Analog Electronics MCQ Quiz - Objective Question with Answer for Analog Electronics - Download Free PDF

An operational amplifier is an integrated circuit that can amplify weak electric signals. An operational amplifier has two input pins and one output pin. Its basic role is to amplify and output the voltage difference between the two input pins.

OP-AMP Integrator:

The expression for the output voltage of OP-AMP integrator as:

\({V_0} = - \frac{1}{{{R_1}{C_f}}}\mathop \smallint \limits_0^t {V_{in}}dt\)

Where,

Vin = Input voltage

Vo  = Output voltage

Cf  = Feedback capacitor

R1 = Input resistance

VA, VB  = Virtual ground potentials

I = Input current

 Additional Information

OP-AMP differentiator:

The expression for the output voltage of OP-AMP differentiator as:

\({V_0} = - {R_f}C\frac{{d{V_{in}}}}{{dt}}\)

Vin = Input voltage

Vo = Output voltage

Rf  = Feedback resistance

C = Input capacitance

VA, VB = Virtual ground points

Important Points

By changing positions of Resistance and Capacitor OP-AMP Integrator converts to Differentiator and vice-versa.

The correct answer is: 2) A circuit that generates a non-sinusoidal wave without any input is called a linear oscillator.

Explanation:

1. Linear Oscillators (Option 1 - TRUE)

   • Generate sinusoidal waveforms (e.g., sine waves).

   • Examples: LC oscillators (Hartley, Colpitts), RC oscillators (Wien Bridge, Phase Shift).

2. Non-Linear Oscillators (Option 2 - FALSE)

   • Generate non-sinusoidal waveforms (e.g., square, triangle, sawtooth waves).

   • Not called linear oscillators—they are called relaxation oscillators or multivibrators.

3. Frequency Determination (Option 3 - TRUE)

   • Oscillator frequency depends on RC (resistor-capacitor) or LC (inductor-capacitor) networks.

4. Multivibrators (Option 4 - TRUE)

   • Used to generate square waves, pulses, or other non-sinusoidal signals.

   • Examples: Astable, Monostable, and Bistable multivibrators.

Given:

• Input voltage: V(t) = sin²(t) + cos²(t) V
• Load resistor: 1 kΩ
• Ideal diode (no voltage drop when conducting)

Key Analysis:

1. Input Voltage Simplification:

   Using the trigonometric identity:

    sin2(t) + cos2(t)  = 1 

   Therefore, V(t) = 1 V (constant DC voltage at all times)

2. Diode Behavior:

   Since the input is always positive (1V):

   • The diode remains continuously forward-biased
   • Acts as a short circuit (0V drop)
3. Output Voltage:

   The entire input appears across the resistor:

   \( V_{out}(t) = V(t) = 1 \text{ V} \)

Average Voltage Calculation:

For a constant DC voltage:

\(V_{avg} = \frac{1}{T}\int_0^T 1\ dt = 1 \text{ V} \)

Final Answer:

4) +1 V

Concept:

The Colpitts oscillator is a type of LC oscillator used to generate high-frequency sinusoidal oscillations, especially in RF applications.

It works based on the principle of LC parallel resonance and uses a combination of capacitors and inductors to determine its frequency of oscillation.

Explanation:

The formula provided in statement 3 is incorrect in the way it's expressed.

The correct frequency of oscillation is given by:

\( \omega = \frac{1}{\sqrt{L \cdot C_{eq}}} \), where \( C_{eq} = \frac{C_1 C_2}{C_1 + C_2} \)

This is the equivalent capacitance of two capacitors in series.

Conclusion:

Statement 3 is incorrect due to the incorrect formula for the oscillation frequency in a Colpitts oscillator.

The Correct Answer is:  4) The transformer Utilisation Factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Explanation:
Transformer Utilisation Factor (TUF):
TUF indicates how efficiently the transformer is used in a rectifier circuit.

Bridge Rectifier TUF ≈ 0.812

Centre-Tapped Full-Wave Rectifier TUF ≈ 0.693

Therefore, a bridge rectifier uses the transformer more efficiently, making option 4 correct.

Additional Information 
1) The PIV of both rectifiers is the same
False — In a bridge rectifier, each diode withstands only V_m (peak voltage),
while in a centre-tapped rectifier, each diode must withstand 2V_m ⇒ PIV is higher.

2) The transformer utilisation factor is the same for both
 False — TUF is better in a bridge rectifier, not the same.

3) A bridge rectifier has double the PIV compared to the centre-tapped rectifier.
False — It's the centre-tapped rectifier that has higher PIV, not the bridge.

Final Answer:
4) The transformer utilisation factor (TUF) is better for a bridge rectifier than for a centre-tapped rectifier.

Concept:

The efficiency of a rectifier is defined as the ratio of dc output power to input power.

The efficiency of a half-wave rectifier will be:

\(\eta = \frac{{{P_{dc}}}}{{{P_{ac}}}}\)

\(\eta= \frac{{\frac{{V_{dc}^2}}{{{R_L}}}}}{{\frac{{V_{rms}^2}}{{{R_L}}}}} \)

V_DC = DC or average output voltage

R_L = Load Resistance

For a half-wave rectifier, the output DC voltage or the average voltage is given by:

\(V_{DC}=\frac{V_m}{\pi}\)

Also, the RMS voltage for a half-wave rectifier is given by:

\(V_{rms}=\frac{V_m}{2}\)

Calculation:

The efficiency for a half-wave rectifier will be:

\(\eta= \frac{{{{\left( {\frac{{{V_m}}}{\pi }} \right)}^2}}}{{{{\left( {\frac{{{V_m}}}{{2 }}} \right)}^2}}} = 40.6\;\% \)

For Half wave rectifier maximum efficiency = 40.6%

Note: For Full wave rectifier maximum efficiency = 81.2%

Concept:

For a transistor, the base current, the emitter current, and the collector current are related as:

IE = IB + IC

where IC = β IB

β = Current gain of the transistor

Typical base-emitter voltages, VBE for both NPN and PNP transistors are as follows:

• If the transistor is made up of a silicon material, the base-emitter voltage VBE will be 0.7 V.
• If the transistor is made up of a germanium material, the base-emitter voltage VBE will be 0.3 V.
   

Application:

From the given figure, Apply KVL

10 - I_B × R_B - V_BE = 0

Let us assume V_BE = 0.7 V

10 - I_B (1 × 10⁶) - 0.7 = 0

I_B = 9.3 μA

We know that,

I_C = β I_B

Where,

I_C  & I_B = collector current and base current

Therefore,

I_C = 100 × 9.3 μA

= 930 μA

= 0.93 mA

≈ 1 mA

• A diode is an electronic device allowing current to move through it only in one direction.
• Current flow is permitted when the diode is forwaforward-biased
• Current flow is prohibited when the diode is reversed-biased.
• The direction of the arrow represents the direction of conventional current flow when the diode is forward biased

• In the figure given above, the symbol represents the circuit symbol of a semiconductor junction diode.
• The ‘P’ side of the diode is always positive terminal and is designated as anode for forward bias.
• Another side that is negative is designated as cathode and is the ‘N’ side of diode.

Concept: 

The working of the Zener diode is explained in the below figures.

Calculation:

Given,

Zener voltage Vz = 10 V

E_in = 6 V ⇒ E_in < V_z

Hence zener will be reverse biased and get open-circuited.

Output voltage E₀ = 0 V

Varactor diode:

• It is represented by a symbol of diode terminated in the variable capacitor as shown below:

• Varactor diode refers to the variable Capacitor diode, which means the capacitance of the diode varies linearly with the applied voltage when it is reversed biased.
• The junction capacitance across a reverse bias pn junction is given by

​           \(C=\frac{A\epsilon}{W}\)

• As the reverse bias voltage increases, the depletion region width increases resulting in a decrease in the junction capacitance.
•  Varactor diodes are used in electronic tuning systems to eliminate the need for moving parts
• Varactor [also called voltacap, varicap, voltage-variable capacitor diode, variable reactance diode, or tuning diode] diodes are the semiconductor, voltage-dependent, variable capacitors
• Varactors are used as voltage-controlled capacitors and it operated in a reverse-biased state

• A limiter circuit is also known as a clipper circuit.
• A clipper is a device that removes either the positive half (top half) or negative half (bottom half), or both positive and negative halves of the input AC signal.
• The clipping (removal) of the input AC signal is done in such a way that the remaining part of the input AC signal will not be distorted
• In the below circuit diagram, the positive half cycles are removed by using the series positive clipper.

​Note: A Clamper circuit can be defined as the circuit that consists of a diode, a resistor, and a capacitor that shifts the waveform to the desired DC level without changing the actual appearance of the applied signal.

Early Effect:

• A large collector base reverse bias is the reason behind the early effect manifested by BJTs.
• As reverse biasing of the collector to base junction increases, the depletion region penetrates more into the base, as the base is lightly doped.
• This reduces the effective base width and hence the concentration gradient in the base increases.
• This reduction in the effective base width causes less recombination of carriers in the base region which results in an increase in collector current. This is known as the Early effect.
• The decrease in base width causes ß to increase and hence collector current increases with collector voltage rather than staying constant.
• The slope introduced by the Early effect is almost linear with IC and the common-emitter characteristics extrapolate to an intersection with the voltage axis VA, called the Early voltage.

This is explained with the help of the following VCE (Reverse voltage) vs IC (Collector current) curve:

Concept:

\(\beta = \frac{\alpha }{{1 - \alpha }}\)

Where β = common-emitter current gain

α = Common base current gain

Calculation:

Common base current gain = α = 0.98

\(\beta = \frac{{0.98}}{{1 - 0.98}} = 49\)

Note: \(\alpha = \frac{{{I_C}}}{{{I_E}}}\) & \(\beta = \frac{{{I_C}}}{{{I_B}}}\)

Where I_C = Collector current

I_E = Emitter current

BJT Amplifier:

• Transistors biasing is done to keep stable DC operating conditions needed for its functioning as an amplifier.
• A properly biased transistor must have it's Q-point (DC operating parameters like IC and VCE) at the center of saturation mode and cut-off mode i.e. active mode.
• In the active mode of transistor operation, the emitter-base junction is forward biased and the collector-base junction is reverse biased.​
• In the cut-off mode of transistor operation, the emitter-base junction is reverse biased and the collector-base junction is reverse biased.​

​

Different modes of BJT operations are: