141. If the input to differentiating circuit is a sawtooth wave, then the output will be ______ wave.
Square
Rectangular
Triangular
Sinusoidal
Answer.2. Rectangular
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
charge exponentially at a rate depending on the RC time constant
charge exponentially at a rate depending on the input voltage
charge when the input voltage is decreasing
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
Of the problem of drift with differentiating circuits
Because of the poor efficiency
Because of the complex circuitry requirements
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:
Square wave
Triangular wave
Parabolic wave
Sine wave
Answer.1. Square 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
Answer: C
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
Answer: B
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
Answer: D
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
Answer: D
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
Answer: C
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
A. Adding feedback capacitor
B. Feedback capacitor and an internal resistor
C. Feedback capacitor and feedback resistor
D. Internal capacitor and internal capacitor
Answer: B
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
Answer: 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
Answer: B
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
Answer: B
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