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How many NAND gates are there in J-K flip-flop?
four NAND gates
J-K Flip-Flop Structure Note that the outputs feed back to the enabling NAND gates. This is what gives the toggling action when J=K=1. While this implementation of the J-K flip-flop with four NAND gates works in principle, there are problems that arise with the timing.
What is the working of J-K flip-flop?
The JK flip flop work as a T-type toggle flip flop when both of its inputs are set to 1. The JK flip flop is an improved clocked SR flip flop. But it still suffers from the “race” problem. This problem occurs when the state of the output Q is changed before the clock input’s timing pulse has time to go “Off”.
What is the output of a NAND gate flip flop?
An important point about NAND gate is that its dominating input is 0 i.e., if any of its input is Logic ‘0’, the output is Logic ‘1’, irrespective of the other input. The output is 0, only if all the inputs are 1. With this in mind, let us see the working of a NAND based RS Flip-Flop.
What happens when you reset the JK flip flop?
When we apply a clock pulse to the J K flip flop and the inputs are J is low and K is high the output of the NAND gate connected to J input becomes 1. Then Q becomes 0. This will reset the flip flop again to its previous state. So the Flip flop will be in RESET state.
How are high inputs connected to JK flip flop?
For high inputs of J K flip flop, only the lower NAND gates are triggered by the outputs that are compliment to each other i.e Q and Q’. So while high inputs are connected to flip – flop, at any instant, one gate is enabled and other gate will be disabled.
Why is the NOR gate forbidden in flip flops?
As ‘1’ at input R makes the output to switch to one of its stable states and resets it to ‘0’, the R input is known as RESET input. This input condition is forbidden as it forces outputs of both NOR Gates to become 0, which is a violation of complementary outputs.