Question 1 |
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 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
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.
25 | |
28 | |
34 | |
30 |
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}
\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
-1V to -2 V | |
-2 V to -4 V | |
+1V to -2 V | |
+2 V to -4 V |
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
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 _____.
100 | |
200 | |
300 | |
400 |
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
\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
\sin \omega t | |
(\sin \omega t + |\sin \omega t|)/2 | |
(\sin \omega t - |\sin \omega t|)/2 | |
0 for all t |
Question 5 Explanation:
For positive half cycle,
V_{WX}=0
For negative half cycle,
Short circuit condition,
V_{WX}=0
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
125 and 125 | |
125 and 250 | |
250 and 125 | |
250 and 250 |
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
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
10 mA | |
9.3 mA | |
6.67 mA | |
6.2 mA |
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
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
Assuming forward voltage drops of the diodes to be 0.7 V, the input-output transfer characteristics of the circuit is
A | |
B | |
C | |
D |
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.
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
If the forward voltage drop of diode is 0.7 V, then the current through collector will be
168mA | |
108mA | |
20.54mA | |
5.36mA |
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.
\;\; =\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
4V | |
5V | |
7.5V | |
12.12V |
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
So, V_0=10 \times \frac{10}{10+10}=5\; V
There are 10 questions to complete.