# Homework Help: How to read SR Latch

1. Nov 27, 2012

### whitehorsey

Using this image, how would you get the second input if the second input has to come from the output of the nor gates?

Last edited by a moderator: May 6, 2017
2. Nov 27, 2012

### Staff: Mentor

I don't understand what you are asking? Do you even understand what you are asking?

The S-R latch is a basic memory cell. S and R cannot* both simultaneously be TRUE. Provided only one of R and S is TRUE, this forces the cell to take up one of two stable memory states. When that input returns to FALSE, meaning both inputs are now FALSE, the cell remembers that last state.

Now, what was your question? http://imageshack.us/a/img402/3247/undecidedg.gif [Broken]

Last edited by a moderator: May 6, 2017
3. Nov 28, 2012

### whitehorsey

Ah my question is because the nor gate needs two inputs what is the other input the one that belongs to the criss crossed line. So if S = 1 what is the other input that is going into the nor gate with it?

Last edited by a moderator: May 6, 2017
4. Nov 28, 2012

### aralbrec

It doesn't matter. The result out of the NOR gate will be zero.

When S and R are both zero, the output is in a stable state with Q and /Q being complements of each other. If for some reason Q and /Q are the same (both 1 or both 0), the state immediately changes to 1/0 with the fastest NOR gate deciding which will be 1 and which will be 0.

5. Nov 28, 2012

### Staff: Mentor

The second input is ¬Q. Maybe it's 1, maybe it's 0. Try both, one at a time.

6. Nov 30, 2012

### CWatters

The short answer is that the other input is Q but its state doesn't matter.

So you asked what happens with S=1.... Follow it around the circuit step by step...

If S=1 then Qbar = 0 whatever the other input is. Draw the truth table for a NOR gate.

Then if R=0 the inputs to the top gate are R=0 and Qbar=0 so the output Q=1.

If Q=1 then the two inputs at the lower gate are S=1 and Q=1 so the output Qbar = 0

If you then set S back to 0 the circuit will stay ("remember") Q=1 and Qbar=0. That's why S stands for "Set". It sets Q=1.

By symetry if you pulse R from 0 to 1 and back you get Q=0, and Qbar = 1 which is why R stands for "Reset".

So the truth table is...

RS Q Qbar
------------
00 Q and Qbar remember their state
01 Q=1, Qbar=0
10 Q=0, Qbar=1

It is possible to have RS=11 in which case Q=0 and Qbar=0 but if RS change back from 11 to 00 at exactly the same time you can't predict how Q and Qbar will end up. If they change at different times the last one remaining at logic 1 "wins".

7. Dec 1, 2012

### whitehorsey

Thank You for your responses!! So then I tried working on some practice problems I found on the internet but there's one that I'm stuck on.

E Qt St Rt Qt+1
0 0 0 0 Qt
.....
1 0 0 0 0
1 0 1 1 undefined
1 0 1 0 1
1 1 0 0 Qt

In this problem, you have to solve for Qt+1.
The answers I got was Qt whenever the Enable was equal to 0. I also got Qt with the second one and fourth and fifth one. Yet, the second one is equal to 0, the fourth one is equal 1 and the fifth one is equal to Qt. (Answers that came with this problem) Why is that?

Also, just to make sure if I understand a part about SR Latch is that whenever S and R = 1 the state would always be undefined?

8. Dec 1, 2012

### Staff: Mentor

Precisely what state are we talking about?

If your question involves what are the gates' outputs when S=R=1, then you should be able to work that out. What is the output of any NOR gate when an input equals 1?