Analog electronics circuits miscellaneous
- Which of the following pairs is/are correctly matched?
Waveform Circuit and Input Signal 1. Triangular wave Integrating circuit and square wave 2. Impulsive wave Differentiating circuit and step signal. 3. Sawtooth wave Differentiating circuit and triangular wave.
Select the correct answer using the codes given below:
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(i) Square wave passes through a integrating circuit → Triangular wave.
(ii) Step input passes through a differentiating circuit → Impulse wave.
(iii) Triangular wave passes through a differentiating circuit → Square wave
Since (i) and (ii) is only correct.
Hence alternative (B) is the correct choice.Correct Option: B
(i) Square wave passes through a integrating circuit → Triangular wave.
(ii) Step input passes through a differentiating circuit → Impulse wave.
(iii) Triangular wave passes through a differentiating circuit → Square wave
Since (i) and (ii) is only correct.
Hence alternative (B) is the correct choice.
- When a square wave pulse fed to the input of the circuit shown in the given figure then VC, the voltage across C will the waveform:
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From given fig. we conclude that output coming from the C.E. amplifier is connected with low pass filter which acts as a integrator, due to C.E. amplifier the phase of the triangular wave become reverse.
Hence alternative (D) is the correct choice.
Correct Option: D
From given fig. we conclude that output coming from the C.E. amplifier is connected with low pass filter which acts as a integrator, due to C.E. amplifier the phase of the triangular wave become reverse.
Hence alternative (D) is the correct choice.
- Which one of the following is the output of the circuit shown in the following figure for the given input?
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For input voltage greater than 3 V the diode will conduct and
V0 = 0 V
and for νin < 3 V diode will not conduct
and V0 = (νin – 3) V
Hence alternative (D) is the correct choice
Correct Option: D
For input voltage greater than 3 V the diode will conduct and
V0 = 0 V
and for νin < 3 V diode will not conduct
and V0 = (νin – 3) V
Hence alternative (D) is the correct choice
- The output waveform for the circuit given below:
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During positive cycle when Vi < 8 V, both diode are off V0 → V0 = Vi
for Vi > 8 V → V0 = 8 V i.e., D1 is on and D2 is off.
During negative cycle when Vi < 6 V, both diode are off, → V0 = Vi
for Vi > 6 V, D2 is on → V0 = – 6 V.
Hence alternative (C) is the correct choice.
Correct Option: C
During positive cycle when Vi < 8 V, both diode are off V0 → V0 = Vi
for Vi > 8 V → V0 = 8 V i.e., D1 is on and D2 is off.
During negative cycle when Vi < 6 V, both diode are off, → V0 = Vi
for Vi > 6 V, D2 is on → V0 = – 6 V.
Hence alternative (C) is the correct choice.
- What will be the output for the circuit given below:
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for νi < 4 V the diode will conduct gives ν0 = 4 V
for νi > 4V diode will not conduct gives ν0 = νi
Hence alternative (D) is the correct choice.
Correct Option: D
for νi < 4 V the diode will conduct gives ν0 = 4 V
for νi > 4V diode will not conduct gives ν0 = νi
Hence alternative (D) is the correct choice.