Sequential Circuits Flip Flops 1 Flip Flops A

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Sequential Circuits Flip Flops 1

Sequential Circuits Flip Flops 1

Flip Flops § A flip-flop is a one bit memory that is edge-triggered or

Flip Flops § A flip-flop is a one bit memory that is edge-triggered or edge-sensitive memory element which means it active only at transitions; i. e. either from 0 1 or 1 0 § The flip-flop is at a “set state” if it forces the output to logic 1 and it is at a “reset state” if it forces the output to logic 0. level positive (rising) edge negative (falling) edge 2

Set Reset Flip-Flop (SR FF) • Two cross coupled NAND gates. • Two inputs

Set Reset Flip-Flop (SR FF) • Two cross coupled NAND gates. • Two inputs S and R. • C is a control input (enable) to determine when the state of the filp-flop can be changed. S R Q C SR FF Q Q C R S Q C S R Qn+1 0 X X No change 1 0 0 No change 1 0 Reset state 1 1 0 1 Set state 1 1 1 Undefined 3

Now we will study the case when we suppose we have two cases for

Now we will study the case when we suppose we have two cases for the present state and then apply the inputs. SR d ≡ don’t care 10 0 d 1 0 d 0 01 Excitation Table Present state Qn The State diagram S R Next state Qn+1 0 0 0 1 1 0 1 1 1 1 0 x 1 1 1 x No change Reset state Set state Not allowed 4

To get the next state Q n +1 R S Qn 0 1 0

To get the next state Q n +1 R S Qn 0 1 0 0 1 1 x x Then Boolean Expression is Q n +1 = S + Qn 5

D flip-flop • D flip-Flop is designed to eliminate the undefined state in the

D flip-flop • D flip-Flop is designed to eliminate the undefined state in the SR FF When (S=R=1). D C D Qn+1 0 X No change 1 0 0 Reset state 1 1 1 Set state Q C D FF Q D Q C Q 6

To study the case when we consider that we have two cases for the

To study the case when we consider that we have two cases for the present state and then apply the inputs: CD 11 0 d, 10 Excitation Table 0 C D Qn Qn+1 0 x 0 0 0 x 1 1 1 0 0 0 1 0 1 1 1 1 1 0 d, 11 1 10 Hold The State diagram Reset Set 7

D C Qn 1 1 1 Qn+1 = CD + 1 Qn When the

D C Qn 1 1 1 Qn+1 = CD + 1 Qn When the FF is enabled, C = 1 Qn+1 = 1. D + 0. Qn = D 8

JK Flip-Flop • It is similar to SR with J replaced by S and

JK Flip-Flop • It is similar to SR with J replaced by S and K replaced by R except for undefined state. J J K Qn+1 0 0 No change 0 1 0 Reset state 1 0 1 Set state 1 1 Complement K Q C JK FF Q Toggle 9

To study the case when we suppose that we have two cases for the

To study the case when we suppose that we have two cases for the present state and then apply the inputs: JK 10, 11 Excitation Table d 0 0 d J K Qn Qn+1 0 0 0 1 1 0 0 1 1 0 0 0 1 1 1 Set state 1 1 0 Toggle 1 0 01, 11 Hold The State diagram Reset state K J Qn 1 1 Qn+1 = 1 1 Qn + J 10

T Flip-Flop It is equivalent to JK FF with input J= K=T T Excitation

T Flip-Flop It is equivalent to JK FF with input J= K=T T Excitation Table T Qn+1 0 Qn 1 Q C No change T FF Q Complement Since it is equivalent to Jk, we can use the same characteristic equation to find the next state: Qn+1 = Qn + J = Qn + T 11

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