Diodes and their Applications

Question 1
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 1 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 2
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 2 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 3
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 3 Explanation: 
For positive half cycle,

V_{WX}=0
For negative half cycle,

Short circuit condition,
V_{WX}=0
Question 4
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 4 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 5
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 5 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 6
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 6 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 7
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 7 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 8
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 8 Explanation: 
Diode D_1 is On and diode D_2 is OFF.
So, V_0=10 \times \frac{10}{10+10}=5\; V
Question 9
In the voltage doubler circuit shown in the figure, the switch 'S' is closed at t=0. Assuming diodes D_1 \; and \; D_2 to be ideal, load resistance to be infinite and initial capacitor voltages to be zero. The steady state voltage across capacitor C_1 \; and \; C_2 will be
A
V_{c1}=10V,V_{c2}=5V
B
V_{c1}=10V,V_{c2}=-5V
C
V_{c1}=5V,V_{c2}=10V
D
V_{c1}=5V,V_{c2}=-10V
GATE EE 2008   Analog Electronics
Question 9 Explanation: 


Forst capacitor charge through Diode (D_1) upto V_{max} (5 \; V) with shown polarities.
\therefore V_{c1}=5\; V
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}
\therefore V_{c2}=-10\; V
Question 10
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 v_0 of the circuit will be
A
A
B
B
C
C
D
D
GATE EE 2008   Analog Electronics
Question 10 Explanation: 


V_P=\frac{10}{10+10}\times 10sin\omega t
\;\; =5sin\omega t
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.

\therefore V_0=\frac{10}{10+10}\times 10sin\omega t=5sin\omega t
There are 10 questions to complete.
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