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
-1V to -2 V | |
-2 V to -4 V | |
+1V to -2 V | |
+2 V to -4 V |
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
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 _____.
100 | |
200 | |
300 | |
400 |
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
\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
\sin \omega t | |
(\sin \omega t + |\sin \omega t|)/2 | |
(\sin \omega t - |\sin \omega t|)/2 | |
0 for all t |
Question 3 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 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
125 and 125 | |
125 and 250 | |
250 and 125 | |
250 and 250 |
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
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 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
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
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 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.
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
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 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.
\;\; =\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
4V | |
5V | |
7.5V | |
12.12V |
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
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
V_{c1}=10V,V_{c2}=5V | |
V_{c1}=10V,V_{c2}=-5V | |
V_{c1}=5V,V_{c2}=10V | |
V_{c1}=5V,V_{c2}=-10V |
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
If such a diode is used in clipper circuit of figure given above, the output voltage v_0 of the circuit will be
A | |
B | |
C | |
D |
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.
