I know the following for two positive edge triggered flip flops in cascade.
Max(combinational logic delay) < Tclk_period + Tskew - Tsetup and
Min(combinational logic delay) > Tskew - Thold
But when I consider positive edge ff followed by negative edge ff how will the equation be modified. Will it be:
Max(combinational logic delay) < Tclk_period + (Tclk_period / 2) + Tskew - Tsetup and
Min(combinational logic delay) > Tskew - Thold
or:
Max(combinational logic delay) < (Tclk_period / 2) + Tskew - Tsetup and
Min(combinational logic delay) > Tskew - Thold
Best Answer
Considering a positive-edge triggered FF followed by a negative-edge triggered FF and 1st FF is triggered at t=0 sec. The triggering edge for 2nd FF will arrive only after \$T_{clk} + T_{skew}\$. But the output of 1st FF takes some time (\$T_{clk\rightarrow Q}\$) to settle to a stable value. Stable value should reach at input to the 2nd FF atleast \$T_{setup}\$ seconds before the triggering edge arrives. ie,
$$\frac{T_{clk}}{2} + T_{skew} > T_{clk\rightarrow Q} + T_{logic,max} + T_{setup} $$ or, $$ T_{logic,max} < \frac{T_{clk}}{2} - T_{clk\rightarrow Q} - T_{setup} + T_{skew}$$
The triggering edge to 2nd FF arrives at \$t = T_{clk} + T_{skew}\$. The next triggering edge for 1st FF will happen at \$t =T_{clk}\$ sec, ie., \$T_{clk}/2-T_{skew}\$ sec after triggering 2nd FF. Hence 2nd FF can have a hold time of \$T_{clk}/2-T_{skew}+ min\ delay\ between\ the\ FF's\$. ie., $$T_{hold} < \frac{T_{clk}}{2} - T_{skew} + T_{clk\rightarrow Q,cd} + T_{logic,min}$$ or, $$ T_{logic,min} > \frac{T_{clk}}{2} + T_{clk\rightarrow Q,cd} + T_{hold} + T_{skew}$$
Where, \$T_{clk\rightarrow Q,cd}\$ is the clock to Q contamination delay defined as time taken to produce the first change in output of the FF after applying clock and \$T_{clk\rightarrow Q}\$ is the time taken to make stable change in the FF output after clock is applied.