Previous Year Questions- Diodes and their Applications | Analog and Digital Electronics - Electrical Engineering (EE) PDF Download

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
V_X = 2 mA × 1kΩ
= 2 V
States: D₁ and D₂ ON  
3V_X = 13
V_X = 13/3 = 4.33 V
I₁ + I_D1 = I₂
I_DI = 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 V_BE. 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)I_B
= 100I_B
⇒ I_B = I_E/100
Apply KVL in loop,
25 - (I_E/100) x 7000 - I_E x 10 x I_E x 20 = 0
⇒ I_E = 0.25 A or 250 mA

Q3: For the circuit shown below with ideal diodes, the output will be (2022)
(a) V_out = V_in for V_in > 0
(b) V_out = V_in for V_in < 0
(c) $Vout=-Vin$V_out= −V_in for V_in > 0
(d) $Vout=-Vin$V_out = −V_in for V_in < 0
Ans: (a)
Sol: Case (i): For positive half cycle of V_i
Diode D₁ & D₂ are in forward biased.
Redraw the circuit:
So, V_o = V_in
Case (ii): For negative half cycle of V_i
Both diodes D₁ & D₂ are in reverse biased.
Redraw the circuit:
Hence, V_o = V_in for V_in > 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 R_S 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 I_L max
I_{s min} = I_zmin + I_L max
For ideal zener diode, I_zmin⁡ = 0
I_L 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 v_i 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
D₂ → conducts
D₁ → OFF
So, V_o = −2V
CASE-II:
−4V ≤ V_i ≤ −2V
Both the diodes will be OFF V_o = V₁′
CASE-III:
V_i ≥ −2V
D₁ → conducts
D₂ → 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 R_L, the minimum value of R_s 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 R_L, the minimum value of R_L 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 = 10³i + 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 I_CBO 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 D₁ is On and diode D₂ is OFF.
So, V₀ =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) V_c1 = 10V, V_c2 = 5V
(b) V_c1 = 10V, V_c2 = −5V
(c) $Vc1=5V,Vc2=10V$V_c1 = 5V, V_c2 = 10V
(d) $Vc1=5V,Vc2=-10V$V_c1 = 5V, V_c2 = −10V
Ans: (d)
Sol: Forst capacitor charge through Diode (D₁) upto V_max(5V) with shown polarities.
∴ V_c1 = 5V  
Now diode D₁ will be reversed biased and D₂ will be forward biased and capacitor C₂ will charge in reverse direction through diode D₂ 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 V_in 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: V₀ = V_Z − V_BE
= 6.6V − 0.7V = 5.9V
V_CE − V_i − V₀
= 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 (D₁ OFF, D₂ OFF)
V_o = 10V
V_i > 10V (D₁ ON, D₂ OFF)
V_o = V_in

Q17: 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) $D1ON,D2OFF,D3ON$D₁ ON, D₂ OFF, D₃ ON
(d) D₁ OFF, D₂ ON, D₃ ON
Ans: (a)
Sol: Analyzing the circuit, we can see that D₁ and D₂ are in forward bias and D₃ is in reverse bias. But no current flow through D₂, bcause current gets shortest path through D₁.

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. D₁ become forward biased and D₂ 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 'D₁' is 'ON' Kirchoff's law is not satisfied
∴ D₁ is OFF and D₂ 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 V_out 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) D₂ON, D₁OFF, V_out = 4V
where V_in < 4V, then V_out = V_in because D₂ also become (OFF).
For negative cycle, when V_in < −4V
D₁ ON, D₂ 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 D_z and diode D shown in figure is 0.7 V, while break down voltage of D_z 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, V₀ = 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:  = V_m/50π