Question 1 |
In the circuit shown below, D_{1} and D_{2} are silicon diodes with cut-in voltage of 0.7 \mathrm{~V}. V_{\text {IN }} and V_{\text {OUT }} are input and output voltages in volts. The transfer characteristic is
A | |
B | |
C | |
D |
Question 1 Explanation:
Case I:
\begin{aligned} V_{\gamma}&=0.7 \mathrm{~V} \\ \text { For the +ve half cycle } &\text { if input } V_{\text {in }} \\ D_{1} & \rightarrow O \mathrm{ON} \text { and } D_{2} \rightarrow \text { OFF } \\ \text { For diode } D_{1}:& \quad V_{\text {in }}-1 V \gt 0.7 \\ V_{\text {in }} & \gt 1.7 \mathrm{~V} \\ V_{0}&=V_{\text {in }}-0.7 \end{aligned}
Case II:
For the + ve half cycle if input V_{\text {in }},
D_{1} \rightarrow \mathrm{OFF} and D_{2} \rightarrow \mathrm{ON}
For diode D_{2}: \quad 1-V_{\text {in }} \gt 0.7
V_{\text {in }} \lt 0.3 \mathrm{~V}
V_{0}=V_{\text {in }}+0.7
Case III:
\begin{aligned} 0.3 \mathrm{~V} & \lt V_{\text {in }} \lt 1.7 \mathrm{~V} \\ D_{1} & \rightarrow \mathrm{OFF} \text { and } D_{2} \rightarrow \mathrm{OFF} \\ V_{0} & =1 \mathrm{~V} \end{aligned}
Transfer characteristics,
\begin{aligned} V_{\gamma}&=0.7 \mathrm{~V} \\ \text { For the +ve half cycle } &\text { if input } V_{\text {in }} \\ D_{1} & \rightarrow O \mathrm{ON} \text { and } D_{2} \rightarrow \text { OFF } \\ \text { For diode } D_{1}:& \quad V_{\text {in }}-1 V \gt 0.7 \\ V_{\text {in }} & \gt 1.7 \mathrm{~V} \\ V_{0}&=V_{\text {in }}-0.7 \end{aligned}
Case II:
For the + ve half cycle if input V_{\text {in }},
D_{1} \rightarrow \mathrm{OFF} and D_{2} \rightarrow \mathrm{ON}
For diode D_{2}: \quad 1-V_{\text {in }} \gt 0.7
V_{\text {in }} \lt 0.3 \mathrm{~V}
V_{0}=V_{\text {in }}+0.7
Case III:
\begin{aligned} 0.3 \mathrm{~V} & \lt V_{\text {in }} \lt 1.7 \mathrm{~V} \\ D_{1} & \rightarrow \mathrm{OFF} \text { and } D_{2} \rightarrow \mathrm{OFF} \\ V_{0} & =1 \mathrm{~V} \end{aligned}
Transfer characteristics,
Question 2 |
A circuit and the characteristics of the diode (D) in it are shown. The ratio of the
minimum to the maximum small signal voltage gain \frac{\partial V_{out}}{\partial V_{in}} is ________ (rounded
off to two decimal places)
0.25 | |
0.55 | |
0.75 | |
0.95 |
Question 2 Explanation:
Replacing the given circuit with small signal equivalent.
Case-I when diode is ON
As r_d(ON)=0, the 2k\Omega resistor in parallel to the diode becomes open circuit.
\therefore \; V_{out}=V_{IN}\times \frac{2}{4}=\frac{V_{in}}{2}
\therefore \; \frac{\partial V_{out}}{\partial V_{in}}|_{max}=\frac{1}{2}\;\;\;...(i)
Case-I: When diode is off:
r_d(off)=\infty \Rightarrow total \; R_{eq}=2+2+2=6k\Omega
\therefore \; V_{out}=\frac{V_{in} \times 4}{6}=\frac{2}{3}V_{in}\Rightarrow \frac{\partial V_{out}}{\partial V_{in}}|_{min}=\frac{2}{3}\;\;\;...(ii)
From (i) and (ii)
\frac{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{min}}{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{max}}=\frac{1/2}{2/3}=0.75
Case-I when diode is ON
As r_d(ON)=0, the 2k\Omega resistor in parallel to the diode becomes open circuit.
\therefore \; V_{out}=V_{IN}\times \frac{2}{4}=\frac{V_{in}}{2}
\therefore \; \frac{\partial V_{out}}{\partial V_{in}}|_{max}=\frac{1}{2}\;\;\;...(i)
Case-I: When diode is off:
r_d(off)=\infty \Rightarrow total \; R_{eq}=2+2+2=6k\Omega
\therefore \; V_{out}=\frac{V_{in} \times 4}{6}=\frac{2}{3}V_{in}\Rightarrow \frac{\partial V_{out}}{\partial V_{in}}|_{min}=\frac{2}{3}\;\;\;...(ii)
From (i) and (ii)
\frac{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{min}}{\left (\frac{\partial V_{out}}{\partial V_{in}}\right )_{max}}=\frac{1/2}{2/3}=0.75
Question 3 |
An asymmetrical periodic pulse train v_{in}
of 10\:V
amplitude with on-time T_{\text{ON}}=1\:ms
and off-time T_{\text{OFF}}=1\:\mu s
is applied to the circuit shown in the figure. The diode D_{1} is ideal.
The difference between the maximum voltage and minimum voltage of the output waveform v_{o} (in integer) is ______________ V.
The difference between the maximum voltage and minimum voltage of the output waveform v_{o} (in integer) is ______________ V.
7 | |
5 | |
12 | |
10 |
Question 3 Explanation:
V_{\text{in}} = 10 V: Diode is ON
\therefore Capacitor charges upto 10 \mathrm{~V},
\begin{aligned} \therefore \qquad V_{C}&=10 \mathrm{~V} \\ V_{\text {in }}&=0 ; \text { Diode is OFF } \end{aligned}
Discharging time constant =R \times C
\begin{aligned} &=10 \mathrm{~m} \mathrm{sec} \\ \tau_{\text {discharging }} &>>\tau_{\mathrm{OFF}} \end{aligned}
Capacitor discharges negligibly
\begin{aligned} \therefore\qquad V_{C}&=10 \mathrm{~V}\\ \text{In steady state},\qquad V_{C} &=10 \mathrm{~V} \\ V_{\text {out }} &=V_{\text {in }}-V_{C}=V_{\text {in }}-10 \mathrm{~V}\\ \text{When }\qquad V_{\text {in }}&=10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out }} &=10-10=0 \mathrm{~V} \\ V_{\text {in }} &=10 \mathrm{~V} \\ \mathrm{~V}_{\text {out }} &=0-10=-10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out(max) }}-V_{\text {out(min) }}&=0-(-10)=10 \mathrm{~V} \end{aligned}
\therefore Capacitor charges upto 10 \mathrm{~V},
\begin{aligned} \therefore \qquad V_{C}&=10 \mathrm{~V} \\ V_{\text {in }}&=0 ; \text { Diode is OFF } \end{aligned}
Discharging time constant =R \times C
\begin{aligned} &=10 \mathrm{~m} \mathrm{sec} \\ \tau_{\text {discharging }} &>>\tau_{\mathrm{OFF}} \end{aligned}
Capacitor discharges negligibly
\begin{aligned} \therefore\qquad V_{C}&=10 \mathrm{~V}\\ \text{In steady state},\qquad V_{C} &=10 \mathrm{~V} \\ V_{\text {out }} &=V_{\text {in }}-V_{C}=V_{\text {in }}-10 \mathrm{~V}\\ \text{When }\qquad V_{\text {in }}&=10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out }} &=10-10=0 \mathrm{~V} \\ V_{\text {in }} &=10 \mathrm{~V} \\ \mathrm{~V}_{\text {out }} &=0-10=-10 \mathrm{~V}\\ \Rightarrow\qquad V_{\text {out(max) }}-V_{\text {out(min) }}&=0-(-10)=10 \mathrm{~V} \end{aligned}
Question 4 |
In the circuit shown below, all the components are ideal and the input voltage is
sinusoidal. The magnitude of the steady-state output V_0 (rounded off to two decimal
places) is _________ V.
325.2 | |
650.4 | |
125.45 | |
752.36 |
Question 4 Explanation:
Voltage Doubles, V_{o}=2\: V_{m}=2\times 230\sqrt{2}\cong 650.4V
Question 5 |
In the circuit shown, V_s is a 10 V square wave of period, T=4 ms with R= 500\Omega and C = 10 \mu F. The capacitor is initially uncharged at t=0, and the diode is assumed to be ideal. The voltage across the capacitor (V_c) at 3 ms is equal to ____ volts (rounded off to one decimal place).
2.8 | |
3.3 | |
4.6 | |
5.4 |
Question 5 Explanation:
\begin{aligned} \tau &=R C=500 \times 10 \times 10-6=5 \mathrm{m}_{8} \\ 0 \lt t \lt \frac{t}{2} \; :\; \; v_{\mathrm{cap}} &=v_{f}+\left(v_{i}-v_{f}\right) e^{-t / \tau} \\ &=10+(0-10) e^{-t / \mathrm{Rc}}\\ \text{At }t&=2 \mathrm{msec,} \\ v_{\mathrm{cap}}&=10-10 e^{\frac{-2 \times 10^{-3}}{5 \times 10^{-3}}}\\ &=3.296 \mathrm{V} \simeq 3.3 \mathrm{V} \end{aligned}
For 2 \mathrm{ms} \lt t \lt 4 \mathrm{ms}, diode is OFF and Capacitor
has no path to discharge. Hence, at t=3 ms V_{\text {cap }}=3.3 \mathrm{V}
There are 5 questions to complete.