For a photodiode amplifier using trans-impedance configuration, increasing the feedback resistance Rf by 10 times will affect the noise figure and gain as:

The correct answer is B: Gain increases 10×, noise figure worsens (increases). Increasing Rf increases gain proportionally (V_out = I_in × Rf). However, thermal noise of Rf (4kTRf/Δf) increases, degrading noise figure.

For a linear time-invariant system with Laplace transform H(s) = 1/(s+2), determine the response to input x(t) = e^(-2t)×u(t):

The correct answer is A: y(t) = t×e^(-2t)×u(t). X(s) = 1/(s+2). Y(s) = H(s)×X(s) = 1/[(s+2)²]. This is a repeated pole, inverse Laplace gives y(t) = t×e^(-2t)×u(t). This is a resonance condition in the system.

What is the primary advantage of a Darlington pair configuration in BJT amplifiers?

The correct answer is A: Higher input impedance and current gain. 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.

In a BJT common-emitter amplifier with emitter bypass capacitor, what is the primary effect on voltage gain?

The correct answer is A: Increases due to reduced emitter degeneration. Bypass capacitor shorts RE to ground at signal frequencies, eliminating degeneration feedback and increasing Av = gm·RC approximately.

The gain-bandwidth product (GBP) of a BJT is approximately equal to:

The correct answer is C: fT (transit frequency) where hFE·fβ = constant. GBW ≈ fT; BJT current gain drops with frequency as β·f ≈ constant, where fT is extrapolated cutoff frequency (~1GHz for silicon BJTs).

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Electronics (ECE)

Analog/digital electronics, communication

135 Q 4 Topics Take Mock Test

Difficulty: All Easy Medium Hard 71–80 of 135

Topics in Electronics (ECE)

All Electronic Devices 100 Digital Electronics 100 Analog Circuits 100 Signals & Systems 100

Q.71 Easy Digital Electronics

In a JK flip-flop, what happens when both J and K inputs are 1?

A Output remains unchanged

B Output toggles on clock pulse

C Output becomes 0

D Output becomes 1

Correct Answer: B. Output toggles on clock pulse

EXPLANATION

When J=1 and K=1, the JK flip-flop enters toggle mode, changing its state with each clock pulse.

Test

Q.72 Easy Digital Electronics

In a 8×4 RAM configuration, what do 8 and 4 respectively represent?

A 8 data lines and 4 address lines

B 8 address locations and 4 bits per location

C 8 bits and 4 bytes of storage

D 8 control signals and 4 chip selects

Correct Answer: B. 8 address locations and 4 bits per location

EXPLANATION

In RAM notation m×n, m represents number of addressable locations and n represents bits per location.

Test

Q.73 Easy Digital Electronics

What is the output of a full adder when A=1, B=1, and Cin=1?

A Sum=1, Cout=0

B Sum=0, Cout=1

C Sum=1, Cout=1

D Sum=0, Cout=0

Correct Answer: C. Sum=1, Cout=1

EXPLANATION

Full adder: A+B+Cin = 1+1+1 = 3 (binary 11). Sum = 1, Carry out = 1.

Test

Q.74 Easy Digital Electronics

What is the truth table output for a D flip-flop when CLK = 1 and D = 1?

A Q = 0, Q' = 1

B Q = 1, Q' = 0

C Q = D, Q' = D'

D Q and Q' remain unchanged

Correct Answer: B. Q = 1, Q' = 0

EXPLANATION

In a D flip-flop, Q takes the value of D on the clock edge. When D = 1, Q becomes 1 and Q' becomes 0.

Test

Q.75 Easy Digital Electronics

Which semiconductor memory loses its data when power is switched off?

A EEPROM

B Flash memory

C DRAM

D ROM

Correct Answer: C. DRAM

EXPLANATION

DRAM (Dynamic RAM) is volatile and requires periodic refreshing. Data is lost without power, unlike non-volatile memories like ROM, EEPROM, and Flash.

Test

Q.76 Easy Digital Electronics

What is the primary purpose of using parity bits in error detection?

A To increase data transmission speed

B To detect single bit errors in data

C To encrypt sensitive information

D To reduce memory requirements

Correct Answer: B. To detect single bit errors in data

EXPLANATION

Parity bits are used for error detection by tracking whether the number of 1s in data is even or odd.

Test

Q.77 Easy Digital Electronics

In a Multiplexer (MUX), how many select lines are required for an 8:1 MUX?

A 2 lines

B 3 lines

C 4 lines

D 8 lines

Correct Answer: B. 3 lines

EXPLANATION

For 2^n inputs, n select lines are needed. For 8 inputs, log₂(8) = 3 select lines.

Test

Q.78 Easy Digital Electronics

What is the Boolean expression for a 2-input NAND gate?

A Y = (A+B)'

B Y = (A.B)'

C Y = A'.B'

D Y = A+B

Correct Answer: B. Y = (A.B)'

EXPLANATION

NAND gate is the complement of AND, so Y = (A.B)' where . represents AND operation.

Test

Q.79 Easy Digital Electronics

Which of the following logic gates has an output of 1 only when both inputs are different?

A XOR gate

B XNOR gate

C AND gate

D NOR gate

Correct Answer: A. XOR gate

EXPLANATION

XOR (Exclusive OR) produces 1 when inputs are different and 0 when they are same.

Test

Q.80 Easy Digital Electronics

In a 3-to-8 decoder, if the binary input is 101, which output line will be activated?

A Line 5

B Line 6

C Line 3

D Line 7

Correct Answer: A. Line 5

EXPLANATION

Binary 101 = 5 in decimal, so output line 5 is activated in a decoder.

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