Question

The n stage Johnson counter will produce a modulus of

• A. n
• B. 2n
• C. \(2^n\)
• D. \(2^{n-1}\)

Hint:

Ring Counter (n stages): Modulus n.
Johnson Counter (n stages): Modulus 2n. Uses complemented feedback.
Binary Ripple/Synchronous Counter (n stages): Modulus \(2^n\).

Answer 1

The Correct Option is B

A Johnson counter (also known as a twisted-ring counter or switch-tail ring counter) is a type of shift register counter where the complement of the output of the last flip-flop is fed back to the input of the first flip-flop. If an n-stage Johnson counter uses 'n' flip-flops:
It cycles through \(2n\) unique states.
Therefore, the modulus of an n-stage Johnson counter is \(2n\). For example: A 2-stage Johnson counter (n=2) has \(2 \times 2 = 4\) states (e.g., 00 \(\rightarrow\) 10 \(\rightarrow\) 11 \(\rightarrow\) 01 \(\rightarrow\) 00...). A 3-stage Johnson counter (n=3) has \(2 \times 3 = 6\) states (e.g., 000 \(\rightarrow\) 100 \(\rightarrow\) 110 \(\rightarrow\) 111 \(\rightarrow\) 011 \(\rightarrow\) 001 \(\rightarrow\) 000...). Compare this to:
A simple ring counter (output of last FF to input of first FF) has 'n' states.
A standard binary counter with 'n' flip-flops has \(2^n\) states. So, for an n-stage Johnson counter, the modulus is \(2n\). \[ \boxed{\text{2n}} \]