1 A latch is a pair of cross-coupled inverters –They can be NAND or NOR gates as shown –Consider their behavior (each step is one gate delay in time) –From R and S to P and Q stable condition is reached in three gate delays Delays in Latches R S Q P R S Q P

2 G acts as a control signal in a gated latch –G = 0 means no writing, G = 1 allows writing In an edge-triggered flip-flop, data is written in flip-flop upon arrival of a clock edge This is achieved by connecting two gated latches as shown Actual implementation may be slightly different, concept is same Delays in Gated-Latch and Flip-Flops G D Q P G DQ P C D G DQ P R S Q P D G R S Q P D G R S Q P D G C D

3 When C is low, second gated latch does not record anything, but the first latch samples the D-input Before the clock changes from low to high –Changes in D propagate through many gates to output. Therefore D should be stable for at least five gate delays –This represents the set up time of a flip flop When clock changes from low to high –The first latch may still sample up to two gate delay time. Therefore, D should remain stable even after clock changes –This is called the hold time of a flip flop Timing Issues in D Flip-flops R S Q P D G R S Q P D G C D

4 The blocks of a Moore machine are shown below Next state logic determines the next state based on current state and next input Output logic determines the output based on current state From rising edge, stored state is stable after four gate delays Next state logic may take 2 or more gate delay worth of time –This is called contamination time, the minimum time taken for output to change after input changes And then input to latches must remain stable for set up time Timing in a State machine Next State Logic INPUTSINPUTS Output Logic OUTPUTSOUTPUTS

5 Consider a four states system State transition table, implementation level state transition table, output table, and implementation level output tables are Using the logic equations below, combination logic takes two gate delay Example S0S1S2S3 L2 = XY+XY = XX := XY+XY L1 = XY+XY = YY := XY+XY = Y L0 = XY+XY+XY+XY = X+X = 1

6 A combination circuit has –Contamination delay (t cd ) -- Minimum delay before output starts to change 2 to 3 gate delays (depends on # of levels) –Propagation delay (t pd ) -- Maximum delay after which all outputs are stable once input change A flip-flop has set up time -- about 5 gate delay A flip-flop has hold time -- minimum time for which input to latch should not change after a clock edge -- about 2 gate delay A flip-flops propagation delay is time from clock edge to time at which its output is stable. Clock low time > set up time Clock high time > FF Prop Clock time > FF Prop + compute + set up time Recap All Delays FF Hold Time FF Propagation Time FF Set up Time Compute Time

7 Computing contamination delay (t cd ) and propagation delay (t pd ) of a combinational circuit –It depends on various paths in the circuit –We need to find the shortest path for contamination and longest path for propagation For the circuit given below –Contamination delay = …. –Propagation delay = ….. Clock cycle time must allow for propagation delay of circuit, set up time of FF, and propagation time of FF Contamination time of combinational circuit must be lower than hold time of FF Clock Cycles T cd = 2 T pd = 5 T cd = 5 T pd = 10 T cd = 2 T pd = 5 T cd = 2 T pd = 5 T cd = 2 T pd = 5 T cd = 2 T pd = 5 T cd = 2 T pd = 8

8 Clock cycles are determined based on timing considerations Circuit runs at clock speed of f Corresponding clock cycle time (period) is T = 1/f Or f = 1/T A frequency of 1MHz gives a clock cycle time of 1 micro second A frequency of 500MHz gives a clock period of 2 nano second Let –T be the clock period –t pd be the propagation time of combinational circuit –t cd be the contamination time of combinational circuit –t rd be the propagation time of FF (register) circuit –t st be the set up time of FF (register) circuit –t ht be the hold time of FF (register) circuit Speed of the circuit and clock frequency

9 What is the minimum clock period? –T = t pd + t rd + t st By how long must any change in external inputs precede the next clock edge? –t pd + t st How long after the clock edge must the external inputs be held valid? –t ht - t cd what is the smallest time after the clock edge that external outputs can be expected to be valid? –t rd + t pd Some Relationships in state machine times