Electronics and Communication questions reward anyone who is comfortable moving between the time domain and the frequency domain. This set covers network theory, analog and digital circuits, signals and systems, control systems, communication systems, and electromagnetics. Numerical solutions keep the units visible at every step, since a dropped factor is the most common reason a correct method still produces a wrong option.
In a Common Source FET amplifier, the gain is primarily dependent on which parameter?
Answer: A
The voltage gain of a CS amplifier is Av = -gm × RD, where gm is transconductance and RD is the drain resistance. This is the fundamental relationship.
Q.82Easy
A BJT amplifier operates in the saturation region. What is the approximate output impedance?
Answer: B
In saturation, the transistor acts almost like a closed switch with minimal voltage drop across it, resulting in very low output impedance.
Q.83Easy
The input impedance of a Common Emitter amplifier is approximately equal to:
Answer: A
The input impedance of a CE amplifier is Zin ≈ β × re, where β is current gain and re is emitter resistance (≈ VT/IE).
Q.84Easy
In a differential amplifier, the Common Mode Rejection Ratio (CMRR) is 80 dB. What is the CMRR in linear form?
A voltage follower (Common Collector) amplifier has a voltage gain of:
Answer: A
The voltage gain of a voltage follower is Av ≈ 1 (or slightly less due to emitter resistance effects). It is used for impedance matching.
Q.86Easy
In a BJT amplifier biased in the active region, what is the relationship between collector current and base current?
Answer: A
In the active region, IC = β × IB where β is the DC current gain (typically 50-300 for silicon BJTs).
Q.87Easy
An op-amp integrator circuit has R = 10 kΩ and C = 100 nF. What is the time constant?
Answer: A
Time constant τ = RC = 10×10³ × 100×10⁻⁹ = 10×10⁻⁴ = 1×10⁻³ s = 1 ms.
Q.88Easy
The gain-bandwidth product (GBW) of a typical general-purpose op-amp (like LM741) is approximately:
Answer: A
LM741 has a GBW product of approximately 1 MHz, which is why it's limited to low-frequency applications. Modern op-amps have higher GBW.
Q.89Easy
The input impedance of an op-amp in an ideal configuration is assumed to be:
Answer: A
Ideal op-amp assumptions include infinite input impedance (no current drawn at inputs), zero output impedance, and infinite open-loop gain.
Q.90Easy
In a Common Emitter BJT amplifier, if the collector resistance is increased while keeping other parameters constant, how does the voltage gain change?
Answer: A
Voltage gain Av = -gm × Rc, where gm is transconductance and Rc is collector resistance. Increasing Rc directly increases voltage gain.
Q.91Easy
A non-inverting amplifier is constructed with an op-amp having open-loop gain A₀ = 100,000. If Rf = 90 kΩ and Rin = 10 kΩ, what is the closed-loop voltage gain?
What is the primary advantage of a Darlington pair configuration in BJT amplifiers?
Answer: A
Darlington pairs consist of two transistors in cascade, providing extremely high current gain (β² product) and very high input impedance due to the two base-emitter junctions.
Q.93Easy
A BJT is operated in the saturation region. Which of the following statements is true?
Answer: B
In saturation, both junctions are forward biased: base-emitter junction is forward biased for current conduction, and base-collector junction is also forward biased, allowing maximum collector current.
Q.94Easy
In a two-stage cascaded amplifier, the first stage has gain A₁ = 50 and bandwidth BW₁ = 100 kHz, while the second stage has A₂ = 20 and BW₂ = 200 kHz. What is the overall voltage gain?
Answer: B
In cascaded amplifiers, overall voltage gain = A₁ × A₂ = 50 × 20 = 1000. The bandwidth is determined by the stage with lowest bandwidth (100 kHz).
Q.95Easy
A voltage-controlled voltage source (VCVS) has transconductance gm = 50 mS. If the input voltage is 10 mV, what is the output current?
Answer: A
Output current Id = gm × Vgs = 50 mS × 10 mV = 50 × 10⁻³ × 10 × 10⁻³ = 0.5 × 10⁻³ A = 0.5 mA
Q.96Easy
What is the typical input impedance of a voltage follower (unity gain buffer) using an ideal op-amp?
Answer: B
An ideal op-amp has infinite input impedance. The voltage follower configuration maintains this high input impedance at the non-inverting input.
Q.97Easy
For maximum power transfer from a source with internal resistance Rs to load RL, what should be the condition?
Answer: B
Maximum power transfer theorem states that maximum power is delivered when load impedance equals the complex conjugate of source impedance. For resistive cases, RL = Rs.
Q.98Easy
What is the input offset voltage of a typical precision op-amp (like OP07)?
Answer: C
Precision op-amps like OP07, OPA2134 are designed with input offset voltages < 1 mV. General purpose op-amps (e.g., 741) have offset voltages of 1-5 mV.
Q.99Easy
What is the slew rate of a standard 741 op-amp?
Answer: A
The 741 op-amp has a slew rate of approximately 0.5 V/μs. This limits the maximum output voltage change rate, causing distortion at high frequencies/amplitudes.
Q.100Easy
What is the gain-bandwidth product (GBW) of a typical 741 op-amp?
Answer: A
The 741 op-amp has a gain-bandwidth product of approximately 1 MHz. This means for a gain of 100 V/V, maximum usable bandwidth is ~10 kHz.
