BASIC ELECTRICAL/ELECTRONICS

EQUIPMENT SERVICING
LEVEL I

Unit of Competence:Troubleshoot AC/DC Power Supply with Single-Phase Input

Module Title: Troubleshoot AC/DC Power Supply with Single-Phase Input

LG Code EEL BEE1 MO13

TTLM CodeEEL BEE13015 v1

Learning Outcome

LO1: Prepare product and work station for troubleshooting

LO2: Diagnose faulty parts of power supply Maintain/repair the power supply unit
LO3: Rewind low-power transformer
LO4: Assemble low-power transformer
LO5: Test and inspect repaired products
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Power Supply Basics

What you´ll learn in Module 1

 Section 1.0 Power Supply Basics.

o Basic functions of a power supply.
o Safety aspects of working on power supplies.
 Section 1.1 Transformers & Rectifiers.
o The Transformer.
o The Rectifier Stage.
 Half Wave.
 Full Wave.
 Bridge.
 Section 1.2 Filter Circuits.
o Reservoir Capacitor.
o Low Pass filter.
 LC Filters
 RC Filters
 Section 1.3 Power Supply Basics Quiz.
 Test your knowledge of basic power supplies

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Parts of a Power Supply

A DC Power Supply Unit (commonly called a PSU) deriving power from the AC
mains (line) supply performs a number of tasks:

1. It changes (in most cases reduces) the level of supply to a value

suitable for driving the load circuit.

2. It produces a DC supply from a pure AC wave.

3. It prevents any AC from appearing at the supply output.

4. It will ensure that the output voltage is kept at a constant level,

independent of changes in:

a. The AC supply voltage at the supply input.

b. The Load current drawn from the supply output.

c. Temperature.

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To do these things the basic PSU has four main stages, illustrated in Fig. 1.0.1

Fig. 1.0.1 Power Supply Block Diagram

Power supplies in recent times have greatly improved in reliability but, because
they have to handle considerably higher voltages and currents than any or most
of the circuitry they supply, they are often the most susceptible to failure of any
part of an electronic system.

Modern power supplies have also increased greatly in their complexity, and can
supply very stable output voltages controlled by feedback systems. Many power
supply circuits also contain automatic safety circuits to prevent dangerous over
voltage or over current situations.

1.1 Transformers & Rectifiers

1.2 Filter Circuits

1.3 Power Supply Basics Quiz

Transformers and Rectifiers

What you´ll learn in Module 1.1

After studying this section, you should be able to:

 Describe the principles of transformers used in basic power supplies.

 Primary and secondary voltages.
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 Isolation.
 Describe the principles of rectification used in basic power supplies.

Half Wave.
Full Wave.
Bridge.

The Transformer

Fig. 1.1.1 Typical Input Transformer

In a basic power supply the input power transformer has its primary winding
connected to the mains (line) supply. A secondary winding, electro-magnetically
coupled but electrically isolated from the primary is used to obtain an AC
voltage of suitable amplitude, and after further processing by the PSU, to drive
the electronics circuit it is to supply.

The transformer stage must be able to supply the current needed. If too small a
transformer is used, it is likely that the power supply's ability to maintain full
output voltage at full output current will be impaired. With too small a

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transformer, the losses will increase dramatically as full load is placed on the
transformer.

As the transformer is likely to be the most costly item in the power supply unit,
careful consideration must be given to balancing cost with likely current
requirement. There may also be a need for safety devices such as thermal fuses
to disconnect the transformer if overheating occurs, and electrical isolation
between primary and secondary windings, for electrical safety.

The Rectifier Stage

Three types of silicon diode rectifier circuit may be used, each having a different
action in the way that the AC input is converted to DC. These differences are
illustrated in Figs. 1.1.2 to 1.1.6

Half Wave Rectification

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Fig. 1.1.2 Half Wave Rectification

Fig. 1.1.3 Full Wave Rectification

Fig. 1.1.4 The Bridge Rectifier

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Fig. 1.1.5 Current Flow on the Positive Half Cycle

Fig. 1.1.6 Current Flow on the Negative Half Cycle

A single silicon diode may be used to obtain a DC voltage from the AC input as
shown in Fig 1.1.2. This system is cheap but is only suitable for fairly non-
demanding uses. The DC voltage produced by the single diode is less than with
the other systems, limiting the efficiency of the power supply, and the amount
of AC ripple left on the DC supply is generally greater.

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The half wave rectifier conducts on only half of each cycle of the AC input wave,
effectively blocking the other half cycle, leaving the output wave shown in Fig.
1.1.2. As the average DC value of one half cycle of a sine wave is 0.637 of the
peak value, the average DC value of the whole cycle after half wave
rectification will be 0.637 divided by 2, because the average value of every
alternate half cycle where the diode does not conduct, will of course be zero.
This gives an output of:

Vpk x 0.318

This figure is approximate, as the amplitude of the half cycles for which the
diode conducts will also be reduced by about 0.6V due to the forward voltage
drop (the depletion layer p.d.) of the silicon rectifier diode. This additional
voltage drop may be insignificant when large voltages are rectified, but in low
voltage power supplies where the AC from the secondary winding of the mains
transformer may be only a few volts, this 0.6V drop across the diode junction
may have to be compensated for, by having a slightly higher transformer
secondary voltage.

Half wave rectification is not very efficient at producing DC from a 50Hz or 60Hz
AC input. In addition the gaps between the 50 or 60Hz diode output pulses
make it more difficult to remove the AC ripple remaining after rectification.

Full Wave Rectification

If a transformer with a centre tapped secondary winding is used, more efficient

full wave rectification can be used. The centre-tapped secondary produces two
anti-phase outputs, as shown in Fig 1.1.3.

If each of these outputs is ‘half wave rectified’ by one of the two diodes, with
each diode conducting on alternate half cycles, two pulses of current occur at
every cycle, instead of once per cycle in half wave rectification. The output
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frequency of the full wave rectifier is therefore twice the input frequency. This
effectively provides twice the output voltage of the half wave circuit,Vpk x
0.637 instead of Vpk x 0.318 as the ‘missing’ half cycle is now rectified,
reducing the power wasted in the half wave circuit. The higher output frequency
also makes the smoothing of any remaining AC ripple easier.

Although this full wave design is more efficient than the half wave, it requires a
centre tapped (and therefore more expensive) transformer.

The Bridge Rectifier

The full wave bridge rectifier uses four diodes arranged in a bridge circuit as
shown in Fig. 1.1.4 to give full wave rectification without the need for a centre-
tapped transformer. An additional advantage is that, as two diodes (effectively
in series) are conducting at any one time, the diodes need only half the reverse
breakdown voltage capability of diodes used for half and conventional full wave
rectification. The bridge rectifier can be built from separate diodes or a
combined bridge rectifier can be used.

The current paths on positive and negative half cycles of the input wave are
shown in Fig. 1.1.5 and Fig. 1.1.6. It can be seen that on each half cycle,
opposite pairs of diodes conduct, but the current through the load remains in
the same polarity for both half cycles.

Filter Circuits

What you´ll learn in Module 1.2

After studying this section, you should be able to:

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 Describe the principles of a reservoir capacitor in basic power supplies.

 Reservoir capacitor action.

 The effect of a reservoir capacitor on the DC component.
 The effect of a reservoir capacitor on the diode current.

 Describe the principles of a low pass filter used in basic power supplies.

 LC filters.
 RC filters.

Filter Components

A typical power supply filter circuit can be best understood by dividing the
circuit into two parts, the reservoir capacitor and the low pass filter. Each of
these parts contributes to removing the remaining AC pulses, but in different
ways.

The Reservoir Capacitor

Fig. 1.2.1 The Reservoir Capacitor

Fig. 1.2.1 shows an electrolytic capacitor used as a reservoir capacitor, so called

because it acts as a temporary storage for the power supply output current. The
rectifier diode supplies current to charge a reservoir capacitor on each cycle of

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the input wave. The reservoir capacitor is a large electrolytic, usually of several
hundred or even a thousand or more microfarads, especially in mains frequency
PSUs. This very large value of capacitance is required because the reservoir
capacitor, when charged, must provide enough DC to maintain a steady PSU
output in the absence of an input current; i.e. during the gaps between the
positive half cycles when the rectifier is not conducting.

The action of the reservoir capacitor on a half wave rectified sine wave is shown
in Fig. 1.2.2. During each cycle, the rectifier anode AC voltage increases
towards Vpk. At some point close to Vpk the anode voltage exceeds the
cathode voltage, the rectifier conducts and a pulse of current flows, charging
the reservoir capacitor to the value of Vpk.

Fig. 1.2.2 Reservoir Capacitor Action

Once the input wave passes Vpk the rectifier anode falls below the capacitor
voltage, the rectifier becomes reverse biased and conduction stops. The load
circuit is now supplied by the reservoir capacitor alone (hence the need for a
large capacitor).

Of course, even though the reservoir capacitor has large value, it discharges as
it supplies the load, and its voltage falls, but not by very much. At some point
during the next cycle of the mains input, the rectifier input voltage rises above

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the vo1tage on the partly discharged capacitor and the reservoir is re-charged
to the peak value Vpk again.

AC Ripple

The amount by which the reservoir capacitor discharges on each half cycle is
determined by the current drawn by the load. The higher the load current, the
more the discharge, but provided that the current drawn is not excessive, the
amount of the AC present in the output is much reduced. Typically the peak-to-
peak amplitude of the remaining AC (called ripple as the AC waves are now
much reduced) would be no more than 10% of the DC output voltage.

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The DC output of the rectifier, without the reservoir capacitor, is either 0.637
Vpk for full wave rectifiers, or 0.317 Vpk for half wave. Adding the capacitor
increases the DC level of the output wave to nearly the peak value of the input
wave, as can be seen from Fig. 1.1.9.

To obtain the least AC ripple and the highest DC level it would seem sensible to
use the largest reservoir capacitor possible. There is a snag however. The
capacitor supplies the load current for most of the time (when the diode is not
conducting). This current partly discharges the capacitor, so all of the energy
used by the load during most of the cycle must be made up in the very short
remaining time during which the diode conducts in each cycle.

The formula relating charge, time and current states that:

Q = It

The charge (Q) on a capacitor depends on the amount of current (I) flowing for a
time (t).

Therefore the shorter the charging time, the larger current the diode must
supply to charge it. If the capacitor is very large, its voltage will hardly fall at all
between charging pulses; this will produce a very small amount of ripple, but
require very short pulses of much higher current to charge the reservoir
capacitor. Both the input transformer and the rectifier diodes must be capable
of supplying this current. This means using a higher current rating for the
diodes and the transformer than would be necessary with a smaller reservoir
capacitor.

There is an advantage therefore in reducing the value of the reservoir capacitor,

thereby allowing an increase in the ripple present, but this can be effectively
removed by using a low pass filter and regulator stages between the reservoir
capacitor and the load.
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This effect of increasing reservoir size on diode and transformer current should
be born in mind during any servicing operations; replacing the reservoir
capacitor with a larger value than in the original design "to reduce mains hum"
may seem like a good idea, but could risk damaging the rectifier diode and/or
the transformer.

With full wave rectification the performance of the reservoir capacitor in

removing AC ripple is significantly better than with half wave, for the same size
of reservoir capacitor, the ripple is about half the amplitude of that in half wave
supplies, because in full wave circuits, discharge periods are shorter with the
reservoir capacitor being recharged at twice the frequency of the half wave
design.

Low Pass Filters

Although a useable power supply can be made using only a reservoir capacitor
to remove AC ripple, it is usually necessary to also include a low pass filter
and/or a regulator stage after the reservoir capacitor to remove any remaining
AC ripple and improve the stabilization of the DC output voltage under variable
load conditions.

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Fig. 1.2.3 LC Filter

Fig. 1.2.4 RC Filter

Either LC or RC low pass filters can be used to remove the ripple remaining after
the reservoir capacitor. The LC filter shown in Fig. 1.2.3 is more efficient and
gives better results than the RC filter shown in Fig. 1.2.4 but for basic power
supplies, LC designs are less popular than RC, as the inductors needed for the
filter to work efficiently at 50 to 120Hz need to be large and expensive
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laminated or toroidal core types. However modern designs using switch mode
supplies, where any AC ripple is at much higher frequencies, much smaller
ferrite core inductors can be used.

The low pass filter passes low frequency, in this case DC (0Hz) and blocks
higher frequencies, whether 50Hz or 120Hz in basic circuits or tens of kHz in
switch mode designs.

The reactance(XC) of the capacitor in the either of the filters is very low
compared with the resistance of resistor R or the reactance of the choke XL at
the ripple frequency. In RC designs the resistance of R must be a fairly low
value as the entire load current, maybe several amperes, must pass through it,
generating a considerable amount of heat. A typical value would therefore be
50 ohms or less, and even at this value, a large wire wound resistor would
normally need to be used. This limits the efficiency of the filter as the ratio
between the resistance of R and the capacitor reactance will not be greater
than about 25:1. This then would be the typical reduction ratio of the ripple
amplitude. By including the low pass filter some voltage is lost across the
resistor, but this disadvantage is offset by the better ripple performance than by
using the reservoir capacitor alone.

The LC filter performs much better than the RC filter because it is possible to
make the ratio between XC and XL much bigger than the ratio between X C and R.
Typically the ratio in a LC filter could be 1:4000 giving much better ripple
rejection than the RC filter. Also, since the DC resistance of the inductor in the
LC filter is much less than the resistance of R in the RC filter, the problem of
heat being generated by the large DC current is very much reduced in LC filters.

With a combined reservoir capacitor and low pass filter it is possible to remove
95% or more of the AC ripple and obtain an output voltage of about the peak
voltage of the input wave. A simple power supply consisting of only transformer,
rectifier, reservoir and low pass filter however, does have some drawbacks.
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Fig. 1.2.5 DC Adaptor

The output voltage of the PSU tends to fall as more current is drawn from the
output. This is due to:

a. The reservoir capacitor being discharged more on each cycle.

b. Greater voltage drop across the resistor or choke in the low pass filter as
current increases.

These problems can be largely overcome by including a regulator stage at the

power supply output as described in Power Supplies Module 2.

The basic power supply circuits described here in Module 1 however, are
commonly used in the common ‘wall wart’ type DC adaptors supplied with
many electronics products. The most common versions comprise a transformer,
bridge rectifier and sometimes a reservoir capacitor. Additional filtering and
regulation/stabilisation being usually performed in the circuit supplied by the
adaptor.

How the output of a basic power supply may be improved by Regulation Circuits
is explained in Power Supplies Module 2

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Power Supply Basics Quiz

Try our quiz, based on the information you can find in Power Supplies Module 1.
Submit your answers and see how many you get right. If you get any answers
wrong. Just follow the hints to find the right answer and learn about Power
Supplies as you go.

1.

Refer to Fig. 1.3.1.What is the function of block B?

a) Rectifier.

 b) Reservoir Capacitor.

 c) Low Pass Filter.

 d) Regulator

2.

Refer to Fig 1.3.1. What is the function of block A ?

 a) Transformer.

 b) Full Wave Rectifier

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 c) Bridge Rectifier

 d) Reservoir Capacitor

3.

Refer to Fig 1.3.1. What will be the approximate value of the DC component of
the waveform at the output of block A?

 a) VPK x 0.318

 b) VPK x 0.5

 c) VPK x 0.637

 d) VPK x 0.707

4.

Refer to Fig 1.3.2. If input B is more positive than input A, which diodes will be
conducting?

 a) D1 and D2

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 b) D2 and D3

 c) D1 and D4

 d) D3 and D4

5.

Refer to Fig 1.3.2. If D4 were to go short circuit, what would be the effect on the
operation of the circuit?

 a) A decrease in the current through D1.

 b) Fuse F1 would blow.

 c) A higher voltage across the load.

 d) A larger peak current through D2 and D3.

6.

What is the action of the reservoir capacitor in a basic power supply circuit?

 a) To de-couple the DC component of the rectifier AC output.

 b) To increase the DC component and reduce the AC component of the

AC wave.

 c) To remove the DC component of the AC wave.

 d) To regulate the AC wave.

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7.

Which of the following is an advantage of using a L C low pass filter rather than
a RC low pass filter in a power supply?

 a) The reactance of L will be much lower than the resistance of R at

mains frequency.

 b) The reactance of L will be much higher than the resistance of R at

mains frequency.

 c) An inductor can dissipate more power than a resistor.

 d) LC filters are less expensive than RC filters.

8.

Refer to Fig 1.3.3. What is the power dissipated in R1?

 a) 5W

 b) 1W

 c) 0.5W
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 d) 0.25W

9.

Refer to Fig 1.3.3. What will be the approximate value of DC across C1?

 a) 3.8V

 b) 7.6V

 c) 10.8V

 d) 14.5V

10.

Refer to Fig 1.3.3. What is the reactance of C2 at the ripple frequency?

 a) 6.4Ω

 b) 0.3Ω

 c) 5.1Ω

 d) 3.2Ω

Regulation &Stabilisation

What you´ll learn in Module 2

 Section 2.0 Regulation &Stabilisation.

o Regulation &Stabilisation.
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o Effects of Poor Regulation.

o The Shunt Regulator.
o The Series Regulator
 Section 2.1 Shunt Voltage Regulators.
o The Zener Diode.
o The Basic Shunt Regulator Circuit.
o Calculate Component Values.
o Limitations of the Basic Shunt Regulator Circuit.
o Improving Current Rating and Line Regulation.
 Section 2.2 Series Voltage Regulators.
o The Simple Series Regulator.
o Using Feedback & Error Amplification.
o Current Limiting.
o Over Voltage Protection.
 Section 2.3 I.C. Regulators.
o 78xx Series of Voltage Regulators
o Decoupling.
o Dropout.
o Reliability.
 Section 2.4 Regulated Power Supplies Quiz.
o Test Your Knowledge of Regulated Power Supplies.

The Regulator/Stabilizer Block

Effects of Poor Regulation

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The effect of poor regulation (or stabilization) of the supply can be seen in Fig.
1.3.1 which shows graphs of output voltage (V DC) for increasing load current (I)
in various versions of a basic power supply.

Notice that the output voltage is substantially higher for full wave designs (red
and yellow) than half wave (green and purple). Also note the slightly reduced
voltage when a LC filter is added, due to the voltage drop across the inductor. In
every case in the basic design the output voltage falls in an almost linear
fashion as more current is drawn from the supply. In addition to this effect, the
extra discharge of the reservoir capacitor also causes the ripple amplitude to
increase.

Fig. 1.3.1 Regulation Curves Compared

Regulator (Stabilizer)

Regulator or Stabilizer?

Strictly speaking, compensating for variations in the mains (line) input voltage
is called REGULATION, while compensating for variations in load current is
called STABILISATION. In practice you will find these terms used rather loosely

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to describe compensation for both effects. In fact most stabilised or regulated

power supplies compensate for both input and output variations and so are both
(at least to some degree) stabilised and regulated power supplies.

In common with much modern usage, the term ‘Regulator’ will be used here to
describe both regulation and stabilisation.

These problems can be largely overcome by including a regulator stage at the

power supply output. The effect of this circuit can be seen in Fig. 1.3.1. as the
black line on the graph, where for any current up to about 200mA the output
voltage (although lower than the absolute maximum provided by the basic
supply) remains constant.

A regulator counteracts the effects of varying load current by automatically

compensating for the reduction in output voltage as current increases.

It is also common in regulated supplies, that the output voltage is automatically

and suddenly reduced to zero as a safety measure if the current demand
exceeds a set limit. This is called Current Limiting.

Regulation requires extra circuitry at the output of a simple power supply. The
circuits used vary greatly in both cost and complexity. Two basic forms of
regulation are used:

1. The shunt regulator.

2. The series regulator.

These two approaches are compared in Fig. 1.3.2 & Fig 1.3.3

The Shunt Regulator

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Fig. 1.3.2 The Shunt Regulator

In the shunt regulator (Fig. 1.3.2) , a circuit is connected in parallel with the
load. The purpose of the regulator is to ensure a stable voltage across the load
at all times; this is achieved by arranging that a current will flow through the
regulator circuit at all times. If the load current increases, then the regulator
circuit reduces its current so that the total supply current I T, (made up of the
load current IL plus the regulator current I S), remains at the same value.
Similarly if the load current decreases, then the regulator current increases to
maintain a steady total current I T. If the total supply current remains the same,
then so will the supply voltage.

The Series Regulator

Fig. 1.3.2 The Series Regulator

In the series regulator (Fig. 1.3.3), the controlling device is in series with the
load. At all times there will be a voltage drop across the regulator. This drop will
be subtracted from the supply voltage to give a voltage V L across the load,
which is the supply voltage VT minus the regulator voltage drop VS. Therefore:

VL = VT - VS

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Series regulators are usually controlled by a sample of the load voltage, using a
negative feedback system. If the load voltage tends to fall, the smaller feedback
causes the controlling device to lessen its resistance, allowing more current to
flow into the load, so increasing the load voltage to its original value. An
increase in load voltage will have the opposite effect. Like shunt regulation, the
action of the series regulator will also compensate for variations in the supply
voltage.

Shunt Voltage Regulators

What you´ll learn in Module 2.1

After studying this section, you should be able to:

 Understand the principles of shunt regulators.

 Regulating action of a zener diode.

 Operation of the basic shunt regulator circuit.
 Calculate component values for a basic shunt regulator circuit.

 Recognise the limitations of the basic shunt regulator.

 Understand methods for increasing current rating.
 Recognise cascading techniques for improving line regulation.

Basic Shunt Regulator Circuits

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Fig.2.1.1 Basic Shunt Regulator

Shunt regulators are widely used because they are cheap, effective and simple.
It is unusual however, to find a shunt regulator used as the main regulating
circuit in a large power supply. Shunt regulation is only really suitable, at
reasonable cost, for relatively small currents and a range of fixed, usually fairly
low voltages. As mentioned in power supplies module 2.0, a regulation current
must always be flowing in addition to the load current. This is wasteful of power
if large currents are involved.

The basic shunt regulator circuit is shown in Fig. 2.1.1 and consists of only two
components; a series resistor R S feeding current to a zener diode, which is
connected in reverse polarity) across the load.

The Zener Diode.

Fig. 2.1.2 Zener Diode Characteristic Curve

The main property of a zener diode is that the voltage across the diode (V Z) will
remain practically steady for a wide range of current (I Z) when the diode is
operated in reverse bias mode, as shown in Fig. 2.0.1.

Fig. 2.1.2 illustrates the characteristic curve of a zener diode where the
operating region (shown in green) of the zener diode is a range of current on
the nearly vertical breakdown region of the curve.

Provided that the reverse current is kept at a value above about 1 to 2mA
(avoiding the "knee" of the reverse characteristic) and does not exceed the safe

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working current for that particular diode type, very little change in reverse
voltage takes place. It is this effect that is used to give the required regulating
effect.

As the power supply operates, two conditions may occur to change the output
voltage;

a. The load current may vary.

b. The supply voltage may vary.

The circuit is arranged so that the total supply current I Sis made up of the
output load current IOUT plus the current in the zener diode IZ:

IS = IZ + IOUT

Provided that the zener diode is working within its allowable range of current,
the voltage VZ will remain practically constant, deviating by only a very small
amount (δV).

Point of Load Regulators

Fig. 2.1.3 Point of Load Regulators

The most common method of using these basic zener diode shunt regulators is
the ‘point of load’ regulation system. This method uses a number of fixed
voltage, relatively low current regulators at various points around the circuit
being supplied. A relatively ineffective main regulator can then be used in the

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power supply unit because each section of the circuit has its own regulator. For
example many complex circuits need different voltage levels for different
electrical and electromechanical parts. Each may have its own regulated supply,
e.g. 9V, 5V or 3.3V, using "point of load" shunt regulators fed from a single
common supply, as shown in Fig. 2.1.3. Each of the voltage regulators will
normally be placed as close to the circuit supplied as physically possible, and
have additional decoupling capacitors to reduce any noise or cross talk between
the individual supply lines.

Variations in Load Current

If the current in the load Iout tends to fall, the voltage across the load would
tend to rise, but because it is connected in parallel with the diode the voltage
will remain constant. What will change is the current (I Z) through the diode. This
will rise by an amount equal to the fall of current in the load. The total supply
current IS being always equal to I Z + IOUT. An increase in load current I OUT will
likewise cause a fall in zener current I Z, again keeping VZ and the output voltage
steady.

Variations in Input Voltage

If the input voltage rises this will cause more supply current I S to flow into the
circuit. Without the zener shunt regulator, this would have the effect of making
the output voltage Vout rise, but any tendency for V OUT to rise will simply cause
the diode to conduct more heavily, absorbing the extra supply current without
any increase in VZ thus keeping the output voltage constant. A fall in the input
voltage would likewise cause a reduction in zener current, again keeping V OUT
steady.

Limitations

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This simple shunt regulator is only suitable for relatively small currents and a
fixed range of voltages. There are a number of limitations to the use of this
circuit:

Output voltage:

The output voltage is equal to the zener voltage of the diode and so is fixed at
one of the available voltage levels.

Output current:

If the output current falls to zero for any reason (the load may become open
circuit due to a fault, or be disconnected from the power supply), all of the load
current must be passed by the zener diode. Therefore the maximum current
available for the load must be no greater than the maximum safe current for
the zener diode alone.

Input voltage:

The input voltage must be higher (usually about 30% higher) than the output
voltage to allow regulation to take place. It should not be too high however, as
this will result in more power being dissipated by the diode.

Power dissipation:

The power dissipated in the diode must be kept within the safe working limits
for the device chosen. Maximum power will be dissipated if the load is allowed
to go open circuit while the input voltage is at its maximum value. This ‘worst
case’ should not exceed the diode‘s maximum power rating.

The above limitations are controlled by the suitable choice of zener diode and
series resistor RS. The design of simple regulator circuits is fairly straightforward
if a few simple steps are followed:

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1. Decide on a zener diode of the required voltage.

2. Select a diode whose maximum current at least matches, and preferably

exceeds the maximum current requirement of the load.

Note:Zener diode ratings usually quote maximum power dissipation rather than
current, so you will need to work out the current rating from the power and
voltage given for the device.

3. Find out the highest voltage likely to occur at the supply input. (V INmax)

4. Calculate the value of a suitable series resistor R S by using the formula:

Where:

VINmax = the highest likely input voltage.

VZ = the zener diode voltage

IOUT max = the maximum output current.

IZmin = the minimum current at which the zener will operate (say about 1 to
2mA).

5. Calculate the power dissipated in the series resistor (R S) by the formula:

Power dissipated by RS = VR x IIN

Note: When calculating power ratings and resistances, your answers will
probably not exactly match the commercially available preferred values.
Therefore choose the closest preferred value, and then put this value into your
calculations to make sure the circuit will operate correctly with the preferred
value. You should then be able to quote:

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A suitable resistor value and its power rating and a suitable zener diode type
number.

Example:

Fig. 2.1.4 Example

Design a simple zener diode (see Fig. 2.1.4) shunt regulator circuit with the
following specifications:

Maximum load current required. 100mA

Output voltage. 12V

Input voltage. 15V nominal, 16V max.

Assuming a minimum zener current of 1mA

The problem can be solved in 4 steps.

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1. Find a value for RS

RS must supply sufficient current to keep I Z at or slightly above 1mA when the
zener diode is passing its minimum current.

As the total circuit current IIN = IOUT + IZ

The minimum zener diode current will occur when the input voltage (V IN) is at its
minimum value and the load current (IOUT) is at its maximum value.

Under these conditions the current (IIN) flowing through RS will be IOUTmax + IZmin

Which is 100mA + 1mA = 101mA

As R = V/I and the voltage across RS = VIN min - VZ

Then RS = (VIN min - VZ)/IIN = (15 -12) / 101exp-3 = 29.7Ω

Therefore a practical value for RS will be the next lowest preferred value27Ω)

2. Calculate the maximum current (IINmax) that will pass through RS.

The maximum current (IINmax) will occur when V IN is at nits maximum value, i.e.
16V

IINmax = (VIN max - VZ) / RS = (16 - 12) / 27 = 4 / 27 = 148mA

3. Calculate the maximum power requirement for RS

This will occur when VIN is at its maximum value).

(VIN max - VZ) x IIN max = (16 - 12) x 148exp-3 = 0.592W

A practical power rating for RS will therefore be the next highest available power
rating = 1W)

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4. Calculate the maximum power (PZmax) that must be dissipated by the

zener diode.

This will be the power that the zener diode would need to dissipate if the load
was disconnected while the input voltage was at maximum, causing the
maximum current (IINmax) of 148mA to flow through the vener diode.

As Power P = I x V, then PZmax = IINmax x VZ = 148exp-3 x 12 = 1.776 = approx.

1.8W

A suitable shunt regulator would therefore consist of a 27Ω 1W resistor

and a 12V zener diode with a power rating of at least 2Watts.

However, it is interesting to compare the power delivered to the load, (12V x

100mA) = 1.2W with the power dissipated in shunt regulator, (0.592W in the
resistor + 1.776W in the diode = 2.368W.

Only about 33% of the total power is in the load with about 66% being
dissipated by the shunt regulator!

The simple zener diode shunt regulator is therefore not very efficient when
handling even these amounts of current. A better option can be to add a
transitor to the circuit to handle larger currents.

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Fig. 2.1.5 Transistor Shunt Regulator

Large Current Shunt Regulators

The current that can be handled by the simple zener diode/resistor shunt
regulator is limited by the maximum current rating of the zener diode, but there
are ways of increasing the maximum current capability of shunt regulators. One
typical method is given in Fig. 2.1.5 where a power transistor, capable of
passing an emitter current much larger than the zener diode is used.

The zener diode now only handles the transistor base current and the main
stabilising current IE is approximately equal to the base current of the transistor,
multiplied by the hfe of the transistor:

IE = IB(1+hfe)

Therefore the regulator can handle I Z (1+hfe) amperes, allowing much larger
load currents to be supplied by the regulator.

Note that the output voltage is no longer equal to V Z but to VZ+VBE. The
transistor will have a VBE of typically 0.7V, so to produce a 5V regulator, a zener
diode of 4.3v would be chosen.

Operation

If the current in a load attached to the regulator output decreases, the voltage
at the right hand end of the series resistor (R) will tend to increase. When this
happens, the base-emitter voltage of the transistor will increase by a similar
amount, as the voltage across the zener diode is constant.

This increase in base-emitter voltage will cause the transistor (Tr1) to conduct
more heavily until the extra current taken by the transistor balances the

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reduction in current taken by the load, the output voltage of the regulator is
therefore returned to its previous ‘normal’ value.

If the load current increases, a similar action takes place, but this time there will
be a reduction in Tr1 base voltage. This will reduce current through the
transistor, thereby balancing the increase in load current. Again, the output
voltage of the regulator remains relatively constant.

Improving Line Regulation

Fig. 2.1.6 Cascaded Shunt Regulators

An effective regulator circuit should compensate for variations in input voltage

as well as variations in output voltage (Output stabilisation). These input
variations may be due to changes in the AC mains (line) supply, hence the
name line regulation, or changes in the rectified DC input to the regulator circuit
caused by variations in current drawn by other parts of the electronic system
sharing the same supply. How well line regulation works can be determined by
comparing any change in the regulator output voltage with the change in
regulator input voltage (assuming a constant load current). This can be
expressed as the line regulation factor, and quoted as a percentage. For

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example if the input voltage to the regulator changes by ±2V but the output
voltage only changes by ±0.2V the line regulation factor is ±10%.

Fig. 2.1.6, shows how line regulation can be improved by using a series of zener
diode regulators connected in cascade, each one will have a lower voltage that
the previous one, but by connecting regulators in cascade, the overall
regulation factor is the product of the factors of the individual circuits. Therefore
two regulators each having a regulation factor of 10% or 0.1 would give an
overall factor of 0.1 x 0.1 = 0.01 or 1%.

Series Voltage Regulators

What you´ll learn in Module 2.2

After studying this section, you should be able to:

Understand the operation of series voltage regulators.

• The simple series regulator.

• Feedback and error amplification.

• Over current protection (current limiting).

• Over voltage protection.

Simple Series Voltage Regulators

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Fig. 2.2.1 Simple Series Regulator

In Fig. 2.2.1 RS and DZ form a simple SHUNT regulator as described in power

supplies module 2.1. In this circuit however, they are used to provide a stable
voltage reference VZ at the base of Tr1. The emitter voltage of Tr1 will be
typically about 0.7V less than the base voltage and V OUT will therefore be at a
lower voltage than the base.

VOUT = VZ - VBE

If the output voltage VOUT falls due to increased current demand by the load, this
will cause VBE to increase and as a result, current through the transistor (from
collector to emitter) will increase. This will provide the extra current required by
the load and thus regulate the output voltage V OUT.

If VOUT tends to rise due to reduced current demand by the load, then this will
reduce VBE as the emitter voltage rises and the base voltage remains stable due
to DZ. This reduction in VBE will tend to turn the transistor off, reducing current
flow, and again regulating the output voltage V OUT.

This regulating effect is due to the base potential of Tr1 being held steady by D Z
so that any variation in emitter voltage caused by varying current flow causes a
change in VBE, varying the conduction of the transistor Tr1, which will usually be
a power transistor. This action counteracts the variation in load current. With

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this simple circuit however, regulation is not perfect, and variations in output do
occur for the following reasons.

Fig. 2.2.2 Zener Diode Operating Region

1. Any increase in load current (I L) causes a small increase in base current by

the ratio IL/hfe. This in turn causes an increase in V BE and because the output
voltage VOUT = VZ - VBE any increase in VBE tends to reduce the output voltage.
The amount of this fall is about 0.25V for a change in output current from 10mA
to 1A.

2. Since base current increases with load, the current through the zener diode
DZ will decrease as more current is taken by the base of Tr1. Because the diode
characteristic has a slope over its operating region as shown in Fig. 2.2.2, a
large change in zener current (ΔI) will cause a very small change in zener
voltage (δV). This in turn will slightly affect VBE and the output voltage.

3. Because of reasons 1 and 2 above, any change in load will result in less than
perfect regulation, therefore any change at the output will slightly change the
loading on the input circuit. As the input is normally taken from an un-regulated
supply, the input voltage will be easily affected by slight changes in load
current, As the input voltage is also the supply for the reference voltage V Z any
change in output current, by affecting the input voltage, can produce a
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noticeable effect on the output voltage, slightly reducing the effectiveness of

the regulation.

Each of the above effects is small, but added together they will provide an
overall effect that is noticeable when the supply is operating under demanding
conditions. Nevertheless this inexpensive circuit is effective enough for many
applications, and is more efficient than the shunt regulator. Also, by using a
suitable power transistor, the series regulator can be used for heavier load
currents than the shunt design.

Feedback and Error Amplification.

Fig. 2.2.3 Series Regulator with Feedback & Error Amplifier

To improve on the simple series regulator a feedback circuit and error amplifier
can be added to the basic series circuit.
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Fig. 2.2.3 shows a block diagram of a series regulator circuit with error
amplification. In this system the reference voltage V Z is compared with a
feedback voltage VF, which is a portion of the actual output voltage. The
difference between the two inputs produces an error voltage that is used to
vary the conduction of the control element, correcting any error in the output
voltage.

Fig. 2.2.4 Circuit Diagram for Fig. 2.2.3

Circuit Diagram.

A circuit diagram for this system is shown in Fig. 2.2.4. Tr1 is the series control
element. It will usually be a power transistor, mounted on a substantial heat
sink to cope with the necessary power dissipation.

A stable reference voltage is provided by R4 & D1 from the un-regulated input

voltage. Tr2 is the error amplifier and its gain is set by the value of its load
resistor R3. Tr2 compares the fraction of the output voltage V F fed back from
the output potential divider R1/R2 with the stable reference voltage V Z across
the zener diode DZ.

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The output voltage VOUT in Fig. 2.2.4 can be expressed as:

VOUT = (VZ + VBE2) + (VOUT - VF)

Where:

VZ is the voltage across DZ

VBE2 is the base/emitter voltage of Tr2

VF is the feedback voltage derived from the slider of VR1

Therefore:

(VZ + VBE2) is the voltage across R2 and the lower portion of VRI

and

(VOUT − VF) is the voltage across R1 and the upper portion of VRI

If the feedback voltage VF is altered by adjusting VR1 potentiometer, the

difference between VF and VZ will change. This will cause a change in the error
voltage controlling Tr1 and a change in the output voltage V OUT. VR1 therefore
provides a variable output voltage, which, once set remains stable at that
setting.

The regulating action of the circuit is governed by the voltage across the
base/emitter junction of Tr2, i.e. the difference between V F and VZ.

If VOUT tends to increase, then VF - VZ also increases. This increases the collector
current of Tr2 and so increases the p.d. across R3 reducing the base voltage,
and therefore the base/emitter voltage of Tr1, reducing the conduction of Tr1,
so reducing current flow to the load.

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The output voltage VOUT is reduced in this way until a balance is reached, as the
feedback portion (VF) of VOUT is also reducing. The overall effect is that the
output is maintained at a level, which depends on the proportion of feedback
set by the variable resistor (part of R1/R2).

If the output voltage tends to decrease, then so does V F. The base/emitter

voltage of Tr2 is reduced due to the stable V Z on the emitter. Tr2 conducts less
and the current through R3 falls, reducing the p.d. across it. Tr1 base voltage
rises, and increases the conduction of the control transistor. This increases
output current and VOUT until VF is once more at the correct level.

Protection Circuits

Over Current Protection (Current Limiting)

Fig. 2.2.5 Series Regulator with Over Current Protection

Fig. 2.2.5 shows how the series stabiliser can be protected against excessive
current being drawn by the load. This will prevent damage to the supply in the
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event of too much current being drawn from the output or even a complete
short circuit across the output terminals.

Two components have been added, Tr3 and R5. The resistor R5 is a very low
value (typically less than 1 ohm).

When the load current rises above a predetermined value, the small voltage
developed across R5 will become sufficient (at about 0.7v) to turn Tr3 on. As Tr3
is connected across the base/emitter junction of the main control transistor Tr1,
the action of turning Tr3 on will reduce the base/emitter voltage of Tr1 by an
amount depending on the amount of excess current. The output current will not
be allowed to increase above a predetermined amount, even if a complete short
circuit occurs across the output terminals. In this case Tr1 base/emitter voltage
will be reduced to practically zero volts, preventing Tr1 from conducting. Under
these conditions the output voltage will fall to zero for as long as the excess
current condition persists, but the supply will be undamaged.

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Fig. 2.2.6 Series Regulator with Over Current & Over Voltage
Protection

Over Voltage Protection.

Where regulated supplies are used, the DC input voltage to the regulator is
often considerably higher than the required output voltage. Therefore if a PSU
fault occurs, it is possible that the regulated output voltage may suddenly rise
to a level that can damage other components. For this reason it is common to
find over voltage protection included in stabilised supplies. The circuit shown in
Fig. 2.2.6 is sometimes called a "crowbar" circuit because when it operates, it
places a complete short circuit across the across the output, a similar effect to
dropping a metal crowbar across the positive and ground output terminals!

Crowbar Circuit Operation.

In Fig. 2.2.6 the zener diode D Z2 has a breakdown voltage slightly less than the
maximum allowed value for VOUT. The remainder of VOUT is developed across R6,
VR2 and R7.

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VR2 is a potentiometer, so that a voltage may be taken from the resistor

network to correctly bias the diode D1. This diode has its cathode held at 0V by
R8, and VR2 is adjusted so that D1 is just out of conduction, i.e. with about 0.5V
on its anode.

Now, if VOUT increases, the voltage across R6, VR2 and R7 will rise by the same
amount, as the voltage across DZ2 will remain the same. Therefore there will be
a substantial rise in the voltage at R7 slider, and at D1 anode. This will cause D1
to conduct, supplying a pulse of current to the gate of thyristor Th1, causing it
to "fire" and conduct heavily until V OUT falls to practically 0v. R9 is included to
limit the resulting current flow through the thyristor to a safe level.

The large current that flows as Th1 fires will cause the current limiter circuit to
come into operation as previously described. This will safely shut down the
supply until the over current caused by Th1 has disappeared, which will of
course happen as soon as VOUTreaches 0V, but should the over voltage still be
present when Th1 switches off and VOUT rises again, the circuit will re-trigger.

I.C. Voltage Regulators

What you´ll learn in Module 2.3

After studying this section, you should be able to:

 Recognise commonly used I.C. Voltage Regulators.

 In Relation to the 78xx series of Voltage Regulators:
o Select appropriate decoupling components.
o Understand the term ‘Dropout’.
o Understand possible causes of IC failure & their prevention.
o Understand methods of producing positive, negative & dual
supplies.

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Buck Converters

Fig. 3.1.1 The Buck Converter

What you´ll learn in Module 3.1

 After studying this section, you should be able to:

 Understand the principles of Buck Converters.

 The Switching Transistor.
 The Flywheel Circuit.
 Recognize the limitations on the output voltage.
 Recognize different input sources.
 Understand the relationship between switching pulse width and output
voltage.

The Buck Converter

The Buck Converter is used in SMPS circuits where the DC output voltage needs
to be lower than the DC input voltage. The DC input can be derived from
rectified AC or from any DC supply. It is useful where electrical isolation is not
needed between the switching circuit and the output, but where the input is
from a rectified AC source, isolation between the AC source and the rectifier
could be provided by a mains isolating transformer.

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The switching transistor between the input and output of the Buck Converter
continually switches on and off at high frequency. To maintain a continuous
output, the circuit uses the energy stored in the inductor L, during the on
periods of the switching transistor, to continue supplying the load during the off
periods. The circuit operation depends on what is sometimes also called a
Flywheel Circuit. This is because the circuit acts rather like a mechanical
flywheel that, given regularly spaced pulses of energy keeps spinning smoothly
(outputting energy) at a steady rate.

AC or DC Input

The buck converter is a form of DC to DC converter that can take an input

directly from a DC source, such as a battery. The input could also be DC derived
from the AC mains (line) as shown in Fig. 3.1.1 via a rectifier/reservoir capacitor
circuit. The AC input to the rectifier circuit could be AC at high voltage directly
from the AC mains supply, or alternatively at a lower voltage via a step down
transformer. However the DC applied to the Buck Converter is obtained, it is
then converted to a high frequency AC, using a switching or ‘chopper’
transistor, driven by a (usually pulse width modulated) square wave. This
results in a high frequency AC wave, which can then be re-converted to DC in a
much more efficient manner than would be possible in the circuits described in
Power Supplies Module 1.

Buck Converter Operation

As shown in Fig. 3.1.1 the buck Converter circuit consists of the switching
transistor, together with the flywheel circuit (Dl, L1 and C1). While the transistor
is on, current is flowing through the load via the inductor L1. The action of any
inductor opposes changes in current flow and also acts as a store of energy. In
this case the switching transistor output is prevented from increasing
immediately to its peak value as the inductor stores energytaken from the

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increasing output; this stored energy is later released back into the circuit as a
back e.m.f. as current from the switching transistor is rapidly switched off.

Fig. 3.1.2 Switching Transistor ‘on’ Period

Transistor Switch ‘on’ Period

In Fig. 3.1.2 therefore, when the switching transistor is switched on, it is

supplying the load with current. Initially current flow to the load is restricted as
energy is also being stored in L1, therefore the current in the load and the
charge on C1 builds up gradually during the ‘on’ period. Notice that throughout
the on period, there will be a large positive voltage on D1 cathode and so the
diode will be reverse biased and therefore play no part in the action.

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Fig. 3.1.2 Switching Transistor ‘off’ Period

Transistor Switch ‘off’ Period

When the transistor switches off as shown in Fig 3.1.3 the energy stored in the
magnetic field around L1 is released back into the circuit. The voltage across
the inductor (the back e.m.f.) is now in reverse polarity to the voltage across L1
during the ‘on’ period, and sufficient stored energy is available in the collapsing
magnetic field to keep current flowing for at least part of the time the transistor
switch is open.

The back e.m.f. from L1 now causes current to flow around the circuit via the
load and D1, which is now forward biased. Once the inductor has returned a
large part of its stored energy to the circuit and the load voltage begins to fall,
the charge stored in C1 becomes the main source of current, keeping current
flowing through the load until the next ‘on’ period begins.

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The overall effect of this is that, instead of a large square wave appearing
across the load, there remains only a ripple waveform, i.e. a small amplitude,
high frequency triangular wave with a DC level of:

VOUT = VIN x (On time of switching waveform (t ON) / periodic time of switching
waveform( T))

or:

Therefore if the switching waveform has a mark to space ratio of 1:1, the output
VOUT from the buck Converter circuit will be V INx(0.5/1) or half of VIN. However if
the mark to space ratio of the switching waveform is varied, any output voltage
between approximately 0V and VIN is possible.

Fig. 3.1.3 Buck Converter Animation

Click play to start.

See the current paths during the on and off periods of the switching transistor.

See the magnetic field around the inductor grow and collapse, and observe the
changing polarity of the voltage across L.

Watch the effect of ripple during the on and off states of the switching
transistor.

Click pause to hold the animation in either the on or the off state.

Click back to reset the animation.

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Fig. 3.1.4 Buck Converter for Negative Supplies

Buck Converter for Negative Supplies

For negative supplies the circuit shown in Fig. 3.1.4 can be used. This involves a
change around in the positions of L1 and D1, and reversing the polarity of C
compared to the circuit in Fig 3.1.2. This variation of the basic buck converter
now inverts the positive DC input to produce a negative supply in the range of
0V to −VIN.

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Fig. 2.3.1 Typical LM78xx Series Packages

The LM78Xxx Integrated Circuit (I.C.) Range

The availability of regulator circuits in I.C. form has greatly simplified power
supply design, and since their introduction, the variety of designs, their power
handling capacity and their reliability has steadily improved. Integrated circuit
regulators are readily available in a variety of current and voltage ratings for
shunt or series applications as well as complete switched mode types. It is now
quite rare to find regulators in the truly discrete forms described in PSU Modules
2.1 to 2.3 but the popular 78Xxx types of regulator (where X indicates the sub
type and xx represents the output voltage) use much the same principles, with
enhanced circuitry, in an integrated form.

There are various ranges, in several package types available from many
component manufacturers, some of which are shown in Fig. 2.4.1. The package
choice is dependent space and performance requirements. Typical ranges are
summarised in Table 1.

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Table 1
Typical
Maximum Maximum
Range Output Voltages (VOUT) Dropout
Current Input Voltage
Voltage
5.0V, 6.2V, 8.2V, 9.0V,
LM78Lxx 100mA 35V VOUT + 1.7V
12V, 15V
LM78Mx
5V, 12V, 15V 500mA 35V VOUT + 2V
x
5.0V, 5.2V, 6.0V, 8.0V, 35 or 40V
LM78xx 8.5V, 9.0V, 12.0V, 15.0V, 1A dependent on VOUT + 2.5V
18.0V, 24.0V type

Dropout Voltage

One of the important pieces of data published in data sheets on linear I.C.
regulators is the devices dropout voltage. With any linear regulator, built with
discrete components or integrated such as the 78 series, the output voltage is
held steady for differing current flows by varying the resistance of the regulator,
(actually by varying the conduction of a transistor as explained in Power
Supplies Module 2.2).

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For this reason two things must be true:

1. The output voltage must always be lower than the input voltage.

2. The greater the difference between the input and output voltages (given the
same current) the more power must be dissipated in the regulator circuit, so the
hotter it will get.

The dropout voltage for any regulator states the minimum allowable difference
between output and input voltages if the output is to be maintained at the
correct level. For example if a LM7805 regulator is to provide 5V at its output,
the input voltage must be no lower than 5v +2.5V =7.5V.

However dropout voltage is not an absolute value, it may vary by about 1V

depending on the current drawn from the output and the temperature at which
the regulator is operating. It seems sensible therefore to allow a comfortable
margin between the lowest possible input voltage and the minimum voltage
allowable (the output voltage + the dropout voltage).

The maximum input voltage listed in table 1 shows that there is plenty of
allowable difference between maximum and minimum input voltage, however it
should be remembered that the higher the input voltage for a given output, the
more power will need to be dissipated by the regulator. Too high an input
voltage and power is wasted, this is bad for battery life in portable equipment
and bad for reliability in high power equipment as more heat means a greater
possibility of faults.

For example a LM7805 delivering 1A at 5V to a load means that the load is

consuming 5Watts. If the input voltage is 8V the current through the regulator is
still 1A, which is 8W; so the regulator is dissipating 8W − 5W = 3W. However if
the input voltage is 20V for example, then the excess power that must be

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dissipated by the regulator is now 20V x 1A = 20W minus the 5W consumed by

the load = 15W.

In modern linear I.C. regulators however, as well as protection against over

current and over voltage protection, as described in PSU Module 2.3 there are
additional thermal shut down circuits to prevent failure due to overheating, so
that if power is excessive, rather than destroying the IC the output will fall to 0V
until the IC has cooled.

Even at more sensible input voltages, regulator I.Cs. do develop considerable

amounts of heat, therefore it is essential that excess heat is dissipated
efficiently by the use of appropriate heat sinks. The criteria for using heat sinks
is the same as that for power transistors, discussed in Amplifiers Module 5.1.

Complementing the 78xx series is the 79 series, which offers I.Cs. for commonly
used negative supply voltages in a similar range of characteristics to the 78
series but with a negative voltage output.

Fig. 2.3.2 Basic PSU Circuit Using a 7805 Linear Regulator I.C.

Reducing AC Ripple

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Fig. 2.3.2 shows a series regulator I.C. and its connections. Notice that C1 and
C2 are much smaller than would be found in discrete component power supply.
A large reservoir capacitor is not necessary, as the regulating action of the I.C.
will reduce the amplitude of any AC ripple (within its maximum input voltage
range) to just a few millivolts at the output.

Ensuring Stability

C2 is no longer a traditional filter capacitor but is there to improve the transient

response, protecting against sudden changes in mains or load conditions e.g.
surges. The use of these capacitors, which with the values shown will be
polarised tantalum types and although not strictly essential in all circuits, are
recommended to ensure maximum stability, preventing any tendency of the I.C.
to oscillate. They should be mounted as close as possible to the regulator, and
the I.C. ground connection should be connected to 0V as physically close as
possible to the ground connection of the load. These issues can be best
accommodated when the regulator I.C. is used as a ‘point of load’ regulator,
rather than (or as well as) a main regulator for the whole system power supply.

Reliability

The use of linear regulator I.Cs. has greatly improved the reliability of power
supplies, but because these I.C.s are often located on plug in sub-panels with a
system, there is a danger that damage may occur to the regulator I.C. (as well
as to other components) if panels are inserted or removed whilst the main
power supply is still live. This may be either because the system is still
connected to the mains supply or because capacitors in the main supply are not
fully discharged.

The reason is that when unplugging or connecting multi-way connectors there is

no guarantee in which order individual pins are connected or disconnected, and

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this can lead to unexpected short circuit or open circuit faults occurring
momentarily during the connection or disconnection process.

To guard against this possibility several additional safeguards can be designed

around the regulator circuitry to safeguard the I.C.

In some circuits electrolytic capacitors may be used for C1 and C2 as an

alternative to using tantalum or polyester capacitors, but in this case the
capacitance use will be considerably greater, 25µF or more. However, in circuits
where C2 is 100µF or more, there is a possibility that if the input is shorted to
ground, either temporarily (or permanently because of a fault) that the charge
on C2 will cause a large current to flow back into the I.C. output terminal,
damaging the I.C. To prevent this, a diode such as a 1N4002 can be connected
across the I.C. as shown in Fig. 2.4.3 so that if, at any time the input terminal is
at a lower potential than the output, the diode will conduct any charge at the
output terminal to ground, rather than allowing the current to flow though the
I.C.

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Fig. 2.3.4 Effect of Open Circuit Gnd Connection on a 7812 IC

If a circuit panel is unplugged whilst the supply is live it is possible that the
ground connection to the I.C. may be disconnected momentarily before the
input as indicated by Fig. 2.4.4. In such event the output terminal can rise to
the voltage level of the unregulated input, which could cause damage to
components being supplied by the regulator. Also if the panel is plugged in with
power already present, the same situation, with the ground connection
momentarily open circuit, then damage to the I.C. is likely.

As voltage regulators are generally fed from the main power supply, they can
be susceptible to any mains borne voltage spikes as well as back e.m.f. voltage
spikes from other parts of the circuit. Any positive voltage spikes greater than
the maximum permitted input voltage (around 35V or 40V) or any negative
spikes greater than -0.8V that have sufficient energy to cause substantial
currents to flow are likely to damage the I.C. Some protection can be given by
using a large value capacitor at the input terminal and/or making sure that
likely causes of transients are minimised by using transient suppressors at the
mains input and preventing back e.m.fs. as described in A.C. Theory Module
3.2.

Dual and Negative Supplies

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Linear regulator I.C.s can also be used to provide regulated negative supplies
using the LM79xx range of regulators are available in a similar range of
voltages as the 78xx series but with negative outputs. They can be used for
regulating negative supply or dual supply rails.

 2.4 Quiz

Regulated Power Supplies Quiz

Try our quiz, based on the information you can find in Power Supplies Module 2.
Submit your answers and see how many you get right. If you get any answers
wrong. Just follow the hints to find the right answer, and learn about Regulated
Power Supplies as you go.

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1.

Refer to Fig. 2.4.1.What would be the likely output voltage V OUT?

a) 4.3V

 b) 4.7V

 c) 5.0V

 d) 3.6V

2. Refer to Fig 2.4.1. Which of the following actions would occur if a load
resistance connected across the output terminals decreased in value?

 a) The output current would increase and Tr1 would conduct less
heavily to maintain a constant output voltage.

 b) The output current would decrease and Tr1 would conduct more
heavily to maintain a constant output voltage.

 c) The output voltage would decrease, reducing the voltage across D Z

causing Tr1 conduction to decrease and the output voltage to rise to norm

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 d) DZ would ensure that the base voltage of Tr1 remains at a constant

voltage so that the output voltage also remains stable.

3.

Refer to Fig 2.4.2. What preferred value of resistance would be required for R in
order to maintain a minimum zener current (I Z) of 1mA?

 a) 330Ω

 b) 270Ω

 c) 220Ω

 d) 180Ω

4. Which of the following methods would be used to improve the regulation

factor of a shunt regulator circuit?

 a) Connect a second shunt regulator in cascade.

 b) Use a point of load regulator.

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 c) Use a transistor shunt regulator.

 d) Use a decoupling capacitor across the regulator output.

5.

Refer to Fig 2.4.3. Which of the following statements correctly describes the
output voltage VOUT?

 a) VOUT = VCB Tr1 + VCETr2

 b) VOUT = VR1 + VDZ

 c) VOUT = VS - VCETr1

 d) VOUT = VR2 + VCETr2 - VDZ

6. Refer to Fig 2.4.3. What will be the effect of moving the slider of VR1
downwards?

 a) The current through Tr2 will increase.

 b) The output voltage will decrease.

 c) The voltage across Tr1 collector/emitter will decrease.

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 d) The current through DZ will be unchanged.

7. Refer to Fig 2.4.3. What is the purpose of Tr2?

 a) Shunt regulator current amplifier transistor.

 b) Over voltage protection transistor.

 c) Over current shut down transistor.

 d) Error amplifier transistor.

8.

Refer to Fig 2.4.4. Which of the following components is responsible for sensing
the current value?

 a) R4

 b) R5

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 c) R8

 d) R9

9. Refer to Fig 2.4.4. What is the purpose of DZ2?

 a) To regulate the output voltage.

 b) To act as a fast switch for the crowbar circuit.

 c) To activate the over current circuit.

 d) To increase the sensitivity of the crowbar circuit.

10.

Refer to Fig. 2.4.5. What is the purpose of D1?

 a) To prevent damage to the I.C. in case of a short circuit input.

 b) To prevent damage to the circuit being supplied, in case of a faulty

I.C.

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 c) To remove positive voltage spikes at the I.C. output.

 d) To provide negative feedback for the I.C.

Switched Mode Power Supplies

What you´ll learn in Module 2

 Section 3.0 SMPS Introduction.

 Switch Mode Topologies.

 Stages in a Switched Mode Power Supply.

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 Voltage Regulation.
 High Frequency Switching.

 Section 3.1 Buck Converters.

 The Switching Transistor

 The Flywheel Circuit
 Buck Converter Operation

 Section 3.2 Boost Converters.

 Boost Converter Operation.

 I.C Boost Converters.
 Output Voltage Range of Boost Converters.

 Section 3.3 Buck-Boost Converters.

 Advantages of Buck-Boost Converters.

 Buck-Boost Converter Operation.

Section 3.4 Push-Pull SMPS.

 Push-Pull Switching.
 Pulse Width Modulation.
 Voltage and Current Limiting.

Section 3.4 SMPS Quiz.

Test your knowledge of Switched Mode Power Supplies

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Introduction

Switched Mode Power Supplies, (often abbreviated to SMPS) are considerably

more complex than the linear regulated power supplies described in Power
Supplies Module 2. The main advantage of this added complexity is that
switched mode operation gives regulated DC supplies that can deliver more
power for a given size, cost and weight of power unit.

Switched Mode Designs

A number of different design types are used. Where the input is the AC mains
(line) supply the AC is rectified and smoothed by a reservoir capacitor before
being processed by what is in effect a DC to DC converter, to produce a
regulated DC output at the required level. Hence a SMPS can be used as an AC
to DC converter, for use in many mains powered circuits, or DC to DC, either
stepping the DC voltage up or down as required, in battery powered systems.

Switched Mode Block Diagram

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Fig. 3.0.1 Typical SMPS Block Diagram

Fig. 3.0.1 shows a block diagram example of a typical SMPS with an AC Mains
(line) input and a regulated DC output. The output rectification and filter are
isolated from the High Frequency switching section by a high frequency
transformer, and voltage control feedback is via an opto isolator. The control
circuit block is typical of specialist ICs containing the high frequency oscillator,
pulse width modulation, voltage and current control and output shut down
sections.

Whatever the purpose of a SMPS, a common feature (after conversion of AC to

DC if required) is the use of a high frequency square wave to drive an electronic
power switching circuit. This circuit is used to convert the DC supply into high
frequency, high current AC, which by various means, depending on the design
of the circuit, is reconverted into a regulated DC output. The reason for this
double conversion process is that, by changing the DC or mains frequency AC to

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a high frequency AC, the components, such as transformers, inductors and

capacitors, needed for conversion back to a regulated DC supply, can be much
smaller and cheaper than those needed to do the same job at mains (line)
frequency.

The high frequency AC produced during the conversion process is a square

wave, which provides a means of controlling the output voltage by means of
pulse width modulation. This allows the regulation of the output to be much
more efficient than is possible in linear regulated supplies.

The combination of a square wave oscillator and switch used in switched mode
supplies can also be used to convert DC to AC. In this way the switched mode
technique also be used as an ‘inverter’ to create an AC supply at mains
potential from DC supplies such as batteries, solar panels etc.

Voltage Regulation

In most switched mode supplies, regulation of both line (input voltage) and load
(output voltage) is normally provided. This is achieved by altering the mark to
space ratio of the oscillator waveform before applying it to the switches. Control
of the mark to space ratio is achieved by comparing voltage feedback from the
output of the supply with a stable reference voltage. By using this feedback to
control the mark to space ratio of the oscillator, the duty cycle and therefore
the average DC output of the circuit can be controlled. In this way, protection
from both over voltage and over current may be provided.

Where it is important to maintain electrical isolation from the mains supply, this
is provided by using a transformer, either at the AC input where it may also be
used to alter the AC voltage prior to rectification, or between the control section
of the power supply and the output section where, as well as providing isolation,
a transformer with multiple secondary windings can produce several different
voltage outputs.

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To provide a well regulated output, a sample of the DC output voltage is

normally fed back to the control circuitry and compared with a stable reference
voltage. Any error produced is used to control the output voltage. To maintain
electrical isolation between input and output, feedback will usually be via a
device such as an opto-isolator.

HF Switching

Using high frequency for the switching drive gives several advantages:

• The transformer will be of a HF type, which is much smaller than a standard

mains transformer.

• The ripple frequency will be much higher (e.g. 100kHz) than in a linear supply,
and so it needs a smaller value of smoothing capacitor.

• Also using a square wave to drive the switching transistors (switched mode
operation) ensures that they dissipate much less power than a conventional
series regulator transistor. Again this means that, for a given amount of power
output, smaller and cheaper transistors can be used, than in similarly rated
linear power supplies.

• The use of smaller transformers and smoothing capacitors makes switched

mode power supplies lighter and less bulky. The added cost of the complex
control circuitry is also offset by the smaller, and therefore cheaper
transformers and smoothing capacitors, making some switched mode designs
less expensive than equivalent linear supplies.

Although linear supplies can provide better regulation and better ripple
rejection at low power levels than switched mode supplies, the above
advantages make the SMPS the most common choice for power supply units in

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any equipment where a stabilised supply is needed to deliver medium to large

amounts of power.

A disadvantage of using such a high frequency square wave in a powerful circuit

such as a SMPS is that many powerful high frequency harmonics are created, so
that without very effective RF screening and filtering, there is a danger of the
SMPS creating RF interference.

Boost Converters

What you´ll learn in Module 3.2

After studying this section, you should be able to:

 Understand the principles of Boost Converters.

o The switching transistor

o The flywheel circuit

 Recognise the limitations on the output voltage.

 Recognise different input sources.
 Understand the relationship between switching pulse width and output
voltage.

Boost Converter

Switched mode supplies can be used for many purposes including DC to DC

converters. Often, although a DC supply, such as a battery may be available, its
available voltage is not suitable for the system being supplied. For example, the
motors used in driving electric automobiles require much higher voltages, in the
region of 500V, than could be supplied by a battery alone. Even if banks of

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batteries were used, the extra weight and space taken up would be too great to
be practical. The answer to this problem is to use fewer batteries and to boost
the available DC voltage to the required level by using a boost converter.
Another problem with batteries, large or small, is that their output voltage
varies as the available charge is used up, and at some point the battery voltage
becomes too low to power the circuit being supplied. However, if this low output
level can be boosted back up to a useful level again, by using a boost converter,
the life of the battery can be extended.

The DC input to a boost converter can be from many sources as well as

batteries, such as rectified AC from the mains supply, or DC from solar panels,
fuel cells, dynamos and DC generators. The boost converter is different to the
Buck Converter in that it’s output voltage is equal to, or greater than its input
voltage. However it is important to remember that, as power (P) = voltage (V) x
current (I), if the output voltage is increased, the available output current must
decrease.

Fig. 3.2.1 Basic Boost Converter Circuit

Fig. 3.2.1 illustrates the basic circuit of a Boost converter. However, in this
example the switching transistor is a power MOSFET, both Bipolar power
transistors and MOSFETs are used in power switching, the choice being
determined by the current, voltage, switching speed and cost considerations.

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The rest of the components are the same as those used in the buck converter
illustrated in Fig. 3.1.2, except that their positions have been rearranged.

Boost converter Operation

Fig. 3.2.2 Boost Converter Operation at Switch On

Fig 3.2.2 illustrates the circuit action during the initial high period of the high
frequency square wave applied to the MOSFET gate at start up. During this time
MOSFET conducts, placing a short circuit from the right hand side of L1 to the
negative input supply terminal. Therefore a current flows between the positive
and negative supply terminals through L1, which stores energy in its magnetic
field. There is virtually no current flowing in the remainder of the circuit as the
combination of D1, C1 and the load represent a much higher impedance than
the path directly through the heavily conducting MOSFET.

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Fig. 3.2.3 Current Path with MOSFET Off

Fig. 3.2.3 shows the current path during the low period of the switching square
wave cycle. As the MOSFET is rapidly turned off the sudden drop in current
causes L1 to produce a back e.m.f. in the opposite polarity to the voltage across
L1 during the on period, to keep current flowing. This results in two voltages,
the supply voltage VIN and the back e.m.f.(VL) across L1 in series with each
other.

This higher voltage (VIN +VL), now that there is no current path through the
MOSFET, forward biases D1. The resulting current through D1 charges up C1 to
VIN +VL minus the small forward voltage drop across D1, and also supplies the
load.

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Fig. 3.2.4 Current Path with MOSFET On

Fig.3.2.4 shows the circuit action during MOSFET on periods after the initial start
up. Each time the MOSFET conducts, the cathode of D1 is more positive than its
anode, due to the charge on C1. D1 is therefore turned off so the output of the
circuit is isolated from the input, however the load continues to be supplied with
VIN +VL from the charge on C1. Although the charge C1 drains away through the
load during this period, C1 is recharged each time the MOSFET switches off, so
maintaining an almost steady output voltage across the load.

The theoretical DC output voltage is determined by the input voltage (V IN)

divided by 1 minus the duty cycle (D) of the switching waveform, which will be
some figure between 0 and 1 (corresponding to 0 to 100%) and therefore can
be determined using the following formula:

Example:

If the switching square wave has a period of 10µs, the input voltage is 9V and
the ON is half of the periodic time, i.e. 5µs, then the output voltage will be:

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VOUT = 9/(1- 0.5) = 9/0.5 = 18V (minus output diode voltage drop)

Because the output voltage is dependent on the duty cycle, it is important that
this is accurately controlled. For example if the duty cycle increased from 0.5 to
0.99 the output voltage produced would be:

VOUT = 9/(1- 0.99) = 9/0.01 = 900V

Before this level of output voltage was reached however, there would of course
be some serious damage (and smoke) caused, so in practice, unless the circuit
is specifically designed for very high voltages, the changes in duty cycle are
kept much lower than indicated in this example.

Fig. 3.2.5 Boost Converter Animation

Click play to start.

See the current paths during the on and off periods of the switching transistor.

See the magnetic field around the inductor grow and collapse, and observe the
changing polarity of the voltage across L.

Watch the effect of ripple during the on and off states of the switching
transistor.

See the input voltage and the back e.m.f. of V L add to give an output voltage
greater than the input voltage.

Click pause to hold the animation in either the on or the off state.

Click back to reset the animation.

I.C. Boost Converter

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Because of the ease with which boost converters can supply large over
voltages, they will almost always include some regulation to control the output
voltage, and there are many I.Cs. manufactured for this purpose A typical
example of an I.C. boost converter is shown in Fig. 3.2.6, in this example the
LM27313 from Texas Instruments. This chip is designed for use in low power
systems such as PDAs, cameras, mobile phones, and GPS devices.

In this circuit, an appropriate fraction of the output voltage (V OUT), dependent on

the ratio of R2:R3 is used as a sample and compared with a reference voltage
within the I.C. This produces an error voltage that is used to alter the duty cycle
of the switching oscillator, enabling a range of automatically regulated boost
voltages between 5V and 28V to be obtained.

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The LM27313 contains an internal oscillator operating at a fixed frequency of

about 1.6MHz. The FET switching transistor is also internal and switches the
current through L1 via the SW terminal. Notice also that a Schottky diode with
an appropriate voltage and current rating is used for D1 to keep losses due to
the forward voltage drop of the diode as small as possible, and to enable high
switching speeds to be achieved. The I.C. also has a shut down (SHDN) facility,
operated by external logic, by which the boost converter may be disabled when
not required, to save battery power.

Protection Circuits

Other safety features provided by the I.C. are over current shut down, which
disables the switch on a cycle-by-cycle basis if too much current is sensed, and
an over temperature shut down facility.

Stability

Another problem facing designers of high frequency boost converters is that of

stability, as at MHz frequencies both negative and positive feedback can occur
simply due to electromagnetic fields radiating between components within the
circuit, especially where the circuit components are in very close proximity as in
surface mount layouts. C2 is therefore added to improve stability and prevent
possible oscillation due to unwanted positive feedback occurring.

Buck-Boost Converters

What you´ll learn in Module 3.3

 After studying this section, you should be able to:

 Understand the need for a choice of DC to DC converter designs.

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 Understand the principles of Buck-Boost Converters.

 • The switching transistors.
 • Switching control systems.
 Understand the relationships between different converter designs.
 • Buck converters.
 • Boost converters.
 • Buck-Boost converters.
 Recognise the limitations on the output voltage.
 Recognise typical commercial I.Cs. using buck boost technology.

Buck-Boost Converters

A Buck-Boost converter is a type of switched mode power supply that combines

the principles of the Buck Converter and the Boost converter in a single circuit.
Like other SMPS designs, it provides a regulated DC output voltage from either
an AC or a DC input.

The Buck converter described in Power Supplies Module 3.1 produces a DC

output in a range from 0V to just less than the input voltage. The boost
converter will produce an output voltage ranging from the same voltage as the
input, to a level much higher than the input.

There are many applications however, such as battery-powered systems, where

the input voltage can vary widely, starting at full charge and gradually
decreasing as the battery charge is used up. At full charge, where the battery
voltage may be higher than actually needed by the circuit being powered, a
buck regulator would be ideal to keep the supply voltage steady. However as
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the charge diminishes the input voltage falls below the level required by the
circuit, and either the battery must be discarded or re-charged; at this point the
ideal alternative would be the boost regulator described in Power Supplies
Module 3.2.

By combining these two regulator designs it is possible to have a regulator

circuit that can cope with a wide range of input voltages both higher or lower
than that needed by the circuit. Fortunately both buck and boost converters use
very similar components; they just need to be re-arranged, depending on the
level of the input voltage.

Fig. 3.3.1 Buck and Boost Converters Combined

In Fig. 3.3.1 the common components of the buck and boost circuits are
combined. A control unit is added, which senses the level of input voltage, then
selects the appropriate circuit action. (Note that in the examples in this section
the transistors are shown as MOSFETs, commonly used in high frequency power
converters, and the diodes shown as Schottky types. These diodes have a low
forward junction voltage when conducting, and are able to switch at high
speeds).

Operation as a Buck Converter

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Fig. 3.3.2 Operation as a Buck Converter During Tr1 ‘on’ Period

The basic operation of the buck boost converter is illustrated in Figs. 3.3.2 to
3.3.5

Fig. 3.3.2 shows the circuit operating as a Buck Converter. In this mode Tr2 is
turned off, and Tr1 is switched on and off by a high frequency square wave from
the control unit. When the gate of Tr1 is high, current flows though L, charging
its magnetic field, charging C and supplying the load. The Schottky diode D1 is
turned off due to the positive voltage on its cathode.

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Fig. 3.3.3 Operation as a Buck Converter During Tr1 ‘off’ Period

Fig 3.3.3 shows the current flow during the buck operation of the circuit when
the control unit switches Tr1 off. The initial source of current is now the inductor
L. Its magnetic field is collapsing, the back e.m.f. generated by the collapsing
field reverses the polarity of the voltage across L, which turns on D1 and current
flows through D2 and the load.

As the current due to the discharge of L decreases, the charge accumulated in C

during the on period of Tr1 now also adds to the current flowing through the
load, keeping VOUT reasonably constant during the off period. This helps keep
the ripple amplitude to a minimum and VOUT close to the value of VS.

Operation as a Boost Converter

Fig. 3.3.4 Operation as a Boost Converter During Tr2 ‘on’ Period

In Boost Converter mode, Tr1 is turned on continually and the high frequency
square wave applied to Tr2 gate. During the on periods when Tr2 is conducting,
the input current flows through the inductor L and via Tr2, directly back to the
supply negative terminal charging up the magnetic field around L. Whilst this is
happening D2 cannot conduct as its anode is being held at ground potential by
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the heavily conducting Tr2. For the duration of the on period, the load is being
supplied entirely by the charge on the capacitor C, built up on previous
oscillator cycles. The gradual discharge of C during the on period (and its
subsequent recharging) accounts for the amount of high frequency ripple on the
output voltage, which is at a potential of approximately V S + VL.

The Off Period

Fig. 3.3.5 Operation as a Boost Converter During Tr2 ‘off’ Period

At the start of the off period of Tr2, L is charged and C is partially discharged.
The inductor L now generates a back e.m.f. and its value that depends on the
rate of change of current as Tr2 switches of and on the amount of inductance
the coil possesses; therefore the back e.m.f can be any voltage over a wide
range, depending on the design of the circuit. Notice particularly that the
polarity of the voltage across L has now reversed, and so adds to the input
voltage VS giving an output voltage that is at least equal to or greater than the
input voltage. D2 is now forward biased and so the circuit current supplies the
load current, and at the same time re-charges the capacitor to V S + VL ready for
the next on period of Tr2.

Circuit Variations
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There are a number of variations of this basic Buck-Boost circuit, some designs
working at lower frequencies or at high voltages may use bipolar transistors
instead of MOSFETs; at low frequencies the higher speed switching of MOSFETs
is less of an advantage. Also, in high voltage designs, silicon diodes may be
used in preference to Schottky types due to the silicon diode’s higher reverse
voltage capabilities. Another variation is to use synchronous switching where,
instead of using diodes that simply respond to the voltage polarity across them,
four synchronised (by the control unit) MOSFETs do all the switching.

The control unit may also carry out over current and over voltage protection, as
well as the normal oscillator and pulse width modulation functions to regulate
the output voltage.

Another commonly used facility is ‘pulse skipping’ where the control unit
prevents charging on one or more oscillator pulses when it senses that the load
current is low. This reduces the overall current drawn from the (typically
battery) supply, prolonging battery life.

Buck-Boost Converter I.Cs.are commonly used to carry out the control unit
functions. These range from very low power, high efficiency I.Cs. for portable
devices such as mobile phones and automotive applications, such as the
TPS63000 series from Texas Instruments, and the LTC3789 from Linear
Technology, to large industrial high power DC-DC converters providing many
kilowatts of output power.

Push-Pull Switched Mode Power Supplies

What you´ll learn in Module 3.4

 After studying this section, you should be able to:

 Understand the operation of a push pull DC to DC converter.

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 Understand the operation of a typical switch mode controller I.C.

 • Power switching transistors.
 • Pulse width modulator.
 • Over current & over voltage protection.
 • Current sensing.
 Recognise components and methods used for output isolation.

Fig. 3.4.1 Push Pull SMPS Block Diagram

The Push Pull SMPS

Fig. 3.4.1 shows a block diagram of a switched mode power supply designed
around a UC3524 Advanced Regulating Pulse Width Modulator by Texas
Instruments.

The circuit is a DC to DC converter using a DC input voltage of 15V to 30V and

produces a regulated 5V output at a current up to about 250mA. The circuit
uses push pull power switching driving a high frequency transformer, which fully

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isolates the output circuit from the input. The output is short circuit protected,
and the output voltage can be manually adjusted. Maximum current can be also
manually set using adjustable current limiting.

Primary Circuit.

A 100kHz oscillator within IC1 (UC3524) generates pulses, which are processed
by the pulse width modulator (within IC1)used to drive the power switching
transistors. The width of the processed drive pulses controls the length of time
for which the power switching transistors conduct, and therefore the amount of
power delivered to the transformer.

The pulse width and therefore the output voltage is controlled by the error
amplifier in IC1. This measures the difference between a sample of the output
voltage, fedback via the opto-isolator, and a reference voltage set by Vr1. When
these two voltages are equal, the circuit output voltage is correct. If there is a
difference, the width of the pulses produced by the pulse width modulator is
increased or decreased to correct the error.

Over current protection is provided to ensure that the supply is shut down in the
event of too high a current demand at the output. The output terminals can
even be shorted together without damaging the supply.

Each pulse of current in the power switching transistors produces a voltage

pulse across the sensing resistor R12. The amplitude of these pulses is
proportional to the current being delivered to the transformer by the switching
circuit. If the peak value of any of these pulses exceeds the DC voltage set by
Vr2 (Current limit) then the output from IC2 will cause pulse width modulator
input to reduce the width of the pulse being produced by the modulator at that
time, momentarily reducing the output voltage. If the over current condition
disappears, the output voltage will be restored to its normal level, but if the
load current remains high, the current limiter will continue to reduce the pulse

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width, dependent on the amount of over current, even down to zero in the case
of a short circuit output.

The Secondary Circuit

The push pull switching transistors create an AC waveform across the

transformer primary winding, and the secondary winding feeds a conventional
full wave rectifier and LC low pass filter to supply the load with a stable 5V at
the output terminals. Negative feedback to the voltage regulating circuit in IC1
is via the opto-isolator (IC3). The higher the output voltage, the brighter the
glow from an LED sealed within the secondary side, and the larger the DC
output voltage derived from pin 4 of IC3 in the primary side of the device. This
voltage is used as a sample at the inverting input of the error amplifier in IC1
where it is compared with a voltage from the ‘set voltage’ control VR1, to
control the pulse width modulator.

Circuit Description

The full schematic diagram for the circuit is shown in Fig. 3.4.2.

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Fig. 3.4.2 Circuit diagram of a push-pull SMPS using the Texas Instruments
UC3524 I.C.

The oscillator within IC1 produces narrow 100kHz (approx) pulses that are used
as clock pulses for the Switch Logic within IC1. The timing components for the
oscillator are R3 and C2. The ramp waveform produced as C2 charges is also
used as an input to the inverting input of the comparator in IC1.

The pulse width modulator comprises the comparator within IC1 and the
switching logic, which consists of a bistable and two three input NOR gates. The
outputs of this block supply variable width pulses to the two transistors Qa and
Qb.

The error amplifier compares a stable reference voltage on pin 1 (set by Vr1
supplied from an internally regulated 5V from pin 16) with a sample of the load

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voltage developed across the opto-isolator emitter resistor, R11. The resulting
error voltage is used as the non-inverting input to the PWM comparator.

The facilities of the UC3524 that are used in this circuit are shown in more detail
in Fig. 3.4.3 (Note: Some unused facilities of the UC3524 have been omitted for
clarity, for more information see the Texas Instruments UC3524 data sheet).

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Pulse Width Modulator

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Fig. 3.4.4 UC3524N and Transformer Primary Waveforms

The action of the pulse width modulator, described by the waveforms shown in
Fig. 3.4 4 is as follows:

Clock pulses (CK) from the oscillator are fed to the Bi-stable (flip-flop), which
produces a square wave with a 1:1 mark/space ratio and a frequency of 50kHz,
(half that of the oscillator) at its Q output, and an inverted version of this wave
at its Q output.

Output Q provides an input to NOR gate a, and output Q (the opposite of Q)

provides an input to NOR gates b. The logic rule for a NOR gate is that its output
will be high, only when all its three inputs are low. Notice that the Q and Q
signals go low at the start of alternate clock pulse low states. The clock signal
also provides an input to both NOR gates.

The third input to each of the NOR gates is provided by the comparator output,
which is a series of variable width low state pulses, produced by comparing the
DC error voltage from the error amplifier in IC1 with the ramp produced by the
oscillator timing capacitor C2.

As each NOR gate output goes high, only when all of its three input signals are
low, alternate high state pulses, whose width depends on the value of the error
voltage, are fed to the bases of the internal transistors Qa and Qb. The lower
the value of the error voltage (due to a higher value of "sample" voltage at pin
1) the narrower the pulses produced. These narrower pulses, when used to turn
on the power switching transistors TR3 and Tr4, will lead to a reduction in power
in the transformer and a reduction in load voltage.

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Power Switching Circuits

Fig. 3.4.5 Speeding up the Power Switch Turn Off

The internal drive transistors Qa and Qb each produces a series of pulses at its
collector, and an a series of anti-phase pulses at its emitter. The emitter signals
a and b drive the power switching transistors Tr3 and Tr4 respectively, and the
collector signals drive the speed up circuits Tr1/Tr2.

The reason for including the speed up circuits is to overcome the delay that
would normally happen because while the power switching transistors Tr3
andTr4 are conducting, their base/emitter junction (which naturally forms a
small capacitor due to the depletion layer between the base and emitter layers
in the transistor) is charged up, and must be discharged before the transistor
will fully turn off.

The power transistor junctions are rapidly discharged by momentarily turning

on Tr1 or Tr2 using a differentiated pulse generated from the rising edge of the
waveform from the collector of Qa or Qb in IC1, which of course happens at
exactly the same instant as Tr3 or Tr4 is turning off, as illustrated in Fig 3.4.5.

Because the transformer primary centre tap is connected to the main (+V IN)
supply, it will always be at the supply potential. The collector voltages of Tr3

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and Tr4 will also be at +VIN during the periods when both transistors are turned
off. During the ‘on’ pulse of Tr3, its collector will be at approximately 0V, and
due to the centre tapping of the transformer primary winding the bottom half of
the primary will be in anti-phase to the top half, so the collector of Tr4 will be
positive at twice the value of +V IN for the period of the Tr3 ‘on’ pulse. This
situation is reversed during the ‘on’ pulse of Tr4. This action produces a stepped
type of waveform with an amplitude of +V IN x 2 across the transformer primary
as shown in Fig 3.4.4.

The Secondary Circuit

The resulting secondary voltage is rectified by D1 and D2, and smoothed by the
low pass filter L1/C10 before being supplied to the load. A sample of the load
voltage is fed back to the LED within opto-isolator IC3 via the LED current
limiting resistor R13.

Fig. 3.4.6 Typical High Frequency Multi-Secondary Transformer

Because of the push pull design used by this circuit, it is a simple matter to
arrange for such a circuit to have multiple outputs. Different (higher or lower)
voltages can be obtained by using a transformer similar to the one illustrated in
Fig. 3.4.6, which has multiple secondary windings with appropriate turns ratios.

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The total current supplied to the multiple outputs however, must not exceed the
maximum current rating of the SMPS. Each supply line will have its own rectifier
and filter system, and may also include some extra point of load regulation. A
voltage sample will normally be taken from only one of the outputs to provide
feedback to the pulse width modulator however, as controlling the power
applied to the transformer primary will control all the voltage outputs.

Current limiting

Current limiting, which is capable of completely shutting down the circuit under
extreme overload conditions is provided by the action of IC2 and the shut down
transistor between pins 9 and 10 within IC1.

Pin 3 of IC2 is provided with a stable reference voltage derived from the shunt
voltage regulator R7/ZD1 via the current limit control Vr2. The non-inverting
input of IC2 is connected to a low resistance current sensing resistor R12 in the
emitter lead common to both switching transistors Tr3/Tr4.

Every time either transistor conducts, the resulting large emitter current
produces a voltage pulse across R12. The peak voltage of this pulse will be
proportional to the emitter current flowing in Tr3/Tr4 and therefore, also to the
output current.

If the peak voltage of any of these pulses applied to the non-inverting input of
IC2 exceeds the stabilised DC voltage at the inverting input, a positive pulse will
be produced at the output, and therefore at the base of Qc within IC1. This will
cause the collector voltage of this transistor to fall, also reducing the error
amplifier output that is controlling the pulse width modulator. This action has
the effect of reducing the width of the pulse presently being produced, thus
instantly reducing output voltage. If the current overload disappears, the pulse
width modulator will return to normal operation. If not, subsequent pulses will
be further reduced until the output voltage falls (if necessary) to zero.

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The action of the current limit circuit is not absolutely instant however, due to
the presence of C4 on the shut down compensation pin (9) of IC1. This capacitor
tends to integrate the voltage changes on the collector of the shut down
transistor so that very rapid (cycle by cycle) variations of the output voltage
during current limiter action are avoided.

Switched Mode Power Supplies Quiz

Try our quiz, based on the information you can find in Power Supplies Module 3.
Submit your answers and see how many you get right. If you get any answers
wrong, just follow the hints to find the right answer, and learn about Switched
Mode Power Supplies as you go.

1. Which answer from the following list is a major advantage of

switched mode power supplies over series regulated supplies?

a) Switched mode supplies create smaller amplitude ripple waveforms than

series regulators.

b) Switched mode supplies dissipate less power in the control element

than series regulators.

c) Switched mode supplies provide better regulation at low power than

series regulators.

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d) Switched mode supplies work digitally, so are more efficient than

series regulators.

2. Which answer from the following list is a disadvantage of switched

mode power supplies compared with series regulated supplies?

a) Switched mode supplies are less suitable for high power applications.

b) Switched mode supplies are more expensive to implement than series

regulated supplies.

c) It is more difficult to prevent high frequency electrical interference in

switched mode supplies.

d) Over voltage protection is not possible in switched mode power

supplies.

Which one of the following acts as an energy store in a DC to DC

Converter?

a) The inductor.

b) The high frequency switching transistor.

c) The load.

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d) The flywheel diode.

4.

Refer to Fig 3.5.1. Which of the following components is not active during the
ON period of the switching waveform?

a) Tr1

b) D1

c) L1

d) C2
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5.

What type of circuit is illustrated in Fig 3.5.2?

a) Buck Converter.

b) Boost Converter.

c) Buck-boost Converter.

d) Flyback Converter

6. Which of the following is one of the main advantages of a push-pull converter

compared to buck or boost DC to DC converters?

a) It works at twice the frequency of other DC to DC converters.

b) It uses pulse width modulation.

c) It provides an output voltage of twice the amplitude of the input .

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d) It uses a transformer to isolate the output from the input.

7. Which of the following statements is true of a Buck-Boost Converter?

a) The output power may be less than, or greater than the input power.

b) The output voltage is always greater than the input voltage.

c) The output current is always greater than the input current.

d) The output voltage may be less than, or greater than the input voltage.

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8.

Refer to Fig. 3.5.3. What is the purpose of Tr1 and Tr2?

a) They act with Tr3 and Tr4 to form two Darlington Pair output stages.

b) They isolate the output transistors from the common ground

connection.

c) They speed up the switch off time of the output transistors.

d) They integrate the square wave signals to the bases of Tr3 and Tr4.

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9. Refer to Fig. 3.5.3. What is the purpose of IC2?

a) It acts as a comparator for the current limiter.

b) It acts as an error amplifier for the current limiter.

c) It acts as a comparator for the over voltage control.

d) It acts as an error amplifier for the over voltage control.

10. Refer to Fig. 3.5.3. What type of signal will be present at pin 4 of IC3.

a) 50kHz pulses with an amplitude proportional to the output voltage.

b) A DC voltage proportional to the output current.

c) 100kHz pulses with an amplitude proportional to the output current.

d) A DC voltage proportional to the output voltage.

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