D Flip-Flop MCQ Quiz in తెలుగు - Objective Question with Answer for D Flip-Flop - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

The given is drawn as:

This is a standard D flip flop circuit.

Analysis:

Step-1:

The characteristic table of the required flipflop i.e D flipflop is:

Step-2:

The excitation table of the given flip-flop i.e. Jk flipflop is:

Step-3:

The K-map for both J and K individually will be:

Step-4:

Now make the connections accordingly to get:

Note: This is a standard flipflop conversion. of making a D flipflop from JK flipflop.

∴ from the above transition diagram only sequence not possible is 2 → 5 → 6

The correct statement about the behavior of a D Flip-Flop when the clock input is NOT transitioning is: 1) The output remains at the last value.

Explanation:

• Edge-Triggered Behavior: A D flip-flop is an edge-triggered device. This means it samples its input (D) and updates its output (Q) only at a specific transition of the clock signal (either the rising edge or the falling edge, depending on its design).

• Latched State: When the clock input is not transitioning (i.e., it's held high, held low, or in between edges), the flip-flop is in a latched state. In this state, its output is isolated from the D input. Any changes in the D input during this period will not affect the output. The output simply holds the data that was present at the last active clock edge.

When \(A = 1\),              \(Y=Q\)

         \(A=0\)               \(Y = \bar Q\)

In the flip-flop \(S = \bar R\). So efficiency the flip – flop will behave like a D flip – flop. So the correct state transition diagram is–

Characteristic equation of normal JK FF is

\(\begin{array}{l} Q\left( {t + 1} \right) = J\bar Q + \bar KQ\\ \therefore Q\left( {t + 1} \right) = {{\bar J}_1}\bar Q + {{\bar K}_1}Q \end{array}\)

In S, R flip flop

S = 1, R = 1 in valid states

So only 2 is true

From the table shown in the explanation of above question, if first state is 010 next State is 011.

 So Total number of distinct outputs is 4