Basic Electronics (GTU)
5-1
Diode Circuits
Chapter 5 : Diode Circuits
Section 5.5 :
Ex. 5.5.2 :
For the series clipper shown in Fig. P. 5.5.2(a), draw the output voltage waveforms and the
transfer characteristics.
.Page No. 5-17.
Fig. P. 5.5.2(a)
Soln. :
This is the series negative clipper circuit. If we assume that the diode is an ideal diode then the
voltage drop across it will be zero in the ON state. Hence it will conduct for the entire positive half cycle
of the input, as shown in Fig. P. 5.5.2(b). The negative half cycle will be completely removed. The
waveforms are as shown in Fig. P. 5.5.2(b).
Transfer characteristics (Ideal Diode) :
1.
2.
3.
The transfer characteristics is the graph of Vin versus Vout of the circuit as shown in
Fig. P. 5.5.2(c).
Fig. P. 5.5.2(c) shows that the output voltage is equal to the output voltage for positive input
voltage.
∴ Vo = Vin
... for Vin ≥ 0
And Vo = 0 for negative values of Vin
∴ Vo = 0
(b)
... for Vin < 0
(c) Transfer characteristics
Fig. P. 5.5.2
Basic Electronics (GTU)
5-2
Diode Circuits
Transfer characteristics (Practical Diode) :
•
If the diode is a practical silicon diode with Vγ = 0.7 V then, the transfer characteristics gets
modified as shown in Fig. P. 5.5.2(d).
•
For Vin negative or less than 0.7 V, the output voltage is zero.
∴
•
... for Vin ≤ 0.7 V
Vo = 0
But as Vin exceeds 0.7 volts, the diode is turned ON and the output voltage is given by,
Vo = Vin – 0.7 ... for Vin > 0.7 V
•
For example if Vin = 2V then Vo = 2 – 0.7 = 1.3 V
Vin = 10 V
•
Table P. 5.5.2 lists the input and output voltages.
Sr.
No.
1.
2.
3.
4.
5.
6.
Input voltage
(volt)
– 10 V to + 0.7 V
2
4
6
8
10
Table P. 5.5.2
Ex. 5.5.3 :
then Vo = 10 – 0.7 = 9.3 V
Output
voltage (volt)
0
1.3
3.3
5.3
7.3
9.3
Fig. P. 5.5.2(d) : Transfer characteristics
for a practical silicon diode
For the circuit shown in Fig. P. 5.5.3(a), explain working of the circuit and draw output
waveform for given input signal. Also draw transfer characteristics.
.Page No. 5-17.
Fig. P. 5.5.3(a)
Soln. :
Working of the circuit :
1.
In the positive half cycle of the input, D2 will be reverse biased. D1 will also be reverse biased as
long as Vin is less than or equal to + 2 V (assuming D1 and D2 to be ideal) and the output voltage
is zero.
Basic Electronics (GTU)
5-3
Diode Circuits
Fig. P. 5.5.3(b) : Waveforms
2.
3.
4.
As soon as Vin > + 2 V , D1 is forward biased and acts as a closed switch. The output voltage is
then given by,
Vo = (Vin – 2)
...(1)
In the negative half cycle of the input, D1 is reverse biased. D2 will also be reverse biased as long
as Vin is between 0 and – 2.5 V and output voltage is zero.
...(2)
∴ Vo = 0 ... – 2.5 V < Vin < 0
As soon as Vin becomes more negative than – 2.5 V, D2 is forward biased and acts as a closed
switch.
...(3)
∴ Vo = Vin + 2.5 V ...for Vin < – 2.5 V
The waveforms are as shown in Fig. P. 5.5.3(b).
Section 5.6 :
Ex. 5.6.1 :
Draw the load voltage waveform for the
clipper circuit shown in Fig. P. 5.6.1(a).
Assume that a sinusoidal voltage Vi = 10
sin 314 t has been applied at the input
and the DC source voltage V = 5 Volts.
The diode D is an ideal diode.
.Page No. 5-21.
Fig. P. 5.6.1(a)
Basic Electronics (GTU)
5-4
Diode Circuits
Soln. :
Refer to the waveforms
shown in Fig. P. 5.6.1(b).
•
•
Fig. P. 5.6.1(b)
Note that the type of circuit is positive parallel clipper. However the load resistance RL has been
included.
The diode will be on if VAB ≥ 5V. And VAB ≥ 5V if Vin ≥ 5.5 Volts.
•
Looking at the waveforms shown in Fig. P. 5.6.1(b) we understand that the circuit operation can
be divided into three modes.
1.
Mode I ( 0 – t1 ) :
In this mode of operation, the input voltage Vi is positive and less than 5.5 volts. Therefore the
diode is reverse biased and acts as an open switch as shown in the equivalent circuit of Fig. P. 5.6.1(c).
The load voltage is given by,
1000
∴ Vo = (1000 + 100) × Vi = 0.9 Vi
(c) Equivalent circuit for mode I
(d) Equivalent circuit for mode II
Fig. P. 5.6.1
2.
Mode II ( t1 – t2 ) :
In this mode, the positive input voltage Vi is higher than 5.5 V. This will turn ON the diode
because then
Vo = 0.9 Vi
hence Vo > 5 V and the diode is forward biased. The dc source get connected across the load as shown
in Fig. P. 5.6.1(d). Thus the output voltage is equal to + 5 V.
3.
Mode III (t2 – 2 π) :
From the instant “t2” onwards, the input voltage is less than V and between “π” to “2π” it is
negative. The diode remains OFF during this interval. Hence the load voltage is given by,
1000 Vi
Vo = 1000 + 100
Basic Electronics (GTU)
5-5
Diode Circuits
The DC source is isolated from the output as shown in the equivalent circuit of Fig. P. 5.6.1(e).
Fig. P. 5.6.1(e) : Equivalent circuit for mode III
The transfer characteristics of the given circuit is as shown in Fig. P. 5.6.1(f). The slope of the
transfer characteristics is equal to 0.9 for Vin < 5.5 V and it is equal to 0 for Vin ≥ 5.5 V.
Fig. P. 5.6.1(f) : Transfer characteristics
Ex. 5.6.2 :
Analyze the circuit shown in Fig. P. 5.6.2(a) and draw
the output voltage waveform for the same, if a
triangular wave is applied at the input with a peak
voltage of 10 Volts. Assume that the diode is a
nonideal silicon diode.
.Page No. 5-22.
Fig. P. 5.6.2(a)
Soln. :
1.
The output voltage is equal to input voltage when the diode is not conducting, i.e. in the positive
half cycle of the input voltage.
∴ Vo = + Vi
...when D is OFF
...(1)
Basic Electronics (GTU)
2.
5-6
Diode Circuits
And the output voltage is negative and equal to one diode drop when the diode is on i.e. in the
negative half cycle of the input voltage.
...when D is ON
...(2)
∴ Vo = – 0.7 volts
The waveforms are as shown in Fig. P. 5.6.2(b).
Fig. P. 5.6.2(b) : Waveforms for the clipper shown in Fig. P. 5.6.2(a)
Section 5.7 :
Ex. 5.7.1 :
The diodes in the circuit shown in Fig. P. 5.7.1(a) are ideal. Sketch the transfer
characteristics for – 20 V ≤ V1 ≤ 20 V. If the diode D2 is reversed, sketch the resulting
characteristics for – 20 V ≤ V1 ≤ 20 V.
.Page No. 5-26.
Fig. P. 5.7.1(a)
Soln. :
Case I : Considering the circuit shown in Fig. P. 5.7.1(a).
1.
Operation for – 20 ≤ V1 ≤ 0 Volts :
When the input voltage V1 is negative, upto – 20 volts, D1 is reverse biased and D2 also is reverse
biased. Both the diodes act as open switches hence the output voltage is equal to zero.
∴ V2 = 0 Volts for – 20 V ≤ V1 < 0
...(1)
Operation for 0 ≤ V1 ≤ 20 V :
Let us calculate the magnitude of Vab required to turn ON diode D2. When diode D2 is turned ON
the output voltage V2 is equal to battery voltage.
2.
Basic Electronics (GTU)
5-7
∴
∴
V2 = 10 Volts
when D2 is ON
R2
V2 = 10V = Vab = R + R × V1
1
2
∴ V1 =
∴
Diode Circuits
...(2)
( R1 + R 2 )
(10 + 5)
× 10 = 10 × 10
R2
V1 = 15 Volts
...(3)
Thus D2 will conduct for V1 ≥ 15 Volts. And when 0 ≤ V1 < 15 Volts the output voltage is
given by,
R2
10
V2 = Vab = ( R + R ) × V1 = 15 × V1 = 0.66 V1
...(4)
1
2
Table P. 5.7.1(a) summarizes the results and Fig. P. 5.7.1(b) shows the transfer characteristics
for case I.
V1 < 15 V
V1 V1 < 0
V1 ≥ 15 V
V2
0
V2 = 0.66 V1
V2 = 10 V
Table P. 5.7.1(a)
Fig. P. 5.7.1(b) : Transfer characteristics for case I
Case II : If D2 in Fig. P. 5.7.1(a) is reversed in direction :
With D2 reversed the circuit is as shown in Fig. P. 5.7.1(c).
Fig. P. 5.7.1(c) : D2 reversed
Fig. P. 5.7.1(d) : Transfer characteristics for case II
Here diode D2 will remain on as long as V1 is less than 15 Volts and Vab is less than 10 Volts.
∴ V2 = 10 Volts for – 20 ≤ V1 < 15
As soon as V1 = 15 V, Vab = 10 V and D2 is reverse biased. The output is given by,
10
V2 = Vab = 15 × V1 = 0.666 V1, for V1 ≥ 15 V
The transfer characteristics for case II is as shown in Fig. P. 5.7.1(d).
V1
– 20 ≤ V1 < 15
V1 ≥ 15 V
V2
V2 = 10 V (D2 ON)
V2 = 0.666 V1
Table P. 5.7.1(b)
...(5)
...(6)
Basic Electronics (GTU)
Ex. 5.7.2 :
5-8
Diode Circuits
The diodes are ideal, plot Vo against Vi indicating all intercepts, slopes and voltage levels for
Fig. P. 5.7.2(a) Sketch Vo if Vi = 40 sin ωt.
.Page No. 5-27.
Fig. P. 5.7.2(a)
Soln. :
This example is similar to Ex. 5.7.1.
∴ Vo = 0 for Vi < 0
When D2 is turned on, Vo = 10 V and D2 is turned on when Vab = 10 V.
10 k
∴ Vo = Vab = 10 =
× Vi
(10 + 10) k
∴ Vi = 20 V
Thus D2 turns on when Vi = 20 Volts
∴
...(1)
...(2)
Vo = 10 V for Vi ≥ 20 Volts
...(3)
For 0 ≤ Vi < 20V, D2 will be OFF and D1 ON
Vo = Vab =
10 k
× Vi = 0.5 Vi
(10 + 10) k
(b) Transfer characteristics
...(4)
(c) Input and output voltages
Fig. P. 5.7.2
Table P. 5.7.2 summarizes the circuit operation, Fig. P. 5.7.2(c) shows the input and output
voltages and Fig. P. 5.7.2(b) gives the transfer characteristics.
Vi
Vi < 0
0 ≤ Vi ≤ 20
Vi > 20
Vo
Vo = 0
Vo = 0.5 Vi
Vo = 10 V
Table P. 5.7.2
Basic Electronics (GTU)
Ex. 5.7.3 :
5-9
Diode Circuits
In the diode clipping circuits shown in Fig. P. 5.7.3, Vi = 20 sin ωt, R = 1 kΩ and VR = 10 V.
This reference voltage is obtained from a tap on a 10 kΩ divider connected to a 100 V
source. Neglect all capacitances. Assume the diode forward resistance 50 Ω with Vo = 0
and reverse resistance = ∞. Sketch the input and output waveforms to scale in both the
cases.
.Page No. 5-27.
(a) Case I
(b) Case II
Fig. P. 5.7.3
Soln. :
Case I :
Step 1 : Let us first obtain the equivalent of the dc source VR. VR has been obtained from a 10 kΩ
divider connected to 100 V source as shown in Fig. P. 5.7.3(c). Thus VR is a 10 V source with
a Thevenin equivalent resistance of ( 1 k || 9 k ) i.e. 900 Ω as shown in Fig. P. 5.7.3(d).
(c)
(d)
Fig. P. 5.7.3
Step 2 : Let us replace the source VR in the original circuit by its equivalent shown in Fig. P. 5.7.3(d)
to redraw the circuit as shown in Fig. P. 5.7.3(e).
Fig. P. 5.7.3(e) : Modified circuit for case I
Basic Electronics (GTU)
5-10
Diode Circuits
Analysis :
1.
For Vi < 0 i.e. when the input is negative diode D will not conduct. The equivalent circuit is as
shown in Fig. P. 5.7.3(f) and the output voltage is equal to input voltage.
∴ Vo = Vi
for Vi < 0
...(1)
(f)
(g)
Fig. P. 5.7.3
2.
For 0 ≤ Vi ≤ 10 V the diode still remains OFF as shown in Fig. P. 5.7.3(g) and output voltage is
equal to input voltage.
∴ Vo = Vi
3.
for 0 ≤ Vi ≤ 10 V
...(2)
For Vi > 10 V, the diode is turned ON and the equivalent circuit is as shown in Fig. P. 5.7.3(h).
I =
Vi – VR
1.9 k
...(3)
And Vo = VR + (I × 900)
∴ I =
And
...(4)
20 sin ωt – 10
3
1.9 × 10
Vo = 10 +
(20 sin ωt – 10)
× 0.9
1.9
Vo = 10 + 9.47 sin ωt – 4.73
∴ Vo = 5.26 + 9.47 sin ωt
At
ωt = π / 2
Vo = 5.26 + 9.47
= 14.73 Volts
... for (π / 6 ≤ ωt ≤ 5π / 6) ...(5)
Basic Electronics (GTU)
5-11
Diode Circuits
The input output voltage waveforms are as shown in Fig. P. 5.7.3(i).
(h)
(i) : Input and output waveforms for case I
Fig. P. 5.7.3
Case II :
Fig. P. 5.7.3(j)
Fig. P. 5.7.3(k) : Input and output voltage
waveforms for case II
The given circuit for case II is redrawn by taking into account the equivalent circuit of VR. This is
as shown in Fig. P. 5.7.3(j).
1.
For – 20 ≤ Vi ≤ 10 V, the diode D will be OFF and the output voltage is equal to VR i.e. 10 V.
2.
For Vi > 10 V the diode will conduct and the output voltage is equal to
Vo = 10 V + Voltage across 1.9 k resistor
where,
= 10 V + (I × 1.9 kΩ)
( Vi – VR )
I =
1.95 kΩ
( Vi – 10 ) × 1.9
= 10 V + 0.974 (Vi – 10 )
1.95
The input and output voltage waveforms are as shown in Fig. P. 5.7.3(k).
∴ Vo = 10 V +
...(6)
Basic Electronics (GTU)
Ex. 5.7.4 :
5-12
Diode Circuits
For the circuit shown in Fig. P. 5.7.4(a) :
1.
The diodes are ideal, write the transfer characteristic equations.
2.
Plot Vo against Vi, indicating all the intercepts, slopes and voltage levels.
(Assume, Vi = 30 sin ωt).
.Page No. 5-27.
Fig. P. 5.7.4(a)
Soln. :
(b) Equivalent circuit for positive half cycle
(c) Equivalent circuit for the negative half cycle
Fig. P. 5.7.4
•
In the positive half cycle of the input, diode D1 is reverse biased and acts like an open switch
while diode D2 is forward biased and acts like a closed switch as shown in Fig. P. 5.7.4(b).
Therefore the output voltage is given by,
1.2 k
1
Vo = (1.2 + 1.2) k × Vi = 2 × Vi
...(1)
•
In the negative half cycle of the input, diode D1 is forward biased and acts like a closed switch
while D2 acts as an open switch as shown in Fig. P. 5.7.4(c). Therefore the output voltage is
given by,
1.2 k × Vi
1
Vo = (1.2 + 1.2) k = 2 × Vi
...(2)
•
Thus in both the half cycles of input, the instantaneous output voltage is equal to half the
instantaneous input voltage.
Vi
...(3)
∴ Vo = 2
•
The input and output voltage waveforms and plot of Vo against Vi are shown in Fig. P. 5.7.4(d).
Basic Electronics (GTU)
5-13
Diode Circuits
Fig. P. 5.7.4(d) : Input output voltage waveforms and transfer characteristics
Ex. 5.7.5 :
Sketch IR and Vo w.r.t time for the network shown in Fig. P. 5.7.5(a), for the input Vi shown
in the same figure.
.Page No. 5-27.
Fig. P. 5.7.5(a)
Assume that both diodes are silicon type with Rf = 0 Ω and Rr = ∞ with Vf = 0.7 V.
Soln. : The waveforms are as shown in Fig. P. 5.7.5(b)
Basic Electronics (GTU)
5-14
Diode Circuits
Fig. P. 5.7.5(b)
Ex. 5.7.6 :
For the circuit shown in Fig. P. 5.7.6, assuming ideal diodes :
1.
Plot transfer characteristics showing all intercepts, slopes and voltage levels.
2.
If an input Vi = 40 sin ωt is applied to the circuit, find the values of ωt at which diode D2
starts and stops conducting for one cycle of AC input.
.Page No. 5-28.
Fig. P. 5.7.6
Basic Electronics (GTU)
5-15
Diode Circuits
Soln. : This example is exactly identical to Ex. 5.7.2.
Part II :
The diode D2 starts conducting when Vi = 20 V.
∴ 20 = 40 sin ωt
sin ωt = 0.5
∴
∴
ωt = sin
–1
0.5 = 30°
...Ans.
Section 5.9 :
Ex. 5.9.2 :
A voltage V = 300 cos100t is applied to a half wave rectifier with RL = 5 kΩ. The rectifier
may be represented by an ideal diode in series with a resistance of 1 kΩ. Calculate :
1.
Im
2.
DC power
3.
AC power
4.
Rectifier efficiency
5.
Ripple factor.
.Page No. 5-42.
Soln. :
From the expression for the input voltage, we can see that Vm = 300 V. As the diode is equivalent
to an ideal diode in series with a 1 kΩ resistor, we can assume that RF = 1 kΩ. As transformer is not
used RS = 0 Ω.
Vm
1.
Peak load current, Im = ( R + R + R )
S
F
L
=
2.
DC power
300
= 50 mA
(0 + 1 kΩ + 5 kΩ)
...Ans.
2
PL dc = IL dc × RL
2
⎡ Im ⎤ × R
=
L
⎣π⎦
2
0.05 ⎤
3
= ⎡
× 5 × 10 = 1.2655 W
⎣ π ⎦
3.
...Ans.
2
AC input power = IS rms ( RS + RF + RL )
= [ Im / 2 ] ×( RF + RL ) = [ 0.05 / 2 ] × 6 × 10
2
2
∴
4.
Pac = 3.75 W
PL dc
1.2665
Rectifier efficiency η = P × 100 % = 3.75 × 100 = 33.77 %
ac
2
5.
2
3
...Ans.
...Ans.
1/2
⎡ IL rms – IL dc ⎤
⎥ × 100
Ripple factor r = ⎢
2
⎣ IL dc
⎦
But
Im
IL rms = 2 = 25 mA
2
and IL dc =
2
Im
π
= 15.91 mA
1/2
⎡ (25) – (15.92) ⎤
⎥ × 100 = 121.13 %
⎣ (15.91)2 ⎦
∴ r = ⎢
...Ans.
Basic Electronics (GTU)
Note :
5-16
Diode Circuits
As shown in the expression for “r” we can calculate “r” from the rms and average values of load
current instead of using the rms and average values of load voltage.
Section 5.10 :
Ex. 5.10.3 : In a centre-tapped full-wave rectifier, the rms half-secondary voltage is 9 V. Assuming ideal
diodes and load resistance. RL = 1 kΩ, find :
1.
Peak current
2.
DC load voltage
3.
RMS current
4.
Ripple factor
5.
Efficiency
Soln. :
Given :
.Page No. 5-51.
VS rms = 9 V, RL = 1 kΩ, RF = RS = 0.
Vm
2 VS rms
Im = R + R + R =
RL
S
F
L
1. Peak current :
∴ Im =
∵ Vm = 2 VS rms
2×9
3 = 12.72 mA
1 × 10
...Ans.
2. DC load voltage ( VL dc ) :
VL dc = IL dc × RL
2 Im
× RL
=
π
2 × 12.72 × 10 × 1 × 10
= 8.1V
π
Im
12.72
=
=
= 8.994 mA
2
2
–3
3
=
3.
4.
rms load current :
Ripple factor :
IL rms
r =
=
5.
Efficiency :
η =
∴ η =
[V
2
2
– VL dc
VL dc
L rms
1/2
]
[I
=
[ (8.994 × 1)2 – (8.1)2 ]
8.1
8 RL
π ( RS + R F + R L )
2
8 RL
π ( RL )
2
=
2
L rms
2
2
RL – VL dc
VL dc
...Ans.
...Ans.
1/2
]
1/2
= 0.4827 or 48.27 %
...Ans.
... [Referring to Equation (5.10.10)]
8
2 = 0.8105 or 81.05 %
π
...Ans.
Ex. 5.10.4 : A full wave rectifier uses a diode with forward resistance of 1 Ω. The transformer secondary
is centre tapped with output 10-0-10 Vrms and has resistance of 5 Ω for each half section.
Calculate :
1.
No-load dc voltage
2.
DC output voltage at 100 mA
3.
% regulation at 100 mA.
.Page No. 5-52.
Basic Electronics (GTU)
Soln. :
5-17
Given : RMS secondary voltage
Diode Circuits
VS rms = 10V,
RS = 5 Ω ,
Secondary resistance
Forward resistance of diode RF = 1 Ω
1.
No-load dc voltage :
2 2 VS rms 2 2 × 10
=
=
= 9 Volts
π
π
π
DC output voltage at IL = 100 mA :
VL dc (at 100 mA) = VL dc (N.L) – IL ( RS + RF )
VL dc =
2.
3.
2 Vm
= 9 – 0.1 (5 + 1) = 8.4 Volts.
% regulation at 100 mA :
VNL – VFL
9 – 8.4
× 100 =
× 100 = 7.142 %
% regulation =
8.4
VFL
...Ans.
...Ans.
...Ans.
Ex. 5.10.5 : What is the necessary ac input power from the transformer secondary used in a half rectifier
to deliver 500 W of dc power to the load ? What would be the ac input power for the same
load in a full wave rectifier ?
.Page No. 5-52.
Soln. :
Given :
PL dc = 500 W
Pac for HWR :
PL dc
Rectifier efficiency η = P
ac
∴ Pac =
For a HWR, η = 0.4
PL dc
η
500
= 0.4 = 1250 W
∴
Pac
∴
500
Pac = 0.812
= 615.76 W
...(1)
...Ans.
Pac for FWR :
For a FWR, η = 0.812
...Ans.
Section 5.11 :
Ex. 5.11.2 : A bridge rectifier is applied with input from a step down transformer having turns ratio 1:8
and input 230 V, 50 Hz. If the diode forward resistance is 1 Ω, secondary resistance is 10 Ω
and load resistance connected is 2 kΩ find :
1.
DC power output
2.
PIV across each diode
3.
% efficiency
4.
% regulation at full load
Soln. :
Given : NP : NS = 8 : 1,
1.
RS = 10 Ω,
RMS secondary voltage,
VS rms
RF = 1 Ω,
RL = 2 kΩ
NS
= N × 230 = 28.75 V
P
.Page No. 5-56.
...(1)
Basic Electronics (GTU)
2.
5-18
Diode Circuits
Peak secondary voltage,
2 VS rms = 2 × 28.75 = 40.65 V
Vm
40.65
= ( R + 2 R + R ) = (10 + 2 + 2000) = 20.2 mA
S
F
L
Vm =
3.
Peak load current,
4.
Im
D.C. load current, IL dc =
2 Im
π
=
...(2)
...(3)
2 × 20.2
= 12.86 mA
π
2
...(4)
(a)
D.C. load power, PL dc = IL dc × RL = (12.86 × 10 ) × 2 × 10 = 331 mW
...Ans.
(b)
PIV across each diode = Vm = 40.65 V
...Ans.
–3 2
3
2
2
AC input power, Pac = IS rms (RS + 2 RF + RL) = [ Im / 2 ] (RS + 2 RF + RL)
(c)
(20.2 × 10 )
× 2012 = 410 mW
2
PL dc
331
∴ % efficiency = P × 100 = 410 × 100 = 80.63 %
ac
–3 2
∴
(d)
Pac =
% regulation =
VNL =
...(5)
...Ans.
VNL – VFL
× 100
VFL
2 Vm
π
=
2 × 40.65
= 25.87 V
π
...(6)
VFL = IL dc × RL = 12.86 × 10 × 2 × 10 = 25.72 V
25.87 – 25.72
× 100 = 0.5832 %
∴ % regulation =
25.72
–3
3
...(7)
...Ans.
Ex. 5.11.3 : Determine the rms value of secondary voltage of a transformer which provides 9 V dc
output voltage when connected to a bridge rectifier. If the secondary winding resistance is
3 Ω and diode forward resistance is 1 Ω what will be the output voltage when 90 Ω load is
connected to the power supply ?
.Page No. 5-56.
Soln. : Given : VL dc = 9 V, RS = 3 Ω, RF = 1 Ω, RL = 90 Ω
1.
To calculate rms secondary voltage VS rms :
Given : VL dc = 9V
Assuming that no load was connected, the expression for average load voltage is,
2 Vm
,
where Vm = Peak secondary voltage
VL dc =
π
2 Vm
∴
= 9
π
9π
∴ Vm = 2 = 14.14 Volt
Vm 14.14
∴ VS rms =
=
≈ 10 V
...Ans.
2
2
2.
This is the rms value of the secondary voltage of the transformer.
To calculate D.C. output voltage ( VL dc ) :
Given :
Vm = 14.14 V, RL = 90 Ω,
RS = 3 Ω, RF = 1 Ω
Basic Electronics (GTU)
5-19
Diode Circuits
Vm
14.14
∴ Peak load current, Im= ( R + 2 R + R ) = (3 + 2 + 90) = 148.8 mA
S
F
L
∴ DC load current, IL dc =
2 Im
π
= 94.75 mA
∴ DC load voltage, VL dc = IL dc × RL = 94.75 × 10
–3
× 90 = 8.528 V
...Ans.
Ex. 5.11.4 : A full wave bridge rectifier is supplied from 230 V, 50 Hz and uses a transformer of turns
ratio of 15 : 1. It uses load resistance of 50 Ω. Calculate load voltage and ripple voltage.
Assume ideal diode and transformer. Assume standard value of ripple factor for full wave
rectifier.
.Page No. 5-56.
Soln. :
Given :
1.NP : NS = 15 : 1
2. RL = 50 Ω
3. RS = 0 Ω and RF = 0 Ω
NS
1
RMS secondary voltage, VS rms = N × 230 = 15 × 230 = 15.33 V
P
2 × VS rms = 2 × 15.33 = 21.68 V
Vm
21.68
= ( R + 2 R + R ) = 50 = 0.4336 A
S
F
L
Peak secondary voltage, Vm =
Peak load current, Im
DC load current,
IL dc =
2 Im
π
= 0.276 A or 276 mA
1.
DC output voltage, VL dc = IL dc × RL = 0.276 × 50 = 13.8 V
2.
Ripple voltage = Ripple factor × VL dc
The standard value of ripple factor is 0.482 for the bridge rectifier.
∴ Ripple voltage = 0.482 × 13.8 = 6.654 Volt
...(1)
...(2)
…(3)
...(4)
...Ans.
...Ans.
Section 5.13 :
Ex. 5.13.2 : A full wave rectifier circuit uses a capacitor input filter with 50 µF capacitor and provides a
load current of 200 mA at 8% ripple. Calculate :
1.
DC voltage across the filter capacitor
2.
The peak rectified voltage obtained from the 50 Hz supply.
.Page No. 5-67.
Soln. :
1.
DC voltage across the filter capacitor ( VL dc ) :
The dc voltage across the filter capacitor is nothing but the dc voltage across the load i.e. VL dc.
1
∴ Ripple factor =
4 3 fC RL
=
∴ r =
1
4 3 fC VL dc / IL dc
IL dc
4 3 fC VL dc
Basic Electronics (GTU)
5-20
∴ VL dc =
Diode Circuits
IL dc
4 3 fC r
200 × 10
–6
4 3 × 50 × 500 × 10 × 0.08
–3
∴ VL dc =
= 14.43 Volts
2.
Peak rectified voltage ( Vm ) :
We know that the average load voltage with capacitor input filter is given by,
IL dc
VL dc = Vm – 4 fC
IL dc
∴ Vm = VL dc + 4 fC
...Ans.
200 × 10
–6
4 × 50 × 500 × 10
–3
= 14.43 +
∴
Vm = 16.43 Volts
...Ans.
Ex. 5.13.3 : Draw the circuit diagram of FWCT rectifier with capacitor shunt filter. If input of the rectifier
is sine wave, draw the nature of the output waveforms for the following :
1.
Filter capacitance only without load resistance.
2.
Capacitor filter with load resistance.
3.
Rectifier without filter.
.Page No. 5-67.
Soln. :
FWCT means full wave rectifier with center tapped transformer. For the circuit diagram of
FWCT rectifier with capacitor shunt filter, refer to Fig. P. 5.13.3. The waveforms are as follows :
Fig. P. 5.13.3
Basic Electronics (GTU)
5-21
Diode Circuits
Ex. 5.13.4 : A bridge rectifier with capacitor filter is fed from 230 V to 50 V step down transformer. If
average dc current in load is 1 Amp. and capacitor filter of 1000 µF. Calculate the load
regulation and ripple factor.
Assume : Power line frequency of 50 Hz. Neglect diode forward resistance and dc
resistance of secondary of transformer.
.Page No. 5-69.
Soln. :
1.
Data given :
C = 1000 µF,
VS rms = 50 Volt,
IL dc = 1 Amp.
f = 50 Hz,
RF = 0,
RS = 0.
Load regulation :
Load regulation =
VNL – VFL
× 100 %
VFL
...(1)
For a bridge rectifier with capacitor filter,
No load voltage = Vm = 2 × VS rms
VNL =
2 × 50 = 70.7 Volts
....(2)
IL dc
Full load voltage, VFL = Vm – 4 fC
= 70.7 –
1
– 6 = 65.71 Volts
4 × 50 × 1000 × 10
....(3)
Substituting Equations (2) and (3) into Equation (1) we get,
Load regulation =
2.
70.7 – 65.71
× 100 % = 7.59 %
65.71
...Ans.
Ripple factor :
Ripple factor =
∴ RF =
1
4 3 fC RL
1
–6
4 3 × 50 × 1000 × 10 × RL
...(4)
We do not know RL in this equation, so let us find it out first.
VFL 65.71
RL = I
= 1 = 65.71 Ω
L dc
...(5)
Substitute this value into Equation (4) to get,
Ripple factor =
1
= 0.04393
–6
4 3 × 50 × 1000 × 10 × 65.71
...Ans.
Basic Electronics (GTU)
5-22
Diode Circuits
Section 5.15 :
Ex. 5.15.2 : Sketch the outputs to the scale for the circuits shown in Fig. P. 5.15.2 and name the circuits.
.Page No. 5-76.
Fig. P. 5.15.2
Ex. 5.15.3 : Explain the performance of the circuit in Fig. P. 5.15.3(a) for the following situations :
1. Diode is connected at A 2. Diode is connected at B 3. Diode is connected at C
.Page No. 5-76.
Basic Electronics (GTU)
5-23
Fig. P. 5.15.3(a)
Soln. :
Fig. P. 5.15.3
Diode Circuits