Power supply unit design

Design examples ranging from very simple unregulated PSU’s to

precise and closely regulated PSU’s

Definition

For the purposes of this article the term “power supply unit” means a unit that takes power from mains AC and produces a DC output.

Throughout this presentation there will be some slides with RED headings. These are more advanced ideas and until you are more familiar with the basic ideas you may want to skip over these slides.

Power supply circuit sections

A power supply unit (psu) comprises:• A transformer to convert mains ac to a

suitable voltage• A rectifier to change AC to DCPSU’s may also have:• Smoothing • Voltage regulation• Current limiting

Basic circuit: Fig 1

L

N

E

Transformer T1 converts the mains voltage to the voltage required. BR1 changes the AC wave into a DC wave.

Its usual to include a fuse (not shown here) in the AC line for circuit protection.

DC wave

Applications

The circuit shown in Fig 1 would be suitable for charging a car battery or operating a dc motor. In these applications the ripple is not important.

However most applications require a smoothed output, and to provide this in the next circuit we will use a capacitor.

Ripple voltageThe input waveform is a sine wave, normally at 50Hz. Its RMS value is the transformer’s output voltage e.g. 12V

The peak value Vpk is 12 √ 2 = 17.0V

The output waveform is a “double hump”. The peak value is still 17V.

This is equivalent to a constant DC level of 12V with an additional ac signal which we call the “ripple voltage”. More on this later.

0V

0V

0V

Vpk

12V

Example design calculationSpecification: we have a 12V dc motor which is to run continuously. It takes 1A in normal operation.

Design:

We choose a 12V transformer rated at 12W (12V * 1A) or a little above.

(The bridge rectifier introduces a voltage drop of 2*0.7=1.4V but the motor will run just fine on 10.6V) We need to calculate the PEAK current and voltage the rectifier will receive.

I(peak) = 1.0 * √2 = 1.414A

V(peak) = 12 * √2 = 17.0V

So we need a bridge rectifier rated over 1.5A, 2A is fine, and a voltage exceeding 2*17V : >50V for safety margin.

Basic circuit with smoothing: Fig 2

Here we have added an electrolytic capacitor to reduce the amount of ripple at the output.

Capacitor smoothing

The output waveform from the rectifier is a “double hump”. The peak value is still 17 V off load, and about 17 - 2*0.7 =15.6V due to the rectifier diodes.As the voltage rises the capacitor “charges up” until the peak is reached. Then it slowly discharges, maintaining the voltage. (red line) When it reaches the next hump it charges again.The amount by which it drops is found from the formula CV =I t where t= 10msec

0V

Time: 0……5…….10……15…..20ms

RMS Ripple voltageAs the ripple is not a sine wave we cant use Vpk/√2 to find the RMS value.

However its almost a sawtooth wave, so instead we can use

Vrms = Vpk-pk/√3

The ripple current flowing in the capacitor is Iripple = Vrms / Xc

And Xc = 1 / (2 *pi * f *C) = 1 / 628 C

Vpk-pk

0V

Design calculations

Specification: A guitar amplifier needs 12V at 6A with a peak – peak ripple of no more than 0.5V.

As before

Transformer: 12V 6A= 72VA+

Bridge rectifier: >50V 10A

Smoothing calculation

Calculate capacitor

CV=I t C * 0.5 = 6 * 0.01

C = 6 * 0.01 / 0.5 = 0.03 Farads = 30,000uF

We need an electrolytic capacitor of 33,000uF * or more, with a voltage rating of over 17V.

*nearest preferred value

Ripple Current rating1: Calculate RMS ripple voltageVpk-pk = 0.5VVrms = 0.5 / √3 = 0.35V

2: now work out RMS ripple currentIripple = Vrms / Xc and Xc = 1 / 2 *pi * f *C = 1 / 628 CXc = 1 / 628 * .03 = 0.053Iripple = 0.35 / 0.053 = 6.6 A

The capacitor must have a ripple current rating of 6.6 A or more.

PSU designer is a simulator package for checking resultshttp://www.duncanamps.com/psud2/

Regulated suppliesIn the previous designs the voltage at the output will vary. With no load it will be a maximum and as the load current is increased the output voltage will fall.

In many applications this is not important, either because the load is constant, or because a very stable voltage is not required.

Load current

Outputvoltage

Vmax

General diagram forseries regulated PSU: Fig 3

Un-regulated supply (basic circuit with smoothing, Fig 2)

AC mains

Series regulator

Voltage reference

0V

V+

Compare and amplify

Positive voltage

General operation of series regulated supply

This uses a series regulator to control the current flowing to the load.

The output voltage is compared with a reference voltage and corrections applied to maintain the output voltage constant.

Series regulated supplies always require a higher voltage at the input to the regulator, which is then reduced to give the required output. This is called insertion loss.

Because of this the series regulator consumes power equal to (Vin - Vout ) * Iout, and so the regulator must be designed to cope with the heat produced.

Simple regulated supply: Fig 4

Vin Vout

Iout

Vref

Circuit explanation

In this simple regulated supply a voltage reference is produced by passing a current through a zener diode. The reference voltage Vref is compared with the emitter voltage Vout. If Vout rises the base-emitter bias is reduced and the voltage restored to its proper value.

Here the transistor is acting as series regulator AND compare/amplify.

Limitations of simple circuit

1. The current to the zener diode is not regulated, so Vref can change

2. The base-emitter voltage of the transistor changes with temperature.

3. The amplification is not very high, so the regulation is not great.

4. The voltage reference circuit uses up to 10% of the available power.

Simple regulated supply: Fig 4aTo deliver 7.5V at 1A

Vin Vout

8.2V+17V max

8.2 - 0.7 = 7.5V

8.2V 1WImax = 100mA

120 ohm

When driving a 1A load the transistor will be dissipating (17V - 7.5V) * 1A = 10W and will need a heat sink.

Limit of regulation of Fig 4a.With no load a current of 17 - 8.2 / 120 = 73mA flows through the zener, establishing a reference voltage of 8.2V.

When a 1A load is applied two things happen; first, a base current flows from R1 through Q1, reducing the current through D1.

Secondly, the unregulated supply now has a load, and its output voltage falls, again reducing the current through D1.

When the current through D1 falls, its voltage also falls.

To improve the regulation we need to stabilise the current through D1.

Three transistor regulated supply: Fig 5

How it works.Q1 and Q2 form a high gain darlington pair so little base current is needed.

The voltage reference is formed by D1 AND the base-emitter junction of Q3; so a 7.5V zener + 0.7V Vbe gives 8.2V

At switch-on the output is zero and Q3 is turned off. Bias current flows through R1 and Q2/Q1 and allows current to flow through these transistors.

When the output reaches 8.2V Q3 starts to turn on and diverts some of the bias current, stabilising the output voltage at 8.2V

Further improvements

The main factors limiting the performance of this supply now are:

1. The voltage reference is Vz + Vbe which are both temperature dependent;

2. The zener diode produces noise. And

3. The current through the reference still changes as the load or input voltage varies.

Zener or avalanche diodes

Voltage reference diodes exhibit both avalanche mode breakdown and Zener breakdown. This is frequently used to provide a stable voltage reference. However avalanche mode has a positive temperature coefficient (tempco). And zener mode a negative tempco.

Avalanche mode predominates above 5.6V and Zener mode below that.

A 5.6V “zener” has virtually zero tempco.

The forward biased junction of a transistor also has a negative tempco, and a combination of a 6.2V “zener” and Vbe, giving 6.9V is again almost zero tempco.

A better circuit• We will use an AD580 bandgap reference IC to provide a

stable reference voltage.• Because this is a fixed value we will use a potential

divider to provide the output voltage required. • We will replace the discrete darlington pair Q1 Q2 with a

power darlington such as a BD681 (100V 10A 70W) • We will use an operational amplifier to compare Vout

with Vref.• Because we are running the op amp from the same

supply the output will never go to zero volts. So we allow it to sit at a higher voltage , and use a zener to bring it down.

Very stable power supply: Fig 6

How it worksThe AD580 is a three terminal device comprising a current source and a stable and accurate voltage reference.

The op amp compares the voltage on the potential divider at C with the reference voltage at D. If its higher a positive output drives current into the base of Q2, draining bias current from the darlington transistor Q1. This reduces the voltage at B, restoring the output to its correct value.

If the output voltage B falls, due for example to a load change, the voltage at C also falls, and as it’s now lower than the reference, the output from the op amp falls, turning off Q2 and allowing more bias current into Q1, thereby restoring the output to its desired value.

Component valuesSpecification; we need a variable supply of 9 -18V 2ADesign: Start with the unregulated supply. To produce an 18V output we need a minimum of about

21V at point A.Transformer – 18V 2A RMS or betterBridge rectifier – 100V 3A RMSThis will give us a peak voltage off-load of 18 √2 = 25.6VMINUS 2 * 0.7 = 1.4 TOTAL: 24VRipple voltage peak-peak must be less than 3VUse CV=It C = 2A * .01 sec / 3V = 6,600 uF rated >32VWe will choose C = 10,000uF so Vmin = 22V

Component values for regulator circuitDarlington transistor: BD681 (100V 10A 70W)

Op Amp: most are suitable

Voltage reference: AD580 (Vmax: 30V)

HFE is 750 min so for a current of 2A we need a base bias of around 3mA minimum.

R1: max V at point F is 18 + 1.4 = 19.4 Min V at point A is 22V

So R1 = (22-19.4)V / 4mA = 650 ohms. (nearest preferred value is 620)

Q2: max current through Q2 occurs when Vout=9V

V(point F) = 9 + 1.4 = 10.4V. Vmax(point A) = 22V

Current in R1 = (22 – 10.4)/620 = 20mA

Maximum voltage across Q2 = 19.4V. Power is 400mW so no heat sink needed

Most npn transistors (eg BC182) will be fine.

ZD1 R2The maximum collector current in Q2 is 20mA. If Q2 has an Hfe > 100 then the maximum base current will be 0.2mA

This is supplied from point A through the op amp via ZD1and R2. If ZD1 is 5.6V the remaining voltage at E is about 21V – 3V - 5.6V = 12.4V.

So the pd across R2 is 12.4 – 0.7 = 11.7V

R2 = 11.7/0.2 = 56k ohms maximum.

We will use a 10k resistor to improve headroom.

The spec is for a variable supply of 9 – 18V. So we need a divider chain that will reduce the voltage at the output to be the same as the reference voltage.

Producing a variable output

R4 will have 2.5V across it. We choose a current of about 10mA in the divider chain. So R4 = 270 ohms. For a 9V output we need a total resistance of Rt / 9 = 270 / 2.5 so Rt = 970 ohmsSo the resistance R3 + VR1 = 970-270 = 700 ohms. (750 is nearest pref value)For 18V output we need a total resistance of Rt / 18 = 270 / 2.5 so Rt = 1940 ohmsSo the resistance R3 + VR1 = 1940 -270 = 1670 ohms. VR1 must be 1670 – 700 = 970 ohms. (1K)

We can adjust the voltages to meet the specification by adding a resistor in parallel with R3 to bring the total value down nearer to 700 ohms.

R4

R3

VR1

B

C

0V

Power and heat calculation

The maximum power in the Darlington occurs when the supply is delivering its maximum current (2A) and its lowest output voltage (9V)

Voltage across transistor = 24V – 9V = 15V

Current = 2A

Power = I * V = 2 *15 = 30W

If we mount the transistor on a heat sink rated 2 degrees C per watt the transistor will heat up to about 20C + 30*2 = 80C.

(assuming the ambient temperature is 20C)

FAQ:Why doesn’t the math work out properly?

Because you can only buy off-the-shelf components in particular “preferred” values.

Also its good practice to leave margins for error.

Where do all these 0.7V values come from?

The voltage across a single forward biased silicon diode junction (rectifier, transistor base-emitter junction etc) is around 0.7V

Can I make this circuit provide a 0 – 30V output?

No because the transistors in the op amp need to be at a voltage of about 2V or more away from the supply rails.

If you really NEED 0V out (WHY?) you will need to provide a –ve supply for the op amp.

Can I play around with the circuit to make it fit my application?

Yes, that’s the whole idea. This basic circuit layout can be used for supplies up to around 30V 10A (with a bigger darlington). You cant go to higher voltages with this layout because the op amp supplies need to be less than +- 18V or so. Also the AD580 has a voltage limit of 30V