\LARGE GATE 2021 Exam with Solutions
Instructor: MITHUN MONDAL
October 22, 2025
An asymmetrical periodic pulse train \(v_{in}\) of 10 V amplitude with on-time \(T_{ON} = 1\,ms\) and off-time \(T_{OFF} = 1\,s\) is applied to the circuit shown in the figure. The diode \(D1\) is ideal. The difference between the maximum voltage and minimum voltage of the output waveform \(v_o\) in integer is V.
\includegraphics{gate2021-question1.png}
Solution 1
In steady state, the capacitor charges to the peak input voltage when the diode is forward biased (ideal diode acts as a short).
Equation
\[V_C = V_{in,peak} = 10\,V\]
The output voltage is given by KVL:
Equation
\[V_{out} = V_{in} - V_C\]
During the ON time (\(T_{ON}\)), \(V_{in} = 10\,V\):
Equation
\[V_{out} = 10 - 10 = 0\,V\]
During the OFF time (\(T_{OFF}\)), \(V_{in} = 0\,V\) (assuming pulse goes 0 to 10V):
Equation
\[V_{out} = 0 - 10 = -10\,V\]
The difference between max and min output voltage:
Equation
\[V_{out,\,max} - V_{out,\,min} = 0 - (-10) = 10\,V\]
A
Final Answer
Correct answer: \(10\,V\).
In the circuit shown below, a square wave of peak-to-peak amplitude 20 V and zero average value is applied. Assuming all components are ideal, what will be the average output voltage \(V_o\) across the resistor?
A. 10 V \quad B. 10 V \quad C. 0 V \quad D. 5 V
\includegraphics{gate2021-question2.png}
Solution 2
A square wave with 20V peak-to-peak and zero average means it swings from -10V to +10V.
The circuit configuration (series capacitor, shunt diode) is a clamper.
- During the negative half cycle (\(V_{in} = -10V\)), the diode conducts and charges the capacitor to \(V_C = 10V\) (polarity: left plate -, right plate +).
- The output voltage is \(V_{out} = V_{in} + V_C\).
- When \(V_{in} = +10V\), \(V_{out} = 10 + 10 = 20V\).
- When \(V_{in} = -10V\), \(V_{out} = -10 + 10 = 0V\).
The output is a square wave swinging from 0V to 20V.
The average value of this 0-20V square wave is:
Equation
\[V_{avg} = \frac{20 + 0}{2} = 10\,V\]
Thus, the average output voltage \(V_o = 10\,V\).
Correct option: A.
In the circuit shown below, assume ideal components. Determine the feedback factor \(\frac{V_f}{V_o}\) based on the configuration.
A. 3.28 \quad B. 2.86 \quad C. 5.55 \quad D. 3.86
\includegraphics{gate2021-question3.png}
Solution 3
MARKS TO ALL AS PER IIT ANSWER KEY.
The feedback factor \(\beta = \frac{V_f}{V_o}\) is determined by the feedback network resistors.
Based on the provided solution text (which seems disjointed, likely specific values were \(R_1, R_2, R_E\), etc.):
Equation
\[V_{f} = V_{out} \times \frac{R_{f2}}{R_{f1} + R_{f2}} \quad \text{(Standard voltage divider feedback)}\]
Using the numbers present in the source text:
Equation
\[V_{in} \approx 5.55\]
(Note: The source text is garbled here, but option C matches the value 5.55).