250+ TOP MCQs on Square Wave Generator & Answers

Linear Integrated Circuit Multiple Choice Questions on “Square Wave Generator”.

1. How are the square wave output generated in op-amp?
A. Op-amp is forced to operate in the positive saturation region
B. Op-amp is forced to operate in the negative saturation region
C. Op-amp is forced to operate between positive and negative saturation region
D. None of the mentioned

Answer: C
Clarification: Square wave outputs are generated where the op-amp is forced to operate in saturated region, that is, the output of the op-amp is forced to swing repetitively between positive saturation, +V_sat and negative saturation, -V_sat.

2. The following circuit represents a square wave generator. Determine its output voltage

A. -13 v
B. +13 v
C. ± 13 v
D. None of the mentioned

Answer: A
Clarification: The differential output voltage V_id = V_in1 – V_in2= 3-7v = -4v.
The output of the op-amp in this circuit depends on polarity of differential voltage V₀= -V_sat ≅ -V_ee = -13 v.

3. Determine the expression for time period of a square wave generator
A. T= 2RC ln×[( R₁+ R₂) / ( R₂)].
B. T= 2RC ln×[( 2R₁+ R₂) / ( R₂)].
C. T= 2RC ln×[( R₁+ 2R₂) / ( R₂)].
D. T= 2RC ln×[( R₁+ R₂) / (2 R₂)].

Answer: B
Clarification: The time period of the output waveform for a square wave generator is T= 2RC ln×[(2R₁+ R₂)/( R₂)].

4. Determine capacitor voltage waveform for the circuit

Answer: C
Clarification: When the op-amp output voltage is at negative saturation, V₁ = [(R₁) / (R₁+ R₂ )] × (-V_sat) = [10kΩ / ( 10 kΩ +11.6 kΩ)] × (-15v) = -7v.
Similarly, when the op-amp’s output voltage is at positive saturation, V₁ = [(R₁) / (R₁+ R₂ )] × (+V_sat) = [10kΩ/ ( 10 kΩ +11.6 kΩ)] × (+15v) = +7v
The time period of the output waveform,T= 2RC ln ×[( 2R₁+ R₂) / ( R₂)] = 2× 10kΩ × 0.05 µF× ln (2×10kΩ + 11.6kΩ) / 11.6kΩ] = 1×10^-3 × ln2.724 = 1ms.
The voltage across the capacitor will be a triangular wave form.

5. What will be the frequency of output waveform of a square wave generator if R₂ = 1.16 R₁?
A. f_o = (1/2RC.
B. f_o = (ln/2RC.
C. f_o = (ln /2 ×√RC.
D. f_o = (ln/√(2 RC.)

Answer: B
Clarification: When R₂= 1.16 R₁, then f_o = 1/2RC× ln[ (2R₁+ R₂) / R₂] = 1/2RC ×ln [(2R₁ + 1.161R₁ )/ (1.161R₁)] = 1/( 2RC×ln2.700)= 1/2RC.

6. What could be the possible output waveform for a free running multivibrator whose op-amp has a supply voltage of ±5v operating at 5khz?

Answer: C
Clarification: In a free running multivibrator, the output is forced to swing repetitively between positive and negative saturation to produce square wave output. Therefore, +V_sat ≅ +V_cc =+5v and -V_sat ≅ -V_cc =-5v.
=> Frequency= 5khz , f =1/t = 0.2ms.

7. Determine the output frequency for the circuit given below
A. 28.77 Hz
B. 31.97 Hz
C. 35.52 Hz
D. 39.47 Hz

Answer: D
Clarification: The output frequency f_o = 1/2RC×ln [ (2R₁+ R₂)/ R₂] = 1 / {(2×33kΩ ×0.33µF)×ln[(2×33kΩ +30kΩ)/30kΩ]} = 1/ (0.02175×ln 32) = 39.47 Hz.

8. The value of series resistance in the square wave generator should be 100kΩ or higher in order to
A. Prevent excessive differential current flow
B. Increase resistivity of the circuit
C. Reduce output offset voltage
D. All of the mentioned

Answer: A
Clarification: In practice, each inverting and non-inverting terminal needs a series resistance to prevent excessive differential current flow because the inputs of the op-amp are subjected to large differential voltages.

9. Why zener diode is used at the output terminal of square wave generator?
A. To reduce both output and capacitor voltage swing
B. To reduce output voltage swing
C. To reduce input voltage swing
D. To reduce capacitor voltage swing

Answer: B
Clarification: A reduced peak-peak output voltage swing can be obtained in the square wave generator by using back to back zener diodes at the output terminal.

10. A square wave oscillator has f_o =1khz. Assume the resistor value to be 10kΩ and find the capacitor value?
A. 3.9 µF
B. 0.3 µF
C. 2 µF
D. 0.05µF

Answer: D
Clarification: Let’s take R₂ = 1.16 R₂, therefore the output frequency f_o = 1/2RC
=> C = 1/2Rf_o = 1/ (2×10kΩ×1khz) = 0.05µF.