Power Supply Notes 1

8/3/2019 Power Supply Notes 1

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Power Supplies

Types of Power Supply

There are many types of power supply. Most are designed to convert highvoltage AC mains electricity to a suitable low voltage supply for electronicscircuits and other devices. A power supply can by broken down into a series ofblocks, each of which performs a particular function.

For example a 5V regulated supply:

Each of the blocks is described in more detail below:

Transformer- steps down high voltage AC mains to low voltage AC. Rectifier- converts AC to DC, but the DC output is varying. Smoothing - smooths the DC from varying greatly to a small ripple. Regulator- eliminates ripple by setting DC output to a fixed voltage.

Power supplies made from these blocks are described below with a circuitdiagram and a graph of their output:

Transformer only Transformer + Rectifier Transformer + Rectifier + Smoothing Transformer + Rectifier + Smoothing + Regulator

Transformer only
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The low voltage AC output is suitable for lamps, heaters and special AC motors.It is not suitable for electronic circuits unless they include a rectifier and asmoothing capacitor.

Transformer + Rectifier

The varying DC output is suitable for lamps, heaters and standard motors. It isnot suitable for electronic circuits unless they include a smoothing capacitor.

Transformer + Rectifier + Smoothing


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The smooth DC output has a small ripple. It is suitable for most electroniccircuits.

Transformer + Rectifier + Smoothing + Regulator

The regulated DC output is very smooth with no ripple. It is suitable for allelectronic circuits.

Transformer


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Transformers convert AC electricity from onevoltage to another with little loss of power.Transformers work only with AC and this isone of the reasons why mains electricity is AC.

Step-up transformers increase voltage, step-down transformers reduce voltage. Most powersupplies use a step-down transformer toreduce the dangerously high mains voltage(230V in UK) to a safer low voltage.

The input coil is called the primary and theoutput coil is called the secondary. There isno electrical connection between the two coils, instead they are linked by analternating magnetic field created in the soft-iron core of the transformer. The twolines in the middle of the circuit symbol represent the core.

Transformers waste very little power so the power out is (almost) equal to thepower in. Note that as voltage is stepped down current is stepped up.

The ratio of the number of turns on each coil, called the turns ratio, determinesthe ratio of the voltages. A step-down transformer has a large number of turns onits primary (input) coil which is connected to the high voltage mains supply, and asmall number of turns on its secondary (output) coil to give a low output voltage.

turns ratio =Vp

=Np

andpower out = power in

Vs Ns Vs Is = Vp Ip

Vp = primary (input) voltageNp = number of turns on primary coilIp = primary (input) current

Vs = secondary (output) voltageNs = number of turns on secondary coilIs = secondary (output) current

Rectifier

There are several ways of connecting diodes to make a rectifier to convert AC toDC. The bridge rectifieris the most important and it produces full-wave varyingDC. A full-wave rectifier can also be made from just two diodes if a centre-taptransformer is used, but this method is rarely used now that diodes are cheaper.A single diode can be used as a rectifier but it only uses the positive (+) parts of

Transformercircuit symbol
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the AC wave to produce half-wave varying DC.

Bridge rectifier

A bridge rectifier can be made using four individual diodes, but it is also availablein special packages containing the four diodes required. It is called a full-waverectifier because it uses all the AC wave (both positive and negative sections).1.4V is used up in the bridge rectifier because each diode uses 0.7V whenconducting and there are always two diodes conducting, as shown in the diagrambelow. Bridge rectifiers are rated by the maximum current they can pass and themaximum reverse voltage they can withstand (this must be at least three timesthe supply RMS voltage so the rectifier can withstand the peak voltages).

Bridge rectifier Alternate pairs of diodes conduct, changing overthe connections so the alternating directions of

AC are converted to the one direction of DC.

Output: full-wave varying DC(using all the AC wave)

Single diode rectifier

A single diode can be used as a rectifier but this produces half-wave varying DCwhich has gaps when the AC is negative. It is hard to smooth this sufficiently wellto supply electronic circuits unless they require a very small current so thesmoothing capacitor does not significantly discharge during the gaps.
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Single diode rectifierOutput: half-wave varying DC(using only half the AC wave)

Smoothing

Smoothing is performed by a large value electrolytic capacitorconnected acrossthe DC supply to act as a reservoir, supplying current to the output when thevarying DC voltage from the rectifier is falling. The diagram shows theunsmoothed varying DC (dotted line) and the smoothed DC (solid line). Thecapacitor charges quickly near the peak of the varying DC, and then dischargesas it supplies current to the output.

Note that smoothing significantly increases the average DC voltage to almost thepeak value (1.4 RMS value). For example 6V RMS AC is rectified to full waveDC of about 4.6V RMS (1.4V is lost in the bridge rectifier), with smoothing thisincreases to almost the peak value giving 1.4 4.6 = 6.4V smooth DC.

Smoothing is not perfect due to the capacitor voltage falling a little as it

discharges, giving a small ripple voltage. For many circuits a ripple which is10% of the supply voltage is satisfactory and the equation below gives therequired value for the smoothing capacitor. A larger capacitor will give less ripple.The capacitor value must be doubled when smoothing half-wave DC.

Smoothing capacitor for 10% ripple, C =5 Io

Vs f
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C = smoothing capacitance in farads (F)Io = output current from the supply in amps (A)Vs = supply voltage in volts (V), this is the peak value of the unsmoothed DCf = frequency of the AC supply in hertz (Hz), 50Hz in the UK

Regulator

Voltage regulator ICs are available with fixed (typically 5, 12 and 15V) or variableoutput voltages. They are also rated by the maximum current they can pass.Negative voltage regulators are available, mainly for use in dual supplies. Mostregulators include some automatic protection from excessive current ('overloadprotection') and overheating ('thermal protection').

Many of the fixed voltage regulator ICs have 3 leads and look like power

transistors, such as the 7805 +5V 1A regulator shown on the right. They includea hole for attaching a heatsink if necessary.

Zener diode regulator

Voltage regulator

zener diodea = anode, k = cathode
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For low current power supplies a simple voltage regulator can be made with aresistor and a zener diode connected in reverse as shown in the diagram. Zenerdiodes are rated by their breakdown voltage Vz and maximum power Pz(typically 400mW or 1.3W).

The resistor limits the current (like an LED resistor). The current through theresistor is constant, so when there is no output current all the current flowsthrough the zener diode and its power rating Pz must be large enough towithstand this.

Choosing a zener diode and resistor:

1. The zener voltage Vz is the output voltage required2. The input voltage Vs must be a few volts greater than Vz

(this is to allow for small fluctuations in Vs due to ripple)3. The maximum current Imax is the output current required plus 10%4. The zener power Pz is determined by the maximum current:

Pz > Vz Imax5. The resistor resistance: R = (Vs - Vz) / Imax6. The resistor power rating: P > (Vs - Vz) Imax

Example:output voltage required is 5V, output current required is 60mA.

1. Vz = 4.7V (nearest value available)

2. Vs = 8V (it must be a few volts greater than Vz)

3. Imax = 66mA (output current plus 10%)

4. Pz > 4.7V 66mA = 310mW, choose Pz = 400mW

5. R = (8V - 4.7V) / 66mA = 0.05k = 50 , choose R = 47

6. Resistor power rating P > (8V - 4.7V) 66mA = 218mW, choose P = 0.5W

Filters


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Overview

The rectifier circuitry takes the initial ac sine wave from the transformer or other source

and converts it to pulsating dc. A full-wave rectifier will produce the waveform shown tothe right, while a half-wave rectifier will pass only every other half-cycle to its output.

This may be good enough for a basic battery charger, although some types of

rechargeable batteries still won't like it. In any case, it is nowhere near good enough for

most electronic circuitry. We need a way to smooth out the pulsations and provide a

much "cleaner" dc power source for the load circuit.

To accomplish this, we need to use a circuit called a filter. In general terms, a filteris any

circuit that will remove some parts of a signal or power source, while allowing other partsto continue on without significant hinderance. In a power supply, the filter must remove

or drastically reduce the ac variations while still making the desired dc available to the

load circuitry.

Filter circuits aren't generally very complex, but there are several variations. Any givenfilter may involve capacitors, inductors, and/or resistors in some combination. Each such

combination has both advantages and disadvantages, and its own range of practical

application. We will examine a number of common filter circuits on this page.

A Single Capacitor


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If we place a capacitor at the output of the full-wave rectifier as shown to the left, thecapacitor will charge to the peak voltage each half-cycle, and then will discharge more

slowly through the load while the rectified voltage drops back to zero before beginning

the next half-cycle. Thus, the capacitor helps to fill in the gaps between the peaks, asshown in red in the first figure to the right.

Although we have used straight lines for simplicity, the decay is actually the normal

exponential decay of any capacitor discharging through a load resistor. The extent to

which the capacitor voltage drops depends on the capacitance of the capacitor and theamount of current drawn by the load; these two factors effectively form the RC time

constant for voltage decay.

As a result, the actual voltage output from this combination never drops to zero, butrather takes the shape shown in the second figure to the right. The blue portion of the

waveform corresponds to the portion of the input cycle where the rectifier provides

current to the load, while the red portion shows when the capacitor provides current to theload. As you can see, the output voltage, while not pure dc, has much less variation (or

ripple, as it is called) than the unfiltered output of the rectifier.

A half-wave rectifier with a capacitor filter will only recharge the capacitor on every

other peak shown here, so the capacitor will discharge considerably more between inputpulses. Nevertheless, if the output voltage from the filter can be kept high enough at all

times, the capacitor filter is sufficient for many kinds of loads, when followed by a

suitable regulator circuit.

RC Filters


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In order to reduce the ripple still more without losing too much of the dc output, we need

to extend the filter circuit a bit. The circuit to the right shows one way to do this. Thiscircuit does cause some dc loss in the resistor, but if the required load current is low, this

is an acceptable loss.

To see how this circuit reduces ripple voltage more than it reduces the dc output voltage,

consider a load circuit that draws 10 mA at 20 volts dc. We'll use 100 f capacitors and a100 resistor in the filter.

For dc, the capacitors are effectively open circuits. Therefore any dc losses will be in that

100 resistor. for a load current of 10 mA (0.01 A), the resistor will drop100 0.01 = 1 volt. Therefore, the dc output from the rectifier must be 21 volts, and thedc loss in the filter resistor amounts to 1/21, or about 4.76% of the rectifier output. This is

generally quite acceptable.

On the other hand, the ripple voltage (in the USA) exists mostly at a frequency of 120 Hz

(there are higher-frequency components, but they will be attenuated even more than the120 Hz component). At this frequency, each capacitor has a reactance of about 13.26.

Thus R and C2 form a voltage divider that reduces the ripple to about 13% of what came

from the rectifier. Therefore, for a dc loss of less than 5%, we have attenuated the rippleby almost 87%. This is a substantial amount of ripple reduction, although it doesn't

remove the ripple entirely.

If the amount of ripple is still too much for the particular load circuit, additional filtering

or a regulator circuit will be required.

LC Filters


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While the RC filter shown above helps to reduce the ripple voltage, it introduces

excessive resistive losses when the load current is significant. To reduce the ripple evenmore without a lot of dc resistance, we can replace the resistor with an inductor as shown

in the circuit diagram to the right.

In this circuit, the two capacitors store energy as before, and attempt to maintain a

constant output voltage between input peaks from the rectifier. At the same time, theinductor stores energy in its magnetic field, and releases energy as needed in its attempt

to maintain a constant current through itself. This provides yet another factor that

attempts to smooth out the ripple voltage.

In some cases, C1 is omitted from this filter circuit. The result is a lower dc outputvoltage, but improved ripple removal. The choice is a trade-off, and must be made

according to the specific requirements in each individual case.

For dc, the inductance has only the resistance of the wire that comprises the coil, which

amounts to a few ohms. Meanwhile, the capacitors still operate as open circuits at dc, sothey do not reduce the dc output voltage. However, at the basic ripple frequency of 120

Hz, a 10 Henry inductance has a reactance of:

XL = 2fL = 7540

At the same time, a 100 f capacitor at the same ripple frequency has a reactance of:

XC = 1/2fC = 13.26

Thus, L and C2 form a voltage divider that drastically reduces the ripple component (toless than 0.2%) while leaving the desired dc output nearly alone. This configuration may

provide sufficiently pure dc for some applications, without the need for any following

regulator at all.

The drawback of this approach is that a 10 Henry inductor is as large as some power

transformers, with a heavy iron core. It takes up a lot of space and is relatively expensive.This is why the RC filter circuit may be preferred to the LC filter, provided the ripple

reduction is sufficient and the power loss in the resistor is not excessive.


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Voltage Dividers

In many circuits, it is necessary to obtain a voltage not available from the

main power source. Rather than have multiple power sources for all needed

voltages, we can derive other voltages from the main power source. In mostcases, the needed voltage is less than the voltage from the main source, so we

can use resistors in an appropriate configuration to reduce the voltage from the

power source, for use in a small circuit.

If we know precisely both the voltage and current required, we can simplyconnect a resistor in series with the power source, with a value calculated in

accordance with Ohm's Law. This resistor will drop some of the source voltage,

leaving the right amount for the actual load, as shown to the right.

Usually, however, this doesn't work too well. The required value of theseries dropping resistor will almost never be a standard value, and the cost of

having special values manufactured for specific circuits is prohibitive. For

example, suppose we have a 9 volt battery as your main power source, and

want to operate a load that requires 5 volts at 3.5 milliamperes. Our seriesresistor, R, must drop 4 volts at 3.5 mA. Using Ohm's Law to calculate the

required resistance value, we find that we need a resistance of

4/0.0035 = 1142.8571 or 1.1428571k. We have a choice between 1.1k and1.2k as standard 5% values, but neither will give us what we want.


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A more practical solution to the problem is to use two resistors in series, and

use the voltage appearing across one of them. This configuration is known as avoltage dividerbecause it divides the source voltage into two parts. The basic

circuit is shown to the right.

In this circuit, the output voltage, VOUT, can be set accurately as a fraction of

the source voltage, E. Using our example above, we want to select R1 and R2such that we will drop 4 volts across R1, leaving 5 volts across R2. Since VOUT is

the voltage across R2, this will give us the voltage we want. But how do we find

the correct values of resistance to do this?

The first step is to note that, with no external connection to VOUT, this is

simply a series connection and the same current must flow through both

resistors. (We'll deal with the load current shortly.) Therefore, in accordance

with Ohm's Law, the ratio of voltage across these resistors will be equal to theratio of the resistance values themselves. In this case, the voltage ratio we want

is 4:5, or 0.8:1. Therefore, we want this resistor ratio as well.

But there are 20 different standard 5% resistance values in each decade

range, so there are lots of possible resistance ratios. Most will be wrong for thispurpose, of course. So how do we find two standard resistance values that will

give the ratio we want? We could do it manually, testing each possible

combination. But a better way is to let the computer do the tedious work andpresent all options to us. Then we can select the values we want from the list of

valid possibilities.

The table shows the significant digits of standard 5% resistor values. When

you type in a ratio, it calculates the corresponding significant digits that wouldbe required to complete that ratio. All you need to do is pick out ratios of valid

significant digits. In this case, the table shows that you can use resistance

values of 15:12, 20:16, or 30:24 to obtain the ratio of 1:0.8. If you had specified1:1.25 (the inverse of 0.8:1), you would have gotten the ratios of 12:15, 16:20,


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and 24:30. Either way, these are workable choices, while all other choices fail

to match standard values.

We can get a VOUT of 5 volts, then, if we set R1 = 1.2k and R2 = 1.5k. Wecan also get the same VOUT if we make R1 = 12k and R2 = 15k. The exact

resistor values don't matter, so long as their ratio is correct.

The one thing we haven't accounted for as yet is the current drawn by theload. This will necessarily upset the resistance balance, since any load current

will flow through R1, but not through R2. As a result, the load will reduce

output voltage of the voltage divider by some amount. Appropriately, this

effect is called loading.

To calculate the effect of loading and its extent in any given instance, wemust realize that the voltage divider circuit behaves in exactly the same way as

a battery of voltage VOUT with a series resistor whose value is equal to the

parallel combination of R1 and R2. The figure to the right shows the equivalentcircuit for our example voltage divider.

Now, we noted earlier that our example load draws 3.5mA at 5 volts. In

accordance with Ohm's Law, this current will drop a voltage of 2.33333 voltsacross that 667 resistor. Thus, our example voltage divider will not be able to

provide +5 volts to this load.

If we reduce the resistors in the voltage divider to 120 and 150 , the

equivalent series resistance is only 66.7 so the voltage drop caused by thisload will be 0.23333 volt. This may be a small enough loss to ignore in a

practical circuit.

The drawback of this is that such low resistance values will draw a

significant amount of current from the original source. This is probablyacceptable if the original source is an electronic power supply, but not if it's an


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actual battery. Thus, this use of a voltage divider is reasonable and appropriate

in some circumstances, but not in all cases.

The voltage divider is a very simple circuit that can be highly accurate if not

loaded down. In many cases it cannot be used directly, as we have seen.

However, in such cases it can either be adapted, or augmented with othercomponents to preserve its operation while avoiding the problems that can

occur. Thus, even in those cases where a voltage divider by itself is not

sufficient to meet the need, it can serve as the basis of a circuit that willperform as required. We'll see any number of examples of this in practical

examples on this Website.


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Power Supply Notes 1

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