Q1: In the given circuit, the diodes are ideal. The current I through the diode D1 in milliamperes is ______ (rounded off to two decimal places). (2024)
(a) 1.12
(b) 2.25
(c) 1.67
(d) 2.92
Ans: (c)
Sol: Assume D2 OFF
Tested: D2 ON
Assume D1 OFF
Tested: D1 ON
VX = 2 mA × 1kΩ
= 2 V
States: D1 and D2 ON
3VX = 13
VX = 13/3 = 4.33 V
I1 + ID1 = I2
IDI = 3 mA - 1.33 mA
= 1.67 mA
Q2: The Zener diode in circuit has a breakdown voltage of 5 V. The current gain β of the transistor in the active region in 99. Ignore baseemitter voltage drop VBE. The current through the 20Ω resistance in milliamperes is ____ (Round off to 2 decimal places). (2023)
(a) 287.36
(b) 145.36
(c) 250
(d) 547.36
Ans: (c)
Sol: Redraw the circuit :
We know,
IE = (1 + β)IB
= 100IB
⇒ IB = IE/100
Apply KVL in loop,
25 - (IE/100) x 7000 - IE x 10 x IE x 20 = 0
⇒ IE = 0.25 A or 250 mA
Q3: For the circuit shown below with ideal diodes, the output will be (2022)
(a) Vout = Vin for Vin > 0
(b) Vout = Vin for Vin < 0
(c) Vout= −Vin for Vin > 0
(d) Vout = −Vin for Vin < 0
Ans: (a)
Sol: Case (i): For positive half cycle of Vi
Diode D1 & D2 are in forward biased.
Redraw the circuit:
So, Vo = Vin
Case (ii): For negative half cycle of Vi
Both diodes D1 & D2 are in reverse biased.
Redraw the circuit:
Hence, Vo = Vin for Vin > 0.
Q4: In the circuit shown, a 5 V Zener diode is used to regulate the voltage across load R0. The input is an unregulated DC voltage with a minimum value of 6 V and a maximum value of 8 V. The value of RS is 6Ω. The Zener diode has a maximum rated power dissipation of 2.5 W. Assuming the Zener diode to be ideal, the minimum value of R0 is _______ Ω. (2021)
(a) 25
(b) 28
(c) 34
(d) 30
Ans: (d)
Sol: To calculate R0 min' we must find IL max
Is min = Izmin + IL max
For ideal zener diode, Izmin = 0
IL max = 1/6 A
R0 min =
Q5: Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage vi must be outside the range (SET-2 (2014))
(a) -1V to -2 V
(b) -2 V to -4 V
(c) +1V to -2 V
(d) +2 V to -4 V
Ans: (b)
Sol: CASE-I:
V
i ≤ −4V
D
2 → conducts
D
1 → OFF
So, V
o = −2V
CASE-II:
−4V ≤ V
i ≤ −2V
Both the diodes will be OFF V
o = V
1′
CASE-III:
V
i ≥ −2V
D
1 → conducts
D
2 → OFF
V
o = −1V
Q6: The sinusoidal ac source in the figure has an rms value of (20/√2) V. Considering all possible values of RL, the minimum value of Rs in Ω to avoid burnout of the Zener diode is _____. (SET-2 (2014))
(a) 100
(b) 200
(c) 300
(d) 400
Ans: (c)
Sol: I
s ≥ I
Z + I
L⇒ R
s ≥ 300Ω
Hence, R
s(min) = 300Ω
Q7: A voltage 1000sin ωt Volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts, is (2013)
(a) sin ωt
(b) (sin ωt + ∣sin ωt∣)/2
(c) (sin ωt − ∣sin ωt∣)/2
(d) 0 for all t
Ans: (d)
Sol: For positive half cycle,
V
WX = 0
For negative half cycle,
Short circuit condition,
V
WX = 0
Q8: In the circuit shown below, the knee current of the ideal Zener dioide is 10 mA. To maintain 5 V across RL, the minimum value of RL in Ω and the minimum power rating of the Zener diode in mW, respectively, are (2013)
(a) 125 and 125
(b) 125 and 250
(c) 250 and 125
(d) 250 and 250
Ans: (b)
Sol: I
load = 50mA − 10mA = 40mA
P
min = 50mA × 5V = 250mW
Q9: The i-v characteristics of the diode in the circuit given below are
The current in the circuit is (2012)
(a) 10 mA
(b) 9.3 mA
(c) 6.67 mA
(d) 6.2 mA
Ans: (d)
Sol: By applying kVL in loop,
10 = 10
3i + V
10 = 2V − 1.4 + V
3V = 11.4
⇒ V = 3.8
⇒ i = (10 − 3.8) × 10
−3 = 6.2 mA
Q10: A clipper circuit is shown below.
Assuming forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit is (2011)
(a) (b) (c) (d) Ans: (c)
Sol: For positive voltage the waveform clips at +5.7 V.
For negative voltage at -0.7 V, the zener diode conducts and clips out.
Q11: The transistor used in the circuit shown below has a β of 30 and ICBO is negligible
If the forward voltage drop of diode is 0.7 V, then the current through collector will be (2011)
(a) 168mA
(b) 108mA
(c) 20.54mA
(d) 5.36mA
Ans: (d)
Sol: = 11.8/2.2k = 5.3636 mA
This is the maximum current possible in any case.
Q12: Assuming that the diodes in the given circuit are ideal, the voltage V0 (2010)
(a) 4V
(b) 5V
(c) 7.5V
(d) 12.12V
Ans: (b)
Sol: Diode D
1 is On and diode D
2 is OFF.
So, V
0 =10 ×(10/(10+10)) = 5V
Q13: In the voltage doubler circuit shown in the figure, the switch 'S' is closed at t = 0. Assuming diode D1 and D2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor C1 and C2 will be (2008)
(a) Vc1 = 10V, Vc2 = 5V
(b) Vc1 = 10V, Vc2 = −5V
(c) Vc1 = 5V, Vc2 = 10V
(d) Vc1 = 5V, Vc2 = −10V
Ans: (d)
Sol: Forst capacitor charge through Diode (D
1) upto V
max(5V) with shown polarities.
∴ V
c1 = 5V
Now diode D
1 will be reversed biased and D
2 will be forward biased and capacitor C
2 will charge in reverse direction through diode D
2 upto 2V
max∴ V
c2 = −10V
Q14: The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure.
If such a diode is used in clipper circuit of figure given above, the output voltage v0 of the circuit will be (2008)
(a) (b) (c) (d) Ans: (a)
Sol: = 5 sinωt
Since maximum voltage across at the point P may be 5 V, hence voltage across the diode always will be less than or equal to zero. So it will be reversed always.
Q15: The three-terminal linear voltage regulator is connected to a 10 Ω load resistor as shown in the figure. If Vin is 10 V, what is the power dissipated in the transistor ? (2007)
(a) 0.6W
(b) 2.4W
(c) 4.2W
(d) 5.4W
Ans: (b)
Sol: V
0 = V
Z − V
BE= 6.6V − 0.7V = 5.9V
V
CE − V
i − V
0= 10 − 5.9 = 4.1V
P
Q = V
CE × I
C= 4.1 × 0.59A = 2.4W
Q16: Assuming the diodes D1 and D2 of the circuit shown in figure to be ideal ones, the transfer characteristics of the circuit will be (2006)
(a) (b) (c) (d) Ans: (a)
Sol: V
i < 10V (D
1 OFF, D
2 OFF)
V
o = 10V
V
i > 10V (D
1 ON, D
2 OFF)
V
o = V
inQ17: What are the states of the three ideal diodes of the circuit shown in figure ? (2006)(a) D1 ON, D2 OFF, D3 OFF
(b) D1 OFF, D2 ON, D3 OFF
(c) D1 ON, D2 OFF, D3 ON
(d) D1 OFF, D2 ON, D3 ON
Ans: (a)
Sol: Analyzing the circuit, we can see that D
1 and D
2 are in forward bias and D
3 is in reverse bias. But no current flow through D
2, bcause current gets shortest path through D
1.
Q18: Assume that D1 and D2 in figure are ideal diodes. The value of current is (2005)
(a) 0 mA
(b) 0.5 mA
(c) 1 mA
(d) 2 mA
Ans: (a)
Sol: Current will pass through its simplest path (or) low resistance path. D
1 become forward biased and D
2 is reverse biased
I = 0 mA
Q19: Assuming that the diodes are ideal in figure, the current in diode D1 is (2004)
(a) 8 mA
(b) 5 mA
(c) 0 mA
(d) -3 mA
Ans: (c)
Sol: From the circuit, if 'D
1' is 'ON' Kirchoff's law is not satisfied
∴ D
1 is OFF and D
2 is ON
∴ I = 0 mA
Q20: The current through the Zener diode in figure is (2004)
(a) 33 mA
(b) 3.3mA
(c) 2 mA
(d) 0 mA
Ans: (c)
Sol: Q21: A voltage signal 10 sin ωt is applied to the circuit with ideal diodes, as shown in figure, The maximum, and minimum values of the output waveform Vout of the circuit are respectively (2003)
(a) +10 V and -10 V
(b) +4 V and -4 V
(c) +7 V and -4 V
(d) +4 V and -7 V
Ans: (d)
Sol: Fpr positive cycle , when V
in > 4V
(V
in > 4V) D
2ON, D
1OFF, V
out = 4V
where V
in < 4V, then V
out = V
in because D
2 also become (OFF).
For negative cycle, when V
in < −4V
D
1 ON, D
2 OFF
but, V
in = −ve
(∵ for minimum output V
in = −10V)
V
out = (-3/10) x 10k − 4 = −7V
Q22: The cut-in voltage of both Zener diode Dz and diode D shown in figure is 0.7 V, while break down voltage of Dz is 3.3 V and reverse breakdown voltage of D is 50 V. the other parameters can be assumed to be the same as those of an ideal diode. The values of the peak output voltage (V0) are (2002)
(a) 3.3 V in the positive half cycle and 1.4 V in the negative half cycle
(b) 4 V in the positive half cycle and 5 V in the negative half cycle
(c) 3.3 V in both positive and negative half cycles
(d) 4 V in both positive and negative half cycles
Ans: (b)
Sol: For positive half cycle, at peak input voltage
So, V
0 = 4V
For negative half cycle, at peak input voltage
So, Vo = 10 × (1/(1+1)) = 5V
Q23: The forward resistance of the diode shown in figure is 5Ω and the remaining parameters are same as those of an ideal diode. The dc component of the source current is (2002)
(a) (b) (c) (d) Ans: (a)
Sol: = Vm/50π