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

A divider circuit can be used to find square roots by connecting both the inputs of the multiplier (in the divider) to the output of an op-amp.

The output of the square root circuit is proportional to the square root of the magnitude of the input voltage.

V_o  = √( V_ref×|V_in|)

Voltage of the given circuit

V_L  = V_o²/ V_ref  = 13²/10 = 16.9v.

∴ I_L  = V_L/R = 16.9/12kΩ = 1.408mA.

The input voltage must be negative or else the op-amp will saturate and lie between the ranges of -1 to -10v.

The circuit in which the output voltage waveform is the integral of the input voltage waveform is called  Integrator.

The integrator circuit produces the output voltage waveform as the integral of the input voltage waveform.

The output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RC_F.

V_o  = (1/R×C_F)×^t∫₀ V_indt+C

Where

C-> Integration constant

C_F-> Feedback capacitor.

A simple low pass RC circuit can work as an integrator when the time constant is very large, which requires the large value of R and C. Due to practical limitations, the R and C cannot be made infinitely large.

In an op-amp integrator, the effective input capacitance becomes C_F×(1-A_v). Where A_v is the gain of op-amp. The gain is infinite for the ideal op-amp. So, the effective time constant of the op-amp integrator becomes very large which results in perfect integration.

In practice, the output of the op-amp never becomes infinite rather the output of the op-amp saturates at a voltage close to the op-amp’s positive or negative power supply depending on the polarity of the input dc signal.

The frequency at which the gain of the integrator becomes zero is f = 1/(2πR₁C_F).

To avoid saturation problems, the feedback capacitor is shunted by a feedback resistance(R_F). The parallel combination of R_F and C_F behaves like a practical capacitor that dissipates power. For this reason, the practical integrator is called a lossy integrator.

The lower frequency limit of integration,

f = 1/(2πR_FC_F)

= 1/(2π×1kΩ×33nF)

= 4.82kHz.

The range of frequency between which the circuit act as an integrator as

[1/(2πR_FC_F)]- [1/(2πR₁C_F)].

Input voltage V_in = 1.2v for 0≤t≤0.4ms. The output voltage at t = 0.4ms is

V_o  = (1/R×C_F)×^t∫₀ V_indt+C = -(1/1) × ^0.4∫₀1.2 dt

= > V_o  = -[^0.1∫₀1.2 dt + ^0.2∫_0.11.2 dt + ^0.3∫_0.21.2 dt + ^0.4∫_0.31.2 dt ]

= -(1.2+1.2+1.2+1.2) = -4.8v

Therefore, the output voltage waveform is a ramp function.

The gain of the lossy integration is

A = (R_F/ R₁)/√[1+(ωR_FC_F)]²

= > A(dB. = 20log{(R_F/R₁)/√[1+(ωR_FC_F)]²}

= > 40db = 20log×[(R_F/R₁)/√1

= > R₁  = R_F/20.

At ω = 20000rad/s, the gain is down by 6db from its peak of 20db and thus is 14db.

The gain at 14db

= > 14db = 20log ×{[ (R_F/ R_F/20)] / [√(1+(200000×0.47×10^-6×R_F)²]}

= > 20log[1+(9.4×10^-3×R_F)²] = 20-14

= > R_F  = √3/9.4×10^-3  = 212.26Ω and

R₁  = 212.26Ω/10 = 21.2Ω.

The input current charging the feedback capacitor produces error voltage at the output of the integrator. Therefore, to reduce error voltages a resistor (R_F) is connected across the feedback capacitor. Hence, R_F minimizes the variation in the output voltage.

The integrator is most commonly used in analog computers mainly for signal waveshaping.

The input signal will be integrated properly if the time period T of the signal is larger than or equal to R_F×C_F (Feedback resistor and capacitor).

If the input frequency is lower than the lower frequency limit of the integrator (i.e. when gain = 0), there will be no integration results, and the circuit act as a simple inverting amplifier.

The mentioned answer gives the exact ranges where the integration starts.