# 250+ TOP MCQs on Clamper Circuit and Answers

Analog Circuits Multiple Choice Questions on “Clamper Circuit”.

1. Which of the following is not true regarding clamper?
A. A positive clamper adds a positive DC voltage
B. A clamper can also be called as a re-inserter
C. To reduce tilt, reduce the RC value
D. Negative clamper will clamp the positive peak of output to the reference voltage

Clarification: Clamper is called a re-inserter since it adds DC voltage to wave, inserts DC. Also, a positive clamper adds a positive DC voltage, shifting the wave up, and vice versa for the negative clamper. However, to reduce the tilt in the output, RC should be large, not small.

2. What is A, B, C and D to design a voltage doubler in the given circuit?

A. A=B=Resistors, C=D=Diodes
B. A=D=Capacitors, B=C=Diodes
C. A=Capacitor, B=D=Diodes, C=Resistor
D. A=C=Diodes, B=D=Capacitors

Clarification: During the negative cycle, D1 conducts and C1 charges up to peak input. When the input is positive, then D2 conducts and C2 charges up to 2Vm. This creates a voltage doubler.

3. Consider the circuit provided. Total discharge time = 0.5 ms. Consider the diode to be an ideal diode, for a square wave input of ± 10 V, what is the percentage tilt? (C = 1MF & R = 50k Ohms)

A. 10%
B. 1%
C. 0.1%
D. 1.1%

Clarification: Total discharge time Td=0.5ms. For ideal diode, no cut-in voltage.
% Tilt = ΔVo/2Vm * 100
Idischarge = 2Vm/R = 2*10/50k = 0.4 mA
ΔVo = Tdischarge*Idischarge/C = 0.5ms*0.4mA/1MF = 0.2
% Tilt = (0.2/20) * 100 = 1 %.

4. Given input for the circuit is f = 1 KHz.

Diode cut-in voltage = 0.7 V

Output waveform:

For the given output, find V1 and discharge time?
A. 4.3 V, 1ms
B. 4V, 0.5ms
C. 4.3V, 0.5ms
D. 5.7V, 1ms

Clarification: Original wave is from +10V to -10V. The final square wave is from +5V to -15V. Negative DC voltage is added, it’s a negative clamper.
Thus, Vo = 5 V when Vin = 10 V. Vcut-in = 0.7 V
Then, Vo = Vin – (V1 + Vcut-in) = 10 – (V1 + 0.7) = 5
Therefore V1 = 4.3 V
Discharge time = To/2 ; To = 1/fo = 1/1000 = 1 ms; Discharge time = 0.5 ms.

5. In the given circuit, given that C=2MF and diode cut-in voltage, Vγ =0. Calculate the average output voltage.

A. 2V
B. 4V
C. -2V
D. -3V

Clarification: Maximum input is Vm = 32 + 5 = 37 V
Capacitor charges up to 37 – 3 = 34 V
Vout = Vin – Vc = 32 + 5 sinωt – 34 = -2 + 5sinωt
Average output voltage = – 2 V.

6. Considering a clamper circuit, where capacitance C, load R, the cut-in voltage of diode are unknown, which is the correct statement?
A. The DC level of the signal changes
B. The peak-to-peak value of signal changes
C. The shape of signal changes
D. The DC level shifts up

Clarification: The shape and peak to peak value of signal remain unchanged in a clamper. A clamper only affects the DC voltage level of the wave, which can be both moved up and down, not simply up.

7. Tilt of output waveform for the circuit is 1%. Given that input is a square wave ± 10 V, f = 2 kHz, diode cut-in voltage = 0.7, what is the relation between C and R?

A. RC=20
B. RC=1/2
C. RC=2
D. RC=1/20

Clarification: To = 1/f = 1 ms; Tdischarge = 0.5 ms
Tilt = 100 *ΔVo/2Vm = 100 *ΔVo/20 = 5 * ΔVo = 1
Thus ΔVo = 0.2 = Idischarge * Tdischarge/C
Idischarge/C = 0.2/Tdischarge = 0.2/0.5 ms = 400
RC = 2Vm/400 = 20/400 = 1/20.

9. Choose the correct option for the relation between the two circuits.

A. Circuit (A. is a voltage doubler circuit, while Circuit (B. does not double the voltage
B. Both are similar and double the voltage to 2Vm
C. Circuit (A. is a doubler circuit with output +2Vm and Circuit (B. is a doubler with output -2Vm
D. Circuit (A. is a doubler circuit, while Circuit (B. is a clipper