Electrical Engineering Tutorials

LM1830 based liquid level indicator circuit

2012-11-01T18:36:00.000+05:30

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LM1830 is a monolithic integrated circuit that can be used in liquid level indicator / control systems. Manufactured by National Semiconductors, the LM1830 can detect the presence or absence of polar fluids . Circuits based on this IC requires minimum number of external components and AC signal is passed through the sensing probe immersed in the fluid. Usage of AC signal for detection prevents electrolysis and this makes the probes long lasting. The IC is capable of driving a LED, high impedance tweeter or a low power relay at its output.

Low liquid level indicator (LED)

The circuit of a low liquid level indicator with LED is shown above. Capacitor Ct sets the frequency of the internal oscillator. With the give value of C1 the frequency will be around 6KHz. Capacitor Cb couples the oscillator output to the probe and it ensures that no DC signal is applied to the probe. The circuit detects the fluid level by comparing the probe to ground resistance with the internal reference resistor Rref.

When the probe to ground resistance goes above the Rref the oscillator output is coupled to the base of the internal output transistor making it conducting. The LED connected to the collector (between pin 12 and Vcc) is driven. Since the base of the transistor is driven using the oscillator, actually the transistor is being switched at the oscillator’s output frequency @50% duty cycle. There is no problem in driving the LED using AC signal and this method is very useful when it comes to use a loud speaker as the indicator. Loud speakers can be driven only by using AC signals and a DC signal will not produce any sound out of the speaker. The circuit diagram of a liquid level indicator using loud speaker at its output is shown below. The circuit is similar to the first circuit except that the LED is replaced by a loud speaker and the load current limiting resistor is changed from 1.2K to 1.5K.

Low liquid level warning (audio)

Low liquid level indicator with external reference resistor.

Some time there may be situations when the fluid resistance (probe to ground) do not match with the 13K internal reference resistor. In such cases there is option to avoid the internal reference resistor and an external reference resistor of your choice (Rx in the circuit) can be employed. The oscillators output is directly available at pin 5 and the Rx connects the oscillator output to the probe through the blocking capacitor. The circuit diagram for the above said scheme is shown below.

Level indicator using external reference resistor

High liquid level activated switch.

A method for activating a relay when the liquid level exceeds a predetermined level is shown here. DC voltage is required for driving the relay and AC voltage cannot be used here just like what we did in the case of LED and speaker. Pin 9 of the IC can be used to solve this problem. A capacitor connected from this pin to ground will keep the internal output transistor steadily ON whenever the probe resistance goes higher than the reference resistor. External transistor Q1 is connected to the collector of the internal transistor. The load that is the relay is connected at the collector of Q1. When the probe is not touching the water it equal to an open circuit situation and surely the probe resistance will be many M ohms and it is greater than the Rref(13K). The internal transistor will be switched ON and Q1 whose base is connected to the collector of the internal transistor will be in OFF condition keeping the relay inactive. When the reverse scenario occurs (fluid level touches the probe) the internal transistor is switched OFF and this in turn makes the transistor Q1 ON resulting in the activation. The load connected through the relay whether pump, lamp, alarm, solenoid valve or anything is driven. Resistor R3 limits the collector current of the internal transistor while resistor R4 provides protection to the IC from transients.

High liquid level activated switch

Probe: The probe used here can be any metal rod with size and shape of your choice. The tank must be made of metal and it should be properly grounded. For non metal tanks fix a small metal contactor at its bottom level and ground it. The probe must be placed at the level you want to monitor.

Notes.

The circuit can be assembled on a Perf board.
I used 12V DC for powering the circuit.
Maximum supply voltage LM1830 can handle is 28V.
The tweeter I used was of a 16 ohm type.
The relay I used is a 200 ohm/12V type.
Maximum load current Q1 (2N2222) can handle is 800mA.
The switching current/voltage ratings of the relay must be according to the load you want to drive using it.
It is recommended to mount the IC on a holder.

Crossover Distortion Summary

2012-10-31T13:03:00.000+05:30

Crossover Distortion

We have seen that one of the main disadvantages of a Class A Amplifier is its low full power efficiency rating. But we also know that we can improve the amplifier and almost double its efficiency simply by changing the output stage of the amplifier to a Class B push-pull type configuration. However, this is great from an efficiency point of view, but most modern Class B amplifiers are transformerless or complementary types with two transistors in their output stage.

This results in one main fundamental problem with push-pull amplifiers in that the two transistors do not combine together fully at the output both halves of the waveform due to their unique zero cut-off biasing arrangement. As this problem occurs when the signal changes or "crosses-over" from one transistor to the other at the zero voltage point it produces an amount of "distortion" to the output wave shape. This results in a condition that is commonly called Crossover Distortion.

Crossover Distortion produces a zero voltage "flat spot" or "deadband" on the output wave shape as it crosses over from one half of the waveform to the other. The reason for this is that the transition period when the transistors are switching over from one to the other, does not stop or start exactly at the zero crossover point thus causing a small delay between the first transistor turning "OFF" and the second transistor turning "ON". This delay results in both transistors being switched "OFF" at the same instant in time producing an output wave shape as shown below.

Crossover Distortion Waveform

In order that there should be no distortion of the output waveform we must assume that each transistor starts conducting when its base to emitter voltage rises just above zero, but we know that this is not true because for silicon bipolar transistors the base voltage must reach at least 0.7v before the transistor starts to conduct thereby producing this flat spot. This crossover distortion effect also reduces the overall peak to peak value of the output waveform causing the maximum power output to be reduced as shown below.

Non-Linear Transfer Characteristics

This effect is less pronounced for large input signals as the input voltage is usually quite large but for smaller input signals it can be more severe causing audio distortion to the amplifier.

Pre-biasing the Output

The problem of Crossover Distortion can be reduced considerably by applying a slight forward base bias voltage (same idea as seen in the Transistor tutorial) to the bases of the two transistors via the centre-tap of the input transformer, thus the transistors are no longer biased at the zero cut-off point but instead are "Pre-biased" at a level determined by this new biasing voltage.

Push-pull Amplifier with Pre-biasing

This type of resistor pre-biasing causes one transistor to turn "ON" exactly at the same time as the other transistor turns "OFF" as both transistors are now biased slightly above their original cut-off point. However, to achieve this the bias voltage must be at least twice that of the normal base to emitter voltage to turn "ON" the transistors. This pre-biasing can also be implemented in transformerless amplifiers that use complementary transistors by simply replacing the two potential divider resistors with Biasing Diodes as shown below.

Pre-biasing with Diodes

This pre-biasing voltage either for a transformer or transformerless amplifier circuit, has the effect of moving the amplifiers Q-point past the original cut-off point thus allowing each transistor to operate within its active region for slightly more than half or 180^o of each half cycle. In other words 180^o + Bias. The amount of diode biasing voltage present at the base terminal of the transistor can be increased in multiples by adding additional diodes in series. This then produces an amplifier circuit commonly called a Class AB Amplifier and its biasing arrangement is given below.

Class AB Output Characteristics

Then to summarise, Crossover Distortion occurs in Class B amplifiers because the amplifier is biased at its cut-off point. This then results in BOTH transistors being switched "OFF" at the same instant in time as the waveform crosses the zero axis. By applying a small base bias voltage either by using a resistive potential divider circuit or diode biasing this crossover distortion can be greatly reduced or even eliminated completely by bringing the transistors to the point of being just switched "ON".

The application of a biasing voltage produces another type or class of amplifier circuit commonly called a Class AB Amplifier. Then the difference between a pure Class B amplifier and an improved Class AB amplifier is in the biasing level applied to the output transistors. One major advantage of using diodes over resistors is that the pn-junctions compensate for variations in the temperature of the transistors. Therefore, we can say the a Class AB amplifier is a Class B amplifier with "Bias" and we can summarise as:

Class A Amplifiers – No Crossover Distortion as they are biased in the centre of the load line.
Class B Amplifiers – Large amounts of Crossover Distortion due to biasing at the cut-off point.
Class AB Amplifiers – Some Crossover Distortion if the biasing level is set too low.

The Class B Amplifier

2012-10-31T13:00:00.000+05:30

To improve the full power efficiency of the previous Class A amplifier by reducing the wasted power in the form of heat, it is possible to design the power amplifier circuit with two transistors in its output stage producing what is commonly termed as a Class B Amplifier also known as a push-pull amplifier configuration.

Push-pull amplifiers use two "complementary" or matching transistors, one being an NPN-type and the other being a PNP-type with both power transistors receiving the same input signal together that is equal in magnitude, but in opposite phase to each other. This results in one transistor only amplifying one half or 180^o of the input waveform cycle while the other transistor amplifies the other half or remaining 180^o of the input waveform cycle with the resulting "two-halves" being put back together again at the output terminal.

Then the conduction angle for this type of amplifier circuit is only 180^o or 50% of the input signal. This pushing and pulling effect of the alternating half cycles by the transistors gives this type of circuit its amusing "push-pull" name, but are more generally known as the Class B Amplifier as shown below.

Class B Push-pull Transformer Amplifier Circuit

The circuit above shows a standard Class B Amplifier circuit that uses a balanced centre-tapped input transformer, which splits the incoming waveform signal into two equal halves and which are 180^o out of phase with each other. Another centre-tapped transformer on the output is used to recombined the two signals providing the increased power to the load. The transistors used for this type of transformer push-pull amplifier circuit are both NPN transistors with their emitter terminals connected together.

Here, the load current is shared between the two power transistor devices as it decreases in one device and increases in the other throughout the signal cycle reducing the output voltage and current to zero. The result is that both halves of the output waveform now swings from zero to twice the quiescent current thereby reducing dissipation. This has the effect of almost doubling the efficiency of the amplifier to around 70%.

Assuming that no input signal is present, then each transistor carries the normal quiescent collector current, the value of which is determined by the base bias which is at the cut-off point. If the transformer is accurately centre tapped, then the two collector currents will flow in opposite directions (ideal condition) and there will be no magnetization of the transformer core, thus minimizing the possibility of distortion.

When an input signal is present across the secondary of the driver transformer T1, the transistor base inputs are in "anti-phase" to each other as shown, thus if TR1 base goes positive driving the transistor into heavy conduction, its collector current will increase but at the same time the base current of TR2 will go negative further into cut-off and the collector current of this transistor decreases by an equal amount and vice versa. Hence negative halves are amplified by one transistor and positive halves by the other transistor giving this push-pull effect.

Unlike the DC condition, these AC currents are ADDITIVE resulting in the two output half-cycles being combined to reform the sine-wave in the output transformers primary winding which then appears across the load.

Class B Amplifier operation has zero DC bias as the transistors are biased at the cut-off, so each transistor only conducts when the input signal is greater than the base-emitter voltage. Therefore, at zero input there is zero output and no power is being consumed. This then means that the actual Q-point of a Class B amplifier is on the Vce part of the load line as shown below.

Class B Output Characteristics Curves

The Class B Amplifier has the big advantage over their Class A amplifier cousins in that no current flows through the transistors when they are in their quiescent state (ie, with no input signal), therefore no power is dissipated in the output transistors or transformer when there is no signal present unlike Class A amplifier stages that require significant base bias thereby dissipating lots of heat - even with no input signal present. So the overall conversion efficiency ( η ) of the amplifier is greater than that of the equivalent Class A with efficiencies reaching as high as 70% possible resulting in nearly all modern types of push-pull amplifiers operated in this Class B mode.

Transformerless Class B Push-Pull Amplifier

One of the main disadvantages of the Class B amplifier circuit above is that it uses balanced centre-tapped transformers in its design, making it expensive to construct. However, there is another type of Class B amplifier called a Complementary-Symmetry Class B Amplifier that does not use transformers in its design therefore, it is transformerless using instead complementary or matching pairs of power transistors. As transformers are not needed this makes the amplifier circuit much smaller for the same amount of output, also there are no stray magnetic effects or transformer distortion to effect the quality of the output signal. An example of a "transformerless" Class B amplifier circuit is given below.

Class B Transformerless Output Stage

The Class B amplifier circuit above uses complimentary transistors for each half of the waveform and while Class B amplifiers have a much high gain than the Class A types, one of the main disadvantages of class B type push-pull amplifiers is that they suffer from an effect known commonly as Crossover Distortion.

Hopefully we remember from our tutorials about Transistors that it takes approximately 0.7 volts (measured from base to emitter) to get a bipolar transistor to start conducting. In a pure class B amplifier, the output transistors are not "pre-biased" to an "ON" state of operation.

This means that the part of the output waveform which falls below this 0.7 volt window will not be reproduced accurately as the transition between the two transistors (when they are switching over from one transistor to the other), the transistors do not stop or start conducting exactly at the zero crossover point even if they are specially matched pairs. The output transistors for each half of the waveform (positive and negative) will each have a 0.7 volt area in which they are not conducting. The resuslt is that both transistors are turned "OFF" at exactly the same time.

A simple way to eliminate crossover distortion in a Class B amplifier is to add two small voltage sources to the circuit to bias both the transistors at a point slightly above their cut-off point. This then would give us what is commonly called an Class AB Amplifier circuit. However, it is impractical to add additional voltage sources to the amplifier circuit so pn-junctions are used to provide the additional bias in the form of silicon diodes.

The Class AB Amplifier

We know that we need the base-emitter voltage to be greater than 0.7v for a silicon bipolar transistor to start conducting, so if we were to replace the two voltage divider biasing resistors connected to the base terminals of the transistors with two silicon Diodes, the biasing voltage applied to the transistors would now be equal to the forward voltage drop of the diode. These two diodes are generally called Biasing Diodes or Compensating Diodes and are chosen to match the characteristics of the matching transistors. The circuit below shows diode biasing.

Class AB Amplifier

The Class AB Amplifier circuit is a compromise between the Class A and the Class B configurations. This very small diode biasing voltage causes both transistors to slightly conduct even when no input signal is present. An input signal waveform will cause the transistors to operate as normal in their active region thereby eliminating any crossover distortion present in pure Class B amplifier designs.

A small collector current will flow when there is no input signal but it is much less than that for the Class A amplifier configuration. This means then that the transistor will be "ON" for more than half a cycle of the waveform but much less than a full cycle giving a conduction angle of between 180 to 360^o or 50 to 100% of the input signal depending upon the amount of additional biasing used. The amount of diode biasing voltage present at the base terminal of the transistor can be increased in multiples by adding additional diodes in series.

Class B amplifiers are greatly preferred over Class A designs for high-power applications such as audio power amplifiers and PA systems. Like the Class A Amplifier circuit, one way to greatly boost the current gain ( A_i ) of a Class B push-pull amplifier is to use Darlington transistors pairs instead of single transistors in its output circuitry.

In the next tutorial about Amplifiers we will look more closely at the effects of Crossover Distortion in Class B amplifier circuits and ways to reduce its effect.

Class A Amplifier

2012-10-30T12:58:00.000+05:30

Common emitter amplifiers are the most commonly used type of amplifier as they have a large voltage gain. They are designed to produce a large output voltage swing from a relatively small input signal voltage of only a few millivolt's and are used mainly as "small signal amplifiers" as we saw in the previous tutorials. However, sometimes an amplifier is required to drive large resistive loads such as a loudspeaker or to drive a motor in a robot and for these types of applications where high switching currents are needed Power Amplifiers are required.

The main function of the power amplifier, which are also known as a "large signal amplifier" is to deliver power, which is the product of voltage and current to the load. Basically a power amplifier is also a voltage amplifier the difference being that the load resistance connected to the output is relatively low, for example a loudspeaker of 4 or 8Ωs resulting in high currents flowing through the collector of the transistor. Because of these high load currents the output transistor(s) used for power amplifier output stages such as the 2N3055 need to have higher voltage and power ratings than the general ones used for small signal amplifiers such as the BC107.

Since we are interested in delivering maximum AC power to the load, while consuming the minimum DC power possible from the supply we are mostly concerned with the "conversion efficiency" of the amplifier. However, one of the main disadvantage of power amplifiers and especially the Class A amplifier is that their overall conversion efficiency is very low as large currents mean that a considerable amount of power is lost in the form of heat. Percentage efficiency in amplifiers is defined as the r.m.s. output power dissipated in the load divided by the total DC power taken from the supply source as shown below.

Power Amplifier Efficiency

Where:
η% - is the efficiency of the amplifier.
Pout - is the amplifiers output power delivered to the load.
Pdc - is the DC power taken from the supply.

For a power amplifier it is very important that the amplifiers power supply is well designed to provide the maximum available continuous power to the output signal.

Class A Amplifier

The most commonly used type of power amplifier configuration is the Class A Amplifier. The Class A amplifier is the most common and simplest form of power amplifier that uses the switching transistor in the standard common emitter circuit configuration as seen previously. The transistor is always biased "ON" so that it conducts during one complete cycle of the input signal waveform producing minimum distortion and maximum amplitude to the output.

This means then that the Class A Amplifier configuration is the ideal operating mode, because there can be no crossover or switch-off distortion to the output waveform even during the negative half of the cycle. Class A power amplifier output stages may use a single power transistor or pairs of transistors connected together to share the high load current. Consider the Class A amplifier circuit below.

Single Stage Amplifier Circuit

This is the simplest type of Class A power amplifier circuit. It uses a single-ended transistor for its output stage with the resistive load connected directly to the Collector terminal. When the transistor switches "ON" it sinks the output current through the Collector resulting in an inevitable voltage drop across the Emitter resistance thereby limiting the negative output capability. The efficiency of this type of circuit is very low (less than 30%) and delivers small power outputs for a large drain on the DC power supply. A Class A amplifier stage passes the same load current even when no input signal is applied so large heatsinks are needed for the output transistors.

However, another simple way to increase the current handling capacity of the circuit while at the same time obtain a greater power gain is to replace the single output transistor with a Darlington Transistor. These types of devices are basically two transistors within a single package, one small "pilot" transistor and another larger "switching" transistor. The big advantage of these devices are that the input impedance is suitably large while the output impedance is relatively low, thereby reducing the power loss and therefore the heat within the switching device.

Darlington Transistor Configurations

The overall current gain Beta (β) or hfe value of a Darlington device is the product of the two individual gains of the transistors multiplied together and very high β values along with high Collector currents are possible compared to a single transistor circuit.

To improve the full power efficiency of the Class A amplifier it is possible to design the circuit with a transformer connected directly in the Collector circuit to form a circuit called a Transformer Coupled Amplifier. The transformer improves the efficiency of the amplifier by matching the impedance of the load with that of the amplifiers output using the turns ratio ( n ) of the transformer and an example of this is given below.

Transformer-coupled Amplifier Circuit

As the Collector current, Ic is reduced to below the quiescent Q-point set up by the base bias voltage, due to variations in the base current, the magnetic flux in the transformer core collapses causing an induced emf in the transformer primary windings. This causes an instantaneous collector voltage to rise to a value of twice the supply voltage 2Vcc giving a maximum collector current of twice Ic when the Collector voltage is at its minimum. Then the efficiency of this type of Class A amplifier configuration can be calculated as follows.

The r.m.s. Collector voltage is given as:

The r.m.s. Collector current is given as:

The r.m.s. Power delivered to the load (Pac) is therefore given as:

The average power drawn from the supply (Pdc) is given by:

and therefore the efficiency of a Transformer-coupled Class A amplifier is given as:

While the transformer improves the efficiency of the amplifier by matching the impedance of the load with that of the amplifiers output impedance, using the turns ratio of an output signal transformer, efficiencies reaching 40% are possible with most commercially available Class-A type power amplifiers being of this type of configuration, but the use of inductive components is best avoided. Also one big disadvantage of this type of circuit is the additional cost and size of the audio transformer required.

The type of "Class" or classification that an amplifier is given really depends upon the conduction angle, the portion of the 360^o of the input waveform cycle, in which the transistor is conducting. In the Class A amplifier the conduction angle is a full 360^o or 100% of the input signal while in other amplifier classes the transistor conducts during a lesser conduction angle.

It is possible to obtain greater power output and efficiency than that of the Class A amplifier by using two complementary transistors in the output stage with one transistor being an NPN or N-channel type while the other transistor is a PNP or P-channel (the complement) type connected in what is called a "push-pull" configuration. This type of configuration is generally called a Class B Amplifier and is another type of audio amplifier circuit that we will look at in the next tutorial.

Amplifier Distortion Navigation

2012-10-29T12:56:00.000+05:30

From the previous tutorials we learnt that for a signal amplifier to operate correctly without any distortion to the output signal, it requires some form of DC Bias on its Base or Gate terminal so that it can amplify the input signal over its entire cycle with the bias "Q-point" set as near to the middle of the load line as possible. This then gave us a "Class-A" type amplification configuration with the most common arrangement being the "Common Emitter" for Bipolar transistors and the "Common Source" for unipolar FET transistors.

We also learnt that the Power, Voltage or Current Gain, (amplification) provided by the amplifier is the ratio of the peak output value to its peak input value (Output ÷ Input). However, if we incorrectly design our amplifier circuit and set the biasing Q-point at the wrong position on the load line or apply too large an input signal to the amplifier, the resultant output signal may not be an exact reproduction of the original input signal waveform. In other words the amplifier will suffer from distortion. Consider the common emitter amplifier circuit below.

Common Emitter Amplifier

Distortion of the output signal waveform may take place because:

1. Amplification may not be taking place over the whole signal cycle due to incorrect biasing.
2. The input signal may be too large, causing the amplifier to be limited by the supply voltage.
3. The amplification may not be linear over the entire frequency range of inputs.

This means then that during the amplification process of the signal waveform, some form of Amplifier Distortion has occurred.

Amplifiers are basically designed to amplify small voltage input signals into much larger output signals and this means that the output signal is constantly changing by some factor or valu, called gain, multiplied by the input signal for all input frequencies. We saw previously that this multiplication factor is called the Beta, β value of the transistor.

Common emitter or even common source type transistor circuits work fine for small AC input signals but suffer from one major disadvantage, the bias Q-point of a bipolar amplifier depends on the same Beta value which may vary from transistors of the same type, ie. the Q-point for one transistor is not necessarily the same as the Q-point for another transistor of the same type due to the inherent manufacturing tolerances. If this occurs the amplifier may not be linear and Amplitude Distortion will result but careful choice of the transistor and biasing components can minimise the effect of amplifier distortion.

Amplitude Distortion

Amplitude distortion occurs when the peak values of the frequency waveform are attenuated causing distortion due to a shift in the Q-point and amplification may not take place over the whole signal cycle. This non-linearity of the output waveform is shown below.

Amplitude Distortion due to Incorrect Biasing

If the bias is correct the output waveform should look like that of the input waveform only bigger, (amplified). If there is insufficient bias the output waveform will look like the one on the right with the negative part of the output waveform "cut-off". If there is too much bias the output waveform will look like the one on the left with the positive part "cut-off".

When the bias voltage is too small, during the negative part of the cycle the transistor does not conduct fully so the output is set by the supply voltage. When the bias is too great the positive part of the cycle saturates the transistor and the output drops almost to zero.

Even with the correct biasing voltage level set, it is still possible for the output waveform to become distorted due to a large input signal being amplified by the circuits gain. The output voltage signal becomes clipped in both the positive and negative parts of the waveform an no longer resembles a sine wave, even when the bias is correct. This type of amplitude distortion is called Clipping and is the result of "Over-driving" the input of the amplifier.

When the input amplitude becomes too large, the clipping becomes substantial and forces the output waveform signal to exceed the power supply voltage rails with the peak (+ve half) and the trough (-ve half) parts of the waveform signal becoming flattened or "Clipped-off". To avoid this the maximum value of the input signal must be limited to a level that will prevent this clipping effect as shown above.

Amplitude Distortion due to Clipping

Amplitude Distortion greatly reduces the efficiency of an amplifier circuit. These "flat tops" of the distorted output waveform either due to incorrect biasing or over driving the input do not contribute anything to the strength of the output signal at the desired frequency. Having said all that, some well known guitarist and rock bands actually prefer that their distinctive sound is highly distorted or "overdriven" by heavily clipping the output waveform to both the +ve and -ve power supply rails. Also, excessive amounts of clipping can also produce an output which resembles a "square wave" shape which can then be used in electronic or digital circuits.

We have seen that with a DC signal the level of gain of the amplifier can vary with signal amplitude, but as well as Amplitude Distortion, other types of distortion can occur with AC signals in amplifier circuits, such as Frequency Distortion and Phase Distortion.

Frequency Distortion

Frequency Distortion occurs in a transistor amplifier when the level of amplification varies with frequency. Many of the input signals that a practical amplifier will amplify consist of the required signal waveform called the "Fundamental Frequency" plus a number of different frequencies called "Harmonics" superimposed onto it. Normally, the amplitude of these harmonics are a fraction of the fundamental amplitude and therefore have very little or no effect on the output waveform. However, the output waveform can become distorted if these harmonic frequencies increase in amplitude with regards to the fundamental frequency. For example, consider the waveform below:

Frequency Distortion due to Harmonics

In the example above, the input waveform consists a the fundamental frequency plus a second harmonic signal. The resultant output waveform is shown on the right hand side. The frequency distortion occurs when the fundamental frequency combines with the second harmonic to distort the output signal. Harmonics are therefore multiples of the fundamental frequency and in our simple example a second harmonic was used. Therefore, the frequency of the harmonic is 2 times the fundamental, 2 x ƒ or 2ƒ. Then a third harmonic would be 3ƒ, a fourth, 4ƒ, and so on. Frequency distortion due to harmonics is always a possibility in amplifier circuits containing reactive elements such as capacitance or inductance.

Phase Distortion

Phase Distortion or Delay Distortion occurs in a non-linear transistor amplifier when there is a time delay between the input signal and its appearance at the output. If we call the phase change between the input and the output zero at the fundamental frequency, the resultant phase angle delay will be the difference between the harmonic and the fundamental. This time delay will depend on the construction of the amplifier and will increase progressively with frequency within the bandwidth of the amplifier. For example, consider the waveform below:

Phase Distortion due to Delay

Any practical amplifier will have a combination of both "Frequency" and "Phase" distortion together with amplitude distortion but in most applications such as in audio amplifiers or power amplifiers, unless the distortion is excessive or severe it will not generally affect the operation of the system.

In the next tutorial about Amplifiers we will look at the Class A Amplifier. Class A amplifiers are the most common type of amplifier output stage making them ideal for use in audio power amplifiers.

The Common Source JFET Amplifier

2012-10-28T12:54:00.000+05:30

So far we have looked at the bipolar type transistor amplifier and especially the common emitter amplifier, but small signal amplifiers can also be made using Field Effect Transistors or FET's for short. These devices have the advantage over bipolar transistors of having an extremely high input impedance along with a low noise output making them ideal for use in amplifier circuits that have very small input signals.

The design of an amplifier circuit based around a junction field effect transistor or "JFET", (N-channel FET for this tutorial) or even a metal oxide silicon FET or "MOSFET" is exactly the same principle as that for the bipolar transistor circuit used for a Class A amplifier circuit we looked at in the previous tutorial.

Firstly, a suitable quiescent point or "Q-point" needs to be found for the correct biasing of the JFET amplifier circuit with single amplifier configurations of Common-source (CS), Common-drain (CD) or Source-follower (SF) and the Common-gate (CG) available for most FET devices. These three JFET amplifier configurations correspond to the common-emitter, emitter-follower and the common-base configurations using bipolar transistors. In this tutorial about FET amplifiers we will look at the popular Common Source JFET Amplifier as this is the most widely used JFET amplifier design.

Consider the Common Source JFET Amplifier circuit configuration below.

Common Source JFET Amplifier

The amplifier circuit consists of an N-channel JFET, but the device could also be an equivalent N-channel depletion-mode MOSFET as the circuit diagram would be the same just a change in the FET, connected in a common source configuration. The JFET gate voltage Vg is biased through the potential divider network set up by resistors R1 and R2 and is biased to operate within its saturation region which is equivalent to the active region of the bipolar junction transistor. Unlike a bipolar transistor circuit, the junction FET takes virtually no input gate current allowing the gate to be treated as an open circuit. Then no input characteristics curves are required. We can compare the JFET to the bipolar junction transistor (BJT) in the following table.

JFET to BJT Comparison

JFET	BJT
Gate, (G)	Base, (B)
Drain, (D)	Collector, (C)
Source, (S)	Emitter, (E)
Gate Supply, (V_G)	Base Supply, (V_B)
Drain Supply, (V_DD)	Collector Supply, (V_CC)
Drain Current, (i_D)	Collector Current, (i_C)

Since the N-Channel JFET is a depletion mode device and is normally "ON", a negative gate voltage with respect to the source is required to modulate or control the drain current. This negative voltage can be provided by biasing from a separate power supply voltage or by a self biasing arrangement as long as a steady current flows through the JFET even when there is no input signal present and Vg maintains a reverse bias of the gate-source pn junction. In this example the biasing is provided from a potential divider network allowing the input signal to produce a voltage fall at the gate as well as voltage rise at the gate with a sinusoidal signal. Any suitable pair of resistor values in the correct proportions would produce the correct biasing voltage so the DC gate biasing voltage Vg is given as:

Note that this equation only determines the ratio of the resistors R1 and R2, but in order to take advantage of the very high input impedance of the JFET as well as reducing the power dissipation within the circuit, we need to make these resistor values as high as possible, with values in the order of 1 to 10MΩ being common.

The input signal, (Vin) of the common source JFET amplifier is applied between the Gate terminal and the zero volts rail, (0v). With a constant value of gate voltage Vg applied the JFET operates within its "Ohmic region" acting like a linear resistive device. The drain circuit contains the load resistor, Rd. The output voltage, Vout is developed across this load resistance. The efficiency of the common source JFET amplifier can be improved by the addition of a resistor, Rs included in the source lead with the same drain current flowing through this resistor. Resistor, Rs is also used to set the JFET amplifiers "Q-point".

When the JFET is switched fully "ON" a voltage drop equal to Rs x Id is developed across this resistor raising the potential of the source terminal above 0v or ground level. This voltage drop across Rs due to the drain current provides the necessary reverse biasing condition across the gate resistor, R2 effectively generating negative feedback. In order to keep the gate-source junction reverse biased, the source voltage, Vs needs to be higher than the gate voltage, Vg. This source voltage is therefore given as:

Then the Drain current, Id is also equal to the Source current, Is as "No Current" enters the Gate terminal and this can be given as:

This potential divider biasing circuit improves the stability of the common source JFET amplifier circuit when being fed from a single DC supply compared to that of a fixed voltage biasing circuit. Both resistor, Rs and the source by-pass capacitor, Cs serve basically the same function as the emitter resistor and capacitor in the common emitter bipolar transistor amplifier circuit, namely to provide good stability and prevent a reduction in the loss of the voltage gain. However, the price paid for a stabilized quiescent gate voltage is that more of the supply voltage is dropped across Rs.

The the value in farads of the source by-pass capacitor is generally fairly high above 100uF and will be polarized. This gives the capacitor an impedance value much smaller, less than 10% of the transconductance, gm (the transfer coefficient representing gain) value of the device. At high frequencies the by-pass capacitor acts essentially as a short-circuit and the source will be effectively connected directly to ground.

The basic circuit and characteristics of a Common Source JFET Amplifier are very similar to that of the common emitter amplifier. A DC load line is constructed by joining the two points relating to the drain current, Id and the supply voltage, Vdd remembering that when Id = 0: ( Vdd = Vds ) and when Vds = 0: ( Id = Vdd/R_L ). The load line is therefore the intersection of the curves at the Q-point as follows.

Common Source JFET Amplifier Characteristics Curves

As with the common emitter bipolar circuit, the DC load line for the common source JFET amplifier produces a straight line equation whose gradient is given as: -1/(Rd + Rs) and that it crosses the vertical Id axis at point A equal to Vdd/(Rd + Rs). The other end of the load line crosses the horizontal axis at point B which is equal to the supply voltage, Vdd. The actual position of the Q-point on the DC load line is generally positioned at the mid centre point of the load line (for class-A operation) and is determined by the mean value of Vg which is biased negatively as the JFET is a depletion-mode device. Like the bipolar common emitter amplifier the output of the Common Source JFET Amplifier is 180^o out of phase with the input signal.

One of the main disadvantages of using Depletion-mode JFET is that they need to be negatively biased. Should this bias fail for any reason the gate-source voltage may rise and become positive causing an increase in drain current resulting in failure of the drain voltage, Vd. Also the high channel resistance, Rds(on) of the junction FET, coupled with high quiescent steady state drain current makes these devices run hot so additional heatsink is required. However, most of the problems associated with using JFET's can be greatly reduced by using enhancement-mode MOSFET devices instead.

MOSFETs or Metal Oxide Semiconductor FET's have much higher input impedances and low channel resistances compared to the equivalent JFET. Also the biasing arrangements for MOSFETs are different and unless we bias them positively for N-channel devices and negatively for P-channel devices no drain current will flow, then we have in effect a fail safe transistor.

JFET Amplifier Current and Power Gains

We said previously that the input current, Ig of a common source JFET amplifier is very small because of the extremely high gate impedance, Rg. A common source JFET amplifier therefore has a very good ratio between its input and output impedances and for any amount of output current, Io the JFET amplifier will have very high current gain Ai. Because of this common source JFET amplifiers are extremely valuable as impedance matching circuits or are used as voltage amplifiers. Likewise, because power = current × voltage, and output voltages are usually several millivolts or even volts, the power gain, Ap is also very high.

In the next tutorial we will look at how the incorrect biasing of the transistor amplifier can cause Distortion to the output signal in the form of amplitude distortion due to clipping and as well as the effect of phase and frequency distortion.

Common Emitter Amplifier

2012-10-28T08:30:00.000+05:30

In the Bipolar Transistor tutorial, we saw that the most common circuit configuration for an NPN transistor is that of the Common Emitter Amplifier circuit and that a family of curves known commonly as the Output Characteristic Curves, relate the transistors Collector current ( Ic ), to the output or Collector voltage ( Vce ), for different values of Base current ( Ib ).

All types of transistor amplifiers operate using AC signal inputs which alternate between a positive value and a negative value so some way of "presetting" the amplifier circuit to operate between these two maximum or peak values is required. This is achieved using a process known as Biasing. Biasing is very important in amplifier design as it establishes the correct operating point of the transistor amplifier ready to receive signals, thereby reducing any distortion to the output signal.

We also saw that a static or DC load line can be drawn onto these output characteristics curves to show all the possible operating points of the transistor from fully "ON" to fully "OFF", and to which the quiescent operating point or Q-point of the amplifier can be found. The aim of any small signal amplifier is to amplify all of the input signal with the minimum amount of distortion possible to the output signal, in other words, the output signal must be an exact reproduction of the input signal but only bigger (amplified).

To obtain low distortion when used as an amplifier the operating quiescent point needs to be correctly selected. This is in fact the DC operating point of the amplifier and its position may be established at any point along the load line by a suitable biasing arrangement. The best possible position for this Q-point is as close to the centre position of the load line as reasonably possible, thereby producing a Class A type amplifier operation, ie. Vce = 1/2Vcc. Consider the Common Emitter Amplifier circuit shown below.

The Common Emitter Amplifier Circuit

The single stage common emitter amplifier circuit shown above uses what is commonly called "Voltage Divider Biasing". This type of biasing arrangement uses two resistors as a potential divider network across the supply with their center point supplying the required Base bias volatge to the transistor. Voltage divider biasing is commonly used in the design of bipolar transistor amplifier circuits. This method of biasing the transistor greatly reduces the effects of varying Beta, ( β ) by holding the Base bias at a constant steady voltage level allowing for best stability. The quiescent Base voltage (Vb) is determined by the potential divider network formed by the two resistors, R1, R2 and the power supply voltage Vcc as shown with the current flowing through both resistors.

Then the total resistance RT will be equal to R1 + R2 giving the current as i = Vcc/RT. The voltage level generated at the junction of resistors R1 and R2 holds the Base voltage (Vb) constant at a value below the supply voltage. Then the potential divider network used in the common emitter amplifier circuit divides the input signal in proportion to the resistance. This bias reference voltage can be easily calculated using the simple voltage divider formula below:

The same supply voltage, (Vcc) also determines the maximum Collector current, Ic when the transistor is switched fully "ON" (saturation), Vce = 0. The Base current Ib for the transistor is found from the Collector current, Ic and the DC current gain Beta, β of the transistor.

Beta is sometimes referred to as h_FE which is the transistors forward current gain in the common emitter configuration. Beta has no units as it is a fixed ratio of the two currents, Ic and Ib so a small change in the Base current will cause a large change in the Collector current. One final point about Beta. Transistors of the same type and part number will have large variations in their Beta value for example, the BC107 NPN Bipolar transistor has a DC current gain Beta value of between 110 and 450 (data sheet value) this is because Beta is a characteristic of their construction and not their operation.

As the Base/Emitter junction is forward-biased, the Emitter voltage, Ve will be one junction voltage drop different to the Base voltage. If the voltage across the Emitter resistor is known then the Emitter current, Ie can be easily calculated using Ohm's Law. The Collector current, Ic can be approximated, since it is almost the same value as the Emitter current.

Example No1

A common emitter amplifier circuit has a load resistance, R_L of 1.2kΩs and a supply voltage of 12v. Calculate the maximum Collector current (Ic) flowing through the load resistor when the transistor is switched fully "ON" (saturation), assume Vce = 0. Also find the value of the Emitter resistor, R_E with a voltage drop of 1v across it. Calculate the values of all the other circuit resistors assuming an NPN silicon transistor.

This then establishes point "A" on the Collector current vertical axis of the characteristics curves and occurs when Vce = 0. When the transistor is switched fully "OFF", their is no voltage drop across either resistor R_E or R_L as no current is flowing through them. Then the voltage drop across the transistor, Vce is equal to the supply voltage, Vcc. This then establishes point "B" on the horizontal axis of the characteristics curves. Generally, the quiescent Q-point of the amplifier is with zero input signal applied to the Base, so the Collector sits half-way along the load line between zero volts and the supply voltage, (Vcc/2). Therefore, the Collector current at the Q-point of the amplifier will be given as:

This static DC load line produces a straight line equation whose slope is given as: -1/(R_L + R_E) and that it crosses the vertical Ic axis at a point equal to Vcc/(R_L + R_E). The actual position of the Q-point on the DC load line is determined by the mean value of Ib.

As the Collector current, Ic of the transistor is also equal to the DC gain of the transistor (Beta), times the Base current (β x Ib), if we assume a Beta (β) value for the transistor of say 100, (one hundred is a reasonable average value for low power signal transistors) the Base current Ib flowing into the transistor will be given as:

Instead of using a separate Base bias supply, it is usual to provide the Base Bias Voltage from the main supply rail (Vcc) through a dropping resistor, R1. Resistors, R1 and R2 can now be chosen to give a suitable quiescent Base current of 45.8μA or 46μA rounded off. The current flowing through the potential divider circuit has to be large compared to the actual Base current, Ib, so that the voltage divider network is not loaded by the Base current flow. A general rule of thumb is a value of at least 10 times Ib flowing through the resistor R2. Transistor Base/Emitter voltage, Vbe is fixed at 0.7V (silicon transistor) then this gives the value of R2 as:

If the current flowing through resistor R2 is 10 times the value of the Base current, then the current flowing through resistor R1 in the divider network must be 11 times the value of the Base current. The voltage across resistor R1 is equal to Vcc - 1.7v (V_RE + 0.7 for silicon transistor) which is equal to 10.3V, therefore R1 can be calculated as:

The value of the Emitter resistor, R_E can be easily calculated using Ohm's Law. The current flowing through R_E is a combination of the Base current, Ib and the Collector current Ic and is given as:

Resistor, R_E is connected between the Emitter and ground and we said previously that it has a voltage of 1 volt across it. Then the value of R_E is given as:

So, for our example above, the preferred values of the resistors chosen to give a tolerance of 5% (E24) are:

Then, our original Common Emitter Amplifier circuit above can be rewritten to include the values of the components that we have just calculated above.

Completed Common Emitter Circuit

Coupling Capacitors

In Common Emitter Amplifier circuits, capacitors C1 and C2 are used as Coupling Capacitors to separate the AC signals from the DC biasing voltage. This ensures that the bias condition set up for the circuit to operate correctly is not effected by any additional amplifier stages, as the capacitors will only pass AC signals and block any DC component. The output AC signal is then superimposed on the biasing of the following stages. Also a bypass capacitor, C_E is included in the Emitter leg circuit.

This capacitor is an open circuit component for DC bias meaning that the biasing currents and voltages are not affected by the addition of the capacitor maintaining a good Q-point stability. However, this bypass capacitor short circuits the Emitter resistor at high frequency signals and only R_L plus a very small internal resistance acts as the transistors load increasing the voltage gain to its maximum. Generally, the value of the bypass capacitor, C_E is chosen to provide a reactance of at most, 1/10th the value of R_E at the lowest operating signal frequency.

Output Characteristics Curves

Ok, so far so good. We can now construct a series of curves that show the Collector current, Ic against the Collector/Emitter voltage, Vce with different values of Base current, Ib for our simple common emitter amplifier circuit. These curves are known as the "Output Characteristic Curves" and are used to show how the transistor will operate over its dynamic range. A static or DC load line is drawn onto the curves for the load resistor RL of 1.2kΩ to show all the transistors possible operating points.

When the transistor is switched "OFF", Vce equals the supply voltage Vcc and this is point B on the line. Likewise when the transistor is fully "ON" and saturated the Collector current is determined by the load resistor, RL and this is point A on the line. We calculated before from the DC gain of the transistor that the Base current required for the mean position of the transistor was 45.8μA and this is marked as point Q on the load line which represents the Quiescent point or Q-point of the amplifier. We could quite easily make life easy for ourselves and round off this value to 50μA exactly, without any effect to the operating point.

Output Characteristics Curves

Point Q on the load line gives us the Base current Q-point of Ib = 45.8μA or 46μA. We need to find the maximum and minimum peak swings of Base current that will result in a proportional change to the Collector current, Ic without any distortion to the output signal. As the load line cuts through the different Base current values on the DC characteristics curves we can find the peak swings of Base current that are equally spaced along the load line. These values are marked as points N and M on the line, giving a minimum and a maximum Base current of 20μA and 80μA respectively.

These points, N and M can be anywhere along the load line that we choose as long as they are equally spaced from Q. This then gives us a theoretical maximum input signal to the Base terminal of 60μA peak-to-peak, (30μA peak) without producing any distortion to the output signal. Any input signal giving a Base current greater than this value will drive the transistor to go beyond point N and into its Cut-off region or beyond point M and into its Saturation region thereby resulting in distortion to the output signal in the form of "clipping".

Using points N and M as an example, the instantaneous values of Collector current and corresponding values of Collector-emitter voltage can be projected from the load line. It can be seen that the Collector-emitter voltage is in anti-phase (-180^o) with the collector current. As the Base current Ib changes in a positive direction from 50μA to 80μA, the Collector-emitter voltage, which is also the output voltage decreases from its steady state value of 5.8v to 2.0v.

Then a single stage Common Emitter Amplifier is also an "Inverting Amplifier" as an increase in Base voltage causes a decrease in Vout and a decrease in Base voltage produces an increase in Vout. In other words the output signal is 180^o out-of-phase with the input signal.

Voltage Gain

The Voltage Gain of the common emitter amplifier is equal to the ratio of the change in the input voltage to the change in the amplifiers output voltage. Then ΔV_L is Vout and ΔV_B is Vin. But voltage gain is also equal to the ratio of the signal resistance in the Collector to the signal resistance in the Emitter and is given as:

We mentioned earlier that as the signal frequency increases the bypass capacitor, C_E starts to short out the Emitter resistor. Then at high frequencies R_E = 0, making the gain infinite. However, bipolar transistors have a small internal resistance built into their Emitter region called R_e. The transistors semiconductor material offers an internal resistance to the flow of current through it and is generally represented by a small resistor symbol shown inside the main transistor symbol.

Transistor data sheets tell us that for a small signal bipolar transistors this internal resistance is the product of 25mV ÷ Ie (25mV being the internal volt drop across the Base/Emitter junction depletion layer), then for our common Emitter amplifier circuit above this resistance value will be equal to:

This internal Emitter leg resistance will be in series with the external Emitter resistor, R_E, then the equation for the transistors actual gain will be modified to include this internal resistance and is given as:

At low frequency signals the total resistance in the Emitter leg is equal to R_E + R_e. At high frequency, the bypass capacitor shorts out the Emitter resistor leaving only the internal resistance R_e in the Emitter leg resulting in a high gain. Then for our common emitter amplifier circuit above, the gain of the circuit at both low and high signal frequencies is given as:

At Low Frequencies

At High Frequencies

One final point, the voltage gain is dependent only on the values of the Collector resistor, R_L and the Emitter resistance, (R_E + R_e) it is not affected by the current gain Beta, β (h_FE) of the transistor.

So, for our simple example above we can now summarise all the values we have calculated for our common emitter amplifier circuit and these are:

	Minimum	Mean	Maximum

Base Current	20μA	50μA	80μA
Collector Current	2.0mA	4.8mA	7.7mA
Output Voltage Swing	2.0V	5.8V	9.3V
Amplifier Gain	-5.32		-218

Common Emitter Amplifier Summary

Then to summarize. The Common Emitter Amplifier circuit has a resistor in its Collector circuit. The current flowing through this resistor produces the voltage output of the amplifier. The value of this resistor is chosen so that at the amplifiers quiescent operating point, Q-point this output voltage lies half way along the transistors load line.

The Base of the transistor used in a common emitter amplifier is biased using two resistors as a potential divider network. This type of biasing arrangement is commonly used in the design of bipolar transistor amplifier circuits and greatly reduces the effects of varying Beta, ( β ) by holding the Base bias at a constant steady voltage. This type of biasing produces the greatest stability.

A resistor can be included in the emitter leg in which case the voltage gain becomes -R_L/R_E. If there is no external Emitter resistance, the voltage gain of the amplifier is not infinite as there is a very small internal resistance, R_e in the Emitter leg. The value of this internal resistance is equal to 25mV/I_E

In the next tutorial about Amplifiers we will look at the Junction Field Effect Amplifier commonly called the JFET Amplifier. Like the transistor, the JFET is used in a single stage amplifier circuit making it easier to understand. There are several different kinds of field effect transistor that we could use but the easiest to understand is the junction field effect transistor, or JFET which has a very high input impedance making it ideal for amplifier circuits.

Nope

2012-10-27T21:41:00.001+05:30

No issue.

Sent using a Sony Ericsson mobile phone

Simple DC power delay circuit

2012-10-27T18:38:00.002+05:30

Description.
The circuit diagram shown here is of a simple DC power delay circuit that is based on an SCR. This circuit is a very handy one and can be employed in many applications. The working of this circuit is very simple. When the input power is applied the capacitor C2 charges through resistor R2 and when the voltage across the capacitor just exceeds the Zener diode D3’s breakdown voltage, it breaks down and the SCR H1 is triggered and the delayed power will be available at the delayed OUT terminal.
Circuit diagram.

Notes.

The circuit must be assembled on a good quality PCB.
The Zener diode must be rated half the input supply voltage.
The current capacity of the circuit depends on the SCR and here it is 4A.

Soldering A Printed Circuit Board (PCB)

2012-10-27T18:33:00.001+05:30

Soldering a PCB is probably the most common soldering task an electronics hobbyist will perform. The basic techniques are fairly easy to grasp but it is a skill that will take a little practice to master. The best way to practice is to buy a simple electronics kit or assemble a simple circuit (such as an LED chaser) on a perf-board. Don't buy that expensive kit or dive into a huge project after only soldering a few joints.

Soldering components onto a PCB involves preparing the surface, placing the components, and then soldering the joint.

Step 1: Surface Preparation:: A clean surface is very important if you want a strong, low resistance solder joint. All surfaces to be soldered should be cleaned well. 3M Scotch Brite pads purchased from the home improvement, industrial supply store or automotive body shop are a good choice as they will quickly remove surface tarnish but will not abrade the PCB material. Note that you will want industrial pads and not the kitchen cleaning pads impregnated with cleaner/soap. If you have particularly tough deposits on your board, then a fine grade of steel wool is acceptable but be very cautious on boards with tight tolerances as the fine steel shavings can lodge between pads and in holes.Once you have cleaned the board down to shiny copper you can use a solvent such as acetone to clean any bits of the cleaning pad that may remain and to remove chemical contamination from the surface of the board. Methyl hydrate is another good solvent and a bit less stinky then acetone. Be aware that both these solvents can remove ink, so if your board is silk screened, test the chemicals first before hosing down the entire board.
A few blasts with compressed air will dry out the board and remove any junk that may have built up in the holes.
It also never hurts to give the component leads a quick wipe down as well, to remove glue or tarnish that may have built up over time.
Step 2: Component Placement: After the component and board have been cleaned, you are ready to place the components onto the board. Unless your circuit is simple and only contains a few components, you will probably not be placing all the components onto the board and soldering them at once. Most likely you will be soldering a few components at a time before turning the board over and placing more. In general it is best to start with the smallest and flattest components (resistors, ICs, signal diodes, etc.) and then work up to the larger components (capacitors, power transistors, transformers) after the small parts are done. This keeps the board relatively flat, making it more stable during soldering. It is also best to save sensitive components (MOSFETs, non-socketed ICs) until the end to lessen the chance of damaging them during assembly of the rest of the circuit.Bend the leads as necessary and insert the component through the proper holes on the board. To hold the part in place while you are soldering, you may want to bend the leads on the bottom of the board at a 45 degree angle. This works well for parts with long leads such as resistors. Components with short leads such as IC sockets can be held in place with a little masking tape or you can bend the leads down to clamp onto the PC board pads.
In the image below, a resistor is ready to solder and is held in place by slightly bent leads.
Step 3: Apply Heat: Apply a very small amount of solder to the tip of the iron. This helps conduct the heat to the component and board, but it is not the solder that will make up the joint. To heat the joint you will lay the tip of the iron so that it rests against both the component lead and the board. It is critical that you heat the lead and the board, otherwise the solder will simply pool and refuse to stick to the unheated item. The small amount of solder you applied to the tip before heating the joint will help make contact between the board and the lead. It normally takes a second or two to get the joint hot enough to solder, but larger components and thicker pads/traces will absorb more heat and can increase this time.If you see the area under the pad starting to bubble, stop heating and remove the soldering iron because you are overheating the pad and it is in danger of lifting. Let it cool, then carefully heat it again for much less time.
Step 4: Apply Solder To The Joint: Once the component lead and solder pad has heated up, you are ready to apply solder. Touch the tip of the strand of solder to the component lead and solder pad, but not the tip of the iron. If everything is hot enough, the solder should flow freely around the lead and pad. You will see the flux melt liquify as well, bubble around the joint (this is part of its cleaning action), flow out and release smoke. Continue to add solder to the joint until the pad is completely coated and the solder forms a small mound with slightly concave sides. If it starts to ball up, you have used too much solder or the pad on the board is not hot enough.

Once the surface of the pad is completely coated, you can stop adding solder and remove the soldering iron (in that order). Don't move the joint for a few seconds as the solder needs time to cool and resolidify. If you do move the joint, you will get what's called a "cold joint". This is recognized by it's characteristic dull and grainy appearance. Many cold joints can be fixed by reheating and applying a small amount of solder, then being allowed to cool without being disturbed.
Step 5: Inspect The Joint and Cleanup: Once the joint is made you should inspect it. Check for cold joints (described a little above and at length below), shorts with adjacent pads or poor flow. If the joint checks out, move on to the next. To trim the lead, use a small set of side cutters and cut at the top of the solder joint.

After you have made all the solder joints, it is good practice to clean all the excess flux residue from the board. Some fluxes are hydroscopic (they absorb water) and can slowly absorb enough water to become slightly conductive. This can be a significant issue in a hostile environment such as an automotive application. Most fluxes will clean up easily using methyl hydrate and a rag but some will require a stronger solvent. Use the appropriate solvent to remove the flux, then blow the board dry with compressed air.

How To Solder?

2012-10-27T18:31:00.001+05:30

Tinning The Soldering Tip

Before use, a new soldering tip, or one that is very dirty, must be tinned. "Tinning" is the process of coating a soldering tip with a thin coat of solder. This aids in heat transfer between the tip and the component you are soldering, and also gives the solder a base from which to flow from.

Step 1: Warm Up The Iron

Warm up the soldering iron or gun thoroughly. Make sure that it has fully come to temperature because you are about to melt a lot of solder on it. This is especially important if the iron is new because it may have been packed with some kind of coating to prevent corrosion.

Step 2: Prepare A Little Space

While the soldering iron is warming up, prepare a little space to work. Moisten a little sponge and place it in the base of your soldering iron stand or in a dish close by. Lay down a piece of cardboard in case you drip solder (you probably will) and make sure you have room to work comfortably.

Step 3: Thoroughly Coat The Tip In Solder

Thoroughly coat the soldering tip in solder. It is very important to cover the entire tip. You will use a considerable amount of solder during this process and it will drip, so be ready. If you leave any part of the tip uncovered it will tend to collect flux residue and will not conduct heat very well, so run the solder up and down the tip and completely around it to totally cover it in molten solder.

Step 4: Clean The Soldering Tip

After you are certain that the tip is totally coated in solder, wipe the tip off on the wet sponge to remove all the flux residue. Do this immediately so there is no time for the flux to dry out and solidify.

Step 5: You're Done!

You have just tinned your soldering tip. This must be done anytime you replace the tip or clean it so that the iron maintains good heat transfer.

Does the hole in a PNP transistor really move…? Holes and electrons are twin brothers??????

2012-10-27T13:08:00.001+05:30

Vidyasagar Sir • October 21, 2012 • 0 Comments

If you consider the internal atomic structure of a PNP transistor, you will see that the emitter layer and the collector layer, together contain a large amount of holes. All these holes are actually within the atoms of trivalent impurity like Aluminium.

Now if you consider this structure, as shown below, then it is so vivid that these holes are immobile and cannot move from one place to another.

But I don’t understand still, since my last 30 years of teaching experience, that how these so called teachers go on telling the students that…

“When a PNP transistor is forward biased, the HOLES, migrate from emitter layer into base layer and then into collector layer…!"

They also say, that…

"Now all the HOLES in collector layer, are then attracted towards the negative terminal of Vbc battery connected to the transistor, and then they come back into emitter layer again…!"

What a false, statement it is…! Shame on them…!

Why they don’t understand that the HOLES are actually IMMOBILE, and CANNOT move from one place to another…!

Introduction to the Amplifier Tutorial

2012-10-27T12:50:00.001+05:30

In "Electronics", signal amplifiers are commonly used devices as they have the ability to amplify a relatively small input signal, for example from a Sensor such as a photo-device, into a much larger output signal to drive a Relay, lamp or loudspeaker for example. There are many forms of electronic circuits classed as amplifiers, from Operational Amplifiers and Small Signal Amplifiers up to Large Signal and Power Amplifiers. Amplifiers can be thought of as a simple box or block containing the amplifying device, such as a Transistor, Field Effect Transistor or Op-amp, which has two input terminals and two output terminals (ground being common) with the output signal being much greater than that of the input signal as it has been "Amplified".

An ideal signal amplifier has three main properties, Input Resistance or ( Rin ), Output Resistance or ( Rout ) and of course amplification known commonly as Gain or ( A ). No matter how complicated an amplifier circuit is, a general amplifier model can still be used to show the relationship of these three properties.

Ideal Amplifier Model

The difference between the input and output signals is known as the Gain of the amplifier and is basically a measure of how much an amplifier "amplifies" the input signal. For example, if we have an input signal of 1 volt and an output of 50 volts, then the gain of the amplifier would be "50". In other words, the input signal has been increased by a factor of 50. This increase is called Gain. Gain is the ratio of output÷input, it has no units but in Electronics it is commonly given the symbol "A", for Amplification. Then the gain of an amplifier is simply calculated as the "output signal divided by the input signal".

Amplifier Gain

The introduction to the amplifier gain can be said to be the relationship that exists between the signal measured at the output with the signal measured at the input. There are three different kinds of amplifier gain which can be measured and these are: Voltage Gain ( Av ), Current Gain ( Ai ) and Power Gain ( Ap ) depending upon the quantity being measured with examples of these different types of gains are given below.

Amplifier Gain of the Input Signal

Voltage Amplifier Gain

Current Amplifier Gain

Power Amplifier Gain

Note that for the Power Gain you can also divide the power obtained at the output with the power obtained at the input. Also when calculating the gain of an amplifier, the subscripts v, i and p are used to denote the type of signal gain being used.

The power Gain or power level of the amplifier can also be expressed in Decibels, (dB). The Bel is a logarithmic unit (base 10) of measurement that has no units. Since the Bel is too large a unit of measure, it is prefixed with deci making it Decibels instead with one decibel being one tenth (1/10th) of a Bel. To calculate the gain of the amplifier in Decibels or dB, we can use the following expressions.

Voltage Gain in dB: a_v = 20 log Av
Current Gain in dB: a_i = 20 log Ai
Power Gain in dB: a_p = 10 log Ap

Note that the DC power gain of an amplifier is equal to ten times the common log of the output to input ratio, where as voltage and current gains are 20 times the common log of the ratio. Note however, that 20dB is not twice as much power as 10dB because of the log scale. Also, a positive value of dB represents a Gain and a negative value of dB represents a Loss within the amplifier. For example, an amplifier gain of +3dB indicates that the amplifiers output signal has "doubled", (x2) while an amplifier gain of -3dB indicates that the signal has "halved", (x0.5) or in other words a loss.

The -3dB point of an amplifier is called the half-power point which is -3dB down from maximum, taking 0dB as the maximum output value.

Example No1

Determine the Voltage, Current and Power Gain of an amplifier that has an input signal of 1mA at 10mV and a corresponding output signal of 10mA at 1V. Also, express all three gains in decibels, (dB).

Amplifier Gain.

in Decibels (dB).

Then the amplifier has a Voltage Gain of 100, a Current Gain of 10 and a Power Gain of 1,000.

Generally, amplifiers can be divide into two distinct types depending upon their power or voltage gain, Small Signal Amplifiers such as pre-amplifiers, instrumentation amplifiers etc, which are designed to amplify very small signal voltage levels of only a few micro-volts (μV) from sensors or audio signals and Large Signal Amplifiers such as audio power amplifiers or switching amplifiers, which are designed to amplify large input voltage signals or switch heavy load currents.

Power Amplifiers

The Small Signal Amplifier is generally referred to as a "Voltage" amplifier because they usually convert a small input voltage into a much larger output voltage. Sometimes an amplifier circuit is required to drive a motor or feed a loudspeaker and for these types of applications where high switching currents are needed Power Amplifiers are required.

As their name suggests, the main job of a "Power Amplifier" (also known as a large signal amplifier), is to deliver power to the load, and as we know from above, is the product of the voltage and current applied to the load with the output signal power being greater than the input signal power. In other words, a power amplifier amplifies the power of the input signal which is why these types of amplifier circuits are used in audio amplifier output stages to drive loudspeakers.

The power amplifier works on the basic principle of converting the DC power drawn from the power supply into an AC voltage signal delivered to the load. Although the amplification is high the efficiency of the conversion from the DC power supply input to the AC voltage signal output is usually poor. The perfect or ideal amplifier would give us an efficiency rating of 100% or at least the power "IN" would be equal to the power "OUT". However, in reality this can never happen as some of the power is lost in the form of heat and also, the amplifier itself consumes power during the amplification process. Then the efficiency of an amplifier is given as:

Amplifier Efficiency

Ideal Amplifier

We can know specify the characteristics for an ideal amplifier from our discussion above with regards to its Gain, meaning voltage gain:

The amplifiers gain, ( A ) should remain constant for varying values of input signal.
Gain is not be affected by frequency. Signals of all frequencies must be amplified by exactly the same amount.
The amplifiers gain must not add noise to the output signal. It should remove any noise that is already exists in the input signal.
The amplifiers gain should not be affected by changes in temperature giving good temperature stability.
The gain of the amplifier must remain stable over long periods of time.

Amplifier Classes

The classification of an amplifier as either a voltage or a power amplifier is made by comparing the characteristics of the input and output signals by measuring the amount of time in relation to the input signal that the current flows in the output circuit. We saw in the Common Emitter transistor tutorial that for the transistor to operate within its "Active Region" some form of "Base Biasing" was required. This small Base Bias voltage added to the input signal allowed the transistor to reproduce the full input waveform at its output with no loss of signal.

However, by altering the position of this Base bias voltage, it is possible to operate an amplifier in an amplification mode other than that for full waveform reproduction. With the introduction to the amplifier of a Base bias voltage, different operating ranges and modes of operation can be obtained which are categorized according to their classification. These various mode of operation are better known as Amplifier Class.

Audio power amplifiers are classified in an alphabetical order according to their circuit configurations and mode of operation. Amplifiers are designated by different classes of operation such as class "A", class "B", class "C", class "AB", etc. These different classes of operation range from a near linear output but with low efficiency to a non-linear output but with a high efficiency. No one class of operation is "better" or "worse" than any other class with the type of operation being determined by the use of the amplifying circuit. There are typical maximum efficiencies for the various types or class of amplifier, with the most commonly used being:

Class A - has low efficiency of less than 40% but good signal reproduction and linearity.
Class B - is twice as efficient as class A amplifiers with a maximum theoretical efficiency of about 70% because the amplifying device only conducts (and uses power) for half of the input signal.
Class AB - has an efficiency rating between that of Class A and Class B but poorer signal reproduction than class A amplifiers.
Class C - is the most efficient amplifier class as only a very small portion of the input signal is amplified therefore the output signal bears very little resemblance to the input signal. Class C amplifiers have the worst signal reproduction.

Class A Operation

Class A Amplifier operation is where the entire input signal waveform is faithfully reproduced at the amplifiers output as the transistor is perfectly biased within its active region, thereby never reaching either of its Cut-off or Saturation regions. This then results in the AC input signal being perfectly "centred" between the amplifiers upper and lower signal limits as shown below.

Class A Output Waveform

In this configuration, the Class A amplifier uses the same transistor for both halves of the output waveform and due to its biasing arrangement the output transistor always has current flowing through it, even if there is no input signal. In other words the output transistors never turns "OFF". This results in the class A type of operation being very inefficient as its conversion of the DC supply power to the AC signal power delivered to the load is usually very low.

Generally, the output transistor of a Class A amplifier gets very hot even when there is no input signal present so some form of heat sinking is required. The DC current flowing through the output transistor (Ic) when there is no output signal will be equal to the current flowing through the load. Then a pure Class A amplifier is very inefficient as most of the DC power is converted to heat.

Class B Operation

Unlike the Class A amplifier mode of operation above that uses a single transistor for its output power stage, the Class B Amplifier uses two complimentary transistors (an NPN and a PNP) for each half of the output waveform. One transistor conducts for one-half of the signal waveform while the other conducts for the other or opposite half of the signal waveform. This means that each transistor spends half of its time in the active region and half its time in the cut-off region thereby amplifying only 50% of the input signal.

Class B operation has no direct DC bias voltage like the class A amplifier, but instead the transistor only conducts when the input signal is greater than the base-emitter voltage and for silicon devices is about 0.7v. Therefore, at zero input there is zero output. This then results in only half the input signal being presented at the amplifiers output giving a greater amount of amplifier efficiency as shown below.

Class B Output Waveform

In a class B amplifier, no DC current is used to bias the transistors, so for the output transistors to start to conduct each half of the waveform, both positive and negative, they need the base-emitter voltage Vbe to be greater than the 0.7v required for a bipolar transistor to start conducting. Then the lower part of the output waveform which is below this 0.7v window will not be reproduced accurately resulting in a distorted area of the output waveform as one transistor turns "OFF" waiting for the other to turn back "ON". The result is that there is a small part of the output waveform at the zero voltage cross over point which will be distorted. This type of distortion is called Crossover Distortion and is looked at later on in this section.

Class AB Operation

The Class AB Amplifier is a compromise between the Class A and the Class B configurations above. While Class AB operation still uses two complementary transistors in its output stage a very small biasing voltage is applied to the Base of the transistor to bias it close to the Cut-off region when no input signal is present.

An input signal will cause the transistor to operate as normal in its Active region thereby eliminating any crossover distortion which is present in class B configurations. A small Collector current will flow when there is no input signal but it is much less than that for the Class A amplifier configuration. This means then that the transistor will be "ON" for more than half a cycle of the waveform. This type of amplifier configuration improves both the efficiency and linearity of the amplifier circuit compared to a pure Class A configuration.

Class AB Output Waveform

The class of operation for an amplifier is very important and is based on the amount of transistor bias required for operation as well as the amplitude required for the input signal. Amplifier classification takes into account the portion of the input signal in which the transistor conducts as well as determining both the efficiency and the amount of power that the switching transistor both consumes and dissipates in the form of wasted heat. Then we can make a comparison between the most common types of amplifier classifications in the following table.

Power Amplifier Classes

Class	A	B	C	AB
Conduction Angle	360^o	180^o	Less than 90^o	180 to 360^o
Position of the Q-point	Centre Point of the Load Line	Exactly on the X-axis	Below the X-axis	In between the X-axis and the Centre Load Line
Overall Efficiency	Poor, 25 to 30%	Better, 70 to 80%	Higher than 80%	Better than A but less than B 50 to 70%
Signal Distortion	None if Correctly Biased	At the X-axis Crossover Point	Large Amounts	Small Amounts

Badly designed amplifiers especially the Class "A" types may also require larger power transistors, more expensive heat sinks, cooling fans, or even an increase in the size of the power supply required to deliver the extra power required by the amplifier. Power converted into heat from transistors, resistors or any other component for that matter, makes any electronic circuit inefficient and will result in the premature failure of the device.

So why use a Class A amplifier if its efficiency is less than 40% compared to a Class B amplifier that has a higher efficiency rating of over 70%. Basically, a Class A amplifier gives a much more linear output meaning that it has, Linearity over a larger frequency response even if it does consume large amounts of DC power.

In this Introduction to the Amplifier tutorial, we have seen that there are different types of amplifier circuit each with its own advantages and disadvantages. In the next tutorial about Amplifiers we will look at the most commonly connected type of transistor amplifier circuit, the Common Emitter Amplifier. Most transistor amplifiers are of the Common Emitter or CE type circuit due to their large gains in voltage, current and power as well as their excellent input/output characteristics.

EFFECT OF A LOAD

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When a load device is connected across the secondary winding of a transformer, current flows through the secondary and the load. The magnetic field produced by the current in the secondary interacts with the magnetic field produced by the current in the primary. This interaction results from the mutual inductance between the primary and secondary windings.
MUTUAL FLUX
The total flux in the core of the transformer is common to both the primary and secondary windings. It is also the means by which energy is transferred from the primary winding to the secondary winding. Since this flux links both windings, it is called MUTUAL FLUX. The inductance which produces this flux is also common to both windings and is called mutual inductance.
Figure 5-11 shows the flux produced by the currents in the primary and secondary windings of a transformer when source current is flowing in the primary winding.
Figure 5-11. - Simple transformer indicating primary- and secondary-winding flux relationship.

When a load resistance is connected to the secondary winding, the voltage induced into the secondary winding causes current to flow in the secondary winding. This current produces a flux field about the secondary (shown as broken lines) which is in opposition to the flux field about the primary (Lenz's law). Thus, the flux about the secondary cancels some of the flux about the primary. With less flux surrounding the primary, the counter emf is reduced and more current is drawn from the source. The additional current in the primary generates more lines of flux, nearly reestablishing the original number of total flux lines.

TURNS AND CURRENT RATIOS
The number of flux lines developed in a core is proportional to the magnetizing force (IN AMPERE-TURNS) of the primary and secondary windings.
The ampere-turn (I X N) is a measure of magnetomotive force; it is defined as the magnetomotive force developed by one ampere of current flowing in a coil of one turn. The flux which exists in the core of a transformer surrounds both the primary and secondary windings. Since the flux is the same for both windings, the ampere-turns in both the primary and secondary windings must be the same.
Therefore:

By dividing both sides of the equation by I_pN_s, you obtain:

Notice the equations show the current ratio to be the inverse of the turns ratio and the voltage ratio. This means, a transformer having less turns in the secondary than in the primary would step down the voltage, but would step up the current. Example: A transformer has a 6:1 voltage ratio.
Find the current in the secondary if the current in the primary is 200 milliamperes.

The above example points out that although the voltage across the secondary is one-sixth the voltage across the primary, the current in the secondary is six times the current in the primary.
The above equations can be looked at from another point of view.
The expression
<figureeq4">
is called the transformer TURNS RATIO and may be expressed as a single factor. Remember, the turns ratio indicates the amount by which the transformer increases or decreases the voltage applied to the primary. For example, if the secondary of a transformer has two times as many turns as the primary, the voltage induced into the secondary will be two times the voltage across the primary. If the secondary has one-half as many turns as the primary, the voltage across the secondary will be one-half the voltage across the primary. However, the turns ratio and the current ratio of a transformer have an inverse relationship. Thus, a 1:2 step-up transformer will have one-half the current in the secondary as in the primary. A 2:1 step-down transformer will have twice the current in the secondary as in the primary.
Example: A transformer with a turns ratio of 1:12 has 3 amperes of current in the secondary. What is the value of current in the primary?

PRODUCING A COUNTER EMF

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When an alternating current flows through a primary winding, a magnetic field is established around the winding. As the lines of flux expand outward, relative motion is present, and a counter emf is induced in the winding. This is the same counter emf that you learned about in the chapter on inductors. Flux leaves the primary at the north pole and enters the primary at the south pole. The counter emf induced in the primary has a polarity that opposes the applied voltage, thus opposing the flow of current in the primary. It is the counter emf that limits exciting current to a very low value.
Q.9 What is meant by "exciting current" in a transformer?
INDUCING A VOLTAGE IN THE SECONDARY
To visualize how a voltage is induced into the secondary winding of a transformer, again refer to figure 5-8. As the exciting current flows through the primary, magnetic lines of force are generated.
During the time current is increasing in the primary, magnetic lines of force expand outward from the primary and cut the secondary. As you remember, a voltage is induced into a coil when magnetic lines cut across it. Therefore, the voltage across the primary causes a voltage to be induced across the secondary.
Q.10 What is the name of the emf generated in the primary that opposes the flow of current in the primary?
Q.11 What causes a voltage to be developed across the secondary winding of a transformer?

PRIMARY AND SECONDARY PHASE RELATIONSHIP
The secondary voltage of a simple transformer may be either in phase or out of phase with the primary voltage. This depends on the direction in which the windings are wound and the arrangement of the connections to the external circuit (load). Simply, this means that the two voltages may rise and fall together or one may rise while the other is falling.
Transformers in which the secondary voltage is in phase with the primary are referred to as LIKE-WOUND transformers, while those in which the voltages are 180 degrees out of phase are called UNLIKE-WOUND transformers.
Dots are used to indicate points on a transformer schematic symbol that have the same instantaneous polarity (points that are in phase).
The use of phase-indicating dots is illustrated in figure 5-9. In part (A) of the figure, both the primary and secondary windings are wound from top to bottom in a clockwise direction, as viewed from above the windings. When constructed in this manner, the top lead of the primary and the top lead of the secondary have the SAME polarity. This is indicated by the dots on the transformer symbol. A lack of phasing dots indicates a reversal of polarity.
Figure 5-9. - Instantaneous polarity depends on direction of winding.

Part (B) of the figure illustrates a transformer in which the primary and secondary are wound in opposite directions. As viewed from above the windings, the primary is wound in a clockwise direction from top to bottom, while the secondary is wound in a counterclockwise direction. Notice that the top leads of the primary and secondary have OPPOSITE polarities. This is indicated by the dots being placed on opposite ends of the transformer symbol. Thus, the polarity of the voltage at the terminals of the secondary of a transformer depends on the direction in which the secondary is wound with respect to the primary.
Q.12 What is the phase relationship between the voltage induced in the secondary of an unlike-wound transformer and the counter emf of the primary winding?
Q.13 Draw dots on the below symbol to indicate the phasing of the transformer.

TRANSFORMER WINDINGS

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As stated above, the transformer consists of two coils called WINDINGS which are wrapped around a core. The transformer operates when a source of ac voltage is connected to one of the windings and a load device is connected to the other. The winding that is connected to the source is called the PRIMARY WINDING. The winding that is connected to the load is called the SECONDARY WINDING. (Note: In this chapter the terms "primary winding" and "primary" are used interchangeably; the term: "secondary winding" and "secondary" are also used interchangeably.)
Figure 5-5 shows an exploded view of a shell-type transformer. The primary is wound in layers directly on a rectangular cardboard form.
Figure 5-5. - Exploded view of shell-type transformer construction.

In the transformer shown in the cutaway view in figure 5-6, the primary consists of many turns of relatively small wire. The wire is coated with varnish so that each turn of the winding is insulated from every other turn. In a transformer designed for high-voltage applications, sheets of insulating material, such as paper, are placed between the layers of windings to provide additional insulation.

Figure 5-6. - Cutaway view of shell-type core with windings.

When the primary winding is completely wound, it is wrapped in insulating paper or cloth. The secondary winding is then wound on top of the primary winding. After the secondary winding is complete, it too is covered with insulating paper. Next, the E and I sections of the iron core are inserted into and around the windings as shown.
The leads from the windings are normally brought out through a hole in the enclosure of the transformer. Sometimes, terminals may be provided on the enclosure for connections to the windings. The figure shows four leads, two from the primary and two from the secondary. These leads are to be connected to the source and load, respectively.

BASIC OPERATION OF A TRANSFORMER

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In its most basic form a transformer consists of:

A primary coil or winding.
A secondary coil or winding.
A core that supports the coils or windings.

Refer to the transformer circuit in figure 5-1 as you read the following explanation: The primary winding is connected to a 60 hertz ac voltage source. The magnetic field (flux) builds up (expands) and collapses (contracts) about the primary winding. The expanding and contracting magnetic field around the primary winding cuts the secondary winding and induces an alternating voltage into the winding. This voltage causes alternating current to flow through the load. The voltage may be stepped up or down depending on the design of the primary and secondary windings.
Figure 5-1. - Basic transformer action.

Q.2 What are, the three basic parts of a transformer?

THE COMPONENTS OF A TRANSFORMER
Two coils of wire (called windings) are wound on some type of core material. In some cases the coils of wire are wound on a cylindrical or rectangular cardboard form. In effect, the core material is air and the transformer is called an AIR-CORE TRANSFORMER. Transformers used at low frequencies, such as 60 hertz and 400 hertz, require a core of low-reluctance magnetic material, usually iron. This type of transformer is called an IRON-CORE TRANSFORMER. Most power transformers are of the iron-core type. The principle parts of a transformer and their functions are:

The CORE, which provides a path for the magnetic lines of flux.
The PRIMARY WINDING, which receives energy from the ac source.
The SECONDARY WINDING, which receives energy from the primary winding and delivers it to the load.
The ENCLOSURE, which protects the above components from dirt, moisture, and mechanical damage.

CORE CHARACTERISTICS
The composition of a transformer core depends on such factors as voltage, current, and frequency. Size limitations and construction costs are also factors to be considered. Commonly used core materials are air, soft iron, and steel. Each of these materials is suitable for particular applications and unsuitable for others. Generally, air-core transformers are used when the voltage source has a high frequency (above 20 kHz). Iron-core transformers are usually used when the source frequency is low (below 20 kHz). A soft-iron-core transformer is very useful where the transformer must be physically small, yet efficient. The iron-core transformer provides better power transfer than does the air-core transformer. A transformer whose core is constructed of laminated sheets of steel dissipates heat readily; thus it provides for the efficient transfer of power. The majority of transformers you will encounter in Navy equipment contain laminated-steel cores. These steel laminations (see figure 5-2) are insulated with a nonconducting material, such as varnish, and then formed into a core. It takes about 50 such laminations to make a core an inch thick. The purpose of the laminations is to reduce certain losses which will be discussed later in this chapter. An important point to remember is that the most efficient transformer core is one that offers the best path for the most lines of flux with the least loss in magnetic and electrical energy.
Figure 5-2. - Hollow-core construction.

SERIES RLC CIRCUITS

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The principles and formulas that have been presented in this chapter are used in all ac circuits. The examples given have been series circuits.
This section of the chapter will not present any new material, but will be an example of using all the principles presented so far. You should follow each example problem step by step to see how each formula used depends upon the information determined in earlier steps. When an example calls for solving for square root, you can practice using the square-root table by looking up the values given.
The example series RLC circuit shown in figure 4-11 will be used to solve for X_L, X_C, X, Z, I_T, true power, reactive power, apparent power, and power factor.
The values solved for will be rounded off to the nearest whole number.
First solve for X_L and X_C.

Figure 4-11. - Example series RLC circuit

Now solve for X

Use the value of X to solve for Z.

This value of Z can be used to solve for total current (I_T).

Since current is equal in all parts of a series circuit, the value of I_T can be used to solve for the various values of power.

The power factor can now be found using either apparent power and true power or resistance and impedance. The mathematics in this example is easier if you use impedance and resistance.

PARALLEL RLC CIRCUITS
When dealing with a parallel ac circuit, you will find that the concepts presented in this chapter for series ac circuits still apply. There is one major difference between a series circuit and a parallel circuit that must be considered. The difference is that current is the same in all parts of a series circuit, whereas voltage is the same across all branches of a parallel circuit. Because of this difference, the total impedance of a parallel circuit must be computed on the basis of the current in the circuit.
You should remember that in the series RLC circuit the following three formulas were used to find reactance, impedance, and power factor:

When working with a parallel circuit you must use the following formulas instead:

NOTE: If no value for E is given in a circuit, any value of E can be assumed to find the values of I_L, I_C, I_X, I_R, and I_Z. The same value of voltage is then used to find impedance.
For example, find the value of Z in the circuit shown in figure 4-12.

The first step in solving for Z is to calculate the individual branch currents.

Figure 4-12. - Parallel RLC circuit.

Using the values for I_R, I_L, and I_C, solve for I_X and I_Z.

Using this value of I_Z, solve for Z.

If the value for E were not given and you were asked to solve for Z, any value of E could be assumed. If, in the example problem above, you assume a value of 50 volts for E, the solution would be:

First solve for the values of current in the same manner as before.

Solve for I_X and I_Z.

Solve for Z.

When the voltage is given, you can use the values of currents, I_R, I_X, and I_Z, to calculate for the true power, reactive power, apparent power, and power factor. For the circuit shown in figure 4-12, the calculations would be as follows.
To find true power,

To find reactive power, first find the value of reactance (X).

To find apparent power,

The power factor in a parallel circuit is found by either of the following methods.

Power Factor

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The POWER FACTOR is a number (represented as a decimal or a percentage) that represents the portion of the apparent power dissipated in a circuit.
If you are familiar with trigonometry, the easiest way to find the power factor is to find the cosine of the phase angle q. The cosine of the phase angle is equal to the power factor.
You do not need to use trigonometry to find the power factor. Since the power dissipated in a circuit is true power, then:

If true power and apparent power are known you can use the formula shown above.
Going one step further, another formula for power factor can be developed. By substituting the equations for true power and apparent power in the formula for power factor, you get:

Since current in a series circuit is the same in all parts of the circuit, I_Requals I_Z. Therefore, in a series circuit,

For example, to compute the power factor for the series circuit shown in figure 4-10, any of the above methods may be used.

Another method:

If you are familiar with trigonometry you can use it to solve for angle

q and the power factor by referring to the tables in appendices V and VI.

NOTE: As stated earlier the power factor can be expressed as a decimal or percentage. In this example the decimal number .6 could also be expressed as 60%.

POWER IN AC CIRCUITS

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You know that in a direct current circuit the power is equal to the voltage times the current, or P = E X I. If a voltage of 100 volts applied to a circuit produces a current of 10 amperes, the power is 1000 watts. This is also true in an ac circuit when the current and voltage are in phase; that is, when the circuit is effectively resistive. But, if the ac circuit contains reactance, the current will lead or lag the voltage by a certain amount (the phase angle). When the current is out of phase with the voltage, the power indicated by the product of the applied voltage and the total current gives only what is known as the APPARENT POWER. The TRUE POWER depends upon the phase angle between the current and voltage. The symbol for phase angle is q (Theta).
When an alternating voltage is impressed across a capacitor, power is taken from the source and stored in the capacitor as the voltage increases from zero to its maximum value. Then, as the impressed voltage decreases from its maximum value to zero, the capacitor discharges and returns the power to the source. Likewise, as the current through an inductor increases from its zero value to its maximum value the field around the inductor builds up to a maximum, and when the current decreases from maximum to zero the field collapses and returns the power to the source. You can see therefore that no power is used up in either case, since the power alternately flows to and from the source. This power that is returned to the source by the reactive components in the circuit is called REACTIVE POWER.
In a purely resistive circuit all of the power is consumed and none is returned to the source; in a purely reactive circuit no power is consumed and all of the power is returned to the source. It follows that in a circuit which contains both resistance and reactance there must be some power dissipated in the resistance as well as some returned to the source by the reactance. In figure 4-9 you can see the relationship between the voltage, the current, and the power in such a circuit. The part of the power curve which is shown below the horizontal reference line is the result of multiplying a positive instantaneous value of current by a negative instantaneous value of the voltage, or vice versa. As you know, the product obtained by multiplying a positive value by a negative value will be negative. Therefore the power at that instant must be considered as negative power. In other words, during this time the reactance was returning power to the source.
Figure 4-9. - Instantaneous power when current and voltage are out of phase.

The instantaneous power in the circuit is equal to the product of the applied voltage and current through the circuit. When the voltage and current are of the same polarity they are acting together and taking power from the source. When the polarities are unlike they are acting in opposition and power is being returned to the source. Briefly then, in an ac circuit which contains reactance as well as resistance, the apparent power is reduced by the power returned to the source, so that in such a circuit the net power, or true power, is always less than the apparent power.

Calculating True Power in AC Circuits
As mentioned before, the true power of a circuit is the power actually used in the circuit. This power, measured in watts, is the power associated with the total resistance in the circuit. To calculate true power, the voltage and current associated with the resistance must be used. Since the voltage drop across the resistance is equal to the resistance multiplied by the current through the resistance, true power can be calculated by the formula:

For example, find the true power of the circuit shown in figure 4-10.
Figure 4-10. - Example circuit for determining power.

Since the current in a series circuit is the same in all parts of the circuit:

IMPEDANCE

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From your study of inductance and capacitance you know how inductive reactance and capacitive reactance act to oppose the flow of current in an ac circuit. However, there is another factor, the resistance, which also opposes the flow of the current. Since in practice ac circuits containing reactance also contain resistance, the two combine to oppose the flow of current. This combined opposition by the resistance and the reactance is called the IMPEDANCE, and is represented by the symbol Z.
Since the values of resistance and reactance are both given in ohms, it might at first seem possible to determine the value of the impedance by simply adding them together. It cannot be done so easily, however. You know that in an ac circuit which contains only resistance, the current and the voltage will be in step (that is, in phase), and will reach their maximum values at the same instant. You also know that in an ac circuit containing only reactance the current will either lead or lag the voltage by one-quarter of a cycle or 90 degrees. Therefore, the voltage in a purely reactive circuit will differ in phase by 90 degrees from that in a purely resistive circuit and for this reason reactance and resistance are rot combined by simply adding them.
When reactance and resistance are combined, the value of the impedance will be greater than either. It is also true that the current will not be in step with the voltage nor will it differ in phase by exactly 90 degrees from the voltage, but it will be somewhere between the in-step and the 90-degree out-of-step conditions. The larger the reactance compared with the resistance, the more nearly the phase difference will approach 90°. The larger the resistance compared to the reactance, the more nearly the phase difference will approach zero degrees.
If the value of resistance and reactance cannot simply be added together to find the impedance, or Z, how is it determined? Because the current through a resistor is in step with the voltage across it and the current in a reactance differs by 90 degrees from the voltage across it, the two are at right angles to each other. They can therefore be combined by means of the same method used in the construction of a right-angle triangle.
Assume you want to find the impedance of a series combination of 8 ohms resistance and 5 ohms inductive reactance. Start by drawing a horizontal line, R, representing 8 ohms resistance, as the base of the triangle. Then, since the effect of the reactance is always at right angles, or 90 degrees, to that of the resistance, draw the line X_L, representing 5 ohms inductive reactance, as the altitude of the triangle. This is shown in figure 4-5. Now, complete the hypotenuse (longest side) of the triangle. Then, the hypotenuse represents the impedance of the circuit.
Figure 4-5. - Vector diagram showing relationship of resistance, inductive reactance, and impedance in a series circuit.

One of the properties of a right triangle is:

Now suppose you apply this equation to check your results in the example given above.

When you have a capacitive reactance to deal with instead of inductive reactance as in the previous example, it is customary to draw the line representing the capacitive reactance in a downward direction. This is shown in figure 4-6. The line is drawn downward for capacitive reactance to indicate that it acts in a direction opposite to inductive reactance which is drawn upward. In a series circuit containing capacitive reactance the equation for finding the impedance becomes:

Figure 4-6. - Vector diagram showing relationship of resistance, capacitive reactance, and impedance in a series circuit.

In many series circuits you will find resistance combined with both inductive reactance and capacitive reactance. Since you know that the value of the reactance, X, is equal to the difference between the values of the inductive reactance, X_L, and the capacitive reactance, X_C, the equation for the impedance in a series circuit containing R, X_L, and X_C then becomes:

In figure 4-7 you will see the method which may be used to determine the impedance in a series circuit consisting of resistance, inductance, and capacitance.
Figure 4-7. - Vector diagram showing relationship of resistance, reactance (capacitive and inductive), and impedance in a series circuit.

Assume that 10 ohms inductive reactance and 20 ohms capacitive reactance are connected in series with 40 ohms resistance. Let the horizontal line represent the resistance R. The line drawn upward from the end of R, represents the inductive reactance, X_L. Represent the capacitive reactance by a line drawn downward at right angles from the same end of R. The resultant of X_L and X_C is found by subtracting X_L from X_C. This resultant represents the value of X.
Thus:

The line, Z, will then represent the resultant of R and X. The value of Z can be calculated as follows:

REACTANCE, IMPEDANCE, AND POWER RELATIONSHIPS IN AC CIRCUITS

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Up to this point inductance and capacitance have been explained individually in ac circuits. The rest of this chapter will concern the combination of inductance, capacitance, and resistance in ac circuits.
To explain the various properties that exist within ac circuits, the series RLC circuit will be used. Figure 4-4 is the schematic diagram of the series RLC circuit. The symbol shown in figure 4-4 that is marked E is the general symbol used to indicate an ac voltage source.
Figure 4 - 4. - Series RLC circuit.

REACTANCE
The effect of inductive reactance is to cause the current to lag the voltage, while that of capacitive reactance is to cause the current to lead the voltage. Therefore, since inductive reactance and capacitive reactance are exactly opposite in their effects, what will be the result when the two are combined? It is not hard to see that the net effect is a tendency to cancel each other, with the combined effect then equal to the difference between their values. This resultant is called REACTANCE; it is represented by the symbol X; and expressed by the equation X = X_L - X_C or X = X_C - X_L. Thus, if a circuit contains 50 ohms of inductive reactance and 25 ohms of capacitive reactance in series, the net reactance, or X, is 50 ohms - 25 ohms, or 25 ohms of inductive reactance.

For a practical example, suppose you have a circuit containing an inductor of 100 mH in series with a capacitor of .001 mF, and operating at a frequency of 4 MHz. What is the value of net reactance, or X?

Now assume you have a circuit containing a 100 -

mH inductor in series with a .0002-mF capacitor, and operating at a frequency of 1 MHz. What is the value of the resultant reactance in this case?

You will notice that in this case the inductive reactance is smaller than the capacitive reactance and is therefore subtracted from the capacitive reactance.
These two examples serve to illustrate an important point: when capacitive and inductive reactance are combined in series, the smaller is always subtracted from the larger and the resultant reactance always takes the characteristics of the larger.

CAPACITIVE REACTANCE

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So far you have been dealing with the capacitor as a device which passes ac and in which the only opposition to the alternating current has been the normal circuit resistance present in any conductor. However, capacitors themselves offer a very real opposition to current flow. This opposition arises from the fact that, at a given voltage and frequency, the number of electrons which go back and forth from plate to plate is limited by the storage ability-that is, the capacitance-of the capacitor. As the capacitance is increased, a greater number of electrons change plates every cycle, and (since current is a measure of the number of electrons passing a given point in a given time) the current is increased.
Increasing the frequency will also decrease the opposition offered by a capacitor. This occurs because the number of electrons which the capacitor is capable of handling at a given voltage will change plates more often. As a result, more electrons will pass a given point in a given time (greater current flow). The opposition which a capacitor offers to ac is therefore inversely proportional to frequency and to capacitance. This opposition is called CAPACITIVE REACTANCE. You may say that capacitive reactance decreases with increasing frequency or, for a given frequency, the capacitive reactance decreases with increasing capacitance. The symbol for capacitive reactance is X_C.

Now you can understand why it is said that the X_C varies inversely with the product of the frequency and capacitance. The formula is:

Where: X_C is capacitive reactance in ohms f is frequency in Hertz C is capacitance in farads p is 6.28 (2 X 3.1416) The following example problem illustrates the computation of X_C.

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CAPACITORS AND ALTERNATING CURRENT
The four parts of figure 4-3 show the variation of the alternating voltage and current in a capacitive circuit, for each quarter of one cycle. The solid line represents the voltage across the capacitor, and the dotted line represents the current. The line running through the center is the zero, or reference point, for both the voltage and the current. The bottom line marks off the time of the cycle in terms of electrical degrees. Assume that the ac voltage has been acting on the capacitor for some time before the time represented by the starting point of the sine wave in the figure.
Figure 4-3. - Phase relationship of voltage and current in a capacitive circuit.

At the beginning of the first quarter-cycle (0° to 90°) the voltage has just passed through zero and is increasing in the positive direction. Since the zero point is the steepest part of the sine wave, the voltage is changing at its greatest rate. The charge on a capacitor varies directly with the voltage, and therefore the charge on the capacitor is also changing at its greatest rate at the beginning of the first quarter-cycle. In other words, the greatest number of electrons are moving off one plate and onto the other plate. Thus the capacitor current is at its maximum value, as part (A) of the figure shows.

As the voltage proceeds toward maximum at 90 degrees, its rate of change becomes less and less, hence the current must decrease toward zero. At 90 degrees the voltage across the capacitor is maximum, the capacitor is fully charged, and there is no further movement of electrons from plate to plate. That is why the current at 90 degrees is zero.
At the end of this first quarter-cycle the alternating voltage stops increasing in the positive direction and starts to decrease. It is still a positive voltage, but to the capacitor the decrease in voltage means that the plate which has just accumulated an excess of electrons must lose some electrons. The current flow, therefore, must reverse its direction. Part (B) of the figure shows the current curve to be below the zero line (negative current direction) during the second quarter-cycle (90° to 180°).
At 180 degrees the voltage has dropped to zero. This means that for a brief instant the electrons are equally distributed between the two plates; the current is maximum because the rate of change of voltage is maximum. Just after 180 degrees the voltage has reversed polarity and starts building up its maximum negative peak which is reached at the end of the third quarter-cycle (180° to 270°). During this third quarter-cycle the rate of voltage change gradually decreases as the charge builds to a maximum at 270 degrees. At this point the capacitor is fully charged and it carries the full impressed voltage. Because the capacitor is fully charged there is no further exchange of electrons; therefore, the current flow is zero at this point. The conditions are exactly the same as at the end of the first quarter-cycle (90°) but the polarity is reversed.
Just after 270 degrees the impressed voltage once again starts to decrease, and the capacitor must lose electrons from the negative plate. It must discharge, starting at a minimum rate of flow and rising to a maximum. This discharging action continues through the last quarter-cycle (270° to 360°) until the impressed-voltage has reached zero. At 360 degrees you are back at the beginning of the entire cycle, and everything starts over again.
If you examine the complete voltage and current curves in part D, you will see that the current always arrives at a certain point in the cycle 90 degrees ahead of the voltage, because of the charging and discharging action. You know that this time and place relationship between the current and voltage is called the phase relationship. The voltage-current phase relationship in a capacitive circuit is exactly opposite to that in an inductive circuit. The current of a capacitor leads the voltage across the capacitor by 90 degrees.
You realize that the current and voltage are both going through their individual cycles at the same time during the period the ac voltage is impressed. The current does not go through part of its cycle (charging or discharging), stop, and wait for the voltage to catch up. The amplitude and polarity of the voltage and the amplitude and direction of the current are continually changing. Their positions with respect to each other and to the zero line at any electrical instant-any degree between zero and 360 degrees-can be seen by reading upwards from the time-degree line. The current swing from the positive peak at zero degrees to the negative peak at 180 degrees is NOT a measure of the number of electrons, or the charge on the plates. It is a picture of the direction and strength of the current in relation to the polarity and strength of the voltage appearing across the plates.
At times it is convenient to use the word "ICE" to recall to mind the phase relationship of the current and voltage in capacitive circuits. I is the symbol for current, and in the word ICE it leads, or comes before, the symbol for voltage, E. C, of course, stands for capacitor. This memory aid is similar to the "ELI" used to remember the current and voltage relationship in an inductor. The phrase "ELI the ICE man" is helpful in remembering the phase relationship in both the inductor and capacitor.
Since the plates of the capacitor are changing polarity at the same rate as the ac voltage, the capacitor seems to pass an alternating current. Actually, the electrons do not pass through the dielectric, but their rushing back and forth from plate to plate causes a current flow in the circuit. It is convenient, however, to say that the alternating current flows "through" the capacitor. You know this is not true, but the expression avoids a lot of trouble when speaking of current flow in a circuit containing a capacitor. By the same short cut, you may say that the capacitor does not pass a direct current (if both plates are connected to a dc source, current will flow only long enough to charge the capacitor). With a capacitor type of hookup in a circuit containing both ac and dc, only the ac will be "passed" on to another circuit.
You have now learned two things to remember about a capacitor: A capacitor will appear to conduct an alternating current and a capacitor will not conduct a direct current.

CAPACITORS IN SERIES AND PARALLEL

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Capacitors may be connected in series or in parallel to obtain a resultant value which may be either the sum of the individual values (in parallel) or a value less than that of the smallest capacitance (in series).
CAPACITORS IN SERIES
The overall effect of connecting capacitors in series is to move the plates of the capacitors further apart. This is shown in figure 3-13. Notice that the junction between C1 and C2 has both a negative and a positive charge. This causes the junction to be essentially neutral. The total capacitance of the circuit is developed between the left plate of C1 and the right plate of C2. Because these plates are farther apart, the total value of the capacitance in the circuit is decreased. Solving for the total capacitance (C_T) of capacitors connected in series is similar to solving for the total resistance(R_T)of resistors connected in parallel.
Figure 3-13. - Capacitors in series.

Note the similarity between the formulas for R_T and C_T:

If the circuit contains more than two capacitors, use the above formula. If the circuit contains only two capacitors, use the below formula:

Note: All values for C_T, C1, C2, C3,... C_n should be in farads. It should be evident from the above formulas that the total capacitance of capacitors in series is less than the capacitance of any of the individual capacitors.
Example: Determine the total capacitance of a series circuit containing three capacitors whose values are 0.01 mF, 0.25 mF, and 50,000 pF, respectively.

The total capacitance of 0.008mF is slightly smaller than the smallest capacitor (0.01mF).
CAPACITORS IN PARALLEL
When capacitors are connected in parallel, one plate of each capacitor is connected directly to one terminal of the source, while the other plate of each capacitor is connected to the other terminal of the source. Figure 3-14 shows all the negative plates of the capacitors connected together, and all the positive plates connected together. C_T, therefore, appears as a capacitor with a plate area equal to the sum of all the individual plate areas. As previously mentioned, capacitance is a direct function of plate area. Connecting capacitors in parallel effectively increases plate area and thereby increases total capacitance.
Figure 3-14. - Parallel capacitive circuit.

For capacitors connected in parallel the total capacitance is the sum of all the individual capacitances. The total capacitance of the circuit may by calculated using the formula:

where all capacitances are in the same units.
Example: Determine the total capacitance in a parallel capacitive circuit containing three capacitors whose values are 0.03 mF, 2.0 mF, and 0.25 mF, respectively.