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 output of each tristate buffer are connected to a common wire bus but the decoder guarantees that only one tristate buffer is active at any one time, due to its enable signal, allowing its data to pass directly onto the common bus while the outputs of the other nonenabled buffers are in a highimpedance state depending on the binary value of the decoders select inputs.
Thus 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 inputs to a single output 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.
]]>Here in part 5 of our series of video tutorials for beginners on power supplies, we will finish off by comparing the different types of power supply we have seen, both Linear and Switching, and see the advantages and drawbacks of both the switching and linear power supply.
Time: 0:00sHello I’m Chris Richardson, and if you made it to part 5 then you probably know by now that I am an electronics engineer focused on power supplies. This is the fifth part in a series of web seminars for power supply enthusiasts, or hobbyists who aren’t necessarily trained as electronics engineers.
So far we have gathered some low cost equipment to test power supplies. Looked at unregulated power supplies. Tested various linear regulators and tested various switching regulators. In this section we will compare examples of these different power supplies and examine them to see which fits best in different applications.
Time: 0:31sThere are many, many different types of power supplies out there. From tiny systems using milliwatts, (mW) or even microwatts, (uW) in areas, like the so called “energy harvesting” field, to megawatts, (MW) in electrical generation and distribution. Selecting the most appropriate device for your application is therefore a critical step in power supply use and design.
Time: 0:48sRegardless of the type of power supply being tested, accurate power efficiency testing requires one ammeter and one voltmeter for the input to your supply, and then another ammeter and another voltmeter for each output. For very low power circuits, generally under onetenth of a watt (1/10W or 0.1W), special equipment is needed because the ammeter and voltmeter always consume some power and would distort those measurements.
I am going to focus on power supplies of at least one watt of output power, since that special equipment is definitely not on my list of affordable devices that I talked about in part one.
Time: 1:18sKelvin Sensing refers to measuring the input voltage and output voltage directly at the inputs and outputs of your power supply. The demo boards I have been using always include test points right next to the input capacitors and also the output capacitors for this purpose.
If you use the voltage readout of a lab power supply or trust the ATX box to give exactly 12 volts or 5 volts, your measurements will be wrong, since voltage is lost due to resistive drops in the connecting cables. The ammeter itself also uses a series resistor and some voltage is lost there too.
Time: 1:48sFor the first efficiency experiment, I am going back to my unregulated power supply here. It is being used with a linear current source set to draw 500mA. The blue multimeter is measuring the input current, (I_{IN}) and the orange multimeter is measuring the input voltage, (V_{IN}).
When switched “ON” the input is drawing 30.8mA at 226 volts rms and now I will switch things around and look at the output current, (I_{OUT}) and output voltage, (V_{OUT}).
Time: 2:16sThe same circuit, but now when I switch it “ON” I am going to measure the output current at 510mA and output voltage at 6.2mA. Remember that this is a semiregulated circuit. In practice if we were going to calculate lots of efficiency points, we would vary the load. However, this linear current source is actually logarithmic in the way it adjusts with this potentiometer so it makes it a little bit more difficult.
Now I am repeating the experiment but instead of using the unregulated power supply, I am using a regulated switching power supply. This is the output current and output voltage and when I turn it “ON”, again 510mA. But it is not actually as well regulated as I expected it to be since its 6.5 volts and here we actually have 7.07 volts. Nonetheless, we can take these two data points and make an efficiency plot.
Now I am testing the input current and the input voltage to my switching dctodc power supply, and we can already see that the input current is much lower whereas the input voltage is almost exactly the same. So we know that the efficiency is going to be much better and now we have two more data points so let’s go ahead and calculate.
Time: 3:26sIn the previous slides and videos, we saw that switching regulators are vastly more efficient than linear regulators in most cases. So it should not be too much of a surprise that linear regulators used in the same conditions of V_{IN}, V_{OUT} and I_{OUT}, dissipate a lot more power and their components get a lot more hotter than equivalent switching regulators.
Still the lower electrical noise, simplicity and low cost of NPN regulators and LDO’s (Low Dropout Regulators) make them my preferred choice whenever it is reasonable to use them. My criteria are the following:
This assumes that you do not have the space or the budget for a heatsink, and in my experience there is really space or money for heatsinks. It might be surprising to you that many heatsinks that can dissipate more than one watt cost more than the power supply control chip.
Time: 4:29sTo talk about heat, I have here my synchronous buck converter. This is probably the most efficient of all the switching power converters and its delivering about 27 or 28 watts. I am powering it from 12 volts from my ATX power supply here. I have almost exactly 5 volts as the output voltage, about 5.5 amps as the output current and that’s thanks to a group of power resistors who’s total resistance is just under 1Ω.
To get an idea, the ambient temperature in the room is somewhere between 27 and 28^{o}C. One of the power resistors is quite hot at close to 50^{o}C. I always think of that as anything over 50^{o}C is uncomfortably to the touch.
If I start to measure the temperature of some of the power components, the switching MOSFET is the component that gets the hottest, and it is barely over 30^{o}C. This is the synchronous power MOSFET, slightly cooler at 29^{o}C.
The power inductor is also an element that can heat up a lot and barely over 30^{o}C. The last thing we will measure is an aluminium electrolytic input capacitor that is barely heating up at all. So that means it should have a nice long lifetime.
Time: 5:57sIf you watched the section on linear regulators, then you will remember that this discrete linear regulator (showing circuit board) has a big, big heatsink and has a control chip and a discrete power transistor. So we will compare this to the buck converter that we just did.
The blue multimeter is the output current, (I_{OUT}) and the orange multimeter is the output voltage, (V_{OUT}), so actually the load here which is altogether 1Ω is drawing so much current it is actually collapsing to output voltage a little bit here. But this is still a good thermal test with the ambient temperature here about 27^{o}C. The only thing that really matters in a linear regulator is the chip itself, which in this case the discrete pass element.
If I put the temperature probe on it it is nice and hot, probably over 100^{o}C. I have the tip of the thermocouple right at the junction of the heatsink and the tab of that discrete power MOSFET. There is a huge difference, but remember that none of the components of the buck regulator were getting to be over 31 or 32^{o}C, or so. I will turn it “OFF” as it is overheating.
Time: 7:16sLinear regulators beat switching regulators without any doubt when it comes to low conductive noise. That goes for their inputs which we see at the top of the screen here, and those may be subject to legal limits, and for their outputs which are often sensitive to noise. For example, most digital circuits are sensitive to noise of certain frequencies.
Three ways to reduce voltage ripple when your power dissipation or voltage transformation forces you to use a switcher are:
Time: 8:26sTo compare power supply voltage ripple, I have both the buck regulator on the left and my LDO with the discrete power transistor on the right. Each one is using +12 volts in from the ATX power supply. Each one has the same load, two 8Ω power resistors in parallel to make a total of a 4Ω load, and on the previous slide you saw the input voltage ripples.
Time: 8:47sNow we are using the oscilloscope to measure the two output voltage ripples. Again, 5 volts output for the LDO and 5 volts output for the buck regulator and if we look at the oscilloscope, the buck regulator ripple in yellow and the LDO ripple is in blue.
At first you may say that they look almost the same, but most of the ripple we see in blue is the result of noise coupling from the buck. If we take out the probe we can see that the LDO noise is much much lower.
Time: 9:20sLinear regulators also radiate far less noise than the switching regulator, even a switcher the processes far less power. When I design power supplies it usually takes me as long or longer to design the filters and the electromagnetic noise reducing circuits as it does to design the switcher itself.
If you know someone who has taken their product for UL laboratories testing in the United States, or CE testing in Europe, then they may have stories of spending days on trial and error fixes for conductive or more commonly radiated noise that exceeded the limits. This is another reason why I use linear regulators whenever I can.
Time: 9:52sTo demonstrate electromagnetic interference (EMI), I have here an AM (Amplitude Modulated) radio tuned to somewhere around 600kHz which is in the range of the switching frequency of this buck regulator. Right now it is not “ON” and we can here some Spanish talk radio. When I turn “ON” the circuit, sure enough noise. When I get the radio closer to the source, which is the switching node and the inductor, the more interference we hear.
Once again for the EMI test I have the AM radio tuned to around 600kHz or so and we saw earlier with the buck regulator that as soon as I turned it “ON”, we got nothing but static noise. When I turn the the linear regulator “ON” we have approximately the same amount of output power, but although there is a little bit of interference, we can hear the AM radio just fine even if we get close to the power supply, and that’s the beauty of linear regulators, there is very little radiated noise.
Time: 10:59sOne more test. I now have the same radio but now it is tuned to FM (Frequency Modulated) 92MHz and sounds good. If I get the antenna (aerial) close then the interference starts again. This happens because not only does the switching inverter operate at maybe 400kHz to 1MHz, but it has a lot of harmonics and also has high frequency noise. Those extend and also interfere with the FM band.
Now we are going to test the linear regulator. I have here my FM radio and with the antenna close, I will switch it “ON” but have to be quick because this gets very hot as you can see it is drawing lots of current and voltage here, and no matter what I do we can continue to hear the Spanish radio.
That concludes part 5 of power supplies for nonEE’s. This is the last video tutorial in the series for now and it has been my pleasure making these videos. On behalf of myself and electronicstutorials.ws I sincerely hope that you have learnt something and thanks again for watching.
End of transcription.
You can find more information and a great tutorial about the different types of power supplies by following this link about: Switch Mode Power Supply.
]]>In part 3 of our series of video tutorials for beginners on power supplies, we looked at testing and using Linear Power Supplies. Here in part 4 of our video tutorial series we will look at testing and using Switching Power Supplies including buck and boost converters which can stepdown (buck) or stepup (boost) the output voltage.
Time: 0:00sHello I’m Chris Richardson, an electronics engineer who focuses on power supplies. This is the fourth in a series of web seminars for people who like power supplies but who aren’t necessarily trained to be electronics engineers.
So far in this series we have gathered some low cost equipment to test power supplies, both unregulated power supplies, tested various linear regulators, and now its time to test some switching regulators. The modern power supply is what dominates the market today.
Time: 0:25sThe “buck” is the simplest switching regulator, and is the easiest to understand. The control switch on top, a bipolar transistor, or more commonly a MOSFET works together with diode D_{1} to make a rectangular wave at the point where the switch, diode and inductor connect.
This point is the switching node and it is the most important voltage to probe in the system. The inductor and capacitor form a low pass filter whose output is then mostly DC (Direct Current) with some AC (Alternating Current) ripple. The average value of that output voltage depends upon the input voltage and upon the duty cycle of the rectangular wave.
Time: 0:56sDuty Cycle is equal to T_{ON} divided by the sum of T_{ON} and T_{OFF} and the higher the duty cycle, the higher the output voltage. This switcher bucksdown the output voltage, hence its name, Buck Regulator.
Much like a linear regulator, the theoretical maximum V_{OUT} is equal to V_{IN}. In practice the maximum V_{OUT} we can achieve is somewhat less than V_{IN}.
Time: 1:17sThe first switching regulator we are going to test today is the buck regulator, and you can see the circuit here. This is the input capacitor, and small capacitors, these are the two control switches. This is a synchronous regulator, meaning that instead of a diode for the low side it has a MOSFET. The power inductor with the loop of wire connected in series is to put a current probe if required, and these are the output capacitors here.
Right now this buck converter is unloaded and I am using the +5 volts from the ATX power supply and here is approximately 5 volts in, and the output I have adjusted to 1.9 volts, approximately. Right now there is no load and these four power resistors here of 8Ω each are connected in parallel to give a load of 2Ω.
When I connect them they give a load of about 1 amp. You can see that in the input voltage drops slightly, and also the output voltage drops slightly but it is still regulating.
Time: 2:13sI have switched things around to show the high efficiency of the switching regulator here. So now I am measuring input voltage here on the blue multimeter, and input current on the orange multimeter here. So I want you to see that when we power the circuit with the load connected, at 5 volts it draws about 380mA (0.380A).
Now I am using the 12 volt input to power it and you can see 12 volts on the blue multimeter, the load is the same and the output voltage is the same, but now the output current has dropped to 210mA (0.210A).
This is an interesting property of switching converters. As the input voltage goes up, the input current goes down. In fact when you are testing a switching regulator, one of the first basic tests is if you have a variable input supply is to watch and make sure that as you increase the input voltage the input current goes down.
Time: 2:59sThis is the backside of the buck regulator circuit. I am measuring the switching node and output voltage with two oscilloscope probes, there is one amp load being powered by the 12 volt input. We can see the switching node in yellow with its duty cycle and the resulting ripple on the output voltage. About 20 to 30mV peaktopeak.
Time: 3:23sThe same experiment again except that this time we are powering it by 5 volts at the input, and now we can see that the duty cycle goes much higher, we are in the buck converter and the duty cycle is approximately equal to the output voltage at 1.9 volts divided by the input voltage of 5 volts, and we also have slightly lower ripple maybe somewhere in the range of 10 to 15mV.
Time: 3:45sA boost regulator is little more than a buck regulator operating in reverse. Just imagine that D_{1} was a MOSFET and that TR_{1} a diode. Where you see V_{IN} there is always a capacitor even though it is not shown in this schematic.
A Boost Regulator is a great circuit for explaining why an inductor is the heart of most switching regulators. To make an output voltage that is higher than the input voltage, the boost regulator stores energy in the inductors magnetic field which develops as the current in L_{1} increases while TR_{1} is “ON”.
The control circuit turns TR_{1} “OFF” while the current is still flowing. The inductor can increase the voltage across it to infinity in theory in order to maintain that current flow. Unfortunately it does not have to go to infinity just increase enough to put D_{1} ON and into forward bias then the current can flow to the output. In theory the boost regulator itself can increase the output voltage to infinity, but in practice it is limited to about 10 times the input voltage (10*V_{IN}).
One last but important note here. The boost regulator can only increase V_{OUT} with respect to V_{IN} which is like the exact opposite of the buck regulator.
Time: 4:45sFrom my example boost regulator, I am using a pcb here that actually has two switching regulators. The top one here is an inverting regulator, which is a very interesting topology but we do not have time here to talk about it so it is disabled.
The bottom part is a boost regulator and the parts are quite small. The input capacitor is actually hidden underneath the input leads. This is the power inductor and the output diode is hidden behind this test fixture which I use to make very precise measurements of the switching node while the output capacitor is also hidden beneath this test fixture and that is also there to make precise measurements for the output voltage.
Right now I have +5 volts in coming from my ATX power supply and about 14.7 volts out. You may be wondering why I would have 14.7 volts, and the honest answer is, this circuit like most of the ones I have shown are ones left over from things I did especially for different kinds of customers. 14.7 volts is not a typical voltage, but these are adjustable regulators so you can get just about any voltage you want out of them.
Here is the boost converter again, but this time I am measuring input current and output current, so when I switch it “ON”, the 5 volts it is drawing about 900mA (0.9A) at the output and 3.5 amps at the input. Remember that that is the opposite of the buck converter and in the buck converter since the output voltage is lower, the input current is always lower than the output current.
In the boost converter it is the reverse. As power efficiency is high, input power is approximately equal to output power. So since input voltage is lower than output voltage, input current is higher than output current.
Time: 6:25sFor the last boost converter test, I have the circuit here powering the same load at about 900mA and 5 volts in to 14.7 volts out and here we can see the switching node in yellow and the output voltage AC coupled in blue.
There are two important things to notice as differences between this and the buck converter. The switching node voltage goes between zero and the output voltage, and the ripple is much higher. That is always true in a boost converter. In the boost converter the input voltage is much smoother while the output voltage has more ripple, and the opposite is true in a buck converter.
Time: 7:02sThe final topology of the three basics is the Buckboost Regulator. As the name implies, it can generate an output voltage whose actual value is higher or lower than that of the input voltage. But, and this is a big but, the polarity of the output voltage is reversed with respect to the input.
The number of modern circuit which need negative voltages (ve) is shrinking. But sensitive amplifiers, sensors and other equipment still use both positive and negative voltages to operate.
Like the boost, the buckboost uses the amazing ability of the inductor to make that negative voltage. In this case the voltage across the inductor reverses polarity in order to maintain current flow, when TR_{1} turns “OFF”. If you inspect the transfer function on the left, in theory the output voltage can go to negative infinity. In practice you get to about minus 10 times the input voltage, (10*V_{IN}).
Time: 7:50sI am sorry to say that I could not find any evaluation board or demo pcb to show off and actual inverting buckboost regulator. But you can usually take almost any buck regulator and turn it into an inverting buckboost by changing the polarity and the connection of the output diode and the output inductor. So perhaps I could do something like this in a future video.
Time: 8:11sNo discussion of switching regulators would be complete the Flyback regulator. In terms of shear volume, the most common switcher in existence is the buck. Just your mobile phone has about five or ten of them, but the flyback is number two. Just about every AC to DC power supply under 50 watts uses this very flexible topology in one form or another.
The flyback regulator is based on the buckboost regulator but it has two windings in its inductor. In fact if you made Nps which is the ratio of the two windings equal to one (1:1), the flyback regulator and the buckboost regulator would have exactly the same transfer function. The ratio of those windings allows V_{OUT} to be equal to V_{IN}, but it could also be much, much higher or much, much lower.
The two windings can also be isolated, and this is great for both electrical safety, as in not electrocuting anyone and also for isolating sensitive circuits from noisy ones. Finally as the flyback output is disconnected by the transformer or coupledinductor, can be positive or negative.
Time: 9:05sWith all the wires disconnected I can show you the most important parts of this supply. So we have the input capacitors here, a discrete power and MOSFET here on the primary side. This is the transformer or better known as a coupled inductor, the output diode and the output capacitors. Notice also that there is this separation between the primary side and the secondary side so this can be an isolated converter, again that is for electrical safety or to get rid of noise.
On the backside we can see again the separation here and these diodes are actually shorting the ground of the primary to the ground of the secondary. So this particular circuit is not isolated, but it could be. If we wanted to isolate it we would get rid of these resistors and use a device called an optocoupler to do the feedback for the control of the power supply.
Time: 9:54sHere I have a small flyback power supply that is designed to operate from 36 volts upto 72 volts. That is typically known as a telecom range. It’s at the very, very limit of what I can do with my ATX power supply. So to get these 22.4 volts, which should be 24 volts, I am actually running from the negative 12 volts (12V) up to the positive 12 volts (+12V) and I can do that because the ATX power supply uses the same common ground for both of these two.
But there is a problem. If I was to try and measure with an oscilloscope probe here. Since this is earth, as soon as I connect and touch this here, my supply turns “OFF” as I have caused a short circuit from the output voltage to earth. So what I am going to do is cheat a little bit and actually use what is called an isolated lab power supply to do the rest of the experiments.
Here is the flyback regulator again, but now it is being powered by 48 volts at the input and this is coming from this triple lab DC power supply. I checked on eBay and these typically cost anywhere from about 100 euros to 200 euros. So they are not free, but they are not, lets say a budget buster.
In any case, a flyback regulator actually has two switching nodes. It consists really of two inductors that are coupled on the same core. So we are looking at the primary side connected to the power MOSFET in yellow, and the secondary side connected to the diode in blue. Also notice that these voltages are much higher than the ones we have been dealing with so far.
So you have seen me in these videos touching the circuits while they are operating, and that’s fine maybe for circuits operating up to 12 volts or so, but I definitely would not touch this circuit while it is operating as 48 volts is enough to give you a nasty shock.
That concludes part 4 of power supplies for nonEE’s. Stay tuned for part 5 were we will compare and contrast the different types of power supplies so far to see which ones work best in which situations.
On behalf of myself and electronicstutorials.ws thanks for watching.
End of transcription.
You can find more information and a great tutorial about buck and boost power supplies by following this link about: Switch Mode Power Supply.
In part 5 and final tutorial of our video tutorial series on power supplies for beginners, we will discuss and compare the different types of power supplies including the switching and linear power supply we have looked at over the previous tutorials.
]]>Previously in part 2 of our series of video tutorials for power supplies for beginners, we explained how to test and use Unregulated Power Supplies and showed how an unregulated power supply has difficulty in controlling its output. Here in part 3 of our video tutorial series we will look at Linear Power Supplies and shows how series and shunt regulators are much better at controlling their output.
Time: 0:00sHello I’m Chris Richardson, and I’m an electronics engineer focused on power supplies. This is the third part of a series of web seminars for power supply enthusiasts who aren’t necessarily trained as power electronics engineers.
So far in parts one and two we gathered the basic equipment for testing power supplies without spending a fortune and then we found and tested some older unregulated power supplies. Now its time to evaluate and test the oldest and most basic type of regulated power supply known as a Linear Regulator.
Time: 0:27sThe diagram on the left is a discrete linear regulator made up of a zener diode and a resistor, R_{S}. All the resistor does is limit the current. If it wasn’t there, the input supply would either melt the zener with too much current, or the input supply itself would run into its own current limit.
Such a circuit is very, very cheap, but the tolerance to the output voltage depends upon the zener voltage, V_{Z} and that depends on the load current, the temperature, and the natural distribution of V_{Z} itself from part to part. To me it’s debatable whether or not this type of supply is regulated, but its a good introduction to the true regulated circuits coming later. Since the active element is in parallel with the load, we say that it “shunts the load”, hence the name Shunt Regulator.
Time: 1:08sOn the right is a similar circuit that uses a genuine integrated circuit for better precision. The TL431 and its variants are everywhere in the power supply world, but are not often used as shunt regulators like we see here. The TL431 is so similar in its function to a true zener diode, that the symbol is often drawn like a zener as I have shown.
Resistor R_{S} still limits current, and still must meet the minimum and maximum limit, but now resistors R_{TOP} and R_{BOTTOM} divide down a portion of V_{OUT} and feed it back the the reference pin. Inside the TL431 are active transistors, and the Ref pin allows V_{OUT} to vary from the reference voltage up to V_{IN} minus about 1 volt. That 1 volt is the so called dropout voltage and we will discuss that in detail in the next slides in the video segments.
Time: 1:53sHere I’m showing the very basic TL431 based shunt regulator solution. Here on this piece of protoboard (prototyping board) I have the actual TL431 and R_{S} current limiting resistor and on the backside is a blue tenturn precision 50kΩ potentiometer.
So this is both, R_{TOP} and R_{BOTTOM}, or R_{A} and R_{B}, so if I was to turn this dial I would adjust the output voltage. I have adjusted it to 5 volts which would be typical for something like an Arduino.
Time: 2:33sThis is the unregulated power supply to a phone that my cat decided to kill, so if I switch it ON, then we can see here we have just under 10 volts input and 5 volts here at the output.
The next test I am going to do is apply a load and show that this regulator still maintains the output voltage under load.
Time: 2:55sHere is the same circuit but now its loaded by 75Ω’s that two 150Ω power resistors placed in parallel. You can see that the input voltage has dropped a little bit but the output voltage is still maintained.
Once again, the shunt regulator based on the TL431 is connected to its 66mA or 75Ω load, and what I want to show here on the screen is the ripple in yellow, that’s the input voltage, and how nice and smooth the same volts per division the output voltage is. So that’s really what the shunt regulator is doing for us.
Another example would be if you are using an Arduino power supply, then this could clean up a voltage that’s both too high and has too much ripple and make it nice and smooth and give the 5 volts that the Arduino would want.
Time: 3:43sHere’s the last test for the shunt regulator and is a good explanation of why shunt regulators aren’t used except for very low power situations. So I have changed things around, now this multimeter is actually measuring the input current. So the unregulated power supply comes in here, gets measured as we are now using this as an ammeter, and goes back into the circuit.
Right now I have the load connected and we can see its drawing about sixty milliamps, (60mA). What I am going to do now is disconnect the load and you can see that there’s a transient but then the current goes straight back to normal. The upper voltage stays the same, but because of this resistor here, the shunt regulator is always drawing load current. So if your circuit is operating at noload you are still using power and some might say wasting power.
Time: 4:33sThese more sophisticated linear power supplies are known as “series regulators”. As the name implies, a transistor operating in its linear active region goes in series with the load. The current limiting resistor R_{S} is not needed here and that saves some power.
Time: 4:48sThe circuit on the left is similar to the zener shunt regulator in that it produces one and only one output voltage value. However, inside the 7809 there are series regulators with fixed output voltages are a pair of feedback voltage divider resistors like R_{1} and R_{2} in the right hand circuit.
In both types, the feedback circuit adjusts the voltage across the active terminals of the transistor. This transistor is often called the pass element since it passes the current from the input to the output. The active voltage is continually adjusted to maintain the desired output voltage, and such a circuit is also called a voltage or a potential divider.
Another way to think about this is to imagine a resistor divider with the top resistor, R_{TOP} is actively adjusted and the bottom resistor, R_{BOTTOM} is the load.
Time: 5:30sHere I am showing the LM317 series linear voltage regulator. It is the same pcb with the device itself. This is a minimum load resistor this device needs between two and three milliamps to regulate properly, but that’s a lot less than the 60mA we were drawing with the shunt regulator, and again here is the 10turn potentiometer to adjust the output voltage, and I have adjusted it to give 5 volts at the output.
It’s finally time to make some actual use of that ATX power supply we turned into a bench power supply. So I am going to use the 12 volt input here, and there is a very noisy fan so you can tell that I am actually using it, and here is 12 volts at the input and 5 volts at the output.
Time: 6:23sOnce again, the LM317 series linear regulator is going to be connected the the four ohms of power resistors here, and if we look at the oscilloscope screen, again in yellow, that’s the input voltage, this is now the output of a switching regulator.
Notice that the time division is much tighter because this is not 100Hz ripple its probably 100kHz ripple and some of the noise gets to the output, but the blue, which is the output, is much smoother.
Time: 6:51sA series regulator is generally capable of a lot more current than a shunt regulator and a LM317 is capable of more than 1 amp. I have switched things around here and now the orange multimeter is measuring my output voltage while the blue multimeter is measuring the output current.
Here I have these two 8Ω power resistors in parallel to give a load of about 4Ω. So when I connect them in, the circuit starts to draw over one amp, and now the control here is ok, but keep in mind that there is no output capacitors here and we are not doing kelvin sensing (4terminal sensing) of the load.
Time: 7:28sI will have to do this experiment quickly as my linear regulator is quickly overheating, so here it is with the load and we can see more clearly how much ripple the switching power supply has and that’s due to the circuit overheating.
Time: 7:40sThe 780x series and LM317 regulators are often called “NPN Regulators” because their pass elements are npn bipolar junction transistors. They are great parts and some of their designs are over 40 years old and still going strong.
But their biggest drawback is their large drop out voltage. That’s the minimum difference needed between the input voltage and the output voltage to keep the circuit regulating properly, and its about 2.5 volts for NPN regulators. An LDO, which stands for “low drop out regulator”, uses PNP transistors or more commonly MOSFET’s to allow the maximum output voltage to get very close to the minimum input voltage. Some parts get close to 100mV’s or less.
That’s perfect for modern circuits where you might want to drop from 1.8 volts down to 1.5 volts. One example is the circuit shown on the left with the simplified block diagram of the inside of the LTC3025 shown on the right. In theory this circuit can regulate down to a drop out voltage which is just above the load current multiplied by the ONresistance of MOSFET, M1 (V = I_{LOAD}*R_{ON}).
Time: 8:43sI am now showing a genuine low dropout voltage regulator built with the LT1575 and this one is not that common as it has the control chip here and actually has a discrete power transistor and then a MOSFET. You can see this giant heatsink clearer when I do the heat experiment and we will compare this to the LM317 that had no heatsink.
Right now there is no load and I have all four of these 8Ω power resistors in parallel to give me a 2Ω load. The input voltage is nominally the 3.3 volts coming from the ATX bench power supply, and I am also using the 12 volt to actually power the control section and that is the way that this chip is actually able to get to such a low drop out voltage as its the higher voltage that actually powers and drives the gate of that nMOSFET.
So what happens when I actually connect the load, (2Ω) we can see that the input voltage collapses. Now that is not because the ATX power supply isn’t capable of giving all the current, the problem is the voltage drop in all these long thin wires.
What I want you to notice is that the output voltage only drops to 2.1 volts. It is suppose to be 2.8 volts as it is with no load, but with a load we can see that the drop out about 100mV’s or so.
Time: 10:09sI have setup my low drop out regulator once again to show that the input current and the output current are related almost directly in a linear regulator. The yellow multimeter is the input current and the blue multimeter is the output current and this device is an LM317 being used as a constant current source.
You can see that there is almost no current being drawn at the output, the input is drawing about 150mA and that’s due to this minimum load resistor down here. However as I start to increase the load on the output, notice that the input current is equal to the output current plus the 150mA, or so. As I turn it all the way up to its maximum, you can see that they track.
Time: 10:50sIn part two, we saw that the bulky line transformers of the unregulated power supplies barely heated up at all even at their maximum loads. With linear regulators heat is a much more immediate concern. The power dissipation is easy to predict as its equal to V_{IN} minus V_{OUT} multiplied by the load current, (V_{IN}V_{out}*I_{L}).
Engineers take the worst case into account, the maximum input voltage, V_{IN(max)} and the maximum load current, I_{L(max)}. As a general rule, semiconductor packages can take about 1 watt before they can get too hot. Anymore than 1 watt, then a heatsink or forced airflow (fan) is needed.
But how hot is too hot but that depends on many factors like ambient temperature, the air flow, the presence of sensitive components nearby, like aluminium electrolytic capacitors, but also something more basic, how long the power supply needs to last.
Time: 11:38sI am running a heat and power dissipation test on my discrete LDO regulator, so I am back to my 2Ω load and I am now using a full 5 volts from my ATX power supply at the input. I have the 2.8 volts out and the output load current is 1.32 amps. The thermocouple is put inside the clip of this big heatsink and reads only 35^{o}C.
I will remove it and stick it on the tab which is the hottest part of the MOSFET to see if I can see what the temperature there is. Now that’s heating up much more quickly. The tab is probably getting to somewhere around 40^{o}C or so, and to me that is fine as long as it doesn’t get to over 50^{o}C then I think it is perfectly safe.
Time: 12:39sThat concludes part three of power supplies for nonEE’s. Stay tuned for part four where we will look at switching power supplies and by far the most exciting topic.
On behalf of myself and ElectronicsTutorials.ws, thanks for watching and how to see you for part four.
End of video tutorial transcription.
You can find more information and a great tutorial about linear power supplies by following this link: Variable Linear Power Supply.
In part 4 of our video tutorial about power supplies for beginners, we will look at using Switching Power Supplies and see how Buck and Boost converters can increase (boost) or decrease (buck) the output voltage.
]]>In part 1 of our series of video tutorials for beginners on power supplies, we explained how you can get set up to test, modify and use power supplies without spending a fortune on expensive equipment. Here in part 2 of our video tutorial series we look at testing and using Unregulated Power Supplies.
Time: 0:00sHello I’m Chris Richardson, and I’m an electronics engineer focused on power supplies. This is the second part in a series of web seminars for power supply enthusiasts who aren’t necessarily trained as Electronics Engineers.
In Part 1 of the series I talked about the basic supplies needed to start testing, now lets look at some power supplies that don’t actively control their outputs. This type of supply, an unregulated supply, is less and less common because its getting more and more affordable to make regulated power supplies, but we can still learn a lot from older unregulated supplies.
Time: 0:31sIn this video we are going to see where the unregulated power supply can still be found, but its more and more rare to actually find them. Then we will look at the heart of most of these supplies which are based on transformers that operate at the AC (Alternating Current) line frequency, either 50Hz or 60Hz depending upon where you live.
Since transformers take an alternating current (AC), or voltage, and also up the alternating current or voltage, just about every supply needs to be rectified, meaning that the AC is turned into DC (Direct Current).
By their nature, unregulated power supplies allow their output voltage to change as their output current changes, so we will test this on some real supplies. Then we will explore the universal challenge of all power supplies, heat!
Time: 1:11sSo here are three power supplies. The first one was salvaged from a telephone, the next from another phone. The first is an unregulated line transformer based power supply and the second is a switching power supply. The first one is 6.5 volts at 500mA and the second one is 6.5 volts at 600mA. So notice the difference, but the big thing is the weight.
If you want to know the difference between a transformer or line transformer based power supply and a standard one, you just have to see how heavy it is. The first one weighs 220 grams, and a device that’s physically smaller and provides actually more power based on a switching regulator weighs only 55 grams. The last one here, is an unregulated power supply that I made myself and its even heavier at 338 grams.
Time: 2:15sI’m getting ready to test my discrete unregulated power supply here, but before I do I want to make a very important note about electrical safety. Here at the AC input we have “Earth”, those are the tabs there in brass, and whenever you test with an oscilloscope or most pieces of lab equipment, the negative connection, the silver there that we see on the oscilloscope is also earth, in fact here is a connection.
So I have my tip there connected and I am going to use the testing function here, so if I go over and touch the tabs, sure enough its electrical earth, and the reason I say this is that we can not connect the oscilloscope to either of the AC inputs, that would be basically shorting earth to line or to neutral. We would cause the differential circuit breaker to trip, or cause a lot of voltage or current across the probe which would go through the oscilloscope and likely damage something.
Time: 3:17sAnother important thing to note, if we wanted to test across both sides of the diode bridge it would not be possible with this oscilloscope and two standard nonisolated probes, because again if we were to connect one probe here and another probe on the other end of the diode bridge we would be short circuiting it.
Time: 3:36sIf you are a conscientious environmentally inclined person then you would take your old unwanted electronics to a recycling centre. But you might still have a bag or a box somewhere with old electronics you have not gotten around to recycling yet. Look in there for the wall adaptors and find the heaviest one.
Time: 3:52sHere I have my unregulated power supply setup, I have got my two multimeters. This blue one is going to measure the input voltage, the orange one is going to measure the output voltage at the output terminals of the transformer. An important note as far as safety goes, is that this multimeter will go up to 750 volts AC rms and the second one is safe up to 250 volts AC rms so we will not destroy anything.
So when I switch it “ON”, we have approximately 230 volts at the input, again AC rms, and the device says 24 volts AC, but we actually got 27 volts, but that’s normal, that’s typical. When we load it to its 12 voltampere rating it should be closer to 24 volts.
Time: 4:49sIn general, the lower the frequency the bigger the transformer will be for a given power level. 50 or 60 Hertz transformers are therefore big and heavy because they need a lot more inductance to operate at such low frequencies.
In comparison, switching power supplies that we will talk about in parts 4 and 5 of this webinar series operate at one thousand to nearly one hundred thousand times higher so their transformers are much smaller and lighter, oh and cheaper.
Time: 5:16sThe same experiment again, but this time the oscilloscope probe is testing the actual DC output voltage. Here is the multimeter reading about 38 volts, and what I have done is to take the oscilloscope and put it into AC coupled mode at only two volts per division. So you can see here that it is very, very smooth and that’s because there is no load.
Now our unregulated power supply is loaded by this 330Ω power resistor. So from the positive output it goes to the resistor then it goes into the blue multimeter, and this one is actually measuring DC current, there we can see 93mA. The other multimeter is measuring the DC output voltage (31.6V) and from the blue multimeter the voltage returns back to the load.
Another important thing to note now that our circuit is under load, we can actually see some ripple. I did have to zoom in and now this is now 500mV per division, but we can definitely see a difference between the loaded and unloaded case.
Time: 6:14sPretty much all modern electronics run with DC, Direct Current, so rectifiers are employed to convert the AC outputs of the line transformers into DC. In general there are three things that destroy microchips or electronics.
Time: 6:28sToo much voltage is the first and most common, a negative voltage connected where a positive voltage should be used is the second cause and that’s where the rectifier comes in. The third cause of death is heat, and we will discuss that towards the end of this webinar and at the end of all the remaining webinars as well.
Discrete diodes and diode bridges come with voltage ratings for the peak or DC voltage they can handle in reverse, and they come with current ratings for the DC or RMS current. In most cases exceeding the voltage by even a little bit destroys the diode almost immediately, whereas too much current causes too much heat. That can destroy the device, but it usually takes longer.
Here is the unregulated discrete component power supply, but now the diode bridge has been taken out of the circuit and replaced by this single rectifier diode here. At noload we still have about 38 volts at the output and we can see with noload the output voltage is very, very smooth.
Time: 7:25sOnce your AC is rectified into DC it still has a lot of ups and downs in terms of output voltage and it gets worse as you apply more and more load. Applying rectified AC directly to most electronics will work sporadically at best and will destroy your electronics at worst because the peaks of the output voltage can easily be too high and cause over voltage.
A large capacitor, usually hundreds of microfarads or millifards absorbs charge while the rectified AC is above the desired output voltage and then supplies that charge to the load when rectified AC is below the desired output voltage. With enough capacitance the output begins to look very smooth.
Time: 8:02sHere is the halfwave rectifier circuit again, but this time with the 330Ω power load connected in. Drawing just under 90mA, the voltage is about 30 volts, and most importantly we can see quite a big difference as there is a lot more ripple at the output voltage now.
Time: 8:20sAfter rectifying and smoothing there is still the drupe or loss of output voltage as more and more load current is drawn. This is why the outside of the wall adaptor says “6.5 volts at 500mA” because it has been tested to have that voltage at that current but the average output voltage will rise at lower loads and sink at higher loads. Whatever device is powered by this adaptor must either have a steady constant current draw or must be able to withstand the changes in that voltage.
Time: 8:48sHere in this experiment, what I am doing is testing the transformer at the maximum voltamperes that it is capable of doing. So on the back of it that we can not see now it says a maximum of 12 voltamperes (12VA). We know that when we connect it to the line we have 230 volts rms.
So 12VA divided by 230V is about 52mA (0.052A) and this is as close as I could get with the type of variable load which I have here called a constant linear current source and you can see this in more detail in part 5 of the series, so I have jumped ahead a little bit to make a point.
When loaded to its maximum we have about 25 and a half volts here, and also when load at the maximum we can see that there is more ripple at the output.
This ripple can be reduced by having more output capacitance but we can see that the approximately 24 volts that was listed as the secondary voltage at maximum load is approximately correct.
Time: 9:43sOur basic unregulated power supply consisted of a transformer, some rectifier diodes and a whole load of capacitors and of these three components, the capacitors are the most sensitive to heat.
To get all the capacitance needed the type of capacitor used is almost always an aluminium electrolytic. These have a liquid or gel inside called the electrolyte and over time it evaporates. Once its gone the capacitor isn’t a capacitor anymore, its just a resistor. Now the hotter the air on the capacitor and the more the capacitor heats up having current pass through it, the shorter its usable lifetime.
A lot of electronics enthusiasts are familiar with the term “recapping” and this refers to saving a piece of electronics and replacing all the dried out aluminium electrolytic capacitors.
Time: 10:25sOne of the last things I am going to do is to test how hot all of the different components in my discrete semiregulated power supply get. So what I have done is connect it to the maximum load, in the previous test I checked to make sure we were drawing 12 voltamperes by watching the current at the input.
Now I am watching the current at the output, about 320mA we can see that the voltage is following approximately the nominal voltage, 25.5 volts in this case, and if you are wondering what was the use of that ATX power supply I turned into a lab supply I am now using it to run this DC fan.
Time: 11:01sNow I am not blowing the air onto the actual power supply, I am blowing the air onto the load. The load again is basically a 10turn precision potentiometer connected to a linear regulator makes me a current source but the watts that I am dissipating here are a lot higher here than what it can normally take, so the air is keeping this part cool.
Here is my thermocouple and the tip is resting in free air right now and you can see it is a pretty hot day here inside my house its almost 30 degrees Celsius. So what I am going to do is grab the tip here and I am going to touch it to the tops of different components. So for example, the actual transformer top is maybe 33.5 degrees or so, so at the maximum load it is not getting very hot.
However what is even more interesting is to look at the aluminium electrolytic capacitor and that device is about 32^{o}C. So overall this power supply is not heating up very much at all. The two critical components are only maybe 2 to 3 degrees, maybe 4 degrees higher than the ambient air temperature and they are also below the 40^{o}C maximum listed on the transformer.
Time: 12:23sThere is one other component in this power supply whose temperature I want to measure and that’s the diode bridge I am looking at here. Again, its about 30^{o}C ambient temperature.
When I do the temperature test here I need to be very careful as the blue and the black wires we see here are connected to the 230 volts AC. So I definitely do not want to short that with my hand. So very carefully put the tip on the component and we can see that we get to almost 39^{o}C, almost 40^{o}C, so this is the hottest component in the circuit.
Time: 13:01sThat concludes part two of power supplies for nonEE’s. Stay tuned for part three were we will look at regulated linear power supplies.
On behalf of myself and ElectronicsTutorials.ws thanks for watching and we hope to see you again for part 3.
End of video tutorial transcription.
You can find more information and a great tutorial about unregulated power supplies by following this link: Unregulated Power Supply.
In part 3 of our video tutorial about power supplies for beginners, we will look at using Linear Power Supplies and see how series and shunt linear regulators are better at controlling their output.
]]>A Video Tutorial for Beginners – Getting set up to test, modify and use power supplies without spending a fortune.
Time: 0:00s Hello I’m Chris Richardson, and I’m an electronics engineer focused on power supplies. This is the first of a series of videos for viewers who aren’t necessarily Electronics Engineers but, want to learn more about test and used power supplies.
If you are a student, hobbyist or someone who needs to modify power supply for pretty much any reason, electronicstutorials.ws and I hope that these videos will get you started.
Time: 0:25sOne important goal of this first video is to show you some basic items that will help to test a power supply, but to do so without spending thousands of dollars or euros or the equivalent wherever you are watching this from. I have put together a list with some of the approximate costs here in Spain where I live and work.
Time: 0:43sHere I’ve gathered some of the basic supplies needed to work with and to test power supplies, so, wire strippers, some clippers, some thin tweezers for grabbing small components. Here two silver box power supplies that I salvaged from an old PC, two different old PC’s. This one is a very old one, it actually has a 20pin connector, and here we can see after modification. I will get into the details of this one later.
Time: 1:13sIf you focus in closely, a silver box power supply will tell you how much power it can provide overall and also how much of the different voltages it gets. Also salvaged from some old PC’s were two DC fans, these run off 12 volts plus they come with a convenient connector also. Seems like something basic, but for the basic plugs here you can turnon and turnoff at the switch, very nice.
Time: 1:53sA soldering iron, one that has a fairly thin tip that will allow us to solder some small components. Some fairly thin solder and of course safety goggles.
As far as electrical tools go, I like to have two multimeters, they come with these kind of tips here. Two multimeters are good for measuring two voltages, but also for measuring either a current and a voltage, and at least one wire that has a banana plug on one end, an alligator or so called grabber clip on the other end.
Time: 2:27sThe last tool here looks a lot like a multimeter but this is actually a thermocouple, I’m going to turn it on. What is does is actually measure temperature, measures the temperature out of the tip here. I am using a plastic workbench here, so this is the kind of thing you can find anywhere, nothing special.
Time: 2:46sIn planning for this series of videos, I debated very seriously the topic of using or not using an oscilloscope. Do a quick search and you’ll find plenty of devices like the one on the screen that attach to your PC and make it into an oscilloscope. In the end I decided that this was better than nothing because actually seeing some power supply voltage waveforms really helps to understand them.
But be aware that 20MHz, even though it sounds high, isn’t enough to see many of the so called transient effects on power supplies. That means things that happen very quickly so, during these videos we will stick to things that happen mostly in steady state.
Time: 3:21sHere we have an oscilloscope that is not the 60 euro model I talked about, its a slightly fancier one. But what I am going to do to make my waveforms that I show in these presentations more realistic and closer to what you would see in the cheaper model you can get off of the internet is to do two things:
One, these aren’t the probes that came with my fancier oscilloscope these are some lower quality probes. A lower quality probe has lower output resistance or impedance and has higher output capacitance. Those are the things that will distort the waveforms.
The other thing I will do is, focusing in here you will see the BW written there stands for Bandwidth. That means that the oscilloscope is bandwidth limited to 20MHz. That’s the same limit that the cheaper oscilloscope has. So that will make the measurements I show closer to what you see if you had the less expensive device.
Time: 4:19sJust about everyone will have and old PC gathering dust in their basement or attic. The floppy disk drive may be useless, but that power supply, the so called silver box may still be good. As the web based tutorial on Electronics Tutorials shows, an ATX Power Supply provides a whole host of different voltages and quite a bit of power.
Also take a moment to remove any fans you find in the case of your old PC. Those will be great later for blowing air and keeping your power supplies and other electronics cool.
Time: 4:48sHere is the pin of an ATX Power Supply. This actually has the 20pins of the older ones and the four extra pins too attached. Its coming of a power supply that was donated to the cause. Of course there are lots of extra wires connected. One thing to keep in mind is that they are colour coded.
Reference: ATX Power Supply Tutorial
Pin  Name  Colour  Description  
1  3.3V  Orange  +3.3 VDC  
2  3.3V  Orange  +3.3 VDC  
3  COMMON  Black  Ground  
4  5V  Red  +5 VDC  
5  COMMON  Black  Ground  
6  5V  Red  +5 VDC  
7  COMMON  Black  Ground  
8  Pwr_Ok  Grey  Power Ok (+5 VDC when power is Ok)  
9  +5VSB  Purple  +5 VDC Standby Voltage  
10  12V  Yellow  +12 VDC  
11  3.3V  Orange  +3.3 VDC  
12  12V  Blue  12 VDC  
13  COMMON  Black  Ground  
14  Pwr_ON  Green  Power Supply On (active low)  
15  COMMON  Black  Ground  
16  COMMON  Black  Ground  
17  COMMON  Black  Ground  
18  5V  White  5 VDC  
19  5V  Red  +5 VDC  
20  5V  Red  +5 VDC 
Time: 5:08sEvery single wire that is Yellow delivers positive 12 volts (+12V). Every wire that is Black is ground or the reference (0V). Every Red wire is 5 volts (+5V). My suggestion is to follow what the tutorials says as the main connector also has some negative voltages so this is the one we will actually cutoff.
Here is the other silver box power supply after I cutoff the main connector and converted into this breakout pcb. You can see here that I have these spring loaded clamps that allow me to connect different wires.
I have soldered a lot of the wires parallel here to give me more power. This particular ATX power supply doesn’t have a switch on the back, so when I want to turn it on I am going to use one of my little independent switches here.
When I do, we don’t hear anything. The fan is not running and that’s because it actually has an onoff switch, that’s the Blue wire here. Switch it “on” and now it makes lots of noise. It’s definitely running and I have switched out the negative lead so that I can test the different voltages with the multimeter.
Negative 12 volts (12V). Standby power, this is always on even if I flick off the switch. Minus 5 volts (5V). The power good signal is a logic level signal that actually tells us whether on not the power supply is operating. Also notice the positive 5 volts (+5V), the positive 12 volts (+12V) and the positive 3.3 (+3.3V) don’t have a particularly great tolerance, and that’s because there’s not much load.
Meaning to say that when they are delivering much current and in this case they are delivering almost no current they are not particularly precise. That will improve once they start to deliver some power.
Time: 7:20sI used some so called perfboard (perforated board) to make the back side of my connector for the ATX power supply here. It lets me put lots of wires in parallel. In this case I used a type which has a 2.54mm or 100 mil pitch and the rows are all connected together in parallel. Here is another kind of perfboard which is good for other types of experiments, also with a pitch of 2.54mm or 100 mil but with each little square separated from its neighbours.
Time: 7:46sEarth in the context refer to the potential of so called Safety Earth, or Protective Earth. That’s the third connection in your wall power outlet. In the European Union (EU) there are little tabs in each electrical socket is the only connection that your finger can touch easily because it is perfectly safe to do so.
In fact if your work space is a plastic or wooden table like the one that I’ll be using, then you want to earth yourself by touching an earth connector regularly especially before handling any semiconductor microchip or anything else that is sensitive to ESD. That’s Electro Static Discharge.
Time: 8:18sSince I am using a plastic workbench here, it could build up electro static discharge or ESD. So, what I want to do is earth myself fairly regularly, especially before I touch any semiconductor chips. I am using the continuity tester, the beeping function of my multimeter here and I am connected to the earthing clip which I can touch with my finger.
The actual power supply is connected through the cable and in theory is a device which the case should be connected to earth. So I can take the other end of my multimeter and test. If I press hard enough to go through the coating I can see it does, but what I want to do is to touch the screws as they are connected to the frame.
So when I go ahead and do any actual testing, regularly I will reach a finger over here and actually touch to discharge any ESD built up on my body before I transfer it to anything sensitive like a semiconductor chip.
Time: 9:18sAs the next video segment shows, a charged capacitor with nothing to drain the voltage out of it can stay charged for a long time. A typical practical joke among Electronics and Electrical Engineers is to charge up a capacitor and then hand it to someone who is not expecting it.
I do not recommend you try that at home, and the charged capacitor phenomenon is why a lot of electronics still recommend that when you need to reset them, you turn them off, wait a while, and then turn them on a again. That’s to allow all the internal capacitors to discharge to zero to be sure that anything digital inside the device is actually turnedoff.
Time: 9:53sTo demonstrate how a capacitor that is not loaded or not connected to anything can hold charge for a long time, I am going to use a laptop here and its charger.
This laptop battery is almost dead, so wants some power and when I turnon the charger, they included a little white LED here that turns on to let us know that it is charging.
Time: 10:13sThe laptop itself is drawing a lot of current, so when I turn it off, the LED begins to fade. When the LED has faded all the way we know that the output capacitor, and there is a lot of output capacitance in a power supply in this laptop adaptor is completely drained.
Now if I disconnect it, and perform the same test, the LED turns on and when I disconnect it, nothing seems to happen. That’s because the LED is barely drawing any current at all and there is a huge amount of capacitance. Millifarads, (mF) that’s to say thousands of microfarads, (uF) of capacitance here. It’s going to take possibly a minute or so for this output capacitance to completely discharge.
Note: 100 seconds of the LED’s intensity reducing intentionally removed from video to save time.
We can see the LED getting dimmer here, but the moral of the story is, whenever a capacitance is charged up and there is no load on it, it may still be charged minutes later so you need to be careful especially if its charged up to a voltage higher than say 30 to 40 volts. That’s enough to give you a nasty shock.
Time: 11:25sThat concludes part one of power supplies for non electrical engineers and hopefully you are now set up to begin testing an actual power supply. Stay tuned for part two were we’ll look at unregulated or semiregulated power supplies, just to be clear, the ATX we have converted is a regulated power supply.
On behalf of myself and ElectronicsTutorials.ws, thanks and see you next time.
End of video tutorial transcription.
You can find more information and a great tutorial about converting an old computer ATX power supply into a bench power supply by following this link: ATX to Bench PSU.
In the next video tutorial about power supplies for beginners, we will look at using Unregulated Power Supplies and see how an unregulated power supply has difficulty controlling its output.
]]>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
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.
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 nΩ.m), whereas the resistivity of a poor conductor (insulator) such as air can be well over 1.5 x 10^{14} or 150 trillion Ω.m.
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.
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