module title : digital & analogue devices & circuits · pdf...

MODULE TITLE : DIGITAL & ANALOGUE DEVICES & CIRCUITS

TOPIC TITLE : D.C. POWER SUPPLIES

LESSON 2 : SMOOTHING

DADC - 3 - 2

© Teesside University 2011

Published by Teesside University Open Learning (Engineering)

School of Science & Engineering

Teesside University

Tees Valley, UK

TS1 3BA

+44 (0)1642 342740

All rights reserved. No part of this publication may be reproduced, stored in a

retrieval system, or transmitted, in any form or by any means, electronic, mechanical,

photocopying, recording or otherwise without the prior permission

of the Copyright owner.

This book is sold subject to the condition that it shall not, by way of trade or

otherwise, be lent, re-sold, hired out or otherwise circulated without the publisher's

prior consent in any form of binding or cover other than that in which it is

published and without a similar condition including this

condition being imposed on the subsequent purchaser.

________________________________________________________________________________________

INTRODUCTION________________________________________________________________________________________

Having described the construction and operation of transformers and rectifiers

in a d.c. power supply system, we now deal with the smoothing that is

necessary to produce a steady and sustainable d.c. supply.

________________________________________________________________________________________

YOUR AIMS________________________________________________________________________________________

On completion of this lesson you should be able to:

• explain the function of a 'reservoir' capacitor in a rectified d.c. supply

circuit

• define ripple factor and carry out simple related calculations

• state the advantages of incorporating an inductive component into a

'smoothing' circuit

________________________________________________________________________________________

STUDY ADVICE________________________________________________________________________________________

See the Appendix for the notation used for alternating quantities.

1

i i i i © T id U i it 2011

________________________________________________________________________________________

SMOOTHING________________________________________________________________________________________

The output from a rectifier circuit cannot be described as continuous direct

current because of the presence of a relatively large alternating component.

This form of output, particularly if resulting from full-wave rectification, is

suitable for some applications such as battery charging. However, in order to

be suitable for electronic equipment, the alternating component must be

reduced to a small percentage of the direct current output. This reduction

process is referred to as smoothing.

Smoothing is usually achieved by introducing a reactive component into the

circuit to 'filter' out most of the alternating component. The most common

method employs a shunt capacitor which is often referred to as a reservoir

capacitor because its function is similar to that of a pneumatic pressure vessel

or hydraulic accumulator.

CAPACITOR SMOOTHING

To achieve the smoothing process, the reservoir capacitor, C, is connected

across the rectifier output in parallel with the load, RL, as shown in FIGURE 1.

2


FIG. 1

The capacitor charges up near the peak of each rectified cycle and discharges

into the load for the remainder of the cycle. For a large part of each cycle,

therefore, the load current is being supplied by the reservoir capacitor whilst

the rectifier 'tops up' the capacitor charge at the peak of each cycle.

This effect is illustrated in the output waveform shown in FIGURE 2.

FIG. 2

Period T

Chargeperiod

Dischargeperiod

TC TD

0t

Loadvoltagev

RL

a.c.

Transformer

Reservoircapacitor

Full-wavebridge rectifier

LoadC

VsVD

VD

VL

3


RIPPLE

As FIGURE 2 shows, the supply of load current causes the capacitor voltage to

fall during each cycle. The resulting unevenness in d.c. output voltage is

termed ripple which appears as an alternating waveform superimposed on the

direct output.

A quantitative measure of the degree of ripple is the ripple factor (KV) which is

defined as:

or

where Vr is the r.m.s. value of the ripple voltage and the average load

voltage.

In order to simplify the calculations relating to ripple factor the following

approximations can be made.

• The output voltage waveforms can be assumed to be composed of

linear sections which produce a triangular or 'sawtooth' ripple

waveform as shown in FIGURE 3*.

________________________________________________________________________________________

*See the Appendix for the notation used for alternating quantities.

V L

KV

VVr

L= ......................................................... 1( )

KV = rms value of the a.c. voltage componentavverage value of the load voltage

4


FIG. 3

• Because the ripple voltage, Vr, will, in practice, be small compared

with the d.c. voltage, the discharge period, TD, can be assumed to

equal the total period, T. This allows approximation of the ripple

waveform to that shown in FIGURE 4.

FIG. 4

Evaluation of the a.c. voltage component can now be based on the rms value

(Vr) of one cycle of this 'ramp' waveform which can be calculated using

FIGURE 5.

time (s)Period T

Vrˆ̂

v

Period T

TD

t0

VL

VL

Vrˆ

ˆ̂

5


FIG. 5

The relationship between the instantaneous voltage, v, and the ripple voltage, Vr, can

be derived using similar triangles. See if you can write down this relationship.

....................................................................................................................................................

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________________________________________________________________________________________

v

t

V

T

v Vt

T

rr

r r

=

∴ =

∨

∧

∨

∧ ........................................ 3( )

VT

v tT

rms d ........................=⌠

⌡⎮⎮

1

0

2 ..... 2( )

T

v

t

Vr

0 t

v

6


Now substituting Equation (3) into Equation (2):

Substitution of Equation (4) into equation (1) gives the simple expression for

the voltage ripple factor:

Two other useful expressions are those for the ripple voltage, , and the

average d.c. output voltage, V L .

Vr∨

∧

K

V

VV

r

L= ∨

∧

3 .................................... 5( )

VT

Vt

Tt

V

Tt

r

T

r

r T

2

0

2

2

30

2

1=⎛⎝⎜

⎞⎠⎟

=

⌠

⌡⎮⎮

⌠

⌡⎮⎮

∨

∧

∨

∧

d

ddt

V

T

tV

T

T

V

r T r

r

=⎡

⎣⎢

⎤

⎦⎥ = ⎛

⎝⎜⎞⎠⎟

∴ =

∨

∧

∨

∧2

3

3

0

2

3

3

3 3

VV

V

V

r

r

r

2

3

3

∨

∧

∨

∧

= .......................................... 4( )

7


Ripple Voltage

During the charging period, the accumulated charge, ΔQ, in the capacitor is

given by the product of capacitance and voltage increase. Hence,

This same charge is transferred to the load, RL, during the discharge of the

capacitor. The transferred charge approximately equates to the product of the

average discharge current and discharge time. The average discharge current is

approximately the same as the average load current, (The validity of this

assumption is demonstrated later in the lesson.) Hence,

Combining Equations (6) and (7):

Substituting where f is the frequency of the original a.c. input,

gives:

VI

f CrL

∨

∧≈

2 .......................................... 8( )

Tf

= 12

CV I T

VI T

C

r L

rL

∨

∧

∨

∧

≈

∴ ≈

ΔQ I TL≈ ............................................ 7( )

I L .

ΔQ CVr=∨

∧ .......................................... 6( )

8


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________________________________________________________________________________________

The basic relationship between periodic time in seconds and frequency in hertz is

Because the frequency of the rectified waveform is twice that of the a.c. input waveform,

the relationship becomes

D.C. Output Voltage

Applying Ohm's law:

V f CR VL L r=∨

∧2 ............................ 10( )

Substituting for from Equation (8) givesI L ::

V I RL L L= .................................... 9( )

Tf

= 12

.

Tf

= 1.

Explain the relationship Tf

= 12

.

9


Example 1

A 24 V power supply delivers 200 mA to a resistive load. The a.c. supply is

240 V, 50 Hz and a bridge rectifier is used in conjunction with a 1000 μF

reservoir capacitor.

Estimate:

(a) the ripple voltage

(b) the transformer secondary voltage if the total diode forward voltage

drop in the rectifier is 1.3 V.

Solution

(a) Ripple voltage is given by:

(b) Reference to FIGURE 3 shows that the peak of the a.c. waveform is

given by:

where is the amplitude of the ripple voV r

∧lltage.

V V

V

V VL L

r

L r

∧ ∨

∧

∧= + = +

2

VI

f CrL

∨

∧=

= ×× × ×

=

2

200 102 50 1000 10

23

6

–

–.V

10


Allowing for the diode forward voltage drop, the peak transformer

secondary voltage, is given by:

Average d.c. Output Voltage of a FW Bridge Rectifier with Smoothing

The previous example (and with reference to FIGURES 1 and3) shows that:

where

Thus

V V V V V

V V V

V V V

L L r L r

L s D

L s D

= =

=

=

∧

∨

∧ ∧ ∧

∧

∧

– –

–

– –

12

2

2122

2

V

V V V V

r

L s D r

∨

∧

∧ ∧∴ = ................– – .............. 11( )

V V V

VV

S L D

SS

∧ ∧

∧

= +

= + = ( )

∴ = =

2

25 1 3 26 3

2

26 3

. .

.

V peak

22

18 6= . .V

V S

∧

∴ = +

∴ =

∧

∧

V

V

L

L

2422

25 V.

11


Peak Reverse Voltage of a Bridge Rectifier with Smoothing

The peak reverse-voltage (PRV) is the maximum reverse voltage that the

bridge circuit can produce across the rectifying diodes. The voltage rating of

the diode must be in excess of the PRV to prevent voltage breakdown

In the case of the bridge rectifier, the PRV can be found by consideration of the

circuit of FIGURE 6 which shows one pair of diodes conducting and the other

pair reverse-biased.

FIG. 6

The PRV is given by the voltage across the points ×-× of the circuit. Denoting

the PRV by the symbol (for repetitive reverse voltage), then is

given by:

V V VRR L D

∧ ∧= +

V RR

∧V RR

∧

Vs

X

X

VL

VD

Vxx

12


For FW rectification:

Note again that is a property of the circuit, not the rating of the diode. To

avoid damage, the diode’s reverse-voltage rating should be somewhat in excess

of .

CAPACITOR RIPPLE CURRENT IN A FW BRIDGE RECTIFIER

The charging current in the smoothing capacitor of FW rectifier consists of a

series of pulses as shown in FIGURE 7. Current only flows into the capacitor

during the brief period for which the rectifier output voltage exceeds the load

voltage, i.e. during the charging period TC. This effect is problematic in

several respects:

• the smoothing capacitor will have to have a high ripple current rating

• the current in the rectifying diodes will also consists of pulses of

current and the diodes will therefore have to have a suitable surge

current rating

V RR

∧

V RR

∧

V VRR s

∧ ∧� ................................................ 13( )

Assuming then:V Vs D

∧>> ,

V V V

V V V

L s D

RR s D

∧ ∧

∧ ∧

=

∴ =

–

–

2

...................................... 12( )

13


• the high peak values of current will cause a large volt drop across the

transformer windings

• the I2R power loss in the transformer will be greater than it would be

for an average current passed over the entire cycle, requiring a

transformer with a higher VA rating.

FIG. 7

During the discharge period the capacitor provides the entire load current.

This is because during discharge the diode anode voltage falls below the

voltage across the capacitor and the diode is reverse-biased. Thus, the

capacitor’s current decays exponentially to the equation

iLRLC

i i

I It

CRI

DC L

L LL

L

=

= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

+∧ ∨ ∨

– exp –

Period T

Chargeperiod

Dischargeperiod

TC TD

0t

Loadvoltage

v/vo

lts

Pulse ofcharging current

I/am

ps

14


This exponential decay of

current is represented in the

lower curve of the sketch of

FIGURE 8.

In practice, because of the

long time constant CRL, this

curve can be, to a very good

approximation, regarded as

being a straight line. Indeed, more often than not, the discharge current can

assumed to be constant, as represented by the upper horizontal dotted line.

This assumption can be seen to be justified the PSpice simulation of a bridge

rectifier with smoothing shown in the circuit of FIGURE 9(a)1. Here the

primary of a 20:1 turns ratio transformer is connected to a 240 V, 50 Hz supply.

Thus the secondary voltage is 10 V (rms). Current and voltage waveforms for

the circuit are plotted in FIGURE 9(b). Note that the diode pairs conduct every

other cycle and the diode current consists of a pulse whilst the capacitor is

charging. For a load of 100 Ω, this pulse has a peak value of over 1.4 amps,

whereas the load current is little more than 100 mA; the bulk of the diode current

is flowing into the capacitor. That this is so can be seen in the waveform for the

capacitor current. This consists of a pulse of current during the charging period

that is IL less than the diode current, i.e. the diode supplies both the charging

capacitor and the load currents. During discharge the capacitor supplies the load

current (this is represented as a negative current in the waveform, as positive

current has been assumed to be flowing into the capacitor).

________________________________________________________________________________________

1Note the following points in relation to the use of PSPICE:

The transformer model used is ideal and PSPICE requires resistance (R2) to be included in the primary

winding to prevent this winding appearing as a short circuit to the voltage source. Similarly for the analysis to

run, the secondary circuit will have to be ‘earthed’, although of course this might not be the case in practice.

The load voltage has been measured using a ‘differential voltage probe’ as a single-ended voltage probe will

measure the voltage with respect to the earth point.

TDC

FIG. 8

IL

IL

ˆˆ

15


For C = 1 mC and RL = 100 Ω, the ripple voltage can be seen to be about

±500 mV.

From the information given in FIGURE 9, estimate the forward volt drop across the

diodes.

FIG. 9(a)

16


FIG. 9(b)

Dio

de c

urre

nt/a

mps

1.4

1.2

1

0.8

0.6

0.4

0.2

0

Time/Secs 2mSecs/dw1 1.002 1.004 1.006 1.008 1.01 1.012 1.014 1.016 1.018


Cap

aici

tor

curr

ent/

amps

1.2

1

0.8

0.6

0.4

0.2

0

–0.1

Seco

ndar

y vo

ltag

e/vo

lts

10

5

0

–5

–10



Rip

ple

volt

age/

volt

s

13

12.8

12.6

12.4

12.2

17


From equation (11):

Thus

Therefore

CHOKE INPUT FILTER

The use of a reservoir capacitor produces minimum ripple and maximum d.c.

output when the output resistance is relatively high. A reduction in load

resistance, therefore, will cause an increase in the ripple factor.

This effect can be reduced by introducing an inductor into the smoothing

circuit. An inductive element in the circuit will reduce the ripple voltage when

the load resistance falls and the ratio of ωL to R increases.

The circuit is shown in FIGURE 10.

FIG. 10

The inclusion of a series inductance will also reduce any 'surge' effects which

may occur and possibly remove the need for any additional surge current

protection.

RL

L

CInputfrom

rectifier

VV V V

V

V

rD

s DC

D

D vo

=

=

=

∧ ∧– –

. – . – .

.

2

14 14 0 5 12 62

0 52 llts

18


________________________________________________________________________________________

NOTES________________________________________________________________________________________

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19


________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1. Explain what is meant by the following terms:

(a) peak-to-peak ripple voltage

(b) rms ripple voltage

(c) ripple factor.

2. A full-wave rectifier has an input voltage of 30 volts (rms) at 50 Hz. If a

500 μF smoothing capacitor is used across a load resistance of 100 Ω,

calculate the following:

(a) the rms ripple voltage

(b) the average voltage across the load

(c) the d.c. load current

(d) the ripple factor

(e) the maximum reverse-bias voltage applied to the diode.

3. A full-wave bridge rectifier with a 1000 μF shunt capacitor. It is required

to supply a load resistance with an average d.c. voltage of 20 volts with a

maximum peak-to-peak ripple voltage of 2 volts. If the supply frequency

is 50 Hz, calculate:

(a) the rms input voltage to the rectifier

(b) the minimum load resistance

(c) the PRV across each diode.

20


________________________________________________________________________________________

NOTES________________________________________________________________________________________

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21


________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1. (a) The peak to peak ripple voltage is the maximum variation in output

voltage and is denoted by

(b) The rms ripple voltage is the value (as would be indicated by an a.c.

voltmeter if connected across the load) which is given by:

(c) The ripple factor is the ratio between the rms ripple voltage and the

average d.c. voltage:

ripple factor = ( )V

Vr rms

DC

VV

V

r rmsr

r

( )

∧∨

∧

= =3 2 3

Vr∨

∧.

22


2. (a)

(b)

VL = 36.92 volts

V V VL C r=

=

∧

∨

∧–

. –.

12

41 038 21

2

Vr∨

∧= 2.37 volts

V V VC s D

∧ ∧=

=

2

42

–

.. – .

.

43 2 0 7

41 03

12

( )

=

=⎛⎝⎜

⎞⎠⎟∨

∧ ∧

volts

Vf CR

VrS L

C

==× × × ×

⎛⎝⎜

⎞⎠⎟

=

12 50 500 10 100

41 03

8 21

6–.

. volts

Vr rrms

rV

( )∨

∧

=

=

2 3

8 21

2 3

.

23


(c)

(d)

(e)

VRR = 41.73volts

PRV, V V VRR s D=

=

∧–

. – .42 43 0 7

Ripple factor 0.064=

Ripple factor =

=

( )V

V

r rms

L

2 3736 92

..

I L = 369 mA

IV

RLL

L

=

= 36 92100

.

24


3. (a)

The rms value of a sinusoidal voltage is equal to its peak value

divided by the square root of two, therefore:

Vs rms( ) = 15.84 volts

VV

s rmss

( )

∧

=

=

2

22 4

2

.

V V V

V V V

L C r

C L r

=

∴ = +⎛⎝⎜

⎞⎠⎟

= +

∧

∨

∧

∧

∨

∧

–12

12

2012

××⎛⎝⎜

⎞⎠⎟

=

=

∴

∧

2

21

2

volts

since:

V V VC s D–

VV V Vs C D= +

= + ×( )

=

∧2

21 2 0 7

22 4

.

. volts

25


(b)

This will equal the minimum value of load resistance because a

smaller RL value will result in a higher ripple voltage

(c)

VRR = 21.7 volts

PRV, V V VRR s D=

=

∧–

. – .22 4 0 7

RL min( ) = 105 Ω

Vf CR

V

Rf CV

rL

C

Lr pk pk

∨

∧ ∧

( )

=⎛⎝⎜

⎞⎠⎟

⇒ =⎛

⎝⎜

⎞

⎠⎟

12

12 –

VV C

∧

=× × ×( ) ×

⎛

⎝⎜⎞

⎠⎟

=

12 50 1000 10 2

21

105

6–

Ω

26


________________________________________________________________________________________

SUMMARY________________________________________________________________________________________

• A filter reduces the output d.c. voltage variation from a rectifier. This

voltage variation is referred to as ripple voltage.

• The simplest type of filter consists of a capacitor connected in parallel

(shunt) across the output terminals of the rectifier – this capacitor can be

referred to as a reservoir capacitor.

• Higher smoothing capacitance results in:

smaller ripple voltage

higher average load d.c. voltage.

• Ripple voltage is also dependant on load resistance, i.e. as a load

resistance decreases, then ripple voltage will increase.

• Average load voltage, V V VL C r=∧

∨

∧–

12

•• Peak reverse voltage (PRV) V VRR =∧

ss DV– 2

• Peak-to-peak ripple voltage Vfr∨

∧=

12 CCR

VL

C

∧

• Ripple factor =V

Vr rms

DC

( )

where Vr rms

rV

( )∨

∧

=2 3

27


________________________________________________________________________________________

APPENDIX________________________________________________________________________________________

Notation for Representing Alternating Voltages and Currents

For a periodic function the mean value is given by:

For a periodic function the r.m.s. value is given by:

VT

V t tT

= ( )( )⌠

⌡⎮⎮

1

0

2d

VT

V t tT

= ( )⌠

⌡⎮⎮

1

0d

Alternating Quantity

Instantaneous value

Average (mean) value

RMS value

Maximum (peak) value

Minimum (valley) value

Peak to valley value(peak to peak)

Examples of notation Comment

v

Va

V

Vm

Vmin

Ve

i

Ia

I

Im

Imin

Ie

V(t)

V

Vrms

V

V

V

I(t)

I

Irms

Î

I

Î ‘e’ for excursionˆ

ˆ

ˆ ˆ

ˆ ˆ

28


Labelling applied to a sinusoid:

Labelling applied to a ripple present in a d.c. voltage:

V

t0

v

V

ˆ

Vmin V̂Vm

ˆ̂

t0

v

v = Vm sin(ωt)

Ve V̂ˆ

V̂Vm

29


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