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OK, converting RS FF (which is with nand gates) to JK FF is simple. We just take K.Q as R and J.Q' as S, right?

But how can we prove it?

2006-11-10 13:56:31 · 1 answers · asked by exodusfe 1 in Science & Mathematics Engineering

1 answers

JK flip-flop consists of two RS flip-flops, the input one called master and the output one slave. The RS inputs of the master are the (controlled by clock) AND gates outputs of the JK inputs while RS inputs of the slave are the outputs of the (controlled by inverse of the same clock) AND gates of the Q and Q' inputs of the master. When the clock is high the input signals are absorbed into the master. When clock goes low JK inputs are disconnected, however, the master RS remembered them as its output (I assume here RS made of two OR gates). At the same time inter-RS gates open and slave RS absorbs the outputs from the master RS. In this way, in one clock cycle, the input JK signals stay remembered as the outputs of JK although the input signal is never directly connected to the output. This is necessary to avoid the race condition in synchronous sequential circuits which use JK flip-flops as a state memories.

2006-11-10 17:44:30 · answer #1 · answered by fernando_007 6 · 0 0

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