11. What happens if the potentiometer Rp is connected to a negative supply?
A. Output waveform below -Vref will be clipped off
B. Output waveform above +Vref will be clipped off
C. Output waveform above -Vref will be clipped off
D. Output waveform below +Vref will be clipped off
Answer: C
If the potentiometer Rp is connected to negative supply -VEE instead of +VCC, the reference voltage Vref will be negative. As a result, the entire output waveform above -Vref is to be clipped off.
12. What happens if the input voltage is higher than the reference voltage in a positive clipper?
A. Output voltage = Reference voltage
B. Output voltage = DC Positive voltage
C. Output voltage = Input voltage
D. All of the mentioned
Answer: A
When the input voltage is higher than the reference voltage, the op-amp operates in an open loop, and the diode becomes reverse biased. Thus, the output voltage will be equal to a reference voltage.
13. A positive small-signal halfwave rectifier can
A. Rectify signals with peak value only
B. Rectify signals with the value of few millivolts only
C. Rectify signals with both peak value and down to a few millivolts
D. None of the mentioned
Answer: C
A positive small-signal halfwave rectifier can rectify signals with peak values down to a few millivolts because the high open-loop gain of the op-amp automatically adjusts the voltage drive to the diode so that the rectified output peak is the same as the input.
14. Diode in small-signal positive halfwave rectifier circuit acts as
A. Ideal diode
B. Clipper diode
C. Clamper diode
D. Rectifier diode
Answer: A
The diode acts as an ideal diode since the voltage across the ON diode is divided by the open-loop gain of the op-amp. As the input voltage starts increasing in the positive direction, the output of the op-amp also increases positively till the diode become forward biased.
15. How to minimize the response time and increase the operating frequency range of the op-amp?
A. Positive halfwave rectifier with two diodes
B. Positive halfwave rectifier with one diode
C. Negative halfwave rectifier with two diodes
D. Negative halfwave rectifier with one diode
Answer: C
A negative halfwave rectifier circuit with two diodes is used so that the output of the op-amp does not saturate. This minimizes the response time and increases the operating frequency range.
16. Why a voltage follower stage is connected at the output of the negative small-signal half-wave rectifier?
A. Due to Non-uniform input resistance
B. Due to Non-uniform output resistance
C. Due to Uniform output voltage
D. None of the mentioned
Answer: B
The output resistance of the circuit is non-uniform as it depends on the state of the diode. That is, the output impedance is low when the diode is on and high when the diode is off.
17. A circuit with a predetermined dc level is added to the output voltage of the op-amp is called
A. Clamper
B. Positive clipper
C. Halfwave rectifier
D. None of the mentioned
Answer: A
A clamper clamps the output to a desired dc level.
18. An op-amp clamper circuit is also referred as
A. DC cutter
B. DC inserter
C. DC lifter
D. DC leveler
Answer: B
In an op-amp clamper circuit, a pre-determined dc level is deliberately inserted at the output voltage. For this reason, the clamper is sometimes called a DC inserter.
19. At what values of Ci and Rd a precision clamping can be obtained in peak clamper when the time period of the input waveform is 0.4s?
A. Ci=0.1µF and Rd=10kΩ
B. Ci=0.47µF and Rd=10kΩ
C. Ci=33µF and Rd=10kΩ
D. Ci=2.5µF and Rd=10kΩ
Answer: A
For precision clamping, Ci and Rd-3 i=0.1µF and Rd=10kΩ.
20. Which circuit can be used as a full-wave rectifier?
A. Absolute vale output circuit
B. Positive clipper with two diodes
C. Negative clipper with two diodes
D. Peak clampers
Answer: A
An absolute value output circuit produces an output signal that swings positively only, regardless of the polarity of the input signal; because of the nature of its output waveform, the circuit is used as a full-wave rectifier.
21. For the circuit shown below find the output voltage
A. Vo (+) = +10 v
B. Vo (+) = +12v
C. Vo (+) = +7v
D. None of the mentioned
Answer: B
The voltage at the terminal
V1 = (Vp -Vd1) /2
V1 = (12-0.7) /2 = 5.65 v (Vd1= voltage drop across diode=0.7)
Similarly, the voltage at the negative terminal
V2 = (Vo -Vd3 ) /2 = (Vo – 0.7) /2
Since Vid ≅ 0v , ∴ V1 = V2
Vo = (5.65 *2 ) + 0.7 = 12v.
22. What is the alternate method to measure the values of non-sinusoidal waveform other than ac voltmeter?
A. Clipper
B. Clamper
C. Peak detector
D. Comparator
Answer: C
A conventional ac voltmeter is designed to measure the RMS value of the pure sine wave whereas, the peak value of the non-sinusoidal waveforms can be a peak detector.
