# Linear Integrated Circuit MCQ [Free PDF] – Objective Question Answer for Linear Integrated Circuit Quiz

141.  If the input to differentiating circuit is a sawtooth wave, then the output will be ______ wave.

1. Square
2. Rectangular
3. Triangular
4. Sinusoidal

The basic operational amplifier differentiator circuit produces an output signal which is the first derivative of the input signal.

If the input to differentiating circuit is a sawtooth wave, then the output will be a Rectangular wave.

142. In an RC differentiator, the capacitor

1. charge exponentially at a rate depending on the RC time constant
2. charge exponentially at a rate depending on the input voltage
3. charge when the input voltage is decreasing
4. charge to approximately on time constant

Answer.1. charge exponentially at a rate depending on the RC time constant

• The input voltage is given to the capacitor and the output is taken at the resistor.
• At low frequencies, the reactance of the capacitor XC = ∞. So, the capacitor blocks DC voltage or behaves like an open circuit.
• At high frequencies, the reactance of the capacitor XC = 0. So, the capacitor allows the varying signals or behaves like a short circuit.

In an RC differentiator, the capacitor charge exponentially at a rate depending on the RC time constant.

143. The operational amplifiers are seldom used for differentiation because

1. Of the problem of drift with differentiating circuits
2. Because of the poor efficiency
3. Because of the complex circuitry requirements
4. The noises are amplified, and this can be significant in the output

Answer.4. The noises are amplified, and this can be significant in the output

The ideal differentiator circuit is as shown below.

The relationship between input and output is obtained by observing that

$${i_s}\left( t \right) = {C_s}\frac{{d{v_s}\left( t \right)}}{{dt}}$$

And $${i_F}\left( t \right) = \frac{{{v_{out}}\left( t \right)}}{{{R_F}}}$$

So that the output of the differentiator circuit is proportional to the derivative of the input:

$${v_{out}}\left( t \right) = – {R_F}{C_s}\frac{{d{v_s}\left( t \right)}}{{dt}}$$

Although mathematically attractive, the differentiation property of this op-amp circuit is seldom used in practice, because differentiation tends to simplify any noise that may be present in a signal.

144. If OP-amp is ideal and Vi is a triangular wave then V0 will be:

1. Square wave
2. Triangular wave
3. Parabolic wave
4. Sine wave

The input signal to the differentiator is applied to the capacitor. The capacitor blocks any DC content so there is no current flow to the amplifier summing point, resulting in zero output voltage.

If we give a triangular wave as input (Vi) to this circuit then it will give a square wave as output after differentiating the input.

145. Differentiation amplifier produces

A. Output waveform as the integration of input waveform
B. Input waveform as the integration of output waveform
C. Output waveform as derivative of the input waveform
D. Input waveform as derivative of the output waveform

A differentiation amplifier or differentiator is a circuit that performs the mathematical operation of differentiation and produces the output waveform as a derivative of the input waveform.

146. Determine the output voltage of the differentiator?

A. VO = RF×C1×[dVin/dt].
B. VO = -RF×C1×[dVin/dt].
C. VO = RF×CF×[dVin/dt].
D. None of the mentioned

The output voltage is equal to the RF×C1 times the negative instantaneous rate of change of the input voltage Vin with time.

147. which factor makes the differentiator circuit unstable?

A. Output impedance
B. Input voltage
C. Noise
D. Gain

The gain of the differentiator circuit (RF / XC1) increases with an increase in frequency at a rate of 20dB/decade. This makes the circuit unstable.

148. The increase in the input frequency of the differentiation amplifier to input impedance creates

A. Component noise
B. External noise
C. Low-frequency noise
D. High-frequency noise

The input impedance of the amplifier decreases with an increase in frequency and makes the circuit susceptible to high-frequency noise such that noise can completely override differential output.

149. Calculate the gain limiting frequency for the circuit

A. 15.64Hz
B. 23.356Hz
C. 33.89Hz
D. None of the mentioned

The gain limiting frequency

fb= 1/(2π×R1×C1).

fb= 1/(2π×10kΩ×0.47µF)

= 1/(2.9516×10-2) = 33.89Hz.

149. The stability and high-frequency noise problem are corrected by

B. Feedback capacitor and an internal resistor
C. Feedback capacitor and feedback resistor
D. Internal capacitor and internal capacitor

The stability and high-frequency noise problem are corrected by the addition of two components to the differentiator:

The internal resistor is in series with the internal capacitor and the Feedback capacitor shunts with the feedback resistor.

150. Select the order in which the frequency should be maintained to enhance the stability of the differentiator? Where fa -> Frequency at which gain =0 ; FB -> Gain limit frequency ; fc -> Unity gain bandwidth.

A. fab c
B. fa > fb > fc
C. fbc > fa
D. fbc a

The value of the internal resistor and capacitor and feedback resistor and capacitor of the differentiator values should be selected such that fab c makes the circuit more stable.

151. Which application uses a differentiator circuit?

A. None of the mentioned
B. FM modulators
C. Wave generators
D. Frequency Shift keying

The differentiators are used in FM modulator as a rate of change detector.

152. Choose the value of RF and C for a 5kHz input signal to obtain good differentiation.

A. RF = 1.6×103Ω, C1 = 33×10-6F
B. RF = 1.6×103Ω, C1 = 0.47×10-6F
C. RF = 1.6×103Ω, C1 = 47×10-6F
D. RF = 1.6×103 Ω, C1 = 10×10-6F

For a good differentiation, the time period of the input signal must be larger than or equal to RF C1 i.e. T ≥ RF×C1

Given f=5kHz

T=1/f = 1/5kHz

= 2×10-4 –> Equ(1)

RF×C1 = 0.4µF×1.6kΩ

=7.52×10-4 –> Equ(2)

Hence Equ(1) ≥ Equ(2).

153. Determine the transfer function for the practical differentiator

A. Vo(s) /V1(s) = -S×RF×C1/(1+R1×C1)2
B. Vo(s) /V1(s) = -S×RF×C1/(1+RF×C1)2
C. Vo(s) /V1(s) = -S×RF×C1/(1+R1×CF)2
D. None of the mentioned

The transfer function for the circuit,

Vo(s) /V1(s)

=-S×RF×C1/{[1+(RF CF)]×[1+(S×R1×C1)]}.

In a practical differentiator, RFCF = R1C1

=> Vo(s) /V1(s)

= -SRF×C1/(1+SRF×CF)2 or Vo(s) /V1’(s) = –

S×RF×C1/(1+R1×C1)2.

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