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

2. 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)

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

4. Id Q y S R y* Q* y** Q** y*** Q*** 0 0 0 0 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 1 0 2 0 0 1 0 0 1 0 1 0 1 3 0 0 1 1 0 0 0 0 0 0 4 0 1 0 0 1 0 1 0 1 0 5 0 1 0 1 1 0 1 0 1 0 6 0 1 1 0 0 0 0 1 0 1 7 0 1 1 1 0 0 0 0 0 0 8 1 0 0 0 0 1 0 1 0 1 9 1 0 0 1 0 0 1 0 1 0 10 1 0 1 0 0 1 0 1 0 1 11 1 0 1 1 0 0 0 0 0 0 12 1 1 0 0 0 0 1 1 0 0 13 1 1 0 1 0 0 1 0 1 0 14 1 1 1 0 0 0 0 1 0 1 15 1 1 1 1 0 0 0 0 0 0 Q y State Transition SR 10 00 11 00 10 SR 11 10 01 11 01 11 01 1000 10 00 01 00 11 State Diagram 01

5. 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). 0 0 0 1 - 1 1 0 1 - PS inputs 00 01 10 11 State table Q(t+1) SR Characteristic Eq Q(t+1) = S(t)+R’(t)Q(t) NS (next state)

6. 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

7. D Flip-Flop (Delay) D CLK Q Q’ Id D Q(t) Q(t+1) 0 0 0 1 1 0 2 1 0 1 3 1 1 - Characteristic Eq Q(t+1) = D(t) 0 0 1 1 0 1 PS D 0 1 State table NS= Q(t+1)

8. JK F-F J CLK Q Q’ Characteristic Eq Q(t+1) = Q(t)K’(t)+Q’(t)J(t) 0 0 0 1 1 1 1 0 1 0 PS JK 00 01 10 11 State table Q(t+1) K

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

10. 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

11. 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