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Understanding Basic S-R Latch |
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| Nov17-12, 10:08 PM | #1 |
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Understanding Basic S-R Latch
I have a question about S-R latch for a specific diagram below (no, this is NOT a homework question).
Is there something, physically, about the way that AND (or even other) gates are built that once they know one input, they know what the output will be? I know, for example, that if R = 0, the output of AND will be 0, because both inputs need to be 1 or high for the output to be 1. I'm guessing the answer is no, the circuit does not know what the output will be with one input because if R = 1, how do we know what the second input will be? What boggles my mind is this: let's pretend that R always gives its output first, because for some reason there is a delay in S. How is this possible? How can there even be an output for either AND gate if the AND gate needs two inputs to make a decision about an output?
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| Nov18-12, 09:40 AM | #2 |
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SpaceCreature,
Ratch |
| Nov19-12, 01:04 AM | #3 |
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A high output simply means a current is flowing through the wires. If the S gate is slower than the R gate, than the R gate will be receiving a low signal from one of its outputs, and will produce a 0.
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| Nov19-12, 08:33 AM | #4 |
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Recognitions:
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Understanding Basic S-R Latch
In a logic gates arrangement, at least those you will be working with, the signal at every point and on every wire is at all times either a 1 or else it's a 0, there are never any "undecided" or "still thinking about it" or "wait, I'm not ready yet" logic levels.
![]() If a 2-input AND gate on one input has a 0, neither you nor the gate needs to take into consideration what is on the other input—the output of that AND gate is already determined to be 0 so the electronics is designed to make the output a 0. |
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