# 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.