Operational Amplifier Application MCQ with Explanation- Objective Question Answer for Operational Amplifier Application Quiz

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.

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 X_C  = ∞. So, the capacitor blocks DC voltage or behaves like an open circuit.
• At high frequencies, the reactance of the capacitor X_C  = 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.

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.

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.

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.

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

V_O  = -R_F×C₁×[dV_in/dt].

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

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.

The gain limiting frequency

f_b = 1/(2π×R₁×C₁).

f_b = 1/(2π×10kΩ×0.47µF)

= 1/(2.9516×10^-2) = 33.89Hz.

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.

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

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

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

Given f = 5kHz

T = 1/f = 1/5kHz

= 2×10^-4 –> Equ(1)

R_F×C₁  = 0.4µF×1.6kΩ

= 7.52×10^-4 –> Equ(2)

Hence Equ(1) ≥ Equ(2).

The transfer function for the circuit,

V_o(s) /V₁(s)

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

In a practical differentiator, R_FC_F  = R₁C₁

= > V_o(s) /V₁(s)

= -SR_F×C₁/(1+SR_F×C_F)² or V_o(s) /V₁’(s) = –

S×R_F×C₁/(1+R₁×C₁)².