Op Amp As a Amplifier MCQ Quiz in తెలుగు - Objective Question with Answer for Op Amp As a Amplifier - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

In ideal op-amp, due to the concept of virtual ground, both the inverting and non-inverting terminals of the op-amp should be at the same voltage i.e. both the input terminals are at the same potential.

• One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
• The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.
• An Op-Amp inverting input (-) is at zero potential (A virtual ground), even though it does not have a galvanic connection to the ground. This is because of feedback due to Rf and the high gain of the Op-Amp.
• The virtual ground is only valid if the Op-Amp gain is much greater than the circuit programmed gain (Rf / R1) in the given figure.

CMRR (Common mode rejection ratio):

It is defined as the ratio of differential-mode voltage gain (Ad) and the common-mode voltage gain (Ac).

Mathematically, this is expressed as:

\(CMRR = \frac{A_d}{A_c}\)

Ad = Differential gain

Ac = Common mode gain

Common-mode gain is the gain when the same output is applied at both input terminals of the op-amp

Ac= (V₁ - V₂) = 0

∴ A_C = 0

∴ CMRR = ∞ (ideally) or very high practically.

Hence, Statement (I) is true but Statement (II) is false 

Consider the CE RC coupled amplifier circuit as shown:

C₁, C₃: coupling capacitors and load capacitors

C₂: bypass capacitor

C_μ, C_π: junction capacitors

Approximate frequency analysis of the amplifier is shown as:

The consideration of the capacitors and their behavior is expressed in the below table.

Note:

1) For mid-frequency itself coupling capacitors are short-circuited, then that should be the perfect short circuit for high frequency.

2) For mid-frequency itself parasitic capacitors are open circuit, then that should be per feet open circuit for low frequency.

Capacitor analysis:

If ‘C₁’ is considered then C₂ & C₃ are short-circuited (C_P1 & C_P2 open circuit)

Low-frequency analysis:

Lower cutoff frequency due to C₁ alone, i.e. C₂, C₃ → short circuit

\({f_L}\;due\;to\;{C_1} = \frac{1}{{2\pi [{R_S} + \left( {{R_B}{\rm{||}}{r_\pi }{\rm{)}}} \right]{C_1}}}\)

Lower cutoff frequency due to ‘C₂’ alone, i.e.C₃, C₁ → short circuit

\({f_L}\;due\;to\;{C_2} = \frac{1}{{2\pi \left[ {\frac{1}{{9m}} + \frac{{({R_B}||{R_S})}}{{\beta + 1}}} \right]{C_2}}}\)

Similarly,

\({f_L}\;\left( {due\;to\;{C_3}} \right) = \frac{1}{{2\pi [\left( {{R_C}{\rm{||}}{r_o}{\rm{)}} + {R_L}} \right]{C_3}}}\)

Conclusion:

The low-frequency analysis is determined by coupling and bypass capacitors

∴ Statement (1) is false.

High-frequency analysis:

The simplified model of the high-frequency model is as shown:

Miller effect:

Calculation of gain

\(A = \frac{{{V_o}}}{{{V_\pi }}} = - {g_m}\left[ {{{\rm{r}}_{\rm{o}}}{\rm{\;}}\left| {\left| {{{\rm{R}}_{\rm{C}}}{\rm{\;}}} \right|} \right|{{\rm{R}}_{\rm{L}}}} \right]\)

Miller's capacitor is dominant.

Conclusion:

1) High-frequency analysis is determined by the junction capacitors.

Statement (II) is false.

2) Gain at the high frequency is reduced due to the Miller effect.

Statement (III) is true.

Gain × bandwidth = Constant

If the gain is increased then bandwidth reduces.

Statement (IV) is true.

Concept:

Precision rectifier

This is another type of rectifier that converts AC to DC but we use op-amp in it.

This type of configuration is obtained with an op-amp in order to behave the circuit like an ideal diode and rectifier.

Calculation:

Considering the given circuit we can determine the characteristics based on the operation of the diode in different biasing conditions.

Redraw the given circuit as shown with the labeling of currents and voltages.

Case (i)

i₁ + i₂ + i₃ = i

\(\frac{{20}}{{4R}} + \frac{{{V_i}}}{R} + \frac{{{V_0}}}{R} = i\)

If V₀ is positive, the diode does not conduct so i = 0

\(\frac{{20}}{{4R}} + \frac{{{V_i}}}{R} + \frac{{{V_0}}}{R} = 0\)

5 + V_i + V₀ = 0

V₀ = - 5 - V_i

At V_i = - 10 V

V₀ = - 5 + 10 = 5 V

At V_i = - 5 V

V₀ = - 5 + 5 = 0 V

Case (ii)

When V_i > - 5V, D1 is OFF and D2 is ON are conducting so V₀ = 0V

Option 2 is correct.

Effect of Negative Feedback:

In the given figure, the output of an op-amp is fed back to its inverting input. This configuration is known as negative feedback.

As V_in increases, V_out will increase in accordance with the differential gain.

However, as V_out increases, that output voltage is fed back to the inverting input, thereby acting to decrease the voltage differential between inputs, which acts to bring the differential input voltage equal to zero.

The input impedance (Z_in) of an OP-AMP is infinite.

Hence, the current flowing into the input terminal of the ideal op-amp is zero.

Both Statement (l) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I).

Hence, option 2 is the correct answer.

Concept:

Most modern audio amplifiers have a flat frequency response as shown above over the whole audio range of frequencies from 20 Hz to 20 kHz. This range of frequencies, for an audio amplifier is called its Bandwidth, (BW) and is primarily determined by the frequency response of the circuit.

Frequency points ƒ_L (f_c1) and ƒ_H (f_c2) relate to the lower corner or cut-off frequency and the upper corner or cut-off frequency points respectively were the circuits gain falls off at high and low frequencies. These points on a frequency response curve are known commonly as the -3dB (decibel) points. So the bandwidth is simply given as:

Bandwidth, (BW) = f_H - f_L

We see from the curve given above that at the two corner or cut-off frequency points, the output drops from 0 dB to -3 dB and continues to fall at a fixed rate. This fall or reduction in gain is known commonly as the roll-off region of the frequency response curve.

These -3 dB corner frequency points define the frequency at which the output gain is reduced to 70.71% of its maximum value. Then we can correctly say that the -3 dB point is also the frequency at which the systems gain has reduced to 0.707 of its maximum value.

- 3 dB = 20 log₁₀ (0.7071)

The -3dB point is also know as the half-power points since the output power at this corner frequencies will be half that of its maximum 0 dB value as shown.

∴ Output power, P = I² R = V² / R

At f_L or f_H,

 V or I = 0.707 of maximum value

Therefore the amount of output power delivered to the load is effectively “halved” at the cut-off frequency and as such the bandwidth (BW) of the frequency response curve can also be defined as the range of frequencies between these two half-power points.

Calculation:

Given-

f_L = 20 Hz, f_H = 20 kHz, 

R = 12 Ω, P = 20 W   

\(P = {V^2 \over R}\)

V² = 20 x 12

V = 15.49 volts

Now rms load voltage at 20 Hz (f_L) can be calculated as

V_rms = 0.707 x V

V_rms = 10.95 volt

Due to virtual short circuit,

V_A = V_B = 0        ---(1)

Apply Nodal Analysis;

\(\frac{{{V_B} - 3}}{{60 }} + \frac{{{V_B} - 2}}{{40}} + \frac{{{V_B} - 1}}{{20 }} + \frac{{{V_B} - V_{O}}}{{60}} = 0\)

\(\frac{{{0} - 3}}{{60 }} + \frac{{{0} - 2}}{{40}} + \frac{{{0} - 1}}{{20 }} + \frac{{{0} - V_{O}}}{{60}} = 0\)

V₀ = -9 V

At low frequencies,  i.e. at f = 0 → capacitor = open-circuited hence output is zero . So it is not a low pass filter.

 At higher frequencies, i.e. at f = ∞ capacitor is short-circuited and output is having Some value. So it is a high pass filter

\(\begin{array}{l} \frac{{0 - {V_{in}}}}{{{R_1} + \frac{1}{{{S_C}}}}} + \frac{{0 - {V_o}}}{{{R_2}}} = 0\\ \frac{{{V_o}}}{{{V_{in}}}} = {R_2}\left[ {\frac{1}{{{R_1} + \frac{1}{{SC}}}}} \right] = \frac{{SC{R_2}}}{{1 + SC{R_1}}} = \frac{{KS}}{{1 + S\tau }} \end{array}\)

∴ τ = R₁C

Cutoff frequency:

\({\omega _0} = \frac{1}{\tau } = \frac{1}{{{R_1}C}}rad/sec\)

The correct answer is Coupling capacitors
Key Points

• Coupling capacitors block the DC component of a signal while allowing the AC component to pass through.
• This prevents the DC bias from one stage of the amplifier from affecting the bias of subsequent stages.
• Signal Isolation: By blocking DC, coupling capacitors isolate different stages of the amplifier from each other,
• ensuring that each stage operates independently in terms of DC biasing.
• Preventing Signal Distortion: DC biasing can introduce distortion in the signal path, particularly in amplifiers with multiple stages.
• Coupling capacitors prevent this distortion by allowing only the AC signal to be amplified."

Additional Information 

• Junction capacitors: Junction capacitors are typically used in semiconductor devices at the junctions between different layers of material.
  They are not commonly used in amplifier circuits for AC analysis.
• Diffusion capacitors: Diffusion capacitors are also specific to semiconductor devices and are typically used in the fabrication process rather
  than being used as components within amplifier circuits.
• Stray capacitors: Stray capacitors refer to unintended or parasitic capacitances that exist in a circuit due to layout, wiring, or the inherent characteristics of components. While stray capacitance can affect circuit behavior, it is not a specific component intentionally considered during AC analysis of amplifier circuits.

Concept:

The voltage gain of an amplifier is defined as the ratio of the output voltage to the input voltage. It is usually denoted by A_v, i.e.

\(A_v=\frac{{{V_{out}}}}{{{V_{in}}}}\)

Calculation:

Given V_out = 1V

V_in = 15 mV

Gain (A_v) will be:

\(A_v=\frac{{{V_{out}}}}{{{V_{in}}}}=\frac{1}{15m}\)

\(A_v=\frac{1000}{{15}}=66.67\)

• In an inverting amplifier configuration, the input signal is applied to the inverting (-) terminal of the operational amplifier, while the non-inverting (+) terminal is typically grounded.
• This configuration causes the output to be 180 degrees out of phase with the input, resulting in phase inversion.
• Therefore, a positive input voltage results in a negative output voltage and vice versa.

Concept:

Inverting amplifier:

In the inverting type of amplifier, the output is precisely 180 degrees out of phase to input. When the +Ve voltage is applied to the circuit, then the output voltage of the circuit will be –Ve.

Once this amplifier is assumed as an ideal, then we have to apply the virtual short concept at the input terminals of the op-amp. So the voltage at the two terminals is equivalent.

On applying KVL

We get  gain  \(\left( {\frac{{{V_{out}}}}{{{V_{in}}}}} \right) = - \frac{{{R_f}}}{{{R_{in}}}}\;\)  < 0.

Additional Information

Non-inverting-Amplifier:

In Non- inverting type of amplifier, the output is exactly in phase to input. When a +Ve voltage is applied to the circuit, then the output voltage  will be positive (i.e. o/p is non-inverted in terms of phase).

Here also as the feedback is negative, the virtual shot is possible

On KVL we get  gain,  \(\left( {\frac{{{V_{out}}}}{{{V_{in}}}}} \right) = 1 + \frac{{{R_2}}}{{{R_1}}}\;\)   ≥ 1