Triangular Wave Generator Using Op-Amp MCQ [Free PDF] – Objective Question Answer for Triangular Wave Generator Using Op-Amp Quiz

1. How a triangular wave generator is derived from a square wave generator?

A. Connect the oscillator at the output
B. Connect the Voltage follower at the output
C. Connect differential at the output
D. Connect integrator at the output

Answer: D

The output waveform of the integrator is triangular if its input is a square wave. Therefore, a triangular wave generator can be obtained by connecting an integrator at the output of the square wave generator.


2. The increase in the frequency of the triangular wave generator.

A. Ramp the amplitude of the triangular wave
B. Increase the amplitude of the triangular wave
C. Decrease the amplitude of the triangular wave
D. None of the mentioned

Answer: A

As the resistor value increase or decreases, the frequency of the triangular wave will decrease or increase, respectively. Therefore, the amplitude of the triangular wave decreases with an increase in its frequency and vice versa.


3. Which among the following op-amp is chosen for generating a triangular wave of relatively higher frequency?

A. LM741 op-amp
B. LM301 op-amp
C. LM1458 op-amp
D. LM3530 op-amp

Answer: B

The frequency of the triangular wave generator is limited by the slew rate of the op-amp. LM301 op-amp has a high slew rate.


4. What is the peak to peak (PP) output amplitude of the triangular wave?

A. VO(pp) = + VRamp + (- VRamp)
B. VO(pp) = – VRamp + (+ VRamp)
C. VO(pp) = + VRamp – (- VRamp)
D. VO(pp) = – VRamp – (+ VRamp)

Answer: C

The peak to the peak output waveform

VO(pp) = + VRamp-(-VRamp)

Where – VRamp –> Negative going ramp ;

+ VRamp–> positive-going ramp.


6. Find the capacitor value for the output frequency, fo = 2kHz & VO(pp) = 7v, in a triangular wave generator. The op-amp is 1458/741 and supply voltage = ±15v. (Take internal resistor=10kΩ)

A. 0.03nF
B. 30nF
C. 0.3nF
D. 3nF

Answer: D

Given, Vsat =15v

∴ VO(pp) = (2R2/R3) × Vsat

=> R2 =(VO(pp) ×R3) / (Vsat×2)

= [7/(2×15)]×R3 = 0.233R3

∵ Internal resistor, R2 = R1= 10kΩ

=> R3 = 0.233×10kΩ = 2.33kΩ.

So, the output frequency

fO = R3 / ( 4×R1 ×C1× R2)

=> 2khz = 2.33khz/ (4×10kΩ ×10kΩ×C1)

=> C1 = 2.33kΩ / (8×10-11)

= 2.9 ×10-9 ≅3nF.


7. Triangular waveform has

A. Rise time < fall time
B. Rise time = fall time
C. Rise time ≥ fall time
D. None of the mentioned

Answer: B

The triangular waveform has the rise time of the triangular wave always equal to its fall time, that is, the same amount of time is required for the triangular wave to swing from -VRamp to +VRamp as from +VRamp to -VRamp.


8. Output of an integrator producing waveforms of unequal rise and fall time is called

A. Triangular waveform
B. Sawtooth waveform
C. Pulsating waveform
D. Spiked waveform

Answer: B

The sawtooth waveform has unequal rise and fall times. It may rise positively many times faster than it falls negatively or vice versa.


9. Depending on the value of input and reference voltage a comparator can be named as

A. Voltage follower
B. Digital to analog converter
C. Schmitt trigger
D. Voltage level detector

Answer: D

A comparator is sometimes called a voltage level detector because, for the desired value of reference voltage, the voltage level of the input can be detected.


10. Why clamp diodes are used in the comparator?

A. To reduce output offset voltage
B. To increase the gain of the op-amp
C. To reduce input offset current
D. To protect op-amp from damage

Answer: D

The diodes protect the op-amp from damage due to excessive input voltage. Because of these diodes, the differential input voltage of the op-amp is clamped to 0.7v or -0.7 v, hence these diodes are clamp diodes.

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