u24c02
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Hi.
I didn't quite catch the following posting's answer from other people.
How could I calculate this questions?
Would you please let me know?
View attachment 113894
Given the above design,reference the figure
1.What are the effective setup and hold times between IN and CLK in the above circuit?
A. Tsetup = 4 ns, Thold = 1 ns
B. Tsetup = 3 ns, Thold = 0 ns
C. Tsetup = 3 ns, Thold = 1ns
D. Tsetup = 2 ns, Thold = 0 ns
2.What is the maximum operating frequency of the above circuit?
A. 250 MHz
B. 80 MHz
C. 125 MHz
D. 166.7 MHz
[FONT=Courier New] ______ ______ ______
nCLK | |______| |______| |
_____ _____
D ----|_____|-------|_____|-------
'2 ' 2' '2 ' 2' setup = 2 ns, hold = 2 ns
removing the clock delay of 1 ns moves the clock edge back 1 ns
removing the data back by 2 ns moves data transition back 2 ns
______ ______ ______
CLK |______| |______| |
_____ _____
IN --|_____|-------|_____|-------
'3 '1' '3 '1'
[/FONT]
Data delay improves setup at the expense of hold time.Code:[FONT=Courier New] ______ ______ ______ nCLK | |______| |______| | _____ _____ D ----|_____|-------|_____|------- '2 ' 2' '2 ' 2' setup = 2 ns, hold = 2 ns removing the clock delay of 1 ns moves the clock edge back 1 ns removing the data back by 2 ns moves data transition back 2 ns ______ ______ ______ CLK |______| |______| | _____ _____ IN --|_____|-------|_____|------- '3 '1' '3 '1' [/FONT]
Clock delay reduces setup time and helps hold time.
so...
Tsu = Tffsu + Txor - Tinv = 2 + 2 - 1 = 3
Th = Tffh - Txor + Tinv = 2 - 2 + 1 = 1
See the for much better graphics than I'm willing to draw right now.
I only glanced at the picture (small phone screen) and didn't notice the clk->Q = 4 ns and didn't pay attention to the feed back path from Q->xor->D. So the frequency is much lower, as you have to add the path delays up along with the setup.
Tperiod = Tsu + Tcq + Txor = 2 + 4 + 2 = 8, so 1/8ns = 125 MHz. so it should have been C.
Data delay improves setup at the expense of hold time.Code:[FONT=Courier New] ______ ______ ______ nCLK | |______| |______| | _____ _____ D ----|_____|-------|_____|------- '2 ' 2' '2 ' 2' setup = 2 ns, hold = 2 ns removing the clock delay of 1 ns moves the clock edge back 1 ns removing the data back by 2 ns moves data transition back 2 ns ______ ______ ______ CLK |______| |______| | _____ _____ IN --|_____|-------|_____|------- '3 '1' '3 '1' [/FONT]
Clock delay reduces setup time and helps hold time.
so...
Tsu = Tffsu + Txor - Tinv = 2 + 2 - 1 = 3
Th = Tffh - Txor + Tinv = 2 - 2 + 1 = 1
See the for much better graphics than I'm willing to draw right now.
I only glanced at the picture (small phone screen) and didn't notice the clk->Q = 4 ns and didn't pay attention to the feed back path from Q->xor->D. So the frequency is much lower, as you have to add the path delays up along with the setup.
Tperiod = Tsu + Tcq + Txor = 2 + 4 + 2 = 8, so 1/8ns = 125 MHz. so it should have been C.
Thanks Sir,
Does "negative slack" mean the same thing as setup violation? I'm confusing maximum delay meaning setup delay or minimum delay mean hold delay.
Especially, I am wondering about if Slack = Required time- Arrival time < 0 why this is setup violation?
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