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JK flip flop function table

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Parth86

Member
please help me to understand function table of JK flip flop positive edge

J K C Q
--------------
0 0 0 ?
0 1 ↑ ?
1 0 1 ?
1 1 ↓ ?

↑ clock is raising
↓ clock is falling
0 clock is low
1 clock is high

what will be the value of Q
 

crutschow

Well-Known Member
Most Helpful Member
For a JK flip-flop:
Q goes to 1 when J is 1 after the appropriate clock edge and stay there.
Q goes to 0 when K is 1 after the appropriate clock edge and stay there.
Q alternates its state when both J and K are 1 after the appropriate clock edge.

You need to know whether the FF changes state on the rising edge or the falling edge of the clock. Knowing that, you can then complete the table from what is stated above.
 

Flyback

Well-Known Member
JK flip flop with J and K tied together is like a D type ff
If one of the inputs is tied to one of the outputs, i forget which, it behaves like a toggle flip flop.
Wiki is good on jk ff's
 

crutschow

Well-Known Member
Most Helpful Member
JK flip flop with J and K tied together is like a D type ff
If one of the inputs is tied to one of the outputs, i forget which, it behaves like a toggle flip flop.
Wiki is good on jk ff's
You tie the D input to the /Q output to get it to toggle.
 

Parth86

Member
For a JK flip-flop:
Q goes to 1 when J is 1 after the appropriate clock edge and stay there.
Q goes to 0 when K is 1 after the appropriate clock edge and stay there.
Q alternates its state when both J and K are 1 after the appropriate clock edge.

You need to know whether the FF changes state on the rising edge or the falling edge of the clock. Knowing that, you can then complete the table from what is stated above.
its jk flip flop with positive edge triggered flip flop
I know the function table of D flip flop when clock is raising the Q will be same as D input other wise no change
but what about JK flip flop
 

crutschow

Well-Known Member
Most Helpful Member
Look again at what I posted. It tells you want the JK FF does after the rising edge of the clock. By "stay there" I mean there's no change for further transitions of the clock.
 

Parth86

Member
For a JK flip-flop:
Q goes to 1 when J is 1 after the appropriate clock edge and stay there.
Q goes to 0 when K is 1 after the appropriate clock edge and stay there.
Q alternates its state when both J and K are 1 after the appropriate clock edge.
.


J K C Q
--------------
0 0 0 ?
0 1 ↑ ? here clock is raising
1 0 1 ?
1 1 ↓ ?

you said Q goes to 0 when K is 1 that means Q become 0 when K =1
 

crutschow

Well-Known Member
Most Helpful Member
............................

you said Q goes to 0 when K is 1 that means Q become 0 when K =1
Yes, "goes to 0" and "becomes 0" mean the same thing.
 

crutschow

Well-Known Member
Most Helpful Member
You tie the D input to the /Q output to get it to toggle.
J K C Q Q0
--------------
0 0 ↑ 0 1
0 1 ↑ Q Q0
1 0 ↑ Q Q0
1 1 ↑ 1 0

is it correct ?
No. My quote above is for a D type FF. It does not apply to a JK FF. Look again at my explanation for the JK FF.
 

Parth86

Member
For a JK flip-flop:
Q goes to 1 when J is 1 after the appropriate clock edge and stay there.
Q goes to 0 when K is 1 after the appropriate clock edge and stay there.
Q alternates its state when both J and K are 1 after the appropriate clock edge.

You need to know whether the FF changes state on the rising edge or the falling edge of the clock. Knowing that, you can then complete the table from what is stated above.
J K C Q Q0
--------------
0 0 ↑
0 1 ↑ 0 Q goes to 0 when K is 1 after the appropriate clock edge and stay there.
1 0 ↑ 1 Q goes to 1 when J is 1 after the appropriate clock edge and stay there.
1 1 ↑
 

crutschow

Well-Known Member
Most Helpful Member
J K C Q Q0
--------------
0 0 ↑
0 1 ↑ 0 Q goes to 0 when K is 1 after the appropriate clock edge and stay there.
1 0 ↑ 1 Q goes to 1 when J is 1 after the appropriate clock edge and stay there.
1 1 ↑
You are correct as far as you go. Here is the completed table

J K C Q Qo
0 0 ↑ Q Qo No change
0 1 ↑ 0 1
1 0 ↑ 1 0
1 1 ↑ Qo Q Changes state (toggles)
 
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