Diodes and their Applications

Question 1
For the circuit shown below with ideal diodes, the output will be

A
V_{out} =V_{in} for V_{in} \gt 0
B
V_{out} =V_{in} for V_{in} \lt 0
C
V_{out} =-V_{in} for V_{in} \gt 0
D
V_{out} =-V_{in} for V_{in} \lt 0
GATE EE 2022   Analog Electronics
Question 1 Explanation: 
Case (i): For positive half cycle of V_i
Diode D_1 & D_2 are in forward biased.
Redraw the circuit:

So, V_o=V_{in}
Case (ii): For negative half cycle of V_i
Both diodes D_1 & D_2 are in reverse biased.
Redraw the circuit:

Hence, V_o=V_{in} \text{ for } V_{in} \gt 0
Question 2
In the circuit shown, a \text{5 V} Zener diode is used to regulate the voltage across load R_{0}. The input is an unregulated DC voltage with a minimum value of \text{6 V} and a maximum value of \text{8 V}. The value of R_{S} is 6\;\Omega. The Zener diode has a maximum rated power dissipation of \text{2.5 W}. Assuming the Zener diode to be ideal, the minimum value of R_{0} is _________________ \Omega.

A
25
B
28
C
34
D
30
GATE EE 2021   Analog Electronics
Question 2 Explanation: 
To calculate R_{0} min' we must find I_{L} max
\begin{aligned} I_{s \min } &=I_{z} \min +I_{L \max } \\ \frac{V_{i \min }-V_{z}}{R_{s}} &=I_{z} \min +I_{L \max } \end{aligned}
For ideal zener diode, I_{z \min }=0
\begin{aligned} \frac{V_{i \min }-V_{z}}{R_{s}} &=I_{L \max } \\ \frac{6-5}{6} &=I_{L \max } \\ I_{L} \max &=\frac{1}{6} \mathrm{~A} \\ R_{0} \min &=\frac{V_{z}}{I_{L \max }}=\frac{5}{1 / 6}=30 \Omega \end{aligned}
Question 3
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
A
-1V to -2 V
B
-2 V to -4 V
C
+1V to -2 V
D
+2 V to -4 V
GATE EE 2014-SET-2   Analog Electronics
Question 3 Explanation: 
CASE-I:

V_i \leq -4 V
D_2\rightarrow conducts
D_1\rightarrow OFF
So,\;\; V_o=-2V

CASE-II:

-4V\leq V_i \leq -2V
Both the diodes will be OFF V_o=V'_1

CASE-III:

V_i\geq -2V
D_1\rightarrow conducts
D_2\rightarrow OFF
V_o=-1V
Question 4
The sinusoidal ac source in the figure has an rms value of \frac{20}{\sqrt{2}}V. Considering all possible values of R_L, the minimum value of R_s \; in \; \Omega to avoid burnout of the Zener diode is _____.
A
100
B
200
C
300
D
400
GATE EE 2014-SET-2   Analog Electronics
Question 4 Explanation: 
I_s \geq I_Z+I_L
\frac{1}{20} \geq \frac{20-5}{R_s} \;\;\;\;\; (Putting \;\;I_L=0)
\Rightarrow R_s \geq 300\Omega
Hence, R_{s(min)}=300\Omega
Question 5
A voltage 1000sin\omega t Volts is applied across YZ. Assuming ideal diodes, the voltage measured across WX in Volts, is
A
\sin \omega t
B
(\sin \omega t + |\sin \omega t|)/2
C
(\sin \omega t - |\sin \omega t|)/2
D
0 for all t
GATE EE 2013   Analog Electronics
Question 5 Explanation: 
For positive half cycle,

V_{WX}=0
For negative half cycle,

Short circuit condition,
V_{WX}=0
Question 6
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 \Omega and the minimum power rating of the Zener diode in mW, respectively, are
A
125 and 125
B
125 and 250
C
250 and 125
D
250 and 250
GATE EE 2013   Analog Electronics
Question 6 Explanation: 


I_s=\frac{10-5}{100}=50\; mA
I_{load}=50\;mA-10\; mA =40\; mA
R_L=\frac{5 V}{40mA}=125\Omega
P_{min}=50 mA \times 5V=250\;mW
Question 7
The i-v characteristics of the diode in the circuit given below are
i=\left\{\begin{matrix} \frac{v-0.7}{500}A, & v\geq 0.7V\\ 0A & v \lt 0.7 V \end{matrix}\right.
The current in the circuit is
A
10 mA
B
9.3 mA
C
6.67 mA
D
6.2 mA
GATE EE 2012   Analog Electronics
Question 7 Explanation: 
By applying kVL in loop,


10=10^3i+V
10=10^3\left ( \frac{V-0.7}{500} \right )+V
10= 2 V-1.4+V
3V=11.4
\Rightarrow V=3.8
\Rightarrow i=(10-3.8) \times 10^{-3}=6.2\; mA
Question 8
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
A
A
B
B
C
C
D
D
GATE EE 2011   Analog Electronics
Question 8 Explanation: 
For positive voltage the waveform clips at +5.7 V.
For negative voltage at -0.7 V, the zener diode conducts and clips out.
Question 9
The transistor used in the circuit shown below has a \beta 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
A
168mA
B
108mA
C
20.54mA
D
5.36mA
GATE EE 2011   Analog Electronics
Question 9 Explanation: 
I_{c(sat)} =\frac{0-(-12)-0.2}{R_c}
\;\; =\frac{11.8}{2.2k}=5.3636 \; mA
This is the maximum current possible in any case.
Question 10
Assuming that the diodes in the given circuit are ideal, the voltage V_0
A
4V
B
5V
C
7.5V
D
12.12V
GATE EE 2010   Analog Electronics
Question 10 Explanation: 
Diode D_1 is On and diode D_2 is OFF.
So, V_0=10 \times \frac{10}{10+10}=5\; V
There are 10 questions to complete.