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A Comparison Between Ring and Johnson Counters Indicates That

By BYJU'S Exam Prep

Updated on: September 25th, 2023

A comparison between Ring and Johnson counters indicates that

  1. A Ring counter has less number of flip-flops but requires more decoding circuitry
  2. A Ring counter has an inverted feedback path
  3. Johnson counter has more flip-flops but less decoding circuitry
  4. Johnson counter has an inverted feedback path

Answer – D. A Johnson counter has an inverted feedback path

A comparison between Ring and Johnson counters indicates that a Johnson counter has an inverted feedback path.

Solution:

The detailed solution to the question A comparison between Ring and Johnson counters indicates that is provided here.

A shift register is said to be a counter if it repeats a particular sequence of states. We have two types of counters based on shift registers. Those are the Ring counter and the Johnson counter. In these two counters, we will use D Flip-Flops.

An n-bit Ring counter consists of ‘n’ D Flip-Flops. The normal output of the last stage D Flip-Flop is connected as the input to the first stage D Flip-Flop in an n-bit Serial-In-Parallel-Out Shift Register. Since this feedback path is forming a closed loop or ring structure and accordingly the name Ring has been given to this counter.

An n-bit Johnson counter consists of ‘n’ D Flip-Flops. The complemented output of the last stage D Flip-Flop is connected as the input to the first stage D Flip-Flop in an n-bit Serial-In-Parallel-Out Shift Register. Since we connected the complemented output of the last stage as the serial input, this counter is known as the Twisted Ring Counter. This circuit is also known as the Johnson counter.

The comparisons between Ring Counter and Johnson Counter are mentioned in the below tabular form.

Parameter

n-bit Ring Counter

n-bit Johnson Counter

Number of D Flip-Flops required

n

n

Number of States Produced

n

2n

Mod Number

n

2n

Number of Unused States

2n-n

2n-2n

Feedback path

Normal

Inverted

Output Frequency

fclk/n

fclk/2n

Therefore, for the given choices, the correct answer will be Option (D).

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