Home Subjects Electronics (ECE) Analog Circuits

Electronics (ECE)
Analog Circuits

Analog/digital electronics, communication

33 Q 4 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 1–10 of 33
Topics in Electronics (ECE)
What is the bandwidth of a first-order low-pass filter with cutoff frequency fc = 1 kHz?
A 500 Hz
B 1 kHz
C 2 kHz
D 4 kHz
Correct Answer:  B. 1 kHz
EXPLANATION

Bandwidth of a low-pass filter is defined by its cutoff frequency (-3dB point). At f = fc, magnitude drops to 0.707 of DC gain. Bandwidth = fc = 1 kHz.

Test
What is the gain-bandwidth product (GBW) of a typical 741 op-amp?
A 1 MHz
B 10 MHz
C 100 MHz
D 1 GHz
Correct Answer:  A. 1 MHz
EXPLANATION

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.

Test
What is the slew rate of a standard 741 op-amp?
A 0.5 V/μs
B 2 V/μs
C 100 V/μs
D 500 V/μs
Correct Answer:  A. 0.5 V/μs
EXPLANATION

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.

Test
What is the input offset voltage of a typical precision op-amp (like OP07)?
A 100 mV
B 10 mV
C Below 1 mV (typically 0.5-2 mV)
D 1 V
Correct Answer:  C. Below 1 mV (typically 0.5-2 mV)
EXPLANATION

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.

Test
For maximum power transfer from a source with internal resistance Rs to load RL, what should be the condition?
A RL = 0 (short circuit)
B RL = Rs (impedance matching)
C RL >> Rs
D RL = Rs/2
Correct Answer:  B. RL = Rs (impedance matching)
EXPLANATION

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.

Test
What is the typical input impedance of a voltage follower (unity gain buffer) using an ideal op-amp?
A Zero ohms
B Infinite (very high)
C Equal to load impedance
D Typically 1 MΩ
Correct Answer:  B. Infinite (very high)
EXPLANATION

An ideal op-amp has infinite input impedance. The voltage follower configuration maintains this high input impedance at the non-inverting input.

Test
A voltage-controlled voltage source (VCVS) has transconductance gm = 50 mS. If the input voltage is 10 mV, what is the output current?
A 0.5 mA
B 0.05 mA
C 5 mA
D 50 mA
Correct Answer:  A. 0.5 mA
EXPLANATION

Output current Id = gm × Vgs = 50 mS × 10 mV = 50 × 10⁻³ × 10 × 10⁻³ = 0.5 × 10⁻³ A = 0.5 mA

Test
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?
A 70
B 1000
C 250
D 150
Correct Answer:  B. 1000
EXPLANATION

In cascaded amplifiers, overall voltage gain = A₁ × A₂ = 50 × 20 = 1000. The bandwidth is determined by the stage with lowest bandwidth (100 kHz).

Test
A BJT is operated in the saturation region. Which of the following statements is true?
A Base-collector junction is reverse biased
B Both base-emitter and base-collector junctions are forward biased
C Collector-emitter voltage equals breakdown voltage
D Transistor acts as an open circuit
Correct Answer:  B. Both base-emitter and base-collector junctions are forward biased
EXPLANATION

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.

Test
Q.10 Easy Analog Circuits
What is the primary advantage of a Darlington pair configuration in BJT amplifiers?
A Higher input impedance and current gain
B Lower output impedance
C Reduced power dissipation
D Higher frequency response
Correct Answer:  A. Higher input impedance and current gain
EXPLANATION

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.

Test
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