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In the sequential circuit shown below, if the initial value of the output $Q_1Q_0$ is $00$. What are the next four values of $Q_1Q_0$?

 

  1. $11$, $10$, $01$, $00$
  2. $10$, $11$, $01$, $00$
  3. $10$, $00$, $01$, $11$
  4. $11$, $10$, $00$, $01$
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10 Answers

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4 votes

$Q_{0Next} = T_{0}\oplus Q_{0} = 1\oplus Q_{0} = Q_{0}'\:;\:$ For every clock

$Q_{1Next} = T_{1} \oplus Q_{1} = 1 \oplus Q_{1} = Q_{1}'\:;\: $ When $Q_{0}:0 \:to\: 1$ (Because clock is positive edge triggered)

$Q_{1}$ $Q_{0}$ $Q_{1Next}$ $Q_{0Next}$
$0$ $0$ $1$ $1$
$0$ $1$ $0$ $0$
$1$ $0$ $0$ $1$
$1$ $1$ $1$ $0$

State diagram:

So, the correct answer is $(A)$.

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2 votes
Despite all the excellent answers explaining the in depth working of the circuit, i will like to mention a trick, that is time saving too (specific to this question only).

Here, we can see that T Flip Flop is used and T flip flop will complement the current state when presented with input one, here the Flip Flop giving the output Q0 is connected to the clock, and is given constant input 1. So, Q0 will alternate with every clock pulse. Only option A satisfies this condition.
1 votes
1 votes

From the timing diagram it is clear that this counter is Mod 3 ripple down counter.

So, counting goes as 00,11,10,01,00 (Option A)

Note : Transitions only occur at the positive edge of the clock as the Flip Flops are edge triggered.

Answer:

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