Diodes and their Applications

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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}

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.