# J K Flip Flop

Discussion in 'Homework Help' started by hitmen, Mar 26, 2009.

1. ### hitmen Thread Starter Active Member

Sep 21, 2008
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My tutorial has this question but there are 2 things I dont understand when I was givven the answer.

1) To calculate fmax, why is hold time not included?

2) Since there are 2 flip flops, why isnt Tmin doubled?

3) Why is the ouput of the second flip flop the inverse of the first?
I thought they should be the same? Eg, when Q1 = 1 J2 =1, this set the flip flop and Q2 should be one. Where have I gone wrong?

I know how to draw Q1 but I am unsure of Q2.

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Last edited: Mar 26, 2009
2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Hold Time is part of the propagation time. In the example, 5ns Hold Time and 45ns other switching makes up the propagation to a stable output.

Setup time is how long the data needs to be valid on J or K before the rising edge of the clock.

Hold Time is how long that same data needs to be present after start of the clock.

Propagation time is how long it takes between the start of the clock and a valid output on Q. Since this time starts when Setup time ends, at the rising edge of the clock, Hold time is the first "slice" of this parameter.

3. ### hitmen Thread Starter Active Member

Sep 21, 2008
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0
Okay I understand but can anyone explain why the output of Q2 is the inverse of Q1. I thought they should be the same since J2 represents set and K2 represents reset. I believe that output Q2 should be the same as J2???

Also, why isnt the t min doubled as this is asynchronous with 2 flip flop??

Last edited: Mar 26, 2009
4. ### hitmen Thread Starter Active Member

Sep 21, 2008
159
0
Answer from tutor: hold time is part of propagation time
Same clock input to 2 flip flop so no need multiply by 2
Q2 output depend on PRIOR output of Q1.

My tutor is more helpful! Yippie.