power supply operation

Operation of the Power Supply Circuit

This section covers the functioning of the circuit. It is organized into several topic headings that are listed below. The professor will cover these questions in class.

Overview

The figure below shows the incomplete schematic circuit. The Circuit can be broken into AC (Alternating Current) and DC (Direct Current) sections as indicated in the figure. The circuit on the Printed Circuit Board (PCB) can also be broken down into the functional areas as indicated in the diagram: a filtered rectifier circuit; fixed 5V regulator circuit; positive variable regulator circuit; and negative variable regulator circuit. The components to the right of the printed circuit board consist of: the power LED, which is on whenever the power-switch, SW1, in on; and, the binding posts for the output voltages that will be installed in the front face of the power supply case.

Depending on the program in which you are enrolled, some students are in first year and others in second year. The second year students should be familiar with all the components in this schematic; whereas first year students will be studying the electronic components such as diodes and regulators in the Electronic Circuits course at the same time as taking this course. The understanding of how such components function in the circuit will be left until later in the course by which time the first year students will have covered them in the Electronic Circuits course. At the beginning of this course, the emphasis will be the functional

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DC Part of Circuit on PCB Board

AC partOf Circuit

Filtered Rectifier

Positive Variable Regulated

Negative Variable Regulated

Fixed 5V Regulated

areas, i.e. as indicated in the diagram, and other more general topics such as safety and electricity.

The format here will be a discussion of the topic headings given below and at the end of each section are questions to test your understanding.

Safety AC Circuit Filtered Rectifiers Fixed 5V Regulator Circuit Positive Variable Regulator Circuit Negative Variable Regulator Circuit

Safety

Safety is the term used to refer to issues that can potentially result in physical harm to people. So what do we mean when speaking about safety in the design of electrical/electronic products such as the power supply that you will be fabricating in this course, other than mechanical considerations such as avoiding sharp edges? Normally, whenever working with electricity the main safety concerns deal with higher voltages. If only low voltages are present in a product such as battery-powered portable devices such as cell phones, then the only real electrical safety issue would likely be the effect of the battery leaking or exploding; there is very little likelihood of personal injury occurring to anyone fabricating/operating/repairing such a product.

So what is considered high voltage? This question does not have a definitive number as an answer but there is little doubt that the supply voltage of 120 Vac to the power supply does pose a potential lethal threat to your life under the right conditions. If it is a dry day and you’re in good physical health and you happen to touch a live wire and a connection to ground with the fingers of the same hand, there is very little threat to your life. The current, even if at a dangerous level, will pass through your hand and possibly result in local cell damage not to mention a painful sensation that will be evident by the possible rapid flow of expletives from your mouth. If however, you touch the live wire with one hand and a connection to ground with the other hand, the resulting current flow may pass through your heart, possibly interrupting the electrical impulses that control the heart…

There certainly have been cases of fatalities due to line voltages of 120 Vac but they are extremely few. The best protection from such an event, even if extremely unlikely while working with the power supply, is knowledge and understanding. In a nutshell the part of the power supply that represents the most significant safety hazard is the alternating current part since it is at 120 Vac and that the PCB part of the assembly is comparatively harmless. The only exception to this statement would be the danger of a capacitor exploding.

Safety Hazards for Power Supply: 120 Vac line voltage

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Exploding capacitors

As for the 120 Vac hazard the student should be careful if they are troubleshooting with the cover removed while still plugged into the wall. If there are any exposed wires at the fuse holder or the power switch you may receive a shock if you come in contact with them. Remember that the only time you will feel anything that could be called an electrical shock is when you provide a path for the current from the live wire, the black one in the power cord, to ground, either through the white also called the common wire in the power cord or the ground wire both of which are connected together at the fuse panel. So what this means is that simply touching the live wire with your finger, without having some other part of you body touching the grounded metal chassis or a common lead inside the chassis will not “give you a shock.” Whenever you suspect a potential high voltage hazard a simply measure to reduce the risk of a current path through your heart is to keep one hand behind your back, thereby the current is unlikely to travel up your arm.

To receive an electrical shock there must be a path to ground

Regarding the exploding capacitors, this can occur when the electrolytic type, which is the type used here, is installed backwards. This is easily avoided for the large ones used as filters in the rectifier circuit by following the template provided for the PCB traces and components in the second Module. The smaller ones of the axial type can be easily installed incorrectly but it is most unlikely that they will explode anyway. Regardless, make sure that you know which way they should be installed at that time.The following questions are given to test your understanding of the material presented above.

1. What is the main safety concern in this project?2. What other safety concerns exist in this project, in particular

regarding a certain type of capacitor?3. Why is it important that the black wire from the AC power cord be

connected to the fuse and not the white wire?4. Explain how grounding the case protects the consumer;

specifically mentioning what could happen if the case were not grounded.

5. Which of the two electrical connections on the fuse holder should be connected to the black wire from the power cord, the one near the end where the fuse is inserted or the one at the other end? Explain. Think of a incandescent light bulb receptacle for comparison.

6. Is there a hazard if someone touched the white wire from the power cord while it was plugged in a wall outlet? Explain.

7. What would happen if the wire between the power switch SW1 and the transformer were to inadvertently touch the metal case (a) assuming the switch is off, (b) assuming the switch is on?

8. What is the purpose of the ground prong of the AC plug?

AC Circuit

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As is typically the case, schematic diagrams are drawn such that the current is flowing from the left to the right the same as how we read text. This implies that like a river the current source is on the left which in this case is 120 Vac and on the right the river ends as the current is absorbed by the vast ocean or “ground” only to be returned one day to feed the river.

Think of the schematic as the river and streams leading to the ocean The diagram below is extracted from the schematic and represents the part where the currents are alternating and some voltages are potentially harmful. We will look at each component in turn.

Figure ?? AC Circuit

Plug – The plug should be a three prong type where the centre prong is the grounding prong. As shown in the diagram where the green wire leads to the chassis symbol, the ground prong should be connected to the chassis (see photo below) thereby ensuring the chassis is always at the same potential as ground that we could call 0 volts.

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Safety Hazard -Live wires 120 Vac

No Safety Hazard

Low Safety Hazard –AC Current

Figure 1 Grounding

There wasn’t always a third prong on plugs and it was added as a safety measure. You might wonder how it provides additional safety. Before there were ground prongs, if a bare portion of a live wire such as the wires from the black wire in the power cord all the way to the transformer were to somehow come in contact with the metal case then the case would effectively be at the same potential as the live wire. This means that if you touched the case it was the same as touching the live wire… The ground prong, being connected both to the case via the transformer mounting flange and to the earth back at the fuse box, in such an event would result in the fuse or circuit breaker in the electrical panel going open circuit due to the massive current flow that would result the instant the live wire was to contact the grounded case.

The other feature of the plug that is not evident in the schematic is the polarity of the two flat prongs. No doubt you have all noticed that the two flat prongs are not the same size, unless the device is old or from outside of this country. One is wider than the other, the narrower one being the live one and connected to the black wire in the power cord. The wider one is connected to the white wire (wider leads to white??)in the power cord and back at the electrical panel it is connected ground which means that the white wire, in the power cord, and the green ground wire are at the same potential. This effectively means that the white wire that comes into the chassis could be left bare and unlike the black wire it can touch the chassis without presenting any safety threat.

Fuse – The next component in the AC circuit is the fuse, rated at 0.5A. The fuse is held in a fuse holder protruding from the back face of the chassis. The fuses function is to prevent excessive currents from flowing through the components of the power supply thereby damaging them. The maximum current according to the rating is 0.5 amperes. This means the maximum power supplied by this device will be the product of the maximum current and the applied voltage, 120 Vac in this case, giving P = 120 x 0.5 = 60 Watts.

The following questions are given to test your understanding of the material presented above.

1. What is the maximum power that can be delivered by this device?

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Green Ground Wire connected to transformer

Black power cord wire connection

2. What is the maximum current that could flow through T11 before the fuse would blow, assuming no losses between the fuse and the output of the regulator?

3. Why are both primary wires on the transformer the same colour?4. Why are there three secondary wires on the transformer? Do all

transformers have three secondary wires?5. How is the green/yellow secondary wire on the transformer

different than the two green wires?6. What is the average voltage between the two green wires? What

name is used for average AC values?7. What is the average voltage between either green wire and the

green/yellow wire?8. Which prong, the wide or the narrower one, of the AC power cord

is connected to the white wire or hot lead? How would verify your answer with a DMM?

9. Can the common lead on the secondary side of the transformer be connected to the white lead on the primary side? When would you not do this, regarding the respective voltages?

10. What would be the effect of changing the turns ratio of the transformer, a) on the output voltages, b) on voltage at TP4?

11. The rated secondary voltage of the transformer is 28Vac, but the measured value with a DMM indicates 33Vac. Explain the difference.

Filtered Rectifiers

The illustration below is extracted from the one at the start of this section. In the labeled ellipse, it shows the part of the schematic that rectifies and filters the AC current into DC current. The diodes are pn-junction diodes and basically their ability to conduct in only one direction is what makes it possible to convert AC to DC. This should have been covered extensively in your Electronic Circuits course, but will be reviewed here to see how well those lessons were learned.

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First the components included in the rectification and filtering will be identified and then their contribution will be discussed. As you already know, the diodes, D1-D4, are what change the AC to DC and the way this is done can be described in several ways. The AC comes from the transformer and is delivered to the printed circuit board at terminal connections T1, T2, and T3, where T2 is connected to the centre-tap of the transformer. At any given instant in time the connection at T1 may be positive or negative WRT the centre-tap and whichever it is the connection at T3 will be the opposite. The potential at T1 changes from positive to negative 60 times per second. When T1 is positive, any current flow will be from T1 to the centre-tap or T3 depending on where a load is present. If T1 is negative, any current flow will similarly be from T3 to the centre-tap or T1.

What happens once on the board is a little harder to follow but it is review of material covered in Electronic Circuits. A key fact to keep in mind is that the diodes separate the AC for the DC but be careful because for D2 and D2 the AC is at the anodes and DC at the cathodes; whereas for D3 and D4 it is the other way around. This is how we end up with positive DC at TP4 and negative at TP5.

The goal is not to repeat everything you have already learned but at least to give you an opportunity to check how well you learned it. An initial question might be to ask what type of rectifier is used in this circuit. If your answer is a bridge rectifier because there are four diodes, you are partly right. Whenever a load is applied between BP1 or BP2 and BP4 current would flow through all four diodes and thus you have a bridge rectifier.

If on the other hand a load is applied between BP1 or BP2 and BP3 then current only flows through D1 and D2 which would mean a full-wave positive rectifier is in effect. Can you comment on the presence of a full-wave negative rectifier and where the load must be for it to function?

Before TL1 and TL2 are connected, that would allow current to flow to the regulators, the output of the rectifiers is unfiltered since the filter capacitors are on the downstream side of the test links. You should recall that the average DC voltage at TP4 should be a function of the peak AC voltage from the transformer. How do we find the peak AC voltage? Once this value is determined and a minor drop of 0.7 volts is applied to the peak as the signal passes across D1 or D2, the average DC voltage is some fraction of the this peak. If it was a half-wave rectifier the peak is divided by a common irrational number that is related to circles to give the average DC voltage. For a full wave, since there is twice the area under the curve of the waveform the value for the half-wave is doubled.

Once the test links are inserted the capacitors, C1 and C2, have a profound affect on the shape of the waveform when viewed on an oscilloscope. It changes from two clearly identifiable humps, as seen in the AC waveform, to what almost appears as a straight line. Upon closer inspection it is apparent that the valleys between the peaks have been filled in as a result of the discharging of the capacitors. The variation in the voltage from the peak value to some slightly smaller value, depending on the magnitude of the capacitor, is referred to as the ripple voltage. A handy rule of thumb to use when looking for the output voltage

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of a filtered rectifier is to assume it is equivalent to the peak AC voltage, which can be found by dividing by 0.707.


1. What type of rectifier(s) is/are used in the power supply?2. List the components that make up the filtered rectifier(s).3. What is the purpose of the capacitors C1 and C2?4. What would be the result, if the value of the capacitors was

reduced to 10µF, in terms of ripple voltages?5. What would happen if C1 was to malfunction and the circuit went

open at this point, specifically to the voltage at TP4 on the schematic?

6. What type of rectifier is in effect when the load is applied across BP1 and BP3?

7. What type of rectifier is in effect when the load is applied across BP2 and BP4? Remember that the voltage in this case is twice the value across BP2 and BP3.

8. How much DC current passes through capacitor C1? 9. On a copy of the completed schematic diagram, using a

highlighter, show the path of the main power, from the power cord to the load, in solid line and in dashed lines show the control signals.

Fixed 5V Regulator Circuit

The illustration below is extracted from the larger one at the start of this section. It shows the portion of the larger circuit that is responsible for producing a consistent 5V at the load, regardless of load variations. This circuit is relatively simple and typically works successfully.

The components included in this function are the 7805 regulator, the small capacitor and the reverse biased diode. The regulator is of course an integrated circuit that includes many tiny components all packed onto a microchip. These components work together to achieve the goal of producing the desired output voltage, relative to the com pin, regardless of what happens to the load. You should recall that a zener diode is also used as a voltage regulator, in fact D5 does exactly that, but it performs relatively poorly when the load varies.

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Now we will look at the diode D10. This diode is there as a protective device. It is a very common feature with power supplies to place reverse biased diodes in parallel with the output. They are almost always reverse biased during normal operation except for when a load is switched on/off (or connected/disconnected). They are often referred to as "fly back", "clipper" or “transient protection” diodes and provide protection for the power supply output and the load. When a load that has some inductive properties is switched on and off, the stored inductive energy can cause an excessive voltage spike of the opposite polarity at the voltage output. This is often explained in an electronic textbook with the use of a speaker as a load. A speaker is of course a large coil of wire (inductor) with a magnet. An inductor…as you may recall…fights against current changes. When an inductor is switched off, it fights against the current change and can develop an extremely high voltage (delta I/delta t = e/L) or ( e= delta I/delta t x L). This phenomenon can often be demonstrated by recalling when you may have unplugged a device without shutting it off and observed a spark at the receptacle plug when it is removed. This spark is caused by the energy stored within the inductive properties of the load and the fact that the magnetic field is collapsing to zero in a very short period of time. The more inductive properties (L) in the load, the larger the spark. The inductor fights against this current change, generates a very high voltage, creates the spark (arc) and dissipates its stored inductive energy in the spark. Also… the voltage regulator switches on and off at a very high speed (many MHz range) to maintain its voltage output at the specified voltage. For example…if you looked at the output of a regulator on a scope you would see a rapid on/off high frequency waveform that varies by a small voltage…say 5 mV to give us an example to work with. So..if it’s a 5 volt regulator the output varies between 5.005 V and 4.995 volts at a very high frequency. This high frequency voltage causes the inductive properties of the load to switch charge/discharge at the same rate. The diode clips the stored inductive energy. Also… a small capacitor is typically placed in parallel with the output to help the regulator out with this rapid switching and regulating effect. Since the LED appears in parallel with the load at the output of the 7805, we will discuss it at this time. The LED is of course installed on the front face of the power supply, just beneath the power switch, SW1, to show when power is on. The LED is in series with the resistor R8 between the output of the 7805 and common.

Whenever power is on and there is 5 volts at the top of the resistor R8 and thus current flows through R8 and then the LED to common. The resistor acts as a current limiting device without which it would be impossible to produce the required 5 volts at the output of the 7805. If there was only the LED, then as for all forward biased diodes the potential drop across it would be fixed, in the case of LEDs at approximately 1.4V, and since the cathode is connected to common the rise

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in voltage toward the anode that would now be connected to the output of the 7805 would be +1.4V.

Can you predict the approximate current flowing through R8 and therefore also the LED? Remembering that the LED intensity is a function of the magnitude of current flowing through it, it should be evident that the size of R8 is so determined.


1. List all the components that are part of the fixed regulator circuit.

2. How do we know the output voltage of the 7805 from the name?3. Does a 7805 always produce 5 volts at the output?4. What would happen to the output voltage at T11 if D10 was

installed backwards?5. What would happen to the voltage at T11 if a 5 ohm resistor was

installed in place of the diode?6. What is the approximate minimum voltage at the input to the

regulator for it to continue outputting 5 V?7. How much current normally flows through the diode D10?8. What is the normal difference in voltage across D10?9. What would be the absolute minimum voltage at the cathode of D10?10. If diode D10 were to go open circuit, would it adversely affect

the operation of the circuit? If so, explain how.11. Explain how D10 protects the regulator through its function as a

waveform clipper.12. What is the purpose of C5?13. What effect would doubling R8 to 400 ohms have on the operation

of the LED and the output of the 7805?

Positive Variable Regulator Circuit (LM317)

This part of the circuit produces a positive regulated output, like the 7805, but the voltage is variable between 0 and 15V. The illustration below shows the components that contribute to this end in the labelled enclosed area. They include the LM317, diodes D6-D8, resistors R2, R3 (Potentiometer) and R4, and capacitor C3 (not show but pointing upwards).

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This part of the circuit requires some discussion. Let’s begin with the regulator as it is the heart of the operation; the other parts simply allow the regulator to do its job. As we saw with the 7805, the diode, D8, and the capacitor, C3, are at the output of the regulator to protect it and improve its effectiveness.

As you saw in Electronic Circuits, the output voltage of the regulator is approximately proportional to the ratio of resistances R3 and R2, i.e. the larger R3 is compared to R2 the larger the output is. The reason R3 is a potentiometer is so that the ratio of resistance can change and thus the output of the LM317. Another way to look at it is that the regulator produces 1.2V more at the output than at the adjust terminal, thus if the adjust was connected directly to common the output voltage would be 0 + 1.2V = 1.2V. If however this was the case it would mean that zero voltage could not be achieved at the output of the regulator.

Since an output of zero volts is required, a voltage less than zero volts is needed at the adjust terminal. This is accomplished by connecting R3 to the series circuit consisting of D6, D7, and R4, between common and -20 V at the output of the negative filtered rectifier. Note that this series circuit would forward bias the diodes and as such they would be conducting with a 0.7 V drop across each one for a total of 2 x -0.7 = -1.4 V at the cathode of D7. With R3 connected at the cathode of D7, instead of at common, the minimum output of the regulator when R3 equals zero will be -1.4 + 1.2 = -0.2 V. The actual output will depend on the actual drops across the diodes D6 and D7 which may be more or less than 0.7 V. Typically, the minimum output of the regulator is approximately -0.05V or negative 50 millivolts.


1. What role do the diodes D6 & D7 have in the operation of the regulator?

2. What is the voltage at the cathode of D7, with respect to the common or TP2?

3. Are these diodes forward or reverse biased? Explain why.4. What would be the result of adding a third diode between D6 and

D7 on the output of the LM317? Explain.5. What would be the effect of reversing the direction of D6 on the

output of the LM317? Explain. Would reversing both D6 and D7 be any different than just one of them?

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6. Approximately how much current flows through D7, which you may assume is the same as through R4?

7. What is the purpose of the resistor R4? 8. What would be the result if R4 were not there, i.e. replace it

with a jumper, in particular on the output of the LM317?9. What would be the output voltage of the regulator when the

potentiometer R3 was set to zero resistance? Show the calculation.

10. What effect would reducing the value of R3 to 500 ohms have on the output of the regulator?

11. Which direction does the diode D8 point? 12. What would happen to the output voltage at T9 if the diode D8 was

installed such that it pointed downward? Hint: Would it be forward or reversed bias and then what must be the drop across a forward/reverse biased diode?

Negative Variable Regulator Circuit (LM337)

This circuit is shown below extracted from the illustration at the beginning of this document. The components that contribute to the operation of this circuit are shown in the dashed enclosure and include the LM337, Op Amp UI741, R1, R5-R7, D5, D9, & C4.

It should be noted that the larger number of components in this circuit compared to the positive regulator is due to the fact that this circuit automatically tracks the output of the positive voltage regulator and ensures the output of the negative regulator is equal in magnitude. Thus the designation of the power supply as a dual tracking power supply.

The operation of this circuit is more complicated than the other two regulated output circuits but not much. Essentially it works the same as the positive regulated circuit in that the output at T6 is always 1.2 V

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greater than whatever the LM337 regulator senses at the Adj terminal. The capacitor C4 and the diode D9 work the same here as with the LM317 circuit. The resistor R7 works the same as R2 in the LM317 circuit; it allows the regulator to compare voltages at the output and the Adj terminals without creating a short. The remaining components: the Op Amp, resistors R1, R5-R6, and zener diode D5 all work toward producing the correct voltage at the Adj terminal of the LM337 regulator to obtain the required voltage at the output of the regulator.

The voltage at the Adj terminal is the same as that produced at the output of the Op Amp. Pins 4 and 7 of the Op Amp are the negative and positive, respectively, power supplies for the Op Amp. The output of the Op Amp can only exist in the range of voltages present at the two power supplies. The voltage at pin 4 will be the same as the input to the LM337 regulator which is approximately -20 V. The voltage at pin 7 will be the same as found at the cathode of the zener diode. Since the nominal zener voltage of D5 is approximately +5 V and the zener is correctly installed in reverse bias the cathode will be at +5 V and therefore the range of voltages available for the output of the Op Amp is necessarily between -20 and +5 V.

The details of the operation of the Op Amp are beyond the scope of this course and will be covered in your future courses. For our purposes we will simplify the operation similar to what was done for the voltage regulators although you should be much more familiar with their operation and at least can relate the explanation to your own experience.

Op Amps can be used essentially in two modes: as an amplifier, as suggested by its name; secondly, it can be used as a comparator. In our case the latter mode is relevant. Using the Op Amp as a comparator means that the input at pins 2 and 3 are constantly being compared and changes here are reflected in the output. These changes are always the result of a change in voltage at pin 2 since pin 3 is connected to common and thus will always be at the same potential.

The voltage at pin 2 is the same as where resistors R5 and R6 are connected. This connection forms a voltage divider between two resistances of the same value, thus the voltage will be one half of the potential difference between the top of R5 and the bottom of R6, which are at the potential of the output of the positive regulator, at T9, and the output of the negative regulator, at T6, respectively. If we suppose that the potential at T9 was +10 V and at T6 it was -10 V the voltage in between the two resistors should be (+10 - -10)/2 + -10 V = 0 V. This is same voltage as pin 3 and since there is no difference between them the output of the Op Amp will be kept at the same value.

If however, by changing the setting at R3 and thereby increasing the output voltage of the LM317 to say 12 V the potential between the two resistors R5 and R6 will become greater than 0 V. Now we have a situation where pin 2 of the Op Amp is at a higher potential than pin 3. This will cause the Op Amp to increase its output. As the output is increased the same happens at the output of the LM337 since it maintains its output at 1.2 V higher than the Adj terminal. Once the LM337 is outputting -12 V the voltage at pin 2 of the Op Amp should have returned to 0 V and everything is held there until such time the two pins 2 and 3 are different again.

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1. Why does this circuit have more components than the positive regulator?

2. What is meant by the expression dual tracking?3. List the components that enable the Op Amp to function.4. What is the voltage at pin 4 of the Op Amp?5. What is the function of pin 4?6. What is the voltage at pin 7 of the Op Amp?7. What is the function of pin 7?8. What is the voltage at pin 3 of the Op Amp?9. What is the function of pin 3?10. What is the function of pin 2?11. When is the voltage at pin 2 not equal to zero?12. When is the voltage at pin 2 equal to zero?13. What happens when the voltage at pin 2 is not 0 V, say it is +1

V?14. What is the purpose of the Op Amp device? 15. Is it being used as an amplifier or as a comparator?16. Which direction does diode D9 point?17. Which direction would current flow through D9?

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power supply operation

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