# WBUT Solved Question Papers EE

## Diode Circuits B Tech Sem Ist

1.5. In a centre tap full wave rectifier, the transformer secondary peak voltage across each half is Vm.

The P1V at each diode is [WBUT 2007(JUNE)]
»V. ») 2Vm c) VJ2 d) 4V„ .
1.6. For full-wave rectifier [WBUT 2007, 2010(DECEMB ER)]
a) one centre-tapped transformer is required
b) two centre-tapped transformers are required
c) more than two centre-tapped transformers are required
d) centre tapped-transformer is not required Answer: (a)
1.7. The PIV for a half-wave rectifier is [WBUT 2007(DECEMBER)]
a) greater than the PIV of bridge rectifier
b) less than the PIV of bridge rectifier
c) equal to the PIV of bridge rectifier .
1.8. In an unbiased negative parallel clipper using an ideal PN junction diode, when the diode is forward biased, the output is [WBUT 2008{JUNE)]
a) Positive half cycle of the input b) negative half cycle of the input
c) Zero d) breakdown voltage of the diode
1.9. If the filtered load current is 10mA, which of the following has a diode current of 10mA? [WBUT 2008(JUNE)]
a) Half-wave rectifier b) Full wave rectifier
c) Bridge rectifier d) Impossible to say
1.10. The ripple factor of a full-wave rectifier is [WBUT 2008(DECEMBER)] a) 0.406 b) 1.21 c) 0.482 d) 1
1.11. A full-wave rectifier has twice the efficiency of a half-wave rectifier because
a) it makes use of a transformer [WBUT 200B(DECEMBER)]
b) its ripple factor is much less .
c) it utilizes both half-cycles of the input
d) its output frequency is double the line frequency Answer: (c)
1.12. The ripple factor for a haff-wave rectifier is rWBUT 2009/JUNEYI
a) 0.482 b) 0.41_ c)1,21 cQ1.11 1
” _ * , L *
1.13. The ripple factor of a power supply is a measure of [WBUT 2009(JUNE)1
a) its filter efficiency b) its voltage regulation
c) diode rating d) purity of power output
1.14. The maximum efficiency of a full-wave rectifier can be
. . [WBUT 2009,2010<JUNE)]
a) 37.2% b) 40.6% c)S3.9% d)81.2%
””aSta D° *°U,Tn?.?!itT*f *Cte [WBUT 2010<JUNE))
a) Rectifier b) Clamper c) Chopper d) Demodulator
1 . v * *■ ” ■ 1.12. If the line frequency is 60 Hz, the output frequency of a bridge rectifier is
a) 30 Hz b) 60 Hz c) 120 Hz d) 240 Hz {WBUT 2010(JUNE))
AUaWvi * / . -■ ’ ‘ . ■ ‘ ‘ r'”

As one diode is employed, the circuit is half-wave rectifier. The peak load current is
/ – K.’ – »>>/2 flldn,
” 0.3 + 2 .
* # ^
(i) The dc load currcnt is = 1.957A, ‘
(ii) The rms ac load current is
r =(r/2 -i2 ‘ – .. • . . .
rms ^ rms dc / ‘ ‘ . . ^
I • ‘ ‘
Where inns – -5 * 3.074A . • –
■ ■ 2 . ■■ ’
… ” i ■ ■ * ■ ■
Hence, = [(3.074)2 -(1.957)2J* = 2.37A ‘ .
– ■ ^F1. ■
2.2. Write short notes on Load lines-static and dynamic. [WBUT 2002, 2003(JUNE)1 Answer:
Since the voltage applied to the diode resistor combination is varying, the process of drawing a load line for the instantaneous values of V, must be repeated at convenient intervals. However, since the load RL is constant all these load lines will be parallel. The intersection of these load lines with the static curve of the diode will give the current in the circuit corresponding to each instantaneous value of input voltage. ■
Instead of using these values of current to obtain the output current waveform, it is more convenient to construct a dynamic curve of I versus Vg as shown in Fig. (b). ‘
Ca)

2Vp-p

(b)
1 mA
Static curve of I vs Vs
Dynamic curve of
I vs Vs
► Vn volts

vra-iv
Fig: (a) Diode circuit to be analyzed for waveforms of i and v0 (b) Showing how to determine the dynamic curve of / vs. Vsfor a given ‘ RL from the static curve of the diode.
Consider when Vs is at its maximum, V,,, = IV. The intersection of the load lin<» corresponding to this condition with the static curve is shown to occur at A, giving a peak current in the circuit of Ira = 9 mA. Drawing a line horizontally from A will intersect a vertical line drawn from Vm at A’. Now take any other value of Vs, say V/ and draw a load line parallel to^ the first. The intersection at B gives the Current in the circuit, i\ corresponding to V/. Drawing a line horizontally from B and vertically from V/ gives
another point B’. Continuing this process for arbitrary values of Vs will yield sufficient points to construct the dynamic curve shown in Fig.(b). •
The horizontal axis in the graph must now be referred to as vf when using the dynamic curve. It must be emphasized that this dynamic curve applies only to the circuit containing the value of load resistance for which it was drawn.
2.3. Write short note on Clamping circuits.
[WBUT 2003, 2006, 2008(JUNE), 2007, 2009(DECEMBER)}
A clamper adds a dc voltage to the signal

Hie clamper shift the ac reference level (normally zero).
This is the actual circuit for the clamper.. During —ve half cycle of the input signal when the diode is in ON state, the capacitor is charged upto the peak value of the signal Vp. In the +ve half cycle, when the diode is in OFF condition, the capacitor tries to discharge through Rl. NOW if the time constant R1^ is made much greater than the time period of the input signal, it remains almost fully charged during this +ve half. In the next —ve half cycle of the input cycle it is again charged to Vp. Thus the capacitor always retains this Vp volts and acts like a dc source of Vp volts over the ac input signal. Thus the output voltage that appears across Rl is from 0 to 2VP instead of -Vp to VP, and we have a positively clamped output signal.
If we consider the cut in voltage of the diode, then output voltage will like shown below.

– V0„ ■
s2V

– 0.7 V
2.4. a) Write down the PiV rating of hatf-wave rectifier.
PIV is the maximum voltage to which the diode is subjected to. This occurs during the negative half cycle when file diode is not conducting.
In the negative half cycle, the point B is +’Vm volts with respect to the point A. Therefore, in half wave circuit, PIV of the diode = Vm volts. .
b) How can you remove the ripples? , . Answer:
By using a capacitor in parallel with load resistance R*.
2.5. Design a fuH wave rectifier.
The figure shows a full-wave rectifier circuit. The transformer secondary has a centre-tap and each half give a peak voltage of Vm. In each half there is one diode i.e., D, and D2. The load resistance RL is common to both halves.
This can be seen to comprise of two half-wave circuits. On the positive half cycle, when the point A is positive with respect to B, the diode, D, conducts and current i, flows through Rl. During this half cycle, the point C is negative with respect to point B and hence the diode D2 does not conduct. Therefore i2 = 0.
On the negative half cycle, the point C is positive with respect to point B. Hence the
diode D2 conducts and current i2 flows through RL. During this half cycle, the point A is
negative with respect to point B and hence the diode D, does not conduct. Therefore ii = 0.
(b

. . ‘ ■
Fig.: Fall-wave rectifier and the current waveforms
Fig. (b) and (c) shows the waveforms of currents it and i2. Since both i, and i2 flow through the load RL, the total current i through RL is i = i, + i2, which is obtained by adding the two waveform and is shown (d).
» *■ *■ “ ■
. ‘ ” r
2.6. What is understood by Peak Inverse Voltage rating of a junction diode?
It is the maximum reverse voltage that can be applied to the pn junction without damage to the junction. If the reverse voltage across the junction exceeds its PIV, the junction may be destroyed due to excessive heat. The peak inverse voltage is of particular importance in rectifier service. A pn junction i.e. a crystal diode is used as a rectifier to
change alternating current into direct current. In such applications, care should be taken that reverse voltage across the diode during negative half-cycle of a. c. does not exceed the PI V of diode. ’
— ■ ” 1 ..
– _ ■
2.7. Write short notes on Ripple factor.
[WBUT 2006<JUNE)12008,2010(DECEMBER>] Answer: h . .
The purpose of a rectifier is to convert a.c. into d.c. But the simple circuit as used iii half wave rectifier does not fully achieve this purpose.
Let Lfc = Average value of the waveform .
i = The value of alternating component of the current waveform
* . – – ‘ ‘ – ’ ■
* ‘ . . ■’ iJF”
I— r.m.s. value of the alternating component i.e., of i .
a 1 ■ . J
Ripple factor, r, is defined as ■ ■
_ r.m.s, value of the alternating component of the wave(output) _ /

\ida+l\~ [da
IK i – 2w i – 7M 0 . .
There are three terms in the equation above and we shall examine them one by one.
(a) The first term i.e.,
i 2K
f *2 ” 2 ■—■ 11 da has the same expression and therefore is equal to 1 ^ .
2ft o •
(b) In the second term, the expression
— f d,OC is nothing but Idc.
\K i ; . .
‘ 1 2 r
So the complete second term —2.1 ^. J i da = —2 — —2./A.2
2 n
o
1
(c) The third term = 2x~0) = 7
Now we assemble all the three terms under the square root
••• >L = JirJ- 21J+Ij = JTJ-ij
The The above expression is independent of current wave shape and is not restricted to half wave aloneRipple factor can be also expressed in terms of voltage (similar to above derived in terms of currents). Thus

Ripple factor, r =

where Vm = the rms value of output voltage .
ar|d Vjc = average value of the output voltage. ■ ‘
. ■ .
2.8. What is your concept about an ideal diode? How does it differ from an actual
one? ; ■■ – . [WBUT 2006,2009(JUNE)1
P-N junction diode conducts current in only one direction. We can think of an ideal diode as a short circuit when forward biased and as an open circuit when reverse biased.
r

Eo„ –

‘HG

W

An equivalent circuit is a combination of elements properly chosen to best represent the
actual terminal characteristics of a device system or such in a particular operating region.
One technique for obtaining an equivalent circuit for a diode is to approximate the
characteristics of the device by straight line segments. This is called piecewise linear equivalent model [Fig. (a) to Fig. (c)].
2.9. A C-T transformer has 230 V primary winding rated at 12*0-12 volts. This transformer is used in the F.W. rectifier circuit with a load resistance of 100 Q. What are the d.c. output voltage, d.c. load current and the rms voltage developed across the diode? Assume the diodes and the transformer to be ideal.
The rms voltage o^ each half of the secondary is 12V.
/.Vm=V2xl2-16.97 volts
Also given that RL = 100Q and Rf — 0 ,
•• ‘ ■’ . ■
2xl6.97K
(a) The d.c. load current /*. = – = — ■ ‘■ ~— = 0.108 Amps.
* RL ^(100)Q V
(b) The d.c. output voltage is – .
K =IJC RL 08AxlOO£i = 10.8 volts
(c) PIV of each diode = 2V = 2×16.97 — 33.94 volts
f/» ■
■ 33.94
Therefore, RMS voltage across each diode = —— 24 volts.
■ 6 .
■ T ■ – – .n,
2.10. Write short notes on clipping circuit.
[WBUT 2007, 2010(JUNE), 2008, 2009, 2010(DECEMBER}]
A clipper is a circuit that removes either positive or R.
negative parts of a waveform. This kind of ^v-
processing is useful for ‘signal shaping, circuit Vi protection etc.
The above circuit removes all the positive parts of the input signal. This is why the output signal has negative half cycle.
Consider the diode cut in voltage is 0.7, so clipping level is not zero (shown below).
Avm * ‘ ‘
•*» t
If we apply an external voltage in series with the diode to change the reference level of a circuit, it becomes biased clipper

– 0,7- V
I
We can combine two biased clipper. The output becomes
2.11. Calculate the maximum conversion efficiency of a half-wave rectifier.
[WBUT 2007, 2Q09(JUNE), 2009<DECEMB4
-Answer: ■ ‘ . _ •’ ■ : ■
Rectification Efficiency . . ~ .
Rectification efficiency is defined as – :;
= d.c. power delivered to the load xlQ0% _■ ^.xl00%
a.c. input power ‘ .
/
Pac is the rms power delivered into the circuit, composing of RL, R*, and Rf.
= inJi(Rf + K
/ 2 R
•** Rectification Efficiency (t) ) — —t———-—* r X100% •
: , +«,+*,.)
i 1 ■ -r
The above expression is valid for all types of rectifier circuits.
” a‘ h
* ‘ J t
‘ 1 ‘
For half wave Rectifier Circuit ^
‘ * – / T
fj V .
ItiA .
(R/ + K, + Kl) = fif-‘j (R,+R, + Rl)
‘• Rectification Efficiency ( T|)
r I ^
m
7t
IJJT2’ Rf+Rs + RL Rf+R
/ \ 4 ” ‘ ‘
If we assume (Rf + R S ) « RL , then 7] =—jxl00% = 0.406×100% —40.6%
This means even under ideal conditions (i.e., Rf and R„ equal to zero), only 40.6% of the a.c. input is converted into d.c. power. What happens to the remaining 59.4% of the a.c. power? Well, it exists in the rectified output, but as the alternating component, i.e., the ripple. In general it can be understood that higher the rectification efficiency, the lower would be the ripple content.
2.12. What are HM advantages and disadvantages of bridge rectifier over full-wave rectifier using two diodes?. [WBUT 2009<DECEMBER)]
i) it can be driven by a single source voltage, such as the untapped secondary of a
transformer or directly .from the power line.
. The peak reverse voltage that can be tolerated is 2x the reverse breakdown of die diodes. . ‘
ii) Only two diodes needed. The forward voltage drop is one diode forward voltage. The center or neutral of the voltage source is common to both AC and DC parts of the circuit. –

i) The AC portion and DC portion must be isolated (you cannot make a common point
between them). Transformers automatically do this, so direct connection to power lines is rare. Also, requires 4 diodes. The output voltage is dropped by 2 diode forward voltages, . ■
ii) Requires a “push-pull” voltage source (such as a center-tapped secondary on a transformer). Max reverse voltage is only one reverse breakdown voltage of a diode
3.1. a) Explain physically how a p-n junction diode can be used as a rectifier. Answer:
The manner in which the output current waveform is obtained using the diode’s dynamic curve (drawn for RL = 50 ohms) is indicated graphically in Fig. die input voltage V, is drawn with its time axis vertically so that the values of voltage for different time intervals ti, t2, etc., may be projected onto the voltage axis of the diode’s dynamic curve. Intersections with the curve are drawn horizontally to give the values of current corresponding to the times ti, t2, etc. Due to the cut-in voltage of ‘ most
semiconductor diodes the output current does not appear
until some time after the input voltage has been applied. Consequently the period of
conduction of the diode is less than half a cycle. – ■
“ _ r r ■ ” . !*
P “ ” ” m ” B ” ■ . T ” * # + r – 1 ^ *
b) A silicon diode having. internal resistance Rp = 30 Q is used for half wave rectification. The input ac voltage is Vi = 6 sin wt volt and load resistance 5000. Find (i) dc output voltage (ii) ac input power and (iii) efficiency of the rectifier.
. [WBUT 2003(JUNE)]
D.C. Output Voltage (V*)
The d.c. (i.e., average) c^utput voltage appearing across RL is
• “… ‘
n
Vm
v t?
y — v 7T_
*■ MR.+RA R.
+ 1
R,
Putting the values we get V- = — — = 1.8 V
JT(30+500)
Rectification Efficiency – .
Rectification efficiency is defined as
d.c. power delivered to the load
T| = : ; Xl00%
a.c. input power
1. ‘2
For half wave Rectifier Circuit Pdc — I^R, — I ■——J
/ (j y ‘
n1=^2(ft/+ff.+Ri)= -f (*,+*.+«*) v
v z /
Rectification Efficiency (rf) ■ . . .
o’ R : –
(L^frVT,xloo%
-f\ (Rf+R3+RL) ■ m f * L . —L+\ ,
2
Putting the values we get Tj ~ 38.23%
3.2. Draw the circuit of a bridge rectifier and compare its performance with-that of a
hill wave rectifier using centre tapped transformer.
[WBUT2002,2003(JUNE), 2009,2010(DECEMBER)]
It is interesting to note that in both the half cycles, the direction of current flow through RL is the same. Thus we obtain full-wave rectification, the waveform being same. Because of this, it has the same expression for: – .
(i) Average current (I*) .
(ii) RMS current (Inns) .
D.C. output voltage (Vdc) ‘ . . ‘
Ripple factor (r) …
Rectification efficiency (11) .
as those of full-wave rectifier circuit, obtained in the previous sections.

and the d.c. output voltage at full load is
2V” -IAt, +2S,)
ft
The terra 2Rf arises because two diodes are conducting in series at any given time.
The other important features of the bridge rectifier circuit are: ^
(a) TUF: The current flow through both the primary and the secondary windings are sinusoidal (i.e.* full cycle). Due to this, the TUF of both the primary and the secondary are 0.812 and the overall TUF is 0<812. This is higher than the TUF of full-wave rectifier (0.693) and hence for the same d.c. output power* a smaller transformer can be used in the bridge circuit-
(b) A centre tap is not required in the transformer secondary. #
(c) The PIV of each diode = Vm, Let us consider the positive half cycle when diodes Di and D3 are conducting and therefore the *voltage drops across each of them is negligible- Therefore the voltage appearing across D2 is equal to Vm, Also the voltage
appearing across D4 is equal to Vm. In a similar way considering the negative half cycle, it can be understood that PIV of Di and D3 are also equal to V,,,. Thus, for a given voltage rating of the diodes, these can be used for high-voltage applications. Bridge circuit requires four diodes. At any given time two of them conduct in series, causing voltage drop due to two diodes. This additional voltage drop cannot be
Type of Rectifier • * . J
Parameter ■ Half-wave ■ Full-wave Bridge
Number of diodes 1 • 2 . ‘ 4 ”
PIV of diodes vm 2Vm v,n
Secondary voltage (rms) V V-O-V V
Secondary voltage (Peak
Vm=j2V). ‘ . vm ‘ . • v„,-o-vm V* •* ■ .
D.C. output voltage, V^, at no-load
p Vm •
K
~ 0.318 Vm 2Vm * it
= 0.636Vm 2Vm
: It
= 0.636Vm
Ripple factor, r 1.21 0.482 0.482 .
Ripple frequency f 2f , 2f
Rectification Efficiency rj 0.406 0.812 0.812 .

3.3, a) For a rectifier circuit using diodes, define (i) rectification efficiency (ii) ripple factor (iii) PiV. [WBUT 2004, 2005(JUNE), 2007{DECEMBER)J
Rectification Efficiency … . .
Rectification efficiency is defined as
d.c. power delivered to the load „ /V
// = ——— xl00% =-^-xl00%
a.c. input power
p*-V …
Pac is the rms power delivered into the circuit, comprising of RL, R„ and Rf.
: 2 *
■ • Rectification Efficiency ( ti) = — — r-xlQ0%
/,~2 (*/+*,+*,.) ,’v :
The above expression is valid for all types of rectifier circuits.
■, ; ■ ■ .
Ripple Factor ^
The purpose of a rectifier is to convert a.c. into d.c. But the simple circuit as used in half wave rectifier does not fully achieve this purpose. ‘ ,
Let 1^ = Average value of the waveform
* = The value of alternating component of the current waveform
= r.m.s. value of the alternating component i.e., of i‘
r , ‘
* “ . ‘ – Ripple factor, r, is defined as
•t
_ r.m.s. value of the alternating component of the wave(output) _ /
rms
average value of the wave(output)
We shall derive a generalized expression for ripple factor (which is valid not only for half wave, but also for other types rectifier circuit i.e., full-wave, bridge etc.)
The total current — D.C. component + a.c. component • –
■/ « r
or,i =1-/*
/
Now we calculate the rms value, I , from the above expression of I
r”=fe JO’2″2Ui*V)da
There are three terms in the above equation and we shall examine them one by one.
(a) The first term i.e.,
,2 M ■ ‘ ■
” I f * 1 , ■ ■ ■. . “2 ”
yi da has the same expression of Iimt .
JT
(b) In tbe second term, the expression 1 2*
— It da is nothing but Idc
■ i 1 | 2x ‘
So the complete second term —2.1^.-— J i da = — 2./^./^ =—2./
2 x
0
(c) The third term 1
Now we assemble all the three terms under the square root * If — ft 2 — 9 / 2 + / 2 — [} 2 _ r 2
•’ 1 rm.< “V’mj ” y 1 ms 1 dc
■ 7’ If 2^J 2 I
The ripple factor r — – = ———= . I
■ – de ^dc Y
The above expression is independent of current wave shape. Ripple factor can be also expressed in terms of voltage (similar to above derived in terms of currents). Thus,

. V
Ripple factor, r =
V
) dc
where = the rms value of output voltage and Vdc = average value of the output voltage.
Peak Inverse Voltage (PIV) ■ •
PIV is the maximum voltage to which the diode is subjected to. This occurs during the negative half cycle when the diode is not conducting.
In the negative half cycle, the point B is +Vm volts with respect to the point A. Therefore, in half wave circuit, PIV of the diode = Vm volts.

h
. L ■ n
b) Each of the two diodes in a full wave rectifier circuit has a forward resistance of 50 ft. The dc voltage drop across a load resistance of 1.2 KQ is 30 V. Find the primary to total secondary turns ratio of the center tapped X former, primary being fed from 220 Vms. [WBUT 2004, 2005(JUNE), 2007( DEC EMBER)]
The rms secondary voltage is Vs = VpIn, where Vp is the rms primary voltage of the transformer and n is the primary-to-secondary turns ratio. Here Vs — 220 x n = 220n V. The secondary voltage from the center tap is 220n/2 = 1 lOn (rms). The corresponding
peak voltage is Vm = 110»>/2.
(i) The dc load current is = 21m 171, where Im is the peak current through the load resistance. We have ■ ‘
7 no^=0124flA,
m Rf+RL 50+1200 .
Hence Idc = —— = 0.079n A.
2L
71 ’ . • ■ ■
(ii) The output voltage is V# = IdcRL = (0.079n)xl200 = 30V
Therefore n=4 – ‘
, . . ■ ‘
3.4. a) Explain the operation of a full-wave Bridge Rectifier with the help of circuit diagram. [WBUT 2007, 2008,2009<DECEMBER)]
*
to the diagrams below, when the input connected to the left comer of the diamond is positive, and the input connected to the right comer is negative, current flows from the upper supply terminal to the right along the red (positive) path to the output, and returns to the lower supply terminal vja the blue (negative) path.

When the input connected to the left comer is negative, and the input connected to the right comer is positive, current flows from the upper supply terminal to die right
along die red (positive) path to the output, and returns to die lower supply terminal via the blue (negative) path.
Full*wave rectification

In each case, the upper right output remains positive and lower1 right output negative Since this is due whether the input is AC or DC, this circuit not only produces a DC output from an AC input, it can also provide what is sometimes called “reverse polarity protection”. That is, it permits normal functioning of DC-powered equipment when batteries have been installed backwards, or when the leads (wires) from a DC
power source have been reversed, and protects the equipment from potential damage caused by reverse polarity. ’
r – V . ‘
b) Obtain a mathematical expression for the efficiency of a full-wave rectifier and show that its riple factor is 0.482. [WBUT 2005(JUNE), 2007,2008(0EC EMBER)]
Rectification efficiency: ,
J r r
The efficiency of ratification is defined to be the ratio of the dc output power (/V) to the input power { P. ) of the rectifier. It is denoted by 7]
7] =—^xlD0%
Pt
77 is also referred to as the theoretical efficiency.
We have= I*RL
2*
V —> instantaneous i/p voltage at time t.
Rf —► forward resistance of the diode.
KirchhofFs voltage law gives
v-it +rl)
If /?, is assumed to be a constant, equation (3) reduces to Rf + R, p.* ,
p‘=r««*»
=(R,+RL)I,
Therefore from equation (1)
1+R,/RL .
For the full wave rectifier the load current II flows in the same direction for the both halves of the i/p voltage cycle. So the frequency of the load current is twice the frequency of the i/p supply. *
Thus, -^-= Lh&*mU
I* 2IJ* 2V2
The ripple factor is /= j^(l.l I)2 — ll = 0.482
81,2
The conversion efficiency is; 7J ~
1 +R,]R,
3.5. a) Find out the expression for efficiency, form factor and rjpple factor for a half-wave rectifier. [WBUT 2009(DECEMBER>]
Rectification efficiency is defined as
d.c. power delivered to the load P.
7 = xl00%–^xl00%
a.c. input power P
Pac is the rms power delivered into the circuit, comprising of RL, R*, and Rf.
I 2R
-*• Rectification Efficiency (r|) = ——*—~ rXl00%
. I^2(Rf+R,+RL)
The above expression is valid for all types of rectifier circuits. For half wave Rectifier Circu

If we assume (jfy + /?,)« , then 7] = -^-xl00% = 0.406×100% = 40.6%
This means even under ideal conditions (i.e., Rf and Rs equal to zero), only 40.6% of die
a c. input is converted into d.c. power. What happens to the remaining 59.4% of the a c.
power? Well, it exists in the rectified output, but as the alternating component, i.e., the
ripple. In general it can be understood that higher the rectification efficiency, the lower would be the ripple content. –
/
Current ripple factor:
The ripple factor, r is given by,
For form factor:
_ , A rms value V 12 nr Form factor — —2—=— — \ 57
Average value Vm In 2 ‘
a,
b) A full-wave rectifier uses a double diode, the forward resistance of each element being 100 ohm. The rectifier supplies current to a load resistance of 1000 ohm. The primary to secondary turns ratio of the centre tapped transformer is 10 : 1. The transformer primary is fed from a supply of 240 V (rms). [WBUT 2009(DECEMBER)]
ii) direct current in each diode
iii) the ripple voltage and ‘
iv) the efficiency of rectification
The rms secondary voltage is Vs = Vp jn where Vp is rms primary voltage of the
transformer and n is primary to secondary turns ratio. Here V — 240x-— = 2400V The
‘ ‘ ‘ 1 ‘ . ,
secondary voltage from the center tap is 2400/2 = 1200V (rms).
The corresponding peak voltage is Vm -1200>/2
i) The dc load current is where Im is the peak current through the load
resistance.

i±wy± = ^ 200+1000

Hence, /* = ^ = 2×1,4142×7 = tAxLAVto
n 22 .22 ‘ ’
ii) The direct current supplied by each diode is 7^/2 = 0.9/2 = 0.45 A
iii) The ripple voltage across the load is RL = Y1 Here I^ ~ IJ V2 — -Jlfypl — 1 Amp
. J*
So, c, RL = J(l2 -0.92)x1000 =0.4358×1000* 435.8 V
3.6. a) What is ripple factor? Evaluate the ripple factor and efficiency of a full-wave bridge rectifier. • [WBUT 2011 (DECEMBER)]
Ripple Factor – . . •
This factor measures how much ac component exists in the rectified dc output component. To get the better performance the value of ripple factor should as small as
possible.
rms value of the alternating component of output
y* — . — 1
. . dc voltage of output
/’ ” •
The ripple factor r = (by definition)
dc

The above expression is independent of current wave shape.
Ripple factor can be also expressed in terms of voltage (similar to above derived in terms
of currents). Thus,
Ripple factor,
r =
where = the rms value of output voltage
and Vdc = average value of the output voltage.
Ripple Factor of Bridge Rectifier
We know that the generalized expression for ripple factor is

r =
dc J
For bridge rectifier, we have seen that
I
m
Inns – /V

r —
-1 =0.482

v ft J
We find that ripple content is much lower compared to the half-wave rectifier. Therefore the output of a bridge rectifier is a better d.c. than that of half-wave.
Rectification Efficiency of Bridge Rectifier
» = -^x 100% 1
P..

J
Rectification Efficiency On)
21
m
(*, + R, + Rt)
2 *

If (R, +
rt = -^-xl00% m 0.812×100% = 81.2%
K
b) A silicon diode with internal resistance RF= 25 £2 is used for hatNwave
rectification. The input a.c. voltage is V -20sin OM and the load resistance is
5000. .
Find,
i) d.c. output voltage –
ii) a.c. input power and
iii) efficiency of the rectifier.
Solution:
D.C. Output Voltage (T*) …
The d.c. (i.e., average) output voltage appearing across RL is
It ‘
m
V* =
KRi
*(R,+Rl) «t+1
RL
■ V2
(Rf+RL) = *

Putting the values we get 7] = 38.567%
average value of the wave(output) At
We shall derive a generalized expression for ripple factor (which is valid not only for half wave, but also for other types rectifier circuit i.e., full-wave, bridge etc.)
The total current = D.C. component + a.c. component