CS 140 Lecture 8 Professor CK Cheng 10/22/02. Part II. Sequential Network 1.Flip-flops SR, D, T, JK, State Table Characteristic Eq. Q(t+1) = f(x(t), Q(t)).

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CS 140 Lecture 8 Professor CK Cheng 10/22/02

Part II. Sequential Network 1.Flip-flops SR, D, T, JK, State Table Characteristic Eq. Q(t+1) = f(x(t), Q(t)). 2. Spec 3. Implementation Ch. 8 (Ch. 4 in Mano)

I. Flip-flop Components S R 1. SR F-F (Set-Reset) Inputs: S, R State: (Q, Y)

Id Q y S R y* Q* y** Q** y*** Q*** Q y State Transition SR SR State Diagram 01

When SR=01, (Q,y) = (0,1) When SR=10, (Q,y) = (1,0) When SR=11, (Q,y) = (0,0) SR = 00 => if (Q,y) = (0,0) or (1,1), the output keeps changing Solutions: 1) SR = (0,0). 2) SR = (1,1) PS inputs State table Q(t+1) SR Characteristic Eq Q(t+1) = S(t)+R’(t)Q(t) NS (next state)

S R y Q Q = (R+y)’ y = (S+Q)’ SR F-F Q Q’ CLK S R SR F-F is the cheapest, but inputs cannot be 11

D Flip-Flop (Delay) D CLK Q Q’ Id D Q(t) Q(t+1) Characteristic Eq Q(t+1) = D(t) PS D 0 1 State table NS= Q(t+1)

JK F-F J CLK Q Q’ Characteristic Eq Q(t+1) = Q(t)K’(t)+Q’(t)J(t) PS JK State table Q(t+1) K

T CLK Q Q’ Characteristic Eq Q(t+1) = Q’(t)T(t) + Q(t)T’(t) PS T 0 1 State table Q(t+1) T Flip-Flop (Toggle)

Using a JK F-F to implement a D and T F-F JKJK Q Q’ D CLK D flip flop JKJK Q Q’ T CLK T flip flop

Latch and Flip-flops (two latch) Latch can be considered as a door CLK = 0, door is closedCLK = 1, door is open A flip-flop is a two door entrance CLK = 1CLK = 0CLK = 1