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Question

Read the following statements :

(a) In a J-K Flip Flop, if J = K, the resulting Flip Flop is referred to as a T-type Flip Flop
(b) In a J-K Flip Flop, if J ≠ K, the resulting Flip Flop is referred to as a D-type Flip Flop
(c) An S-R Flip Flop cannot be converted into a T-type Flip Flop since S = R = 1 is not allowed.

Which is correct ?

This question was previously asked in
UGC NET 2015 Paper 3 History Question Paper (28-Jun-2015)
The correct answer is

(a) and (b)

The J-K flip-flop is the universal one: tie its inputs together in the two possible ways and the other two types fall out.

(a) J tied to K gives a T flip-flop — correct. With a single input T driving both,

\(Q_{n+1}=T\overline{Q_{n}}+\overline{T}Q_{n}=T\oplus Q_{n}\)

so T = 0 holds the state and T = 1 toggles it — exactly the T flip-flop, the building block of every counter.

(b) J and K made complementary gives a D flip-flop — correct. Setting \(J=D\) and \(K=\overline{D}\) in the characteristic equation,

\(Q_{n+1}=D\overline{Q_{n}}+\overline{\overline{D}}Q_{n}=D\overline{Q_{n}}+DQ_{n}=D\)

The output simply follows the input at each clock edge — the D or "data" flip-flop, made from a J-K with a single inverter. The statement's "J ≠ K" is loose phrasing for this complementary connection, which is the only way two binary inputs can differ.

So (a) and (b) hold, giving option 1.

(c) is the questionable statement. It is true that S = R = 1 is forbidden in an S-R flip-flop, and true that this is why a bare S-R cannot toggle. But "cannot be converted" is too strong: the conversion is standard and uses the outputs as feedback,

\(S=T\overline{Q}\qquad R=TQ\)

Because Q and \(\overline{Q}\) are always complementary, S and R can never both be 1, and with T = 1 exactly one of them is active — precisely the one that flips the state. That is in fact how the J-K flip-flop itself is built. Since a conversion does exist, (c) is best read as false, and the answer is flagged for confirmation because a key that reads (c) as a statement about the bare S-R would mark it true.

ConversionConnectionResult
J-K → TJ = K = T\(Q_{n+1}=T\oplus Q_{n}\)
J-K → DJ = D, K = D̄\(Q_{n+1}=D\)
S-R → TS = TQ̄, R = TQToggles, invalid state impossible

The general method for any conversion is the same: write the excitation table of the available flip-flop, the required next-state behaviour of the target, and use a K-map to express the available inputs in terms of the target's input and the present state.

Hence, the correct statements are (a) and (b).

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Similar Questions

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    C  DQn+1
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    $\uparrow$00
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  3. Assertion (A) : Asynchronous sequential circuit is also called event driven circuit.

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  6. Under which input condition, the J-K Flip-Flop toggles ?

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Important Questions from Sequential Circuits

  1. A basic memory storage element in a digital system is:

  2. The basic sequential logic building block in which the output follows the data input as long as the ENABLE input is active, is

  3. What can be the maximum clock frequency of a 10-bit ripple counter which will not cause a count to skip, considering 10 ns propagation delay for each of the edge-triggered flip flops?

  4. The output of a sequential circuit depends on

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