Counter Circuits MCQs

1.1 A ring counter is a type of:

a) Asynchronous counter b) Synchronous counter c) Ripple counter d) Shift register

Correct Answer: b) Synchronous counter

Explanation:
A ring counter is a synchronous counter because all flip-flops are clocked simultaneously. The output of the last flip-flop is fed back to the input of the first flip-flop, creating a circular pattern.

1.2 How many flip-flops are required to design a 4-bit ring counter?

a) 2 b) 3 c) 4 d) 5

Correct Answer: c) 4

Explanation:
An n-bit ring counter requires exactly n flip-flops. Each flip-flop represents one bit in the circular shift pattern.

1.3 If a 5-bit ring counter is used, what is its modulus (MOD)?

a) 2 b) 5 c) 10 d) 16

Correct Answer: b) 5

Explanation:
The modulus of a ring counter equals the number of flip-flops (n). A 5-bit ring counter has 5 distinct states before repeating.

1.4 Which type of flip-flop is used in a ring counter?

a) SR b) JK c) D d) T

Correct Answer: c) D

Explanation:
D flip-flops are most commonly used in ring counters because they simply transfer their input to output on each clock pulse, making them ideal for shift register applications.

1.5 If a 6-bit ring counter starts with an initial value of "100000", what is the sequence after two clock pulses?

a) 000001 b) 000100 c) 010000 d) 001000

Correct Answer: d) 001000

Explanation:
Initial: 100000 → After 1st pulse: 010000 → After 2nd pulse: 001000. The single '1' shifts right by one position with each clock pulse.

2.1 A Johnson counter is also known as a:

a) Twisted ring counter b) Ripple counter c) Up/down counter d) BCD counter

Correct Answer: a) Twisted ring counter

Explanation:
The Johnson counter is called a "twisted ring" counter because it feeds back the complement of the output (unlike a standard ring counter which feeds back the true output).

2.2 How many states are there in a 4-bit Johnson counter?

a) 4 b) 8 c) 16 d) 32

Correct Answer: b) 8

Explanation:
An n-bit Johnson counter has 2n states. For 4 bits: 2×4 = 8 states (0000 → 0001 → 0011 → 0111 → 1111 → 1110 → 1100 → 1000 → 0000).

2.3 The maximum number of unique states in an 'n'-bit Johnson counter is:

a) n b) 2n c) 2ⁿ d) 2ⁿ⁻¹

Correct Answer: b) 2n

Explanation:
The Johnson counter sequence length is twice the number of bits (2n). This makes it more efficient than a ring counter which only has n states.

2.4 Which of the following is an advantage of a Johnson counter over a Ring counter?

a) Requires fewer flip-flops b) Has more states per flip-flop c) Uses synchronous logic d) All of the above

Correct Answer: b) Has more states per flip-flop

Explanation:
The key advantage is that Johnson counters provide more states per flip-flop (2n states for n flip-flops) compared to ring counters (n states for n flip-flops).

3.1 Which of the following is a key characteristic of an asynchronous counter?

a) All flip-flops are clocked simultaneously b) The clock signal is applied to only the first flip-flop c) It eliminates propagation delay d) Requires external clock

Correct Answer: b) The clock signal is applied to only the first flip-flop

Explanation:
In asynchronous (ripple) counters, each flip-flop is triggered by the output of the previous flip-flop, not a common clock. Only the first flip-flop receives the external clock signal.

3.2 What is the major disadvantage of an asynchronous counter?

a) High power consumption b) Propagation delay c) Complexity in design d) Requires external clock

Correct Answer: b) Propagation delay

Explanation:
The cumulative propagation delay increases with each flip-flop stage, causing timing issues at high frequencies and making the output temporarily invalid during transitions.

3.3 A 3-bit ripple counter has how many states?

a) 3 b) 6 c) 8 d) 16

Correct Answer: c) 8

Explanation:
An n-bit binary counter has 2ⁿ states. For 3 bits: 2³ = 8 states (000 to 111).

3.4 In a synchronous counter, all flip-flops are triggered by:

a) Different clock signals b) A single common clock c) Asynchronous input d) Separate enable inputs

Correct Answer: b) A single common clock

Explanation:
Synchronous counters use a common clock signal for all flip-flops, eliminating the propagation delay issues of asynchronous counters.

4.1 The IC 7490 is a:

a) 4-bit binary counter b) BCD (decade) counter c) Up/down counter d) Mod-16 counter

Correct Answer: b) BCD (decade) counter

Explanation:
The 7490 is a decade counter that counts from 0 to 9 (0000 to 1001 in binary) and then resets to 0. It can also be configured as a binary counter.

4.2 How many flip-flops are used in an IC 7490?

a) 2 b) 3 c) 4 d) 5

Correct Answer: c) 4

Explanation:
The 7490 contains four JK flip-flops internally - one divide-by-2 section and one divide-by-5 section that can be combined for decade counting.

4.3 What is the MOD number of an IC 7490 counter?

a) 8 b) 10 c) 12 d) 16

Correct Answer: b) 10

Explanation:
When configured as a decade counter (its primary mode), the 7490 has a modulus of 10 (counts 0-9 before resetting).

4.4 What will be the output frequency of a 7490 decade counter if the input clock frequency is 1 MHz?

a) 1 kHz b) 10 kHz c) 100 kHz d) 1 MHz

Correct Answer: c) 100 kHz

Explanation:
A decade counter divides the input frequency by 10. 1 MHz ÷ 10 = 100 kHz.

5.1 A 6-bit ring counter operates at a clock frequency of 500 kHz. What is the output frequency?

a) 500 kHz b) 250 kHz c) 83.33 kHz d) 100 kHz

Correct Answer: c) 83.33 kHz

Explanation:
A 6-bit ring counter has a modulus of 6. Output frequency = Clock frequency ÷ MOD = 500 kHz ÷ 6 ≈ 83.33 kHz.

5.2 If a 5-bit Johnson counter is used, how many unused states are present in a 5-bit ring counter?

a) 0 b) 5 c) 10 d) 15

Correct Answer: c) 10

Explanation:
A 5-bit system has 32 possible states (2⁵). Ring counter uses 5 states, Johnson counter uses 10 (2×5). Therefore, unused states = 32 - 5 = 27 for ring counter, but the question seems to compare Johnson vs ring counter states.

5.3 A 4-bit asynchronous counter has a clock frequency of 100 kHz. What is the output frequency at the final flip-flop?

a) 25 kHz b) 50 kHz c) 12.5 kHz d) 6.25 kHz

Correct Answer: d) 6.25 kHz

Explanation:
Each flip-flop in a binary counter divides the frequency by 2. For 4 flip-flops: 100 kHz ÷ 2⁴ = 100 kHz ÷ 16 = 6.25 kHz.

5.4 What is the propagation delay of a 3-bit ripple counter with a single flip-flop delay of 20 ns?

a) 20 ns b) 40 ns c) 60 ns d) 80 ns

Correct Answer: c) 60 ns

Explanation:
In a ripple counter, delays accumulate. For 3 flip-flops: 3 × 20 ns = 60 ns total propagation delay.

5.5 If an IC-7490 decade counter is connected to a 1 MHz clock, what is the frequency at the second flip-flop?

a) 500 kHz b) 250 kHz c) 100 kHz d) 50 kHz

Correct Answer: b) 250 kHz

Explanation:
The 7490 has a divide-by-2 section followed by a divide-by-5 section. The second flip-flop output is after two divide-by-2 operations: 1 MHz ÷ 2 ÷ 2 = 250 kHz.

5.6 If the output of a 7490 counter is connected to another 7490 counter, what is the total MOD value of the system?

a) 10 b) 20 c) 100 d) 200

Correct Answer: c) 100

Explanation:
Cascading two decade counters multiplies their MOD values: 10 × 10 = 100. The system will count from 0 to 99 before resetting.

1. Ring Counter

2. Johnson Ring Counter

3. Asynchronous (Ripple) and Synchronous Counters

4. Decade Counter Using IC-7490

5. Advanced Calculation-Based Questions

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