Diodes and their Applications

 Question 1
The Zener diode in circuit has a breakdown voltage of $5 \mathrm{~V}$. The current gain $\beta$ of the transistor in the active region in $99$. Ignore baseemitter voltage drop $\mathrm{V}_{\mathrm{BE}}$. The current through the $20 \Omega$ resistance in milliamperes is ____ (Round off to 2 decimal places).

 A 287.36 B 145.36 C 250 D 547.36
GATE EE 2023   Analog Electronics
Question 1 Explanation:
Redraw the circuit :

We know,
\begin{aligned} \mathrm{I}_{\mathrm{E}} & =(1+\beta) \mathrm{I}_{\mathrm{B}} \\ & =100 \mathrm{I}_{\mathrm{B}} \\ \Rightarrow \quad \mathrm{I}_{\mathrm{B}} & =\frac{\mathrm{I}_{\mathrm{E}}}{100} \end{aligned}
Apply KVL in loop,
\begin{aligned} & 25-\frac{\mathrm{I}_{E}}{100} \times 7000-I_{E} \times 10-I_{E} \times 20=0 \\ & \Rightarrow \quad \mathrm{I}_{\mathrm{E}}=0.25 \mathrm{~A} \text { or } 250 \mathrm{~mA} \end{aligned}
 Question 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 Explanation:
$I_s \geq I_Z+I_L$
$\frac{1}{20} \geq \frac{20-5}{R_s} \;\;\;\;\; (Putting \;\;I_L=0)$
$\Rightarrow R_s \geq 300\Omega$
$Hence, R_{s(min)}=300\Omega$

There are 5 questions to complete.