I was looking at the SN74HC164N shift register datasheet and I noticed something that intrigued me:

There are 2 input signals A and B, but they are NAND'd together to produce the final output. My question is why? I suspect it's so the shift register can have 2 different input sources, though I'm not sure why they would embed that on-chip.

If that's the case and I only have 1 input source, can I just tie the two together and have a 3-wire configuration with DATA_IN, CLR, and CLK?


1 Answer 1


The gated serial (A and B) inputs permit complete control over incoming data; a low at either input inhibits entry of the new data and resets the first flip-flop to the low level at the next clock (CLK) pulse. A high-level input enables the other input, which then determines the state of the first flip-flop.

It's essentially an input enable. There's no point in defining one as Data In and the other as Enable since they can perform either function.

To answer your question, you can do a couple different things if you don't care about using an enable: Holding A or B high or tying A and B together will make the input always enabled.


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