# 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.