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).


287.36 | |
145.36 | |
250 | |
547.36 |
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}

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


V_{out} =V_{in} for
V_{in} \gt 0 | |
V_{out} =V_{in} for
V_{in} \lt 0 | |
V_{out} =-V_{in} for
V_{in} \gt 0 | |
V_{out} =-V_{in} for
V_{in} \lt 0 |
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
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.


25 | |
28 | |
34 | |
30 |
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}
\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


-1V to -2 V | |
-2 V to -4 V | |
+1V to -2 V | |
+2 V to -4 V |
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
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 _____.


100 | |
200 | |
300 | |
400 |
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
\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.