Rectifiers are one of the basic building blocks of AC power conversion with halfwave or fullwave rectification generally performed by semiconductor diodes. Diodes allow alternating currents to flow through them in the forward direction while blocking current flow in the reverse direction creating a fixed DC voltage level making them ideal for rectification.
However, direct current which has been rectified by diodes is not as pure as that obtained from say, a battery source, but has voltage changes in the form of ripples superimposed on it as a result of the alternating supply.
But for single phase rectification to take place, we need an AC sinusoidal waveform of a fixed voltage and frequency as shown.
AC waveforms generally have two numbers associated with them. The first number expresses the degree of rotation of the waveform along the xaxis by which the alternator has rotated from 0to360^{o}. This value is known as the period (T) which is defined as the interval taken to complete one full cycle of the waveform. Periods are measured in units of degrees, time, or radians. The relationship between a sine waves periods and frequency is defined as: T = 1/ƒ.
The second number indicates the amplitude of the value, either current or voltage, along the yaxis. This number gives the instantaneous value from zero to some peak or maximum value ( A_{MAX}, V_{MAX} or I_{MAX} ) indicating the sine waves greatest amplitude before returning back to zero again. For a sinusoidal waveform there are two maximum or peak values, one for the positive and one for the negative halfcycles.
But as well as these two values, there are two more which are of interest to us for rectification purposes. One is the waveforms Average Value and the other is its RMS Value. The average value of a waveform is obtained by adding the instantaneous values of voltage (or current) over one halfcycle and is found as: 0.6365*V_{P}. Note that the average value over one complete cycle of a symmetrical sine wave is zero.
The RMS, root mean squared or effective value of a sinusoid (a sinusoid is another name for a sine wave) delivers the same amount of energy to a resistance as does a DC supply of the same value. The root mean square (rms) value of a sinusoidal voltage (or current) is defined as: 0.7071*V_{P}.
All single phase rectifiers use solid state devices as their primary ACtoDC converting device. Single phase uncontrolled halfwave rectifiers are the simplest and possibly the most widely used rectification circuit for small power levels as their output is heavily affected by the reactance of the connected load.
For uncontrolled rectifier circuits, semiconductor diodes are the most commonly used device and are so arranged to create either a halfwave or a fullwave rectifier circuit. The advantage of using diodes as the rectification device is that by design they are unidirectional devices having an inbuilt oneway pnjunction.
This pnjunction converts the bidirectional alternating supply into a oneway unidirectional current by eliminating onehalf of the supply. Depending upon the connection of the diode, it could for example pass the positive half of the AC waveform when forwardbiased, while eliminating the negative halfcycle when the diode becomes reversebiased.
The reverse is also true by eliminate the positive half or the waveform and passing the negative half. Either way, the output from a single diode rectifier consists of only one half of the 360^{o} waveform as shown.
The singlephase halfwave rectifier configuration above passes the positive half of the AC supply waveform with the negative half being eliminated. By reversing the direction of the diode we can pass negative halves and eliminate the positive halves of the AC waveform. Therefore the output will be a series of positive or negative pulses.
Thus there is no voltage or current applied to the connected load, R_{L} for half of each cycle. In other words, the voltage across the load resistance, R_{L} consists of only half waveforms, either positive or negative, as it operates during only onehalf of the input cycle, hence the name of halfwave rectifier.
Hopefully we can see that the diode allows current to flow in one direction only producing an output which consists of halfcycles. This pulsating output waveform not only varies ON and OFF every cycle, but is only present 50% of the time and with a purely resistive load, this high voltage and current ripple content is at its maximum.
This pulsating DC means that the equivalent DC value dropped across the load resistor, R_{L} is therefore only one half of the sinusoidal waveforms average value. Since the maximum value of the waveforms sine function is 1 ( sin(90^{o}) ), the Average or Mean DC value taken over onehalf of a sinusoid is defined as: 0.637 x maximum amplitude value.
So during the positive halfcycle, A_{AVE} equals 0.637*A_{MAX}. However as the negative halfcycles are removed due to rectification by the diode, the average value during this period will be zero as shown.
So for a halfwave rectifier, 50% of the time there is an average value of 0.637*A_{MAX} and 50% of the time there is zero. If the maximum amplitude is 1, the average or DC value equivalent seen across the load resistance, R_{L} will be:
Thus the corresponding expressions for the average value of voltage or current for a halfwave rectifier is given as:
V_{AVE} = 0.318*V_{MAX}
I_{AVE} = 0.318*I_{MAX}
Note that the maximum value, A_{MAX} is that of the input waveform, but we could also use its RMS, or root mean squared value to find the equivalent DC output value of a single phase halfwave rectifier. To determine the average voltage for a halfwave rectifier, we multiply the RMS value by 0.9 (form factor) and divide the product by 2, that is multiplying it by 0.45 giving:
V_{AVE} = 0.45*V_{RMS}
I_{AVE} = 0.45*I_{RMS}
Then we can see that a halfwave rectifier circuit converts either the positive or negative halves of an AC waveform into a pulsed DC output that has a value of 0.318*A_{MAX} or 0.45*A_{RMS} as shown.
A single phase halfwave rectifier is connected to a 50V RMS 50Hz AC supply. If the rectifier is used to supply a resistive load of 150 Ohms. Calculate the equivalent DC voltage developed across the load, the load current and power dissipated by the load. Assume ideal diode characteristics.
First we need to convert the 50 volts RMS to its peak or maximum voltage equivalent.
V_{M} = 1.414*V_{RMS} = 1.414*50 = 70.7 volts
V_{DC} = 0.318*V_{M} = 0.318*70.7 = 22.5 volts
I_{L} = V_{DC} ÷ R_{L} = 22.5/150 = 0.15A or 150mA
P_{L} = V*I or I^{2}*R_{L} = 22.5*0.15 = 3.375W ≅ 3.4W
In practice, V_{DC} would be slightly less due to the forward biased 0.7 volt voltage drop across the rectifying diode.
One of the main disadvantages of a singlephase halfwave rectifier is that there is no output during half of the available input sinusoidal waveform resulting in a low average value as we have seen. One way to overcome this is to use more diodes to produce a fullwave rectifier.
Unlike the previous halfwave rectifier, the fullwave rectifier utilises both halves of the input sinusoidal waveform to provide a unidirectional output. This is because the fullwave rectifier basically consists of two halfwave rectifiers connected together to feed the load.
The single phase fullwave rectifier does this by using four diodes arranged in a bridge arrangement passing the positive half of the waveform as before but inverting the negative half of the sine wave to create a pulsating DC output. Even though the the voltage and current output from the rectifier is pulsating, it does not reverse direction using the full 100% of the input waveform and thus providing fullwave rectification.
This bridge configuration of diodes provides fullwave rectification because at any time two of the four diodes are forward biased while the other two are reverse biased. Thus there are two diodes in the conduction path instead of the single one for the halfwave rectifier. Therefore there will be a difference in voltage amplitude between V_{IN} and V_{OUT} due to the two forward voltage drops of the serially connected diodes. Here as before, for simplicity of the maths we will assume ideal diodes.
So how does the single phase fullwave rectifier work. During the positive half cycle of V_{IN}, diodes D_{1} and D_{4} are forward biased while diodes D_{2} and D_{3} are reverse biased. Then for the positive half cycle of the input waveform, current flows along the path of: D_{1} – A – R_{L} – B – D_{4} and back to the supply.
During the negative half cycle of V_{IN}, diodes D_{3} and D_{2} are forward biased while diodes D_{4} and D_{1} are reverse biased. Then for the negative half cycle of the input waveform, current flows along the path of: D_{3} – A – R_{L} – B – D_{2} and back to the supply.
In both cases the positive and negative halfcycles of the input waveform produce positive output peaks regardless of polarity of input waveform and as such the load current, i always flows in the SAME direction through the load, R_{L} between points or nodes A and B. Thus the negative halfcycle of the source becomes a positive halfcycle at load.
So whichever set of diodes are conducting, node A is always more positive than node B. Therefore the load current and voltage are unidirectional or DC giving us the following output waveform.
Although this pulsating output waveform uses 100% of the input waveform, its average DC voltage (or current) is not at the same value. We remember from above that the average or mean DC value taken over onehalf of a sinusoid is defined as: 0.637 x maximum amplitude value. However unlike halfwave rectification above, fullwave rectifiers have two positive halfcycles per input waveform giving us a different average value as shown.
Here we can see that for a fullwave rectifier, for each positive peak there is an average value of 0.637*A_{MAX} and as there are two peaks per input waveform, this means there are two lots of average value summed together. Thus the DC output voltage of a fullwave rectifier is twice that of the previous halfwave rectifier. If the maximum amplitude is 1, the average or DC value equivalent seen across the load resistance, R_{L} will be:
Thus the corresponding expressions for the average value of voltage or current for a fullwave rectifier is given as:
V_{AVE} = 0.637*V_{MAX}
I_{AVE} = 0.637*I_{MAX}
As before, the maximum value, A_{MAX} is that of the input waveform, but we could also use its RMS, or root mean squared value to find the equivalent DC output value of a single phase fullwave rectifier. To determine the average voltage for a fullwave rectifier, we multiply the RMS value by 0.9 giving:
V_{AVE} = 0.9*V_{RMS}
I_{AVE} = 0.9*I_{RMS}
Then we can see that a fullwave rectifier circuit converts BOTH the positive or negative halves of an AC waveform into a pulsed DC output that has a value of 0.637*A_{MAX} or 0.9*A_{RMS} as shown.
Four diodes are used to construct a single phase fullwave bridge rectifier circuit which is required to supply a purely resistive load of 1kΩ at 220 volts DC. Calculate the RMS value of the input voltage required, the total load current drawn from the supply, the load current passed by each diode and the total power dissipated by the load. Assume ideal diode characteristics.
V_{DC} = 0.9*V_{RMS} therefore: V_{RMS} = V_{DC} ÷ 0.9 = 220/0.9 = 244.4 V_{RMS}
I_{L} = V_{DC} ÷ R_{L} = 220/1000 = 0.22A or 220mA
The load current is supplied by two diodes per cycle, thus:
I_{D} = I_{L} ÷ 2 = 0.22/2 = 0.11A or 110mA
P_{L} = V*I or I^{2}*R_{L} = 220*0.22 = 48.4W
Fullwave rectification has many advantages over the simpler halfwave rectifier, such as the output voltage is more consistent, has a higher average output voltage, the input frequency is doubled by the process of rectification, and requires a smaller capacitance value smoothing capacitor if one is required. But we can improve on the design of the bridge rectifier by using thyristors instead of diodes in its design.
By replacing the diodes within a single phase bridge rectifier with thyristors, we can create a phasecontrolled ACtoDC rectifier for converting the constant AC supply voltage into a controlled DC output voltage. Phase controlled rectifiers either halfcontrolled or fully controlled, have many applications in variable voltage power supplies and motor control.
The single phase bridge rectifier is what is termed an “uncontrolled rectifier” in that the applied input voltage is passed directly to the output terminals providing a fixed average DC equivalent value. To convert an uncontrolled bridge rectifier into a single phase halfcontrolled rectifier circuit we just need to replace two of the diodes with thyristors (SCR’s) as shown.
In the halfcontrolled rectifier configuration, the average DC load voltage is controlled using two thyristors and two diodes. As we learnt in our tutorial about Thyristors, a thyristor will only conduct (“ON” state) when its Anode, (A) is more positive than its Cathode, (K) and a firing pulse is applied to its Gate, (G) terminal. Otherwise it remains inactive.
We also learnt that once “ON”, a thyristor is only turned “OFF” again when its gate signal is removed and the anode current has fallen below the thyristors holding current, I_{H} as the AC supply voltage reverse biases it. So by delaying the firing pulse applied to the thyristors gate terminal for a controlled period of time, or angle (α), after the AC supply voltage has passed the zerovoltage crossing of the anodetocathode voltage, we can control when the thyristor starts to conduct current and hence control the average output voltage.
During the positive half cycle of the input waveform, current flows along the path of: SCR_{1} and D_{2}, and back to the supply. During the negative half cycle of V_{IN}, conduction is through SCR_{2} and D_{1} and back to the supply.
It is clear then that one thyristor from the top group (SCR_{1} or SCR_{2}) and its corresponding diode from the bottom group (D_{2} or D_{1}) must conduct together for for any load current to flow.
Thus the average output voltage, V_{AVE} is dependant on the firing angle α for the two thyristors included in the halfcontrolled rectifier as the two diodes are uncontrolled and pass current whenever forward biased. So for any gate firing angle, α, the average output voltage is given by:
Note that the maximum average output voltage occurs when α = 1 but is still only 0.637*V_{MAX} the same as for the single phase uncontrolled bridge rectifier.
We can take this idea of controlling the average output voltage of the bridge one step further by replacing all four diodes with thyristors giving us a Fullycontrolled Bridge Rectifier circuit.
Single phase fullycontrolled bridge rectifiers are known more commonly as ACtoDC converters. Fullycontrolled bridge converters are widely used in the speed control of DC machines and is easily obtained by replacing all four diodes of a bridge rectifier with thyristors as shown.
In the fullycontrolled rectifier configuration, the average DC load voltage is controlled using two thyristors per halfcycle. Thyristors SCR_{1} and SCR_{4} are fired together as a pair during the positive halfcycle, while thyristors SCR_{3} and SCR_{4} are also fired together as a pair during the negative halfcycle. That is 180^{o} after SCR_{1} and SCR_{4}.
Then during continuous conduction mode of operation the four thyristors are constantly being switched as alternate pairs to maintain the average or equivalent DC output voltage. As with the halfcontrolled rectifier, the output voltage can be fully controlled by varying the thyristors firing delay angle (α).
Thus the expression for the average DC voltage from a single phase fullycontrolled rectifier in its continuous conduction mode is given as:
with the average output voltage varying from V_{MAX}/π to V_{MAX}/π by varying the firing angle, α from π to 0 respectively. So when α < 90^{o} the average DC voltage is positive and when α > 90^{o} the average DC voltage is negative. That is power flows from the DC load to the AC supply.
Then we have seen here in this tutorial about single phase rectification that single phase rectifiers can take on many forms to convert AC voltage to DC voltage from uncontrolled single diode halfwave rectifiers to fullycontrolled fullwave bridge rectifiers using four thyristors.
The advantages of the halfwave rectifier are its simplicity and low cost as it requires only one diode. However, it is not very efficient as only half of the input signal is used producing a low average output voltage.
The fullwave rectifier is more efficient than the halfwave rectifier as it uses both halfcycles of the input sine wave producing a higher average or equivalent DC output voltage. A disadvantage of the fullwave bridge circuit is that is that it requires four diodes.
Phase controlled rectification uses combinations of diodes and thyristors (SCR’s) to convert the AC input voltage into a controlled DC output voltage. Fullycontrolled rectifiers use four thyristors in their configuration, whereas halfcontrolled rectifiers use a combination of both thyristors and diodes.
Then no matter how we do it, the conversion of a sinusoidal AC waveform to a steady state DC supply is called Rectification.
]]>The switching and routing of digital and analogue signals (both voltage and current) can easily be done using mechanical relays and their contacts, but these can be slow and costly. The obvious choice is to use much faster acting solid state electronic switches which use metal oxide semiconductor (MOS) analogue gates to route the signal currents from their input to their output, with the wellknown CMOS 4016B bilateral switch being the most common example.
MOS technology uses both NMOS and PMOS devices to perform the logical switching functions, thus allowing a digital computer or logic circuit to control the operation of these analogue switches. CMOS devices where both NMOS and PMOS transistors are fabricated into the same gate circuit, can pass (closedcondition) or block (opencondition) an analogue or digital signal, depending on the digital logic level that controls it.
The type of solidstate switch which allows for a signal or data transfer in both directions is called a Transmission Gate, or TG. But first lets consider the operation of a Field Effect Transistor, or FET as a basic analogue switch.
Both Bipolar Junction Transistors (BJTs) and Field Effect Transistors (FETs) can be used as a singlepole electronic switch in a wide variety of different applications. The main advantages of MOSFET, or metaloxidesemiconductor FET, technology over bipolar devices is that its gate terminal is insulated from the main conducting channel by a thin layer of metal oxide, and the main MOSFET channel used for switching is purely resistive.
Consider the basic Nchannel and Pchannel enhancement MOSFET (eMOSFET) configurations below.
Then we can see that for the nchannel (NMOS) and pchannel (PMOS) enhancement MOSFET to operate as an open (OFF) or closed (ON) device the following conditions must be true:
Note that a MOSFETs Threshold Voltage, V_{T} is the minimum voltage applied to the gate terminal for the main channel between the drain and source terminals to start conducting. Also, since the eMOSFET is used mainly as a switching device, it generally operates between its cutoff and saturation regions thus V_{GS} acts as an ON/OFF control voltage for the MOSFET.
An ideal analogue switch would create a shortcircuit condition when closed and an opencircuit condition when open, in a similar fashion to a mechanical switch.
However, solidstate analogue switches are not ideal as there is always some loss associated with the conducting channel due to its resistive value when ON.
We would like to think that if we applied a signal to its input pin this would result in the signal being identical and without loss at the output pin, and vice versa. However, while CMOS switches do make excellent transmission gates, their “ON” state resistance, R_{ON} can be several ohms creating an I^{2}*R power loss, while their “OFF” state resistance can be several thousand ohms allowing pico amperes of current to still flow through the channel.
Nevertheless, the ability of complementary metaloxide semiconductor FETs to perform as analogue switches and transmission gates remains high, and MOSFET devices, in particular the enhancement MOSFET which requires a voltage to be applied to the gate to turn it “ON” and zero voltage to turn it “OFF” are the most commonly used switching transistor.
The Nchannel metaloxide semiconductor (NMOS) transistor can be used as a transmission gate for the passing of analogue signals. Assuming that the drain and source terminals are identical, the input is connected to the Drain terminal and the control signal to the gate terminal as shown.
When the control voltage, V_{C} on the gate is zero (LOW), the gate terminal will not be positive with respect to either input terminal (drain) or the output terminal (source), thus the transistor is in its cutoff region and the input and output terminals are isolated from each. Then the NMOS is acting an open switch so any voltage at the input will not be passed to the output.
When there is a positive control voltage +V_{C} at the gate terminal, the transistor is turned “ON” and in its saturation region acting as a closed switch. If the input voltage, V_{IN} is positive and greater than V_{C} current will flow from the drain terminal to the source terminal, thus connecting V_{OUT} to V_{IN}.
If however V_{IN} becomes zero (LOW) while the gates control voltage is still positive, the transistor channel is still open but the draintosource voltage, V_{DS} is zero, so no drain current flows through the channel and thus the output voltage is zero.
Therefore, as long as the gate control voltage, V_{C} is HIGH, the NMOS transistor passes the input voltage to the output. If it is LOW, the NMOS transistor is turned “OFF”, and the output terminal is disconnected from the input. Thus, the control voltage, V_{C} at the gate determines whether the transistor is an “open” or “closed” as a switch.
One issue here with the NMOS switch is that the gatetosource voltage, V_{GS} must be significantly greater than the channel threshold voltage to turn it fullyON or there will be a voltage reduction through the channel. Thus the NMOS device can only transmit a “weak” logic “1” (HIGH) level but a strong logic “0” (LOW) without loss.
The Pchannel metaloxide semiconductor (PMOS) transistor is similar but opposite in polarity to the previous NMOS device with current flowing in the opposite direction, from source to drain. Then for a PMOS device, the input is connected to the Source terminal and the control signal to the gate terminal as shown.
For the PMOS FET, when the control voltage, V_{C} on the gate is zero and is thus more negative with respect to either input terminal (source) or the output terminal (drain), the transistor is “ON” and in its saturation region acting as a closed switch. If the input voltage, V_{IN} is positive and greater than V_{C} current will flow from the source terminal to the drain terminal, that is I_{D} flows out of the drain thus connecting V_{IN} to V_{OUT}.
If the input voltage, V_{IN} becomes zero (LOW) while the gates control voltage is still zero or negative, the PMOS channel is still open but the sourcetodrain voltage, V_{SD} is zero, so no current flows through the channel and thus the voltage at the output (drain) is zero.
When there is a positive control voltage +V_{C} at the gate terminal, the channel of the PMOS transistor is turned “OFF” and in its cutoff region acting as a open switch. Thus no drain current, I_{D} flows through the conducting channel.
Therefore, as long as the gate control voltage, V_{C} is LOW (or negative), the PMOS transistor will pass the input voltage to the output. If it is HIGH, the PMOS transistor is turned “OFF”, and the output terminal is disconnected from the input. Thus as with the previous NMOS device, the control voltage, V_{C} at the gate determines whether the transistor is an “open” or “closed” as a switch.
The problem with the PMOS switch is that the gatetosource voltage, V_{GS} must be significantly less than the channel threshold voltage to turn it fullyOFF or current will still flow through the channel. Thus the PMOS device can transmit a “strong” logic “1” (HIGH) level without loss but a weak logic “0” (LOW).
So we can see that for an NMOS device a positive gatetosource voltage causes current to flow in one direction from DraintoSource, while for the PMOS device, a negative gatetosource voltage will result in current flowing in the reverse direction from SourcetoDrain.
However, the NMOS device only passes a strong “0” but a weak “1”, while the PMOS device passes a strong “1” but a weak “0”. Thus by combining the characteristics of the NMOS and the PMOS devices, it is possible to transmit both a strong logic “0” or a strong logic “1” value in either direction without any degradation. This then forms the basis of a Transmission Gate.
Connecting PMOS and NMOS devices together in parallel we can create a basic bilateral CMOS switch, known commonly as a “Transmission Gate”. Note that transmission gates are quite different from conventional CMOS logic gates as the transmission gate is symmetrical, or bilateral, that is, the input and output are interchangeable. This bilateral operation is shown in the transmission gate symbol below which shows two superimposed triangles pointing in opposite directions to indicate the two signal directions.
Two MOS transistors are connected backtoback in parallel with an inverter used between the gate of the NMOS and PMOS to provide the two complementary control voltages. When the input control signal, V_{C} is LOW, both the NMOS and PMOS transistors are cutoff and the switch is open. When V_{C} is high, both devices are biased into conduction and the switch is closed.
Thus the transmission gate acts as a “closed” switch when V_{C} = 1, while the gate acts as an “open” switch when V_{C} = 0 operating as a voltagecontrolled switch. The bubble of the symbol indicating the gate of the PMOS FET.
As with traditional logic gates, we can define the operation of a transmission gate using both a truth table and boolean expression as follows.
Symbol  Truth Table  
Transmission Gate

Control  A  B 
1  0  0  
1  1  1  
0  0  HiZ  
0  1  HiZ  
Boolean Expression B = A.Control  Read as A AND Cont. gives B 
We can see from the above truth table, that the output at B relies not only the logic level of the input A, but also on the logic level present on the control input. Thus the logic level value of B is defined as both A AND Control giving us the boolean expression for a transmission gate of:
B = A.Control
Since the boolean expression of a transmission gate incorporates the logical AND function, it is therefore possible to implement this operation using a standard 2input AND gate with one input being the data input while the other is the control input as shown.
One other point to consider about transmission gates, a single NMOS or a single PMOS on its own can be used as a CMOS switch, but the combination of the two transistors in parallel has some advantages. An FET channel is resistive so the ONresistances of both transistors are effectively connected in parallel.
As a FETs Onresistance is a function of the gatetosource voltage, V_{GS}, as one transistor becomes less conducting due to the gate drive, the other transistor takes over and becomes more conducting. Thus the combined value of the two ONresistances (as low as 2 or 3Ω) stays more or less constant than would be the case for a single switching transistor on its own.
When can demonstrate this in the following diagram.
We have seen here that connecting a Pchannel FET (PMOS) with an Nchannel FET (NMOS) we can create a solidstate switch which is digitally controlled using logic level voltages and is commonly called a “transmission gate”.
The Transmission Gate, (TG) is a bilateral switch where either of its terminals can be the input or the output. As well as the input and output terminals, the transmission gate has a third connection called the control, where the control input determines the switching state of the gate as an open or closed (NO/NC) switch.
This input is typically driven by a digital logic signal that toggles between ground (0V) and a set DC voltage, usually VDD. When the control input is low (Cont. = 0), the switch is open, and when the control input is HIGH Cont. = 1) the switch is closed.
Transmission gates act like voltagecontrolled switches, and being switches, CMOS transmission gates can be used for switching both analogue and digital signals passing the full range of voltages (from 0V to V_{DD}) in either direction, which as discussed is not possible with a single MOS device.
The combination of an NMOS and a PMOS transistor together within a single gate means that the NMOS transistor will transfer a good logic “0” but a poor logic “1”, while the PMOS transistor transfers a good logic “1” but a poor logic “0”. Therefore, connecting an NMOS transistor with a PMOS transistor in parallel provides a single bilateral switch which offers efficient output drive capability for CMOS logic gates controlled by a single input logic level.
]]>Thus their name “transceiver” is a portmanteau word coming from the amalgamation of the two words transmitter and receive (transmitter/receiver). Transceivers are also known by the names of: send/receive or driver/receiver devices.
In the Digital Buffer Tutorial, we saw that a buffer performs no inversion or decision making capabilities, unlike digital logic gates with two or more inputs, but instead produces an output condition which matches exactly that of its input. Thus a buffer is a “noninverting” device producing the Boolean expression of: Q = A.
A Digital Buffer like the one shown on the left, is a unidirectional device, that is the signal passes through them in one direction only, from input “A” to the output at “Q“.
Thus, when input A is at logic “1”, output Q is at logic “1”, and when input A is at logic “0”, output Q is at logic “0” for a positive logic device such as the CMOS 74HC4050 Hex Buffer Gate.
Buffers can be used to isolate other gates or circuit stages from each other preventing the impedance or operation of one circuit from affecting the impedance or operation of another. Also on their own, buffers can be used as drivers for high current loads such as transistor switches because their output drive capability (fanout) is generally much higher than their input signal requirements. For example, the TTL 74LS07 Hex buffer/driver with open collector, highvoltage (30 volts) outputs.
The digital noninverting buffer function can also be made using spare logic AND, or logic OR gates or by using pairs of NOT gates (inverters) as shown.
One of the disadvantages of a single input digital buffer is that the output at Q will always be at the same logic level as the input possibly affecting whatever circuit or device is connected to the buffers output terminal. One way to overcome this is to turn the basic buffer into a 3State Buffer, more commonly known as a Tristate Buffer.
A Tristate Buffer is another type of buffer circuit which can be used to control the passage of a logic signal from its input to its output. The tristate buffer is a combinational device whose output can be electronically turned “ON” or “OFF” by means of an external “Control” or “Enable” (EN) signal input allowing them to be used in busorientated systems.
As their name implies, the output at “Q” for a Tristate Buffer can take on one of three possible states, logic “0”, logic “1”, and HighZ (high impedance), that is, an open circuit, rather than the standard “0” and “1” states.
The buffers enable or control signal can be either a logic “0” or a logic “1” level signal with the output being inverting and noninverting as the digital signal passes through it. The two most commonly used tristate buffer IC’s being the TTL 74LS125 and the TTL 74LS126.
Thus a tristate buffer requires two inputs. One being the data input (A) and the other being the control or Enable input (EN) as shown.
The tristate buffer’s symbol is very similar to the standard buffer symbol above but with the addition of a second input representing the enable/disable control function. When the enable (EN) input is at a logic level “1” (for positive logic), it acts as a normal buffer allowing the input signal, A to pass directly to the output at Q. Whether it is a logic “0” or a logic “1”.
When the enable input is at logic “0”, the tristate buffer is activated into its third state and disables or turns “OFF” its output producing an open circuit condition. This third condition is neither at a logic “1” (high) or logic “0” (low), but instead gives an output state that is at a very high impedance, HighZ, more commonly written as: HiZ.
Thus a tristate buffer has two logic state inputs, “0” or a “1” but can produce three different output states, “0”, “1” or “HiZ” which is why it is called a “Tri” or “3state” device. Note that this third state is NOT equal to a logic level “0” or a “1”, but is an high impedance state as its output is electrically disconnected.
Then we can correctly state for a positively enabled tristate buffer that:
and we can present the truth table for a tristate buffer as:
Symbol  Truth Table  
Tristate Buffer

Enable  IN  OUT 
0  0  HiZ  
0  1  HiZ  
1  0  0  
1  1  1 
Tristate Buffers are available in integrated form as quad, hex or octal buffer/drivers such as the TTL 74LS244 as shown.
Notice that the eight buffers are configured into two groups of four with the first group (A1 to A4) being controlled by enable input, CA, and the second group (A5 to A8) being controlled by the enable input, CB. The 74LS244 has very high sink and source current capabilities if required to switch transistor loads.
So what can we use a 3state or tristate buffer for. Tristate buffers can allow multiple devices to share a common output wire or bus by having only one tristate device drive the wire bus at any one time while all other buffers remain in their HiZ state. Consider the circuit below.
The outputs from each tristate buffer are connected to a common wire bus but their enable inputs are connected to a binary decoder. The decoder guarantees that only one tristate buffer will be active at any one time, due to its enable signal. This allows the data of the active buffer to pass directly onto the common bus while the outputs of the other nonenabled buffers are effectively disconnected and in their highimpedance state. Thus which buffer is connected to the common line will depending on the binary value of the decoders select inputs.
Therefore, no more than one tristate buffer can be in an “active state” at any given time. You may have noticed that the possible combination of different data inputs connected to a single output line above resembles that of a 4to1 line multiplexer, and you would be right, multiplexer circuits can be easily constructed using tristate buffers.
Any tristate buffer element can easily be converted into a normal digital buffer by simply connecting their enable (EN) input directly to +Vcc or ground, depending on the tristate buffer used. Thus, the output is permanently enabled so any input signal present at “A” will pass straight through the buffer to the output at “Q“.
We have seen thus far, that we can use tristate buffers to send information in a unidirectional way onto a common wire or bus. But how could we use them to send data in both directions, that is, to send data too and receive data from a common wire bus.
It is also possible to connect Tristate Buffers “backtoback” (inverse parallel) to produce what is called a Bidirectional Buffer or transceiver circuit. By using an additional inverter, one tristate buffer is as an “activehigh buffer”, while the other operates as an “activelow buffer”, as shown.
Here, the two tristate buffers are connected in parallel but in reverse from “A” to “B” with the enable control input, EN acting more like a directional control signal thus allowing data to be both read “from” and transmitted “to” the same data terminal.
So in this simple example, when the enable input is HIGH, (EN equals logic “1”) data is allowed to pass from A to B via buffer 1, and when the enable input is LOW, (EN equals logic “0”) data passes from B to A via buffer 2.
Thus the enable input “EN” acts as direction control allowing data to flow in either direction depending upon the logic status of this control input. In this type of application a tristate buffer with bidirectional switching capability such as the TTL 74LS245 or the inverting CMOS 74ALS620 can be used producing what is called a Bus Transceiver.
Bus transceivers are tristate bidirectional devices which allow the flow of data between two points making them compatible with busoriented systems or the bidirectional (input or output) control of interface circuitry. Bus transceivers can be inverting, the TTL 74LS242 or noninverting, the TTL 74LS243 devices.
Thus we can use an 8line octal transceiver to interface any input/output device to an 8bit data bus with the most common bus transceiver IC being used to both send and receive data is the TTL 74LS245 given below.
The TTL 74LS245 is an octal bus transceiver (Transmitter/Receiver) designed for asynchronous twoway communication between two data buses or input/output device. The transceiver allows for the transmission of data from the terminals A to terminals B or the reverse depending on the logic level at the directioncontrol (DIR) input, (pin 1).
So for example, if the directioncontrol input is HIGH at logic level “1”, then data will pass from terminals A to terminals B. If the directioncontrol input is LOW at logic level “0”, then data will pass in the reverse direction from terminals B to terminals A. When held HIGH at logic level “1”, the output chipenable (CE) input, (pin 19) can be used to disable the device so that the terminals, and therefore any connected data buses are effectively isolated from each other.
]]>But as well as each digit being ten times bigger than the previous number as we move from righttoleft, each digit can also be ten times smaller than its neighbouring number as we move along in the opposite direction from lefttoright.
However, once we reach zero (0) and the decimal point, we do not need to just stop, but can continue moving from lefttoright along the digits producing what are generally called Fractional Numbers.
Here in this decimal (or denary) number example, the digit immediately to the right of the decimal point (number 5) is worth one tenth (1/10 or 0.1) of the digit immediately to the left of the decimal point (number 4) which as a multiplication value of one (1).
Thus as we move through the number from lefttoright, each subsequent digit will be one tenth the value of the digit immediately to its left position, and so on.
Then the decimal numbering system uses the concept of positional or relative weighting values producing a positional notation, where each digit represents a different weighted value depending on the position occupied either side of the decimal point.
Thus mathematically in the standard denary numbering system, these values are commonly written as: 4^{0}, 3^{1}, 2^{2}, 1^{3} for each position to the left of the decimal point in our example above. Likewise, for the fractional numbers to right of the decimal point, the weight of the number becomes more negative giving: 5^{1}, 6^{2}, 7^{3} etc.
So we can see that each digit in the standard decimal system indicates the magnitude or weight of that digit within the number. Then the value of any decimal number will be equal to the sum of its digits multiplied by their respective weights, so for our example above: N = 1234.567_{10} in the weighted decimal format this will be equal too:
1000 + 200 + 30 + 4 + 0.5 + 0.06 + 0.007 = 1234.567_{10}
or it could be written to reflect the weighting of each denary digit:
(1×1000) + (2×100) + (3×10) + (4×1) + (5×0.1) + (6×0.01) + (7×0.001) = 1234.567_{10}
or even in polynomial form as:
(1×10^{3}) + (2×10^{2}) + (3×10^{1}) + (4×10^{0}) + (5×10^{1}) + (6×10^{2}) + (7×10^{3}) = 1234.567_{10}
We can also use this idea of positional notation where each digit represents a different weighted value depending upon the position it occupies in the binary numbering system. The difference this time is that the binary number system (or simply binary numbers) is a positional system, where the different weighted positions of the digits are to the power of 2 (base2) instead of 10.
The binary numbering system is a base2 numbering system which contains only two digits, a “0” or a “1”. Thus each digit of a binary number can take the “0” or the “1” value with the position of the 0 or 1 indicating its value or weighting. But we can also have binary weighting for values of less than 1 producing what are called unsigned fractional binary numbers.
Similar to decimal fractions, binary numbers can also be represented as unsigned fractional numbers by placing the binary digits to the right of the decimal point or in this case, binary point. Thus all the fractional digits to the right of the binary point have respective weightings which are negative powers of two, creating a binary fraction. In other words, the powers of 2 are negative.
So for the fractional binary numbers to the right of the binary point, the weight of each digit becomes more negative giving: 2^{1}, 2^{2}, 2^{3}, 2^{4}, and so on as shown.
etc, etc.
Thus if we take the binary fraction of 0.1011_{2} then the positional weights for each of the digits is taken into account giving its decimal equivalent of:
For this example, the decimal fraction conversion of the binary number 0.1011_{2} is 0.6875_{10}.
Now lets suppose we have the following binary number of: 1101.0111_{2}, what will be its decimal number equivalent.
1101.0111 = (1×2^{3}) + (1×2^{2}) + (0×2^{1}) + (1×2^{0}) + (0×2^{1}) + (1×2^{2}) + (1×2^{3}) + (1×2^{4})
= 8 + 4 + 0 + 1 + 0 + 1/4 + 1/8 + 1/16
= 8 + 4 + 0 + 1 + 0 + 0.25 + 0.125 + 0.0625 = 13.4375_{10}
Hence the decimal equivalent number of 1101.0111_{2} is given as: 13.4375_{10}
So we can see that fractional binary numbers, that is binary numbers that have a weighting of less than 1 (2^{0}), can be converted into their decimal number equivalent by successively dividing the binary weighting factor by the value of two for each decrease in the power of 2, remembering also that 2^{0} is equal to 1, and not zero.
0.11 = (1×2^{1}) + (1×2^{2}) = 0.5 + 0.25 = 0.75_{10}
11.001 = (1×2^{1}) + (1×2^{0}) + (1×2^{3}) = 2 + 1 + 0.125 = 3.125_{10}
1011.111 = (1×2^{3}) + (1×2^{1}) + (1×2^{0}) (1×2^{1}) + (1×2^{2}) + (1×2^{3})
= 8 + 2 + 1 + 0.5 + 0.25 + 0.125 = 11.875_{10}
The conversion of a decimal fraction to a fractional binary number is achieved using a method similar to that we used for integers. However, this time multiplication is used instead of division with the integers instead of remainders used with the carry digit being the binary equivalent of the fractional part of the decimal number.
When converting from decimal to binary, the integer (positive sequence righttoleft) part and the fractional (negative sequence from lefttoright) part of the decimal number are calculated separately.
For the integer part of the number, the binary equivalent is found by successively dividing (known as successive division) the integer part of the decimal number repeatedly by 2 (÷2), noting the remainders in reverse order from the least significant bit (LSB) to the most significant bit (MSB), until the value becomes “0” producing the binary equivalent.
So to find the binary equivalent of the decimal integer: 118_{10}
118 (divide by 2) = 59 plus remainder 0 (LSB)
59 (divide by 2) = 29 plus remainder 1 (↑)
29 (divide by 2) = 14 plus remainder 1 (↑)
14 (divide by 2) = 7 plus remainder 0 (↑)
7 (divide by 2) = 3 plus remainder 1 (↑)
3 (divide by 2) = 1 plus remainder 1 (↑)
1 (divide by 2) = 0 plus remainder 1 (MSB)
Then the binary equivalent of 118_{10} is therefore: 1110110_{2} ← (LSB)
The fractional part of the number is found by successively multiplying (known as successive multiplication) the given fractional part of the decimal number repeatedly by 2 (×2), noting the carries in forward order, until the value becomes “0” producing the binary equivalent.
So if the multiplication process produces a product greater than 1, the carry is a “1” and if the multiplication process produces a product less than “1”, the carry is a “0”.
Note also that if the successive multiplication processes does not seem to be heading towards a final zero, the fractional number will have an infinite length or until the equivalent number of bits have been obtained, for example 8bits. or 16bits, etc. depending on the degree of accuracy required.
So to find the binary fraction equivalent of the decimal fraction: 0.8125_{10}
0.8125 (multiply by 2) = 1.625 = 0.625 carry 1 (MSB)
0.625 (multiply by 2) = 1.25 = 0.25 carry 1 (↓)
0.25 (multiply by 2) = 0.50 = 0.5 carry 0 (↓)
0.5 (multiply by 2) = 1.00 = 0.0 carry 1 (LSB)
Thus the binary equivalent of 0.8125_{10} is therefore: 0.1101_{2} ← (LSB)
We can double check this answer using the procedure above to convert a binary fraction into a decimal number equivalent: 0.1101 = 0.5 + 0.25 + 0.0625 = 0.8125_{10}
Find the binary fraction equivalent of the following decimal number: 54.6875
First we convert the integer 54 to a binary number in the normal way using successive division from above.
54 (divide by 2) = 27 remainder 0 (LSB)
27 (divide by 2) = 13 remainder 1 (↑)
13 (divide by 2) = 6 remainder 1 (↑)
6 (divide by 2) = 3 remainder 0 (↑)
3 (divide by 2) = 1 remainder 1 (↑)
1 (divide by 2) = 0 remainder 1 (MSB)
Thus the binary equivalent of 54_{10} is therefore: 110110_{2}
Next we convert the decimal fraction 0.6875 to a binary fraction using successive multiplication.
0.6875 (multiply by 2) = 1.375 = 0.375 carry 1 (MSB)
0.375 (multiply by 2) = 0.75 = 0.75 carry 0 (↓)
0.75 (multiply by 2) = 1.50 = 0.5 carry 1 (↓)
0.5 (multiply by 2) = 1.00 = 0.0 carry 1 (LSB)
Thus the binary equivalent of 0.6875_{10} is therefore: 0.1011_{2} ← (LSB)
Hence the binary equivalent of the decimal number: 54.6875_{10} is 110110.1011_{2}
We have seen here in this tutorial about Binary Fractions that to convert any decimal fraction into its equivalent binary fraction, we must multiply the decimal fractional part, and only the decimal fractional part by 2 and record the digit that appears to the left of the binary point. This binary digit which is the carry digit will ALWAYS be either a “0” or a “1”.
We must then multiply the remaining decimal fraction by 2 again repeating the above sequence using successive multiplication until the fraction is reduced to zero or the required amount of binary bits has been completed for a repeating binary fraction. Fractional numbers are represented by negative powers of 2.
For mixed decimal numbers we must perform two separate operations. Successive division for the integer part to the left of the decimal point and successive multiplication for the fractional part to the right of the decimal point.
Note that the integer part of a mixed decimal number will always have an exact binary number equivalent but the decimal fractional part may not, since we could get a repeating fraction resulting in an infinite number of binary digits if we wanted to represent the decimal fraction exactly.
]]>In other words the algebraic sum of ALL the potential differences around the loop must be equal to zero as: ΣV = 0. Note here that the term “algebraic sum” means to take into account the polarities and signs of the sources and voltage drops around the loop.
This idea by Kirchhoff is commonly known as the Conservation of Energy, as moving around a closed loop, or circuit, you will end up back to where you started in the circuit and therefore back to the same initial potential with no loss of voltage around the loop. Hence any voltage drops around the loop must be equal to any voltage sources met along the way.
So when applying Kirchhoff’s voltage law to a specific circuit element, it is important that we pay special attention to the algebraic signs, (+ and ) of the voltage drops across elements and the emf’s of sources otherwise our calculations may be wrong.
But before we look more closely at Kirchhoff’s voltage law (KVL) lets first understand the voltage drop across a single element such as a resistor.
For this simple example we will assume that the current, I is in the same direction as the flow of positive charge, that is conventional current flow.
Here the flow of current through the resistor is from point A to point B, that is from positive terminal to a negative terminal. Thus as we are travelling in the same direction as current flow, there will be a fall in potential across the resistive element giving rise to a IR voltage drop across it.
If the flow of current was in the opposite direction from point B to point A, then there would be a rise in potential across the resistive element as we are moving from a  potential to a + potential giving us a +IR voltage drop.
Thus to apply Kirchhoff’s voltage law correctly to a circuit, we must first understand the direction of the polarity and as we can see, the sign of the voltage drop across the resistive element will depend on the direction of the current flowing through it. As a general rule, you will loose potential in the same direction of current across an element and gain potential as you move in the direction of an emf source.
The direction of current flow around a closed circuit can be assumed to be either clockwise or anticlockwise and either one can be chosen. If the direction chosen is different from the actual direction of current flow, the result will still be correct and valid but will result in the algebraic answer having a minus sign.
To understand this idea a little more, lets look at a single circuit loop to see if Kirchhoff’s Voltage Law holds true.
Kirchhoff’s voltage law states that the algebraic sum of the potential differences in any loop must be equal to zero as: ΣV = 0. Since the two resistors, R_{1} and R_{2} are wired together in a series connection, they are both part of the same loop so the same current must flow through each resistor.
Thus the voltage drop across resistor, R_{1} = I*R_{1} and the voltage drop across resistor, R_{2} = I*R_{2} giving by KVL:
We can see that applying Kirchhoff’s Voltage Law to this single closed loop produces the formula for the equivalent or total resistance in the series circuit and we can expand on this to find the values of the voltage drops around the loop.
Three resistor of values: 10 ohms, 20 ohms and 30 ohms, respectively are connected in series across a 12 volt battery supply. Calculate: a) the total resistance, b) the circuit current, c) the current through each resistor, d) the voltage drop across each resistor, e) verify that Kirchhoff’s voltage law, KVL holds true.
R_{T} = R_{1} + R_{2} + R_{3} = 10Ω + 20Ω + 30Ω = 60Ω
Then the total circuit resistance R_{T} is equal to 60Ω
Thus the total circuit current I is equal to 0.2 amperes or 200mA
The resistors are wired together in series, they are all part of the same loop and therefore each experience the same amount of current. Thus:
I_{R1} = I_{R2} = I_{R3} = I_{SERIES} = 0.2 amperes
V_{R1} = I x R_{1} = 0.2 x 10 = 2 volts
V_{R2} = I x R_{2} = 0.2 x 20 = 4 volts
V_{R3} = I x R_{3} = 0.2 x 30 = 6 volts
Thus Kirchhoff’s voltage law holds true as the individual voltage drops around the closed loop add up to the total.
We have seen here that Kirchhoff’s voltage law, KVL is Kirchhoff’s second law and states that the algebraic sum of all the voltage drops, as you go around a closed circuit from some fixed point and return back to the same point, and taking polarity into account, is always zero. That is ΣV = 0
The theory behind Kirchhoff’s second law is also known as the law of conservation of voltage, and this is particularly useful for us when dealing with series circuits, as series circuits also act as voltage dividers and the voltage divider circuit is an important application of many series circuits.
]]>Gustav Kirchhoff’s Current Law is one of the fundamental laws used for circuit analysis. His current law states that for a parallel path the total current entering a circuits junction is exactly equal to the total current leaving the same junction. This is because it has no other place to go as no charge is lost.
In other words the algebraic sum of ALL the currents entering and leaving a junction must be equal to zero as: Σ I_{IN} = Σ I_{OUT}.
This idea by Kirchhoff is commonly known as the Conservation of Charge, as the current is conserved around the junction with no loss of current. Lets look at a simple example of Kirchhoff’s current law (KCL) when applied to a single junction.
Here in this simple single junction example, the current I_{T} leaving the junction is the algebraic sum of the two currents, I_{1} and I_{2} entering the same junction. That is I_{T} = I_{1} + I_{2}.
Note that we could also write this correctly as the algebraic sum of: I_{T}  (I_{1} + I_{2}) = 0.
So if I_{1} equals 3 amperes and I_{2} is equal to 2 amperes, then the total current, I_{T} leaving the junction will be 3 + 2 = 5 amperes, and we can use this basic law for any number of junctions or nodes as the sum of the currents both entering and leaving will be the same.
Also, if we reversed the directions of the currents, the resulting equations would still hold true for I_{1} or I_{2}. As I_{1} = I_{T}  I_{2} = 5  2 = 3 amps, and I_{2} = I_{T}  I_{1} = 5  3 = 2 amps. Thus we can think of the currents entering the junction as being positive (+), while the ones leaving the junction as being negative ().
Then we can see that the mathematical sum of the currents either entering or leaving the junction and in whatever direction will always be equal to zero, and this forms the basis of Kirchhoff’s Junction Rule, more commonly known as Kirchhoff’s Current Law, or (KCL).
Let’s look how we could apply Kirchhoff’s current law to resistors in parallel, whether the resistances in those branches are equal or unequal. Consider the following circuit diagram:
In this simple parallel resistor example there are two distinct junctions for current. Junction one occurs at node B, and junction two occurs at node E. Thus we can use Kirchhoff’s Junction Rule for the electrical currents at both of these two distinct junctions, for those currents entering the junction and for those currents flowing leaving the junction.
To start, all the current, I_{T} leaves the 24 volt supply and arrives at point A and from there it enters node B. Node B is a junction as the current can now split into two distinct directions, with some of the current flowing downwards and through resistor R_{1} with the remainder continuing on through resistor R_{2} via node C. Note that the currents flowing into and out of a node point are commonly called branch currents.
We can use Ohm’s Law to determine the individual branch currents through each resistor as: I = V/R, thus:
For current branch B to E through resistor R_{1}
For current branch C to D through resistor R_{2}
From above we know that Kirchhoff’s current law states that the sum of the currents entering a junction must equal the sum of the currents leaving the junction, and in our simple example above, there is one current, I_{T} going into the junction at node B and two currents leaving the junction, I_{1} and I_{2}.
Since we now know from calculation that the currents leaving the junction at node B is I_{1} = 3 amps and I_{2} equals 2 amps, the sum of the currents entering the junction at node B must equal 3 + 2 = 5 amps. Thus Σ_{IN} = I_{T} = 5 amperes.
In our example, we have two distinct junctions at node B and node E, thus we can confirm this value for I_{T} as the two currents recombine again at node E. So, for Kirchhoff’s junction rule to hold true, the sum of the currents into point F must equal the sum of the currents flowing out of the junction at node E.
As the two currents entering junction E are 3 amps and 2 amps respectively, the sum of the currents entering point F is therefore: 3 + 2 = 5 amperes. Thus Σ_{IN} = I_{T} = 5 amperes and therefore Kirchhoff’s current law holds true as this is the same value as the current leaving point A.
We can use Kirchhoff’s current law to find the currents flowing around more complex circuits. We hopefully know by now that the algebraic sum of all the currents at a node (junction point) is equal to zero and with this idea in mind, it is a simple case of determining the currents entering a node and those leaving the node. Consider the circuit below.
In this example there are four distinct junctions for current to either separate or merge together at nodes A, C, E and node F. The supply current I_{T} separates at node A flowing through resistors R_{1} and R_{2}, recombining at node C before separating again through resistors R_{3}, R_{4} and R_{5} and finally recombining once again at node F.
But before we can calculate the individual currents flowing through each resistor branch, we must first calculate the circuits total current, I_{T}. Ohms law tells us that I = V/R and as we know the value of V, 132 volts, we need to calculate the circuit resistances as follows.
Thus the equivalent circuit resistance between nodes A and C is calculated as 1 Ohm.
Thus the equivalent circuit resistance between nodes C and F is calculated as 10 Ohms. Then the total circuit current, I_{T} is given as:
Giving us an equivalent circuit of:
Therefore, V = 132V, R_{AC} = 1Ω, R_{CF} = 10Ω’s and I_{T} = 12A.
Having established the equivalent parallel resistances and supply current, we can now calculate the individual branch currents and confirm using Kirchhoff’s junction rule as follows.
Thus, I_{1} = 5A, I_{2} = 7A, I_{3} = 2A, I_{4} = 6A, and I_{5} = 4A.
We can confirm that Kirchoff’s current law holds true around the circuit by using node C as our reference point to calculate the currents entering and leaving the junction as:
We can also double check to see if Kirchhoffs Current Law holds true as the currents entering the junction are positive, while the ones leaving the junction are negative, thus the algebraic sum is: I_{1} + I_{2}  I_{3}  I_{4}  I_{5} = 0 which equals 5 + 7 – 2 – 6 – 4 = 0.
So we can confirm by analysis that Kirchhoff’s current law (KCL) which states that the algebraic sum of the currents at a junction point in a circuit network is always zero is true and correct in this example.
Find the currents flowing around the following circuit using Kirchhoff’s Current Law only.
I_{T} is the total current flowing around the circuit driven by the 12V supply voltage. At point A, I_{1} is equal to I_{T}, thus there will be an I_{1}*R voltage drop across resistor R_{1}.
The circuit has 2 branches, 3 nodes (B, C and D) and 2 independent loops, thus the I*R voltage drops around the two loops will be:
Since Kirchhoff’s current law states that at node B, I_{1} = I_{2} + I_{3}, we can therefore substitute current I_{1} for (I_{2} + I_{3}) in both of the following loop equations and then simplify.
We now have two simultaneous equations that relate to the currents flowing around the circuit.
Eq. No 1 : 12 = 10I_{2} + 4I_{3}
Eq. No 2 : 12 = 4I_{2} + 16I_{3}
By multiplying the first equation (Loop ABC) by 4 and subtracting Loop ABD from Loop ABC, we can be reduced both equations to give us the values of I_{2} and I_{3}
Eq. No 1 : 12 = 10I_{2} + 4I_{3} ( x4 ) ⇒ 48 = 40I_{2} + 16I_{3}
Eq. No 2 : 12 = 4I_{2} + 16I_{3} ( x1 ) ⇒ 12 = 4I_{2} + 16I_{3}
Eq. No 1 – Eq. No 2 ⇒ 36 = 36I_{2} + 0
Substitution of I_{2} in terms of I_{3} gives us the value of I_{2} as 1.0 Amps
Now we can do the same procedure to find the value of I_{3} by multiplying the first equation (Loop ABC) by 4 and the second equation (Loop ABD) by 10. Again by subtracting Loop ABC from Loop ABD, we can be reduced both equations to give us the values of I_{2} and I_{3}
Eq. No 1 : 12 = 10I_{2} + 4I_{3} ( x4 ) ⇒ 48 = 40I_{2} + 16I_{3}
Eq. No 2 : 12 = 4I_{2} + 16I_{3} ( x10 ) ⇒ 120 = 40I_{2} + 160I_{3}
Eq. No 2 – Eq. No 1 ⇒ 72 = 0 + 144I_{3}
Thus substitution of I_{3} in terms of I_{2} gives us the value of I_{3} as 0.5 Amps
As Kirchhoff’s junction rule states that : I_{1} = I_{2} + I_{3}
The supply current flowing through resistor R_{1} is given as : 1.0 + 0.5 = 1.5 Amps
Thus I_{1} = I_{T} = 1.5 Amps, I_{2} = 1.0 Amps and I_{3} = 0.5 Amps and from that information we could calculate the I*R voltage drops across the devices and at the various points (nodes) around the circuit.
We could have solved the circuit of example two simply and easily just using Ohm’s Law, but we have used Kirchhoff’s Current Law here to show how it is possible to solve more complex circuits when we can not just simply apply Ohm’s Law.
]]>We always measure electrical resistance in Ohms, where Ohms is denoted by the Greek letter Omega, Ω. So for example: 50Ω, 10kΩ or 4.7MΩ, etc. Conductors (e.g. wires and cables) generally have very low values of resistance (less than 0.1Ω) and thus we can neglect them as we assume in circuit analysis calculations that wires have zero resistance. Insulators (e.g. plastic or air) on the other hand generally have very high values of resistance (greater than 50MΩ) and we can ignore them also for circuit analysis as their value is too high.
But the electrical resistance between two points can depend on many factors such as the conductors length, its crosssectional area, the temperature, as well as the actual material from which it is made. For example, let’s assume we have a piece of wire (a conductor) that has a length L, a crosssectional area A and a resistance R as shown.
The electrical resistance, R of this simple conductor is a function of its length, L and the conductors area, A. Ohms law tells us that for a given resistance R, the current flowing through the conductor is proportional to the applied voltage as I = V/R. Now suppose we connect two identical conductors together in a series combination as shown.
Here by connecting the two conductors together in a series combination, that is end to end, we have effectively doubled the total length of the conductor (2L), while the crosssectional area, A remains exactly the same as before. But as well as doubling the length, we have also doubled the total resistance of the conductor, giving 2R as: 1R + 1R = 2R.
Therefore we can see that the resistance of the conductor is proportional to its length, that is: R ∝ L. In other words, we would expect the electrical resistance of a conductor (or wire) to be proportionally greater the longer it is.
Note also that by doubling the length and therefore the resistance of the conductor (2R), to force the same current, i to flow through the conductor as before, we need to double (increase) the applied voltage as now I = (2V)/(2R). Next suppose we connect the two identical conductors together in parallel combination as shown.
Here by connecting the two conductors together in a parallel combination, we have effectively doubled the total area giving 2A, while the conductors length, L remains the same as the original single conductor. But as well as doubling the area, by connecting the two conductors together in parallel we have effectively halved the total resistance of the conductor, giving 1/2R as now each half of the current flows through each conductor branch.
Thus the resistance of the conductor is inversely proportional to its area, that is: R 1/∝ A, or R ∝ 1/A. In other words, we would expect the electrical resistance of a conductor (or wire) to be proportionally less the greater is its crosssectional area.
Also by doubling the area and therefore halving the total resistance of the conductor branch (1/2R), for the same current, i to flow through the parallel conductor branch as before we only need half (decrease) the applied voltage as now I = (1/2V)/(1/2R).
So hopefully we can see that the resistance of a conductor is directly proportional to the length (L) of the conductor, that is: R ∝ L, and inversely proportional to its area (A), R ∝ 1/A. Thus we can correctly say that resistance is:
But as well as length and conductor area, we would also expect the electrical resistance of the conductor to depend upon the actual material from which it is made, because different conductive materials, copper, silver, aluminium, etc all have different physical and electrical properties. Thus we can convert the proportionality sign (∝) of the above equation into an equals sign simply by adding a “proportional constant” into the above equation giving:
Where: R is the resistance in ohms (Ω), L is the length in metres (m), A is the area in square metres (m^{2}), and where the proportional constant ρ (the Greek letter “rho”) is known as Resistivity.
The electrical resistivity of a particular conductor material is a measure of how strongly the material opposes the flow of electric current through it. This resistivity factor, sometimes called its “specific electrical resistance”, enables the resistance of different types of conductors to be compared to one another at a specified temperature according to their physical properties without regards to their lengths or crosssectional areas. Thus the higher the resistivity value of ρ the more resistance and vice versa.
For example, the resistivity of a good conductor such as copper is on the order of 1.72 x 10^{8} ohm metre (or 17.2 nΩm), whereas the resistivity of a poor conductor (insulator) such as air can be well over 1.5 x 10^{14} or 150 trillion Ωm.
Materials such as copper and aluminium are known for their low levels of resistivity thus allowing electrical current to easily flow through them making these materials ideal for making electrical wires and cables. Silver and gold have much low resistivity values, but for obvious reasons are more expensive to turn into electrical wires.
Then the factors which affect the resistance (R) of a conductor in ohms can be listed as:
Calculate the total DC resistance of a 100 metre roll of 2.5mm^{2} copper wire if the resistivity of copper at 20^{o}C is 1.72 x 10^{8} Ω metre.
Given: resistivity of copper at 20^{o}C is 1.72 x 10^{8}, coil length L = 100m, the crosssectional area of the conductor is 2.5mm^{2} giving an area of: A = 2.5 x 10^{6} metres^{2}.
That is 688 milliohms or 0.688 Ohms.
We said previously that resistivity is the electrical resistance per unit length and per unit of conductor crosssectional area thus showing that resistivity, ρ has the dimensions of ohms metre, or Ω.m as it is commonly written. Thus for a particular material at a specified temperature its electrical resistivity is given as.
While both the electrical resistance (R) and resistivity (or specific resistance) ρ, are a function of the physical nature of the material being used, and of its physical shape and size expressed by its length (L), and its sectional area (A), Conductivity, or specific conductance relates to the ease at which electric current con flow through a material.
Conductance (G) is the reciprocal of resistance (1/R) with the unit of conductance being the siemens (S) and is given the upside down ohms symbol mho, ℧. Thus when a conductor has a conductance of 1 siemens (1S) it has a resistance is 1 ohm (1Ω). So if its resistance is doubled, the conductance halves, and viceversa as: siemens = 1/ohms, or ohms = 1/siemens.
While a conductors resistance gives the amount of opposition it offers to the flow of electric current, the conductance of a conductor indicates the ease by which it allows electric current to flow. So metals such as copper, aluminium or silver have very large values of conductance meaning that they are good conductors.
Conductivity, σ (Greek letter sigma), is the reciprocal of the resistivity. That is 1/ρ and is measured in siemens per metre (S/m). Since electrical conductivity σ = 1/ρ, the previous expression for electrical resistance, R can be rewritten as:
Then we can say that conductivity is the efficiency by which a conductor passes an electric current or signal without resistive loss. Therefore a material or conductor that has a high conductivity will have a low resistivity, and vice versa, since 1 siemens (S) equals 1Ω^{1}. So copper which is a good conductor of electric current, has a conductivity of 58.14 x 10^{6} siemens per metre.
A 20 metre length of cable has a crosssectional area of 1mm^{2} and a resistance of 5 ohms. Calculate the conductivity of the cable.
Given: DC resistance, R = 5 ohms, cable length, L = 20m, and the crosssectional area of the conductor is 1mm^{2} giving an area of: A = 1 x 10^{6} metres^{2}.
That is 4 megasiemens per metre length.
We have seen in this tutorial about resistivity, that resistivity is the property of a material or conductor that indicates how well the material conducts electrical current. We have also seen that the electrical resistance (R) of a conductor depends not only on the material from which the conductor is made from, copper, silver, aluminium, etc. but also on its physical dimensions.
The resistance of a conductor is directly proportional to its length (L) as R ∝ L. Thus doubling its length will double its resistance, while halving its length would halve its resistance. Also the resistance of a conductor is inversely proportional to its crosssectional area (A) as R ∝ 1/A. Thus doubling its crosssectional area would halve its resistance, while halving its crosssectional area would double its resistance.
We have also learnt that the resistivity (symbol: ρ) of the conductor (or material) relates to the physical property from which it is made and varies from material to material. For example, the resistivity of copper is generally given as: 1.72 x 10^{8} Ω.m. The resistivity of a particular material is measured in units of OhmMetres (Ω.m) which is also affected by temperature.
Depending upon the electrical resistivity value of a particular material, it can be classified as being either a “conductor”, an “insulator” or a “semiconductor”. Note that semiconductors are materials where its conductivity is dependent upon the impurities added to the material.
Resistivity is also important in power distribution systems as the effectiveness of the earth grounding system for an electrical power and distribution system greatly depends on the resistivity of the earth and soil material at the location of the system ground.
Conduction is the name given to the movement of free electrons in the form of an electric current. Conductivity, σ is the reciprocal of the resistivity. That is 1/ρ and has the unit of siemens per metre, S/m. Conductivity ranges from zero (for a perfect insulator) to infinity (for a perfect conductor). Thus a super conductor has infinite conductance and virtually zero ohmic resistance.
]]>A thermistor is basically a twoterminal solid state thermally sensitive transducer made from sensitive semiconductor based metal oxides with metallised or sintered connecting leads onto a ceramic disc or bead. This allows it to change its resistive value in proportion to small changes in temperature. In other words, as its temperature changes, so too does its resistance and as such its name, “Thermistor” is a combination of the words THERMally sensitive resISTOR.
While the change in resistance due to heat is generally undesirable in standard resistors, this effect can be put to good use in many temperature detection circuits. Thus being nonlinear variableresistance devices, thermistors are commonly used as temperature sensors having many applications to measure the temperature of both liquids and ambient air.
Also, being a solid state device made from highly sensitive metal oxides, they operate at the molecular level with the outermost (valence) electrons becoming more active and producing a negative temperature coefficient, or less active producing a positive temperature coefficient as the temperature of the thermistor is increased. This means that they can have very good reproducible resistance verses temperature characteristics allowing them to operate up to temperatures of about 200^{o}C.
Typical Thermistor
While the primarily used of thermistors are as resistive temperature sensors, being resistive devices belonging the the resistor family, they can also be used in series with a component or device to control the current flowing through them. In other words, they can also be used as currentlimiting devices.
Thermistors are available in a range of types, materials and sizes depending on the response time and operating temperature. Also, hermetically sealed thermistors eliminate errors in resistance readings due to moisture penetration while offering high operating temperatures and a compact size. The three most common types are: Bead thermistors, Disk thermistors, and Glass encapsulated thermistors.
These heatdependent resistors can operate in one of two ways, either increasing or decreasing their resistive value with changes in temperature. Then there are two types of thermistors available: negative temperature coefficient (NTC) of resistance and positive temperature coefficient (PTC) of resistance.
Negative temperature coefficient of resistance thermistors, or NTC thermistors for short, reduce or decrease their resistive value as the operating temperature around them increases. Generally, NTC thermistors are the most commonly used type of temperature sensors as they can be used in virtually any type of equipment where temperature plays a role.
NTC temperature thermistors have a negative electrical resistance versus temperature (R/T) relationship. The relatively large negative response of an NTC thermistor means that even small changes in temperature can cause significant changes in its electrical resistance. This makes them ideal for accurate temperature measurement and control.
We said previously that a thermistor is an electronic component whose resistance is highly dependent on temperature so if we send a constant current through the thermistor and then measure the voltage drop across it, we can thus determine its resistance and temperature.
NTC thermistors reduce in resistance with an increase in temperature and are available in a variety of base resistances and curves. They are usually characterised by their base resistance at room temperature, that is 25^{o}C, (77^{o}F) as this provides a convenient reference point. So for example, 2k2Ω at 25^{o}C, 10kΩ at 25^{o}C or 47kΩ at 25^{o}C, etc.
Another important characteristic is the “B” value. The B value is a material constant which is determined by the ceramic material from which it is made and describes the gradient of the resistive (R/T) curve over a particular temperature range between two temperature points. Each thermistor material will have a different material constant and therefore a different resistance versus temperature curve.
Then the B value will define the thermistors resistive value at the first temperature or base point, (which is usually 25^{o}C), called T1, and the thermistors resistive value at a second temperature point, for example 100^{o}C, called T2. Therefore the B value will define the thermistors material constant between the range of T1 and T2. That is B_{T1/T2} or B_{25/100} with typical NTC thermistor B values given anywhere between about 3000 and about 5000.
Note however, that both the temperature points of T1 and T2 are calculated in the temperature units of Kelvin where 0^{0}C = 273.15 Kelvin. Thus a value of 25^{o}C is equal to 25^{o} + 273.15 = 298.15K, and 100^{o}C is equal to 100^{o} + 273.15 = 373.15K, etc.
So by knowing the B value of a particular thermistor (obtained from manufacturers datasheet), it is possible to produce a table of temperature versus resistance to construct a suitable graph using the following normalised equation:
A 10kΩ NTC thermistor has a B value of 3455 between the temperature range of 25 to 100^{o}C. Calculate its resistive value at 25^{o}C and at 100^{o}C.
Data given: B = 3455, R1 = 10kΩ at 25^{o}. In order to convert the temperature scale from degrees Celsius, ^{o}C to degrees Kelvin add the mathematical constant 273.15
The value of R1 is already given as its 10kΩ base resistance, thus the value of R2 at 100^{o}C is calculated as:
Giving the following two point characteristics graph of:
Note that in this simple example, only two points were found, but generally thermistors change their resistance exponentially with changes in temperature so their characteristic curve is nonlinear, therefore the more temperature points are calculated the more accurate will be the curve.
Temperature (^{o}C) 
10  20  25  30  40  50  60  70  80  90  100  110  120 
Resistance (Ω) 
18476  12185  10000  8260  5740  4080  2960  2188  1645  1257  973  765  608 
and these points can be plotted as shown to give a more accurate characteristics curve for the 10kΩ NTC Thermistor which has a Bvalue of 3455.
Notice that it has a negative temperature coefficient (NTC), that is its resistance decreases with increasing temperatures.
So how can we use a thermistor to measure temperature. Hopefully by now we know that a thermistor is a resistive device and therefore according to Ohms law, if we pass a current through it, a voltage drop will be produced across it. As a thermistor is an active type of a sensor, that is, it requires an excitation signal for its operation, any changes in its resistance as a result of changes in temperature can be converted into a voltage change.
The simplest way of doing this is to use the thermistor as part of a potential divider circuit as shown. A constant voltage is applied across the resistor and thermistor series circuit with the output voltage measured across the thermistor.
If for example we use a 10kΩ thermistor with a series resistor of 10kΩ, then the output voltage at the base temperature of 25^{o}C will be half the supply voltage.
When the resistance of the thermistor changes due to changes in temperature, the fraction of the supply voltage across the thermistor also changes producing an output voltage that is proportional to the fraction of the total series resistance between the output terminals.
Thus the potential divider circuit is an example of a simple resistance to voltage converter where the resistance of the thermistor is controlled by temperature with the output voltage produced being proportional to the temperature. So the hotter the thermistor gets, the lower the voltage.
If we reversed the positions of the series resistor, R_{S} and the thermistor, R_{TH}, then the output voltage will change in the opposite direction, that is the hotter the thermistor gets, the higher the output voltage.
We can use ntc thermistors as part of a basic temperature sensing configuration using a bridge circuit as shown. The relationship between resistors R_{1} and R_{2} sets the reference voltage, V_{REF} to the value required. For example, if both R_{1} and R_{2} are of the same resistive value, the reference voltage will be equal to half of the supply voltage. That is Vs/2.
As the temperature and therefore the resistance of the thermistor changes, the voltage at V_{TH} also changes either higher or lower than that at V_{REF} producing a positive or negative output signal to the connected amplifier.
The amplifier circuit used for this basic temperature sensing bridge circuit could act as a differential amplifier for high sensitivity and amplification, or a simple Schmitttrigger circuit for ONOFF switching.
The problem with passing a current through a thermistor in this way, is that thermistors experience what is called selfheating effects, that is the I^{2}.R power dissipation could be high enough to create more heat than can be dissipated by the thermistor affecting its resistive value producing false results.
Thus it is possible that if the current through the thermistor is too high it would result in increased power dissipation and as the temperature increases, its resistance decreases causing more current to flow, which increases the temperature further resulting in what is known as Thermal Runaway. In other words, we want the thermistor to be hot due to the external temperature being measured and not by itself heating up.
Then the value for the series resistor, R_{S} above should be chosen to provide a reasonably wide response over the range of temperatures for which the thermistor is likely to be used while at the same time limiting the current to a safe value at the highest temperature.
One way of improving on this and having a more accurate conversion of resistance against temperature (R/T) is by driving the thermistor with a constant current source. The change in resistance can be measured by using a small and measured direct current, or DC, passed through the thermistor in order to measure the voltage drop produced.
We have seen that thermistors are primarily used as resistive temperature sensitive transducers, but the resistance of a thermistor can be changed either by external temperature changes or by changes in temperature caused by an electrical current flowing through them, as after all, they are resistive devices.
Ohm’s Law tells us that when an electrical current passes through a resistance R, as a result of the applied voltage, power is consumed in the form of heat due to the I^{2}R heating effect. Because of the selfheating effect of current in a thermistor, a thermistor can change its resistance with changes in current.
Inductive electrical equipment such as motors, transformers, ballast lighting, etc, suffer from excessive inrush currents when they are first turnedon. But series connected thermistors can be used to effectively limit these high initial currents to a sfe value. NTC thermistors with low values of cold resistance (at 25^{o}C) are generally used for current regulation.
Inrush current suppressors and surge limiters are types of series connected thermistor whose resistance drops to a very low value as it is heated by the load current passing through it. At the initial turnon, the thermistors cold resistance value (its base resistance) is fairly high controlling the initial inrush current to the load.
As a result of the load current, the thermistor heats up and reduces its resistance relatively slowly to the point were the power dissipated across it is sufficient to maintain its low resistance value with most of the applied voltage developed across the load.
Due to the thermal inertia of its mass this heating effect takes a few seconds during which the load current increases gradually rather than instantaneously, so any high inrush current is restricted and the power it draws reduces accordingly. Because of this thermal action, inrush current suppression thermistors can run very hot in the lowresistance state so require a cooldown or recovery period after power is removed to allow the resistance of the NTC thermistor to increase sufficiently to provide the required inrush current suppression the next time it is needed.
Thus the speed of response of a current limiting thermistor is given by its time constant. That is, the time taken for its resistance to change by by 63% (i.e. 1 to 1/e) of the total change. For example, suppose the ambient temperature changes from 0 to 100^{o}C, then the 63% time constant would be the time taken for the thermistor to have a resistive value at 63^{o}C.
Thus NTC thermistors provide protection from undesirably high inrush currents, while their resistance remains negligibly low during continuous operation supplying power to the load. The advantage here is that they able to effectively handle much higher inrush currents than standard fixed current limiting resistors with the same power consumption.
We have seen here in this tutorial about thermistors, that a thermistor is a two terminal resistive transducer which changes its resistive value with changes in surrounding ambient temperature, hence the name thermalresistor, or simply “thermistor”.
Thermistors are inexpensive, easilyobtainable temperature sensors constructed using semiconductor metal oxides, and are available with either a negative temperature coefficient, (NTC) of resistance or a positive temperature coefficient (PTC) of resistance. The difference being that NTC thermistors reduce their resistance as the temperature increases, while PTC thermistors increase their resistance as the temperature increases.
NTC thermistors are the most commonly used (especially the 10KΩ ntc thermistor) and along with an addition series resistor, R_{S} can be used as part of a simple potential divider circuit so that changes to its resistance due to changes in temperature, produces a temperaturerelated output voltage.
However, the operating current of the thermistor must be kept as low as possible to reduce any selfheating effects. If they pass operating currents which are too high, they can create more heat than can be quickly dissipated from the thermistor which may cause false results.
Thermistors are characterised by their base resistance and their B value. The base resistance, for example, 10kΩ, is the resistance of the thermistor at a given temperature, usually 25^{o}C and is defined as: R_{25}. The B value is a fixed material constant that describes the shape of the slope of the resistive curve over temperature (R/T).
We have also seen that thermistors can be used to measure an external temperature or can be used control a current as a result of the I^{2}R heating effect caused by the current flowing through it. By connecting an NTC thermistor in series with a load, it is possible to effectively limit the high inrush currents.
]]>A passive RC differentiator is nothing more than a capacitance in series with a resistance, that is a frequency dependant device which has reactance in series with a fixed resistance (the opposite to an integrator). Just like the integrator circuit, the output voltage depends on the circuits RC time constant and input frequency.
Thus at low input frequencies the reactance, Xc of the capacitor is high blocking any d.c. voltage or slowly varying input signals. While at high input frequencies the capacitors reactance is low allowing rapidly varying pulses to pass directly from the input to the output.
This is because the ratio of the capacitive reactance (Xc) to resistance (R) is different for different frequencies and the lower the frequency the less output. So for a given time constant, as the frequency of the input pulses increases, the output pulses more and more resemble the input pulses in shape.
We saw this effect in our tutorial about Passive High Pass Filters and if the input signal is a sine wave, an rc differentiator will simply act as a simple high pass filter (HPF) with a cutoff or corner frequency that corresponds to the RC time constant (tau, τ) of the series network.
Thus when fed with a pure sine wave an RC differentiator circuit acts as a simple passive high pass filter due to the standard capacitive reactance formula of Xc = 1/(2πƒC).
But a simple RC network can also be configured to perform differentiation of the input signal. We know from previous tutorials that the current through a capacitor is a complex exponential given by: i_{C} = C(dVc/dt). The rate at which the capacitor charges (or discharges) is directly proportional to the amount of resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC differentiator circuit is the time interval that equals the product of R and C. Consider the basic RC series circuit below.
For an RC differentiator circuit, the input signal is applied to one side of the capacitor with the output taken across the resistor, then V_{OUT} equals V_{R}. As the capacitor is a frequency dependant element, the amount of charge that is established across the plates is equal to the time domain integral of the current. That is it takes a certain amount of time for the capacitor to fully charge as the capacitor can not charge instantaneously only charge exponentially.
We saw in our tutorial about RC Integrators that when a single step voltage pulse is applied to the input of an RC integrator, the output becomes a sawtooth waveform if the RC time constant is long enough. The RC differentiator will also change the input waveform but in a different way to the integrator.
We said previously that for the RC differentiator, the output is equal to the voltage across the resistor, that is: V_{OUT} equals V_{R} and being a resistance, the output voltage can change instantaneously.
However, the voltage across the capacitor can not change instantly but depends on the value of the capacitance, C as it tries to store an electrical charge, Q across its plates. Then the current flowing into the capacitor, that is i_{t} depends on the rate of change of the charge across its plates. Thus the capacitor current is not proportional to the voltage but to its time variation giving: i = dQ/dt.
As the amount of charge across the capacitors plates is equal to Q = C x Vc, that is capacitance times voltage, we can derive the equation for the capacitors current as:
Therefore the capacitor current can be written as:
As V_{OUT} equals V_{R} where V_{R} according to ohms law is equal too: i_{R} x R. The current that flows through the capacitor must also flow through the resistance as they are both connected together in series. Thus:
Thus the standard equation given for an RC differentiator circuit is:
Then we can see that the output voltage, V_{OUT} is the derivative of the input voltage, V_{IN} which is weighted by the constant of RC. Where RC represents the time constant, τ of the series circuit.
When a single step voltage pulse is firstly applied to the input of an RC differentiator, the capacitor “appears” initially as a short circuit to the fast changing signal. This is because the slope dv/dt of the positivegoing edge of a square wave is very large (ideally infinite), thus at the instant the signal appears, all the input voltage passes through to the output appearing across the resistor.
After the initial positivegoing edge of the input signal has passed and the peak value of the input is constant, the capacitor starts to charge up in its normal way via the resistor in response to the input pulse at a rate determined by the RC time constant, τ = RC.
As the capacitor charges up, the voltage across the resistor, and thus the output decreases in an exponentially way until the capacitor becomes fully charged after a time constant of 5RC (5T), resulting in zero output across the resistor. Thus the voltage across the fully charged capacitor equals the value of the input pulse as: V_{C} = V_{IN} and this condition holds true so long as the magnitude of the input pulse does not change.
If now the input pulse changes and returns to zero, the rate of change of the negativegoing edge of the pulse pass through the capacitor to the output as the capacitor can not respond to this high dv/dt change. The result is a negative going spike at the output.
After the initial negativegoing edge of the input signal, the capacitor recovers and starts to discharge normally and the output voltage across the resistor, and therefore the output, starts to increases exponentially as the capacitor discharges.
Thus whenever the input signal is changing rapidly, a voltage spike is produced at the output with the polarity of this voltage spike depending on whether the input is changing in a positive or a negative direction, as a positive spike is produced with the positivegoing edge of the input signal, and a negative spike produced as a result of the negativegoing input signal.
Thus the RC differentiator output is effectively a graph of the rate of change of the input signal which has no resemblance to the square wave input wave, but consists of narrow positive and negative spikes as the input pulse changes value.
By varying the time period, T of the square wave input pulses with respect to the fixed RC time constant of the series combination, the shape of the output pulses will change as shown.
Then we can see that the shape of the output waveform depends on the ratio of the pulse width to the RC time constant. When RC is much larger (greater than 10RC) than the pulse width the output waveform resembles the square wave of the input signal. When RC is much smaller (less than 0.1RC) than the pulse width, the output waveform takes the form of very sharp and narrow spikes as shown above.
So by varying the time constant of the circuit from 10RC to 0.1RC we can produce a range of different wave shapes. Generally a smaller time constant is always used in RC differentiator circuits to provide good sharp pulses at the output across R. Thus the differential of a square wave pulse (high dv/dt step input) is an infinitesimally short spike resulting in an RC differentiator circuit.
Lets assume a square wave waveform has a period, T of 20mS giving a pulse width of 10mS (20mS divided by 2). For the spike to discharge down to 37% of its initial value, the pulse width must equal the RC time constant, that is RC = 10mS. If we choose a value for the capacitor, C of 1uF, then R equals 10kΩ.
For the output to resemble the input, we need RC to be ten times (10RC) the value of the pulse width, so for a capacitor value of say, 1uF, this would give a resistor value of: 100kΩ. Likewise, for the output to resemble a sharp pulse, we need RC to be one tenth (0.1RC) of the pulse width, so for the same capacitor value of 1uF, this would give a resistor value of: 1kΩ, and so on.
So by having an RC value of one tenth the pulse width (and in our example above this is 0.1 x 10mS = 1mS) or lower we can produce the required spikes at the output, and the lower the RC time constant for a given pulse width, the sharper the spikes. Thus the exact shape of the output waveform depends on the value of the RC time constant.
We have seen here in this RC Differentiator tutorial that the input signal is applied to one side of a capacitor and the the output is taken across the resistor. A differentiator circuit is used to produce trigger or spiked typed pulses for timing circuit applications.
When a square wave step input is applied to this RC circuit, it produces a completely different wave shape at the output. The shape of the output waveform depending on the periodic time, T (an therefore the frequency, ƒ) of the input square wave and on the circuit’s RC time constant value.
When the periodic time of the input waveform is similar too, or shorter than, (higher frequency) the circuits RC time constant, the output waveform resembles the input waveform, that is a square wave profile. When the periodic time of the input waveform is much longer than, (lower frequency) the circuits RC time constant, the output waveform resembles narrow positive and negative spikes.
The positive spike at the output is produced by the leadingedge of the input square wave, while the negative spike at the output is produced by the fallingedge of the input square wave. Then the output of an RC differentiator circuit depends on the rate of change of the input voltage as the effect is very similar to the mathematical function of differentiation.
]]>In Electronics, the basic series connected resistorcapacitor (RC) circuit has many uses and applications from basic charging/discharging circuits to highorder filter circuits. This two component passive RC circuit may look simple enough, but depending on the type and frequency of the applied input signal, the behaviour and response of this basic RC circuit can be very different.
A passive RC network is nothing more than a resistor in series with a capacitor, that is a fixed resistance in series with a capacitor that has a frequency dependant reactance which decreases as the frequency across its plates increases. Thus at low frequencies the reactance, Xc of the capacitor is high while at high frequencies its reactance is low due to the standard capacitive reactance formula of Xc = 1/(2πƒC), and we saw this effect in our tutorial about Passive Low Pass Filters.
Then if the input signal is a sine wave, an rc integrator will simply act as a simple low pass filter (LPF) with a cutoff or corner frequency that corresponds to the RC time constant (tau, τ) of the series network and whose output is reduced above this cutoff frequency point. Thus when fed with a pure sine wave an RC integrator acts as a passive low pass filter.
As we have seen previously, the RC time constant reflects the relationship between the resistance and the capacitance with respect to time with the amount of time, given in seconds, being directly proportional to resistance, R and capacitance, C.
Thus the rate of charging or discharging depends on the RC time constant, τ = RC. Consider the circuit below.
For an RC integrator circuit, the input signal is applied to the resistance with the output taken across the capacitor, then V_{OUT} equals V_{C}. As the capacitor is a frequency dependant element, the amount of charge that is established across the plates is equal to the time domain integral of the current. That is it takes a certain amount of time for the capacitor to fully charge as the capacitor can not charge instantaneously only charge exponentially.
Therefore the capacitor current can be written as:
This basic equation above of i_{C} = C(dVc/dt) can also be expressed as the instantaneous rate of change of charge, Q with respect to time giving us the following standard equation of: i_{C} = dQ/dt where the charge Q = C x Vc, that is capacitance times voltage.
The rate at which the capacitor charges (or discharges) is directly proportional to the amount of the resistance and capacitance giving the time constant of the circuit. Thus the time constant of a RC integrator circuit is the time interval that equals the product of R and C.
Since capacitance is equal to Q/Vc where electrical charge, Q is the flow of a current (i) over time (t), that is the product of i x t in coulombs, and from Ohms law we know that voltage (V) is equal to i x R, substituting these into the equation for the RC time constant gives:
Then we can see that as both i and R cancel out, only T remains indicating that the time constant of an RC integrator circuit has the dimension of time in seconds, being given the Greek letter tau, τ. Note that this time constant reflects the time (in seconds) required for the capacitor to charge up to 63.2% of the maximum voltage or discharge down to 36.8% of maximum voltage.
We said previously that for the RC integrator, the output is equal to the voltage across the capacitor, that is: V_{OUT} equals V_{C}. This voltage is proportional to the charge, Q being stored on the capacitor given by: Q = VxC.
The result is that the output voltage is the integral of the input voltage with the amount of integration dependent upon the values of R and C and therefore the time constant of the network.
We saw above that the capacitors current can be expressed as the rate of change of charge, Q with respect to time. Therefore, from a basic rule of differential calculus, the derivative of Q with respect to time is dQ/dt and as i = dQ/dt we get the following relationship of:
Q = ∫idt (the charge Q on the capacitor at any instant in time)
Since the input is connected to the resistor, the same current, i must pass through both the resistor and the capacitor (i_{R} = i_{C}) producing a V_{R} voltage drop across the resistor so the current, (i) flowing through this series RC network is given as:
therefore:
As i = V_{IN}/R, substituting and rearranging to solve for V_{OUT} as a function of time gives:
So in other words, the output from an RC integrator circuit, which is the voltage across the capacitor is equal to the time Integral of the input voltage, V_{IN} weighted by a constant of 1/RC. Where RC represents the time constant, τ.
Then assuming the initial charge on the capacitor is zero, that is V_{OUT} = 0, and the input voltage V_{IN} is constant, the output voltage, V_{OUT} is expressed in the time domain as:
So an RC integrator circuit is one in which the output voltage, V_{OUT} is proportional to the integral of the input voltage, and with this in mind, lets see what happens when we apply a single positive pulse in the form of a step voltage to the RC integrator circuit.
When a single step voltage pulse is applied to the input of an RC integrator, the capacitor charges up via the resistor in response to the pulse. However, the output is not instant as the voltage across the capacitor cannot change instantaneously but increases exponentially as the capacitor charges at a rate determined by the RC time constant, τ = RC.
We now know that the rate at which the capacitor either charges or discharges is determined by the RC time constant of the circuit. If an ideal step voltage pulse is applied, that is with the leading edge and trailing edge considered as being instantaneous, the voltage across the capacitor will increase for charging and decrease for discharging, exponentially over time at a rate determined by:
Capacitor Charging
Capacitor Discharging
So if we assume a capacitor voltage of one volt (1V), we can plot the percentage of charge or discharge of the capacitor for each individual R time constant as shown in the following table.
Time Constant 
Capacitor Charging 
Capacitor Discharging 
τ  % Charged  % Discharged 
0.5  39.4%  60.6% 
0.7  50%  50% 
1  63.2%  36.7% 
2  86.4%  13.5% 
3  95.0%  4.9% 
4  98.1%  1.8% 
5  99.3%  0.67% 
Note that at 5 time constants or above, the capacitor is considered to be 100 percent fully charged or fully discharged.
So now lets assume we have an RC integrator circuit consisting of a 100kΩ resistor and a 1uF capacitor as shown.
The time constant, τ of the RC integrator circuit is therefore given as: RC = 100kΩ x 1uF = 100ms.
So if we apply a step voltage pulse to the input with a duration of say, two time constants (200mS), then from the table above we can see that the capacitor will charge to 86.4% of its fully charged value. If this pulse has an amplitude of 10 volts, then this equates to 8.64 volts before the capacitor discharges again back through the resistor to the source as the input pulse returns to zero.
If we assume that the capacitor is allowed to fully discharge in a time of 5 time constants, or 500mS before the arrival of the next input pulse, then the graph of the charging and discharging curves would look something like this:
Note that the capacitor is discharging from an initial value of 8.64 volts (2 time constants) and not from the 10 volts input.
Then we can see that as the RC time constant is fixed, any variation to the input pulse width will affect the output of the RC integrator circuit. If the pulse width is increased and is equal too or greater than 5RC, then the shape of the output pulse will be similar to that of the input as the output voltage reaches the same value as the input.
If however the pulse width is decreased below 5RC, the capacitor will only partially charge and not reach the maximum input voltage resulting in a smaller output voltage because the capacitor cannot charge as much resulting in an output voltage that is proportional to the integral of the input voltage.
So if we assume an input pulse equal to one time constant, that is 1RC, the capacitor will charge and discharge not between 0 volts and 10 volts but between 63.2% and 38.7% of the voltage across the capacitor at the time of change. Note that these values are determined by the RC time constant.
So for a continuous pulse input, the correct relationship between the periodic time of the input and the RC time constant of the circuit, integration of the input will take place producing a sort of ramp up, and then a ramp down output. But for the circuit to function correctly as an integrator, the value of the RC time constant has to be large compared to the inputs periodic time. That is RC ≫ T, usually 10 times greater.
This means that the magnitude of the output voltage (which was proportional to 1/RC) will be very small between its high and low voltages severely attenuating the output voltage. This is because the capacitor has much less time to charge and discharge between pulses but the average output DC voltage will increase towards one half magnitude of the input and in our pulse example above, this will be 5 volts (10/2).
We have seen above that an RC integrator circuit can perform the operation of integration by applying a pulse input resulting in a rampup and rampdown triangular wave output due to the charging and discharging characteristics of the capacitor. But what would happen if we reversed the process and applied a triangular waveform to the input, would we get a pulse or square wave output?
When the input signal to an RC integrator circuit is a pulse shaped input, the output is a triangular wave. But when we apply a triangular wave, the output becomes a sine wave due to the integration over time of the ramp signal.
There are many ways to produce a sinusoidal waveform, but one simple and cheap way to electronically produce a sine waves type waveform is to use a pair of passive RC integrator circuits connected together in series as shown.
Here the first RC integrator converts the original pulse shaped input into a rampup and rampdown triangular waveform which becomes the input of the second RC integrator. This second RC integrator circuit rounds off the points of the triangular waveform converting it into a sine wave as it is effectively performing a double integration on the original input signal with the RC time constant affecting the degree of integration.
As the integration of a ramp produces a sine function, (basically a roundoff triangular waveform) its periodic frequency in Hertz will be equal to the period T of the original pulse. Note also that if we reverse this signal and the input signal is a sine wave, the circuit does not act as an integrator, but as a simple low pass filter (LPF) with the sine wave, being a pure waveform does not change shape, only its amplitude is affected.
We have seen here that the RC integrator is basically a series RC lowpass filter circuit which when a step voltage pulse is applied to its input produces an output that is proportional to the integral of its input. This produces a standard equation of: Vo = ∫Vidt where Vi is the signal fed to the integrator and Vo is the integrated output signal.
The integration of the input step function produces an output that resembles a triangular ramp function with an amplitude smaller than that of the original pulse input with the amount of attenuation being determined by the time constant. Thus the shape of the output waveform depends on the relationship between the time constant of the circuit and the frequency (period) of the input pulse.
An RC integrators time constant is always compared to the period, T of the input, so a long RC time constant will produce a triangular wave shape with a low amplitude compared to the input signal as the capacitor has less time to fully charge or discharge. A short time constant allows the capacitor more time to charge and discharge producing a more typical rounded shape.
By connecting two RC integrator circuits together in parallel has the effect of a double integration on the input pulse. The result of this double integration is that the first integrator circuit converts the step voltage pulse into a triangular waveform and the second integrator circuit converts the triangular waveform shape by rounding off the points of the triangular waveform producing a sine wave output waveform with a greatly reduced amplitude.
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