CONT(MUX) PD(MUX) 11.4 : Clock enabled edge-triggered flip-flop

Question Circuit Analysis Comb. Logic ANDD FF Q CLK There are 2 options – t su is positive (as we’ve learned in class) and t su is negative (meaning the flip flops starts looking at its D port AFTER the clock rise). Each solution consists of a maximum of 2 options, which are drawn on the slide. “Logic” means the stability of the comb. gate.

Question Circuit Analysis CLK case #1 – negative setup time t i-1 titi CiCi Logic Q D Not stableStable

Question Circuit Analysis CLK case #1 – negative setup time t i-1 titi CiCi Logic Q D Not stableStable

Question Circuit Analysis CLK case #2 – positive setup time t i-1 titi CiCi Logic Q D Not stableStable CiCi

Question Circuit Analysis CLK case #2 – positive setup time t i-1 titi CiCi Logic Q D Not stableStable CiCi

The Marvelous Toy

Toy Design Identifying system states Identifying state transitions and deciding on Moore or Mealy model Detailing the state machine transition and output functions The combinational circuits The Canonic circuit Clock rate calculation

Toy System States Only the three switching elements keep state. Each has a binary state: Left or Right We can model the state of every switch by a single bit. Convention: 0=Left, 1=Right The total number of states: 2 3 = 8

State Diagram 000

State Diagram /0 1/0 X is Left Z is Left Y is Left Enter from Left Out from Left Swap X Enter from Right Out from Left Swap Y & Z

State Diagram /0 1/ /0 1/0

State Diagram /0 1/ /0 1/ /0 1/1 Enter Right Out Right Swap Y&Z

State Diagram /0 1/ /0 1/ /0 1/1 0/11/1

State Diagram /0 1/ /0 1/ /0 1/1 0/11/1 0/1 1/1

State Diagram /0 1/ /0 1/ /0 1/1 0/11/1 0/1 1/1 0/0 1/1

State Diagram /0 1/ /0 1/ /0 1/1 0/11/1 0/1 1/1 0/0 1/1 0/0 1/1

State Diagram /0 1/ /0 1/ /0 1/1 0/11/1 0/1 1/1 0/0 1/1 0/0 1/1 0/0 1/1

Output Function Y=0 I=0 Y=0 I=1 Y=1 I=1 Y=1 I=0 X=0 Z= X=0 Z= X=1 Z= X=1 Z=0 0010

Output Function Output = YI + XZ + ZI (This is λ) This circuit has 3 AND(2) in parallel, and then an OR(3) No NOT gates. Delay = D(AND)+2*D(OR) –Assuming we use OR(2) only

The Next State Function of X /0 1/ /0 1/ /0 1/1 0/11/1 0/1 1/1 0/0 1/1 0/0 1/1 0/0 1/1

Next State Function for X Y=0 I=0 Y=0 I=1 Y=1 I=1 Y=1 I=0 X=0 Z= X=0 Z= X=1 Z= X=1 Z=0 0110

X Next State Function X = X’I’+XI (This is part of δ) This circuit has: –2 negations in parallel –2 AND(2) in parallel, –and then an OR(2) Delay = D(NOT)+D(AND)+D(OR) Similar to this we find functions to Y,Z

The Canonic Circuit State Register Next State Circuit δ Output Circuit λ Input {0,1} Next State {0,1} 3 State {0,1} 3 Output {0,1}

Stripping away the Flip-Flops Next State Circuit δ Output Circuit λ Input {0,1} Next State {0,1} 3 State {0,1} 3 Output {0,1} D-portQ-port

Attaching Delay Next State Circuit pd(δ) Output Circuit pd(λ) Input {0,1} Next State {0,1} 3 State {0,1} 3 Output {0,1} D-portQ-port t pd pd(IN) setup(OUT) t su

Finding the Clock Rate Next State Circuit pd(δ) Output Circuit pd(λ) Input {0,1} Next State {0,1} 3 State {0,1} 3 Output {0,1} D-portQ-port t pd pd(IN) setup(OUT) t su

The Clock Rate We are done!

Question 4: Synchronous Circuit 1) answer B 2+3) the circuit is drawn for n=4 in the next slide. Extension to n>4 is obvious. Description of a “משוון“ is needed in answer and not drawn… 4) After running the algorithm for calculating asymptotical clock period, we find that it is max{O(logn),O(logk)} (משוון = O(logk), wt(n)=O(logn), adder=O(log(log( )+1)) ) n2n2

MUX FF(K) MUX FF(K) MUX FF(K) MUX FF(K) 01 x0x0 reset x3x3 1 x2x2 1 x1x1 משוון K x3x3 x2x2 x1x1 x0x0 משקל מחרוזת בינארית (wt(4)) pad zeroes Adder FF MUX 0 reset במוצא אנחנו מורידים את הביט הLSB בשביל לחלק ב2!