Question

The following circuit works like a

• A. D Flip-flop
• B. T Flip-flop
• C. Frequency divider
• D. Inverter

Hint:

A D flip-flop with feedback \[ D=\overline{Q} \] always behaves as a T flip-flop. The output toggles on every clock pulse and divides the clock frequency by 2.

Answer 1

The Correct Option is B

Concept: The circuit consists of a D flip-flop with feedback from the output \(Q\) to the input \(D\) through an inverter. A D flip-flop follows the relation \[ Q_{n+1}=D. \] Therefore, the next state of the flip-flop depends entirely on the signal applied at the \(D\)-input.

Step 1: Determine the feedback connection.
From the circuit, \[ D=\overline{Q}. \] This is because the output \(Q\) is passed through an inverter before being fed back to the D-input.

Step 2: Write the next-state equation.
Since a D flip-flop satisfies \[ Q_{n+1}=D, \] substituting \[ D=\overline{Q_n}, \] we obtain \[ Q_{n+1}=\overline{Q_n}. \]

Step 3: Analyze the state transition.
If \[ Q_n=0, \] then \[ Q_{n+1}=1. \] If \[ Q_n=1, \] then \[ Q_{n+1}=0. \] Thus the output toggles at every active clock edge.

Step 4: Compare with T flip-flop operation.
The characteristic equation of a T flip-flop is \[ Q_{n+1}=T\oplus Q_n. \] For \[ T=1, \] \[ Q_{n+1}=\overline{Q_n}. \] This is exactly the behavior obtained from the given circuit. Therefore the circuit functions as a \[ \boxed{\text{T Flip-Flop}} \]

Step 5: Additional observation.
Since the output changes state on every clock pulse, the output frequency becomes half of the clock frequency. \[ f_{out}=\frac{f_{clk}}{2}. \] Hence the circuit can also act as a frequency divider, but its fundamental equivalent flip-flop realization is a T flip-flop. \[ \boxed{\text{Correct Option (B)}} \]