When two asynchronous active low inputs PRESET and CLEAR are applied to a J-K flip flop the output will be
When two asynchronous active low inputs PRESET and CLEAR are applied to a J-K flip flop the output will be Correct Answer Undefined
The PRESET and CLEAR inputs of the JK Flip-Flop are asynchronous, which means that they will have an immediate effect on the Q and Q’ outputs regardless of the state of the clock and / or the J and K inputs.[ alt="F1 Neha.B 30-03-21 Savita D1" src="//storage.googleapis.com/tb-img/production/21/03/F1_Neha.B_30-03-21_Savita_D1.png" style="width: 250px; height: 124px;">
1.When the preset input is activated, the flip-flop will be set (Q=1, not-Q=0) regardless of any of the synchronous inputs or the clock.
2.When the clear input is activated, the flip-flop will be reset (Q=0, not-Q=1), regardless of any of the synchronous inputs or the clock.
3.When preset and clear inputs are activated we get an invalid state on the output, where Q and not-Q go to the same state.
Important Points-
JK Flip-Flop Truth Table- From truth table it can be seen that the CLEAR (CLR) and PRESET inputs are active at a low logic level and put on the Q output of the Flip-Flop, a high logic level regardless of the state of the clock and / or the state of the J and K inputs.
|
|
Input |
Output |
|||||
|
|
Preset |
Clear |
CLK |
J |
K |
Q |
Q̅ |
|
Invalid |
0 |
0 |
✕ |
✕ |
✕ |
1* |
1* |
|
Preset |
0 |
1 |
✕ |
✕ |
✕ |
1 |
0 |
|
Clear |
1 |
0 |
✕ |
✕ |
✕ |
0 |
1 |
|
No change |
1 |
1 |
✕ |
✕ |
✕ |
Q0 |
Q̅0 |
|
No change |
1 |
1 |
↓ |
0 |
0 |
Q0 |
Q̅0 |
|
Reset |
1 |
1 |
↓ |
0 |
1 |
0 |
1 |
|
Set |
1 |
1 |
↓ |
1 |
0 |
1 |
0 |
|
Toggle |
1 |
1 |
↓ |
1 |
1 |
Q̅0 |
Q0 |
