Home Subjects Electronics (ECE) Digital Electronics

Electronics (ECE)
Digital Electronics

Analog/digital electronics, communication

50 Q 4 Topics Take Mock Test
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Topics in Electronics (ECE)
What is the main advantage of a Carry Lookahead Adder (CLA) over a Ripple Carry Adder (RCA)?
A Lower cost
B Reduced propagation delay and faster operation
C Simpler design
D Lower power consumption
Correct Answer:  B. Reduced propagation delay and faster operation
EXPLANATION

CLA reduces propagation delay from O(n) to O(log n) by calculating carry signals in parallel, making it much faster than RCA.

Test
A 4-bit Ripple Carry Adder has a total propagation delay of 12 ns. If each full adder stage has a propagation delay of 3 ns, what is the delay for carry propagation?
A 3 ns
B 9 ns
C 12 ns
D 15 ns
Correct Answer:  B. 9 ns
EXPLANATION

In a 4-bit ripple carry adder, the carry must propagate through 3 stages (from FA1 to FA4), giving delay = 3 × 3 ns = 9 ns.

Test
A 32×8 ROM has how many address lines and data lines respectively?
A 5 address lines, 8 data lines
B 8 address lines, 5 data lines
C 10 address lines, 8 data lines
D 32 address lines, 8 data lines
Correct Answer:  A. 5 address lines, 8 data lines
EXPLANATION

For 32 memory locations, we need log2(32) = 5 address lines. The data width is 8 bits, hence 8 data lines.

Test
In Gray code, how many bits change when transitioning from 5 (0101) to 6 (0110)?
A 1 bit
B 2 bits
C 3 bits
D 4 bits
Correct Answer:  A. 1 bit
EXPLANATION

Gray code for 5 is 0111 and for 6 is 0101. Only 1 bit changes between consecutive numbers, which is the main advantage of Gray code.

Test
In a 3-to-8 decoder, if the enable pin is active low and all select lines are 0, what will be the output?
A All outputs are 1
B All outputs are 0
C Output 0 is 0, all others are 1
D Output 0 is 1, all others are 0
Correct Answer:  C. Output 0 is 0, all others are 1
EXPLANATION

In a 3-to-8 decoder with active low enable and select lines = 000, only output 0 is activated (becomes 0), while all others remain 1.

Test
What is the setup time of a flip-flop?
A Time required for the output to stabilize after clock edge
B Minimum time the input data must be stable before the active clock edge
C Time for which data must remain stable after the clock edge
D Time required to reset the flip-flop
Correct Answer:  B. Minimum time the input data must be stable before the active clock edge
EXPLANATION

Setup time (t_su) is the minimum duration that the input signal must be stable and valid before the active clock edge for reliable operation.

Test
Which logic gate is used in a full adder to generate the carry output?
A AND gate only
B OR gate only
C Combination of AND and OR gates
D XOR gate only
Correct Answer:  C. Combination of AND and OR gates
EXPLANATION

The carry output of a full adder is: C_out = (A·B) + (B·C_in) + (A·C_in), which requires both AND and OR gates.

Test
In a Johnson counter with 4 flip-flops, what is the maximum count sequence length?
A 4
B 8
C 15
D 16
Correct Answer:  B. 8
EXPLANATION

A Johnson counter (twisted ring counter) has a sequence length of 2n, where n is the number of flip-flops. For 4 flip-flops: 2×4 = 8 states.

Test
In a synchronous counter with n flip-flops, what is the maximum frequency at which the counter can operate?
A f_max = 1/(n × t_pd)
B f_max = 1/t_pd, where t_pd is the propagation delay of one flip-flop
C f_max = n/t_pd
D f_max = 1/(2 × t_pd)
Correct Answer:  B. f_max = 1/t_pd, where t_pd is the propagation delay of one flip-flop
EXPLANATION

In synchronous counters, all flip-flops are clocked simultaneously, so the maximum frequency is limited only by the propagation delay of a single flip-flop, not the cumulative delay.

Test
In a state machine with 7 states, how many flip-flops are needed for the state register?
A 2 flip-flops
B 3 flip-flops
C 7 flip-flops
D 4 flip-flops
Correct Answer:  B. 3 flip-flops
EXPLANATION

For N states, minimum flip-flops required = ⌈log₂(N)⌉. For 7 states: ⌈log₂(7)⌉ = ⌈2.807⌉ = 3 flip-flops (can represent 2³=8 states)

Test
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