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How is a JK flip made to toggle?
How is a J-K flip-flop made to toggle? Explanation: When j=k=1 then the race condition is occurs that means both output wants to be HIGH. Explanation: The flip flop is sensitive only to the positive or negative edge of the clock pulse. So, the flip-flop toggles whenever the clock is falling/rising at edge.
What are the limitations of JK flip flop?
JK flip-flop has a drawback of timing problem known as “RACE”. The condition of RACE arises if the output Q changes its state before the timing pulse of the clock input has time to go in OFF state. The timing pulse period (T) should be kept as short as possible to avoid the problem of timing.
Are JK flip flop is in a no change condition when?
A J-K flip-flop is in a “no change” condition when ________. 24. A correct output is achieved from a master-slave J-K flip-flop only if its inputs are stable while the: A….Exercise :: Flip-Flops – General Questions.
| A. | The output toggles if one of the inputs is held HIGH. |
|---|---|
| C. | Only one of the inputs can be HIGH at a time. |
Why we use J-K flip-flop?
A J-K flip-flop is nothing more than an S-R flip-flop with an added layer of feedback. This feedback selectively enables one of the two set/reset inputs so that they cannot both carry an active signal to the multivibrator circuit, thus eliminating the invalid condition.
How to reset a counter in a JK flip flop?
A J-K flip flop will count (toggle) when both J and K = 1. We can make a free-running counter by just using J, tying K high. To reset Q in a J-K flip flop we must set J=0 and K=1. If we make RESET active low, then the circuit below does that.
How many flip flops are there in a JK binary counter?
It actually has eleven states, 0000 through 1010, but as soon as the last state is reached, the NAND gate immediately (asychronously) resets the flip-flops to the 0000 state. In a synchronous counter, all of the flip-flops would share a common clock, and you’d control the sequence of states by driving their J and K inputs.
When to reset a JK binary counter from 0 to 9?
When both bits are one, and that only occurs when the output is 10 if you stay in the 0..9 range, the NAND output goes low and pulls low all the negative reset inputs of the FFs, resetting your counter. When you want to design a counter which counts 0000-1001 (0 to 9), You have to reset when it reaches 1010 (10).
Why is the NAND gate connected to the second flip flop?
In a JK Binary Counter from 0 to 9, why is the NAND gate connected to the second and fourth J-K flip flop and not the first and fourth? In a binary counter design using 4 J-K flip-flops, that counts from 0 to 9, the flip flops are reset when the output from the 2nd flip flop NAND the 4th flipflop equals to 0.