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SR Latch with NOR

D

David Taylor

Jan 1, 1970
0
http://www.ffldusoe.edu/Faculty/Den...Pounds/Section26-30_files/NORLatchCircuit.gif
I'll refer to Q and Q' as Q and QN, as they are different outputs and
one is not necessarily the complement of the other.

I get that R means reset and Set means set.
But I don't understand the meaning of the second output QN. Is it
just an implementaion thing required only to ensure Q has the right
value?
Would QN be discarded, for example the circuit for Prog+ and Prog- on
a remote control or the circuit for a burgler alarm just gives 1
output.

More importantly, if I feed in previous values of Q and QN, I don't
get the right result, I should find that
when R=1 S=0 then Q should equal 0
also
when R=0 S=1 then Q should equal 1
And I'd have thought the previous values of Q and QN are irrelevant.

I'll describe the results i'm getting-
suppose R=0 S=1 and the previous outputs now going in are Q=0 QN=1
I expect a result of Q=1 as R means reset, and S means set.

In the top NOR gate, (R NOR QN)=(0 NOR 1)= 0 so new Q=0
This is not right, Q should be Set, Q should = 1.

Im the bottom NOR gate, new QN=0.

If I had not fed in previous inputs and just said that as R=0 NOR
can't be evaluated yet. S=1 so QN=0 Feeding in R with the new QN, R
NOR QN gives 1 so Q=1. If I do that, then I get the right result.

Another thing i don't understand is that I keep reading that when S=0
and R=0 then Q and QN should retain their original values. I find
this only to be true when Q=0 and QN=1 or Q=1 and QN=0. Clearly,
when Q=0 and QN=0, you feed in
R NOR QN that's 0 NOR 0 =1 so Q=1
you feed in that old Q
S NOR Q so QN=1
Clearly Q and QN have flipped, and not retained their original values.
Similarly, with R=0 S=0 and Q=1 QN=1, the NOR gates see the 1s coming
in, evaluate to 0, and so Q=0 QN=0, so they've flipped too.
Definitely not retainig their values as they should be.

An additional problem I have, is that the definition of sequential
logic circuit "one whose output depends not only on its inputs but
also on its past sequence of inputs" Surely this NOR SR latch is
feeding in past outputs rather than past inputs. I don't see any old
Rs and Ss going in, only an old Q and QN.

Thanks in advance
 
D

David Taylor

Jan 1, 1970
0
Gajski on P.213 implies that Q is input not QN, and also, the input Q
only has an effect on the output Q when S=0 and R=0. Also, when S=1
and R=1
it is something designers try to avoid as it doesn't make sense to
reset and set the latch simultaneously. This clears everything up,
leaving 2 question of minor significance.

1 How electronically does it work out that Q is inputted and QN isn't?
If you look at the circuit diagram, it clearly shows QN inputted.

2 "Sequential logic circuit – outputs depend on its inputs and past
sequence of inputs. " (e.g. prog up and down on a remote control)
Wakerly p.529


note: A far easier to understand boolean algebra expression is given
in Gajski p.226 Q(next)=S+R'Q Q(next) meaning new Q (the Q
outputted), and, Q meaning old Q (the Q inputted).
 
P

petrus bitbyter

Jan 1, 1970
0
David Taylor said:
http://www.ffldusoe.edu/Faculty/Den...Pounds/Section26-30_files/NORLatchCircuit.gif
I'll refer to Q and Q' as Q and QN, as they are different outputs and
one is not necessarily the complement of the other.

This is a very basic logic building block treated in every book a ever saw
on logic systems. You call it a latch. It's often called "trigger" or
"flipflop" as well but it has no use to start a discussion about that.
I get that R means reset and Set means set.
Correct.

But I don't understand the meaning of the second output QN. Is it
just an implementaion thing required only to ensure Q has the right
value?

The circuit will not function without it. There may be or not be a use for
QN outside this circuit but inside you cannot do without it.

Would QN be discarded, for example the circuit for Prog+ and Prog- on
a remote control or the circuit for a burgler alarm just gives 1
output.

It may be hidden in a box or even in a chip but this circuit requires QN.
More importantly, if I feed in previous values of Q and QN, I don't
get the right result, I should find that
when R=1 S=0 then Q should equal 0
also
when R=0 S=1 then Q should equal 1
And I'd have thought the previous values of Q and QN are irrelevant.

You're mixing logic with levels. The outputs are inverted with respect to
the inputs.
I'll describe the results i'm getting-
suppose R=0 S=1 and the previous outputs now going in are Q=0 QN=1
I expect a result of Q=1 as R means reset, and S means set.

In the top NOR gate, (R NOR QN)=(0 NOR 1)= 0 so new Q=0
This is not right, Q should be Set, Q should = 1.

Im the bottom NOR gate, new QN=0.

If I had not fed in previous inputs and just said that as R=0 NOR
can't be evaluated yet. S=1 so QN=0 Feeding in R with the new QN, R
NOR QN gives 1 so Q=1. If I do that, then I get the right result.

As I said the outputs are inverted. You will have the "correct" results by
inverting them again using inverters between the outputs of the gates and
the external signals. In a practical situation you will simply exchange Q
and QN.
Another thing i don't understand is that I keep reading that when S=0
and R=0 then Q and QN should retain their original values. I find
this only to be true when Q=0 and QN=1 or Q=1 and QN=0. Clearly,
when Q=0 and QN=0, you feed in
R NOR QN that's 0 NOR 0 =1 so Q=1
you feed in that old Q
S NOR Q so QN=1
Clearly Q and QN have flipped, and not retained their original values.
Similarly, with R=0 S=0 and Q=1 QN=1, the NOR gates see the 1s coming
in, evaluate to 0, and so Q=0 QN=0, so they've flipped too.
Definitely not retainig their values as they should be.

Q is complementary to QN by definition. They both remember whether the set
or the reset has been "true" the last time. (Although the level has
inverted.) You can make both Q and QN low by driving both inputs high but
that is a forbidden input combination. When this happens the circuit losts
his memory. It is up to the designer to prevent this situation. As an
alternative you can design an RS-element with an overriding set or an
overriding reset.
An additional problem I have, is that the definition of sequential
logic circuit "one whose output depends not only on its inputs but
also on its past sequence of inputs" Surely this NOR SR latch is
feeding in past outputs rather than past inputs. I don't see any old
Rs and Ss going in, only an old Q and QN.

The Qs *are* the old Rs and Ss. The output *does* not depend on the input
only. Q reminds which of the inputs has been true the last time even if the
signal itself has disappeared already. It's all a simple memory element like
this *can* do. It's up to the designer once more to use this type of
elements as building blocks for more complated sequential machines.
Thanks in advance

You're welcome.

petrus
 
S

Steve

Jan 1, 1970
0
I think this examination of an SR flip-flop is akin to using a tea
spoon to dig the channel tunnel.

Or perhaps it's like using a 200 tonne drag line excavator to repair a
divot on a golf green... I'm not sure which is more appropriate.

Either way, such a detailed examination of an SR flip flop would seem
to me to be a waste of time. Just accept that it works, build one if
you have to satisfy your curiosity; then get on with more important
stuff.

niftydog
 
J

John Fields

Jan 1, 1970
0
I think this examination of an SR flip-flop is akin to using a tea
spoon to dig the channel tunnel.

Or perhaps it's like using a 200 tonne drag line excavator to repair a
divot on a golf green... I'm not sure which is more appropriate.

---
Neither one of them is an example of anything which has anything to do
with explaining regeneration or memory, so they're both stupid
analogies.
---
Either way, such a detailed examination of an SR flip flop would seem
to me to be a waste of time. Just accept that it works, build one if
you have to satisfy your curiosity; then get on with more important
stuff.

---
A worse waste of time is your complaining about something which,
obviously, has nothing to do with you. This is sci.electronics.basics,
and a subject which may seem mundane to you may be conceptually
difficult for someone else. This is just the right place for them to
ask whatever they want to without having to worry about whether someone
else thinks it's stupid, or a waste of time, or whatever.
 
P

petrus bitbyter

Jan 1, 1970
0
Steve said:
I think this examination of an SR flip-flop is akin to using a tea
spoon to dig the channel tunnel.

Or perhaps it's like using a 200 tonne drag line excavator to repair a
divot on a golf green... I'm not sure which is more appropriate.

Either way, such a detailed examination of an SR flip flop would seem
to me to be a waste of time. Just accept that it works, build one if
you have to satisfy your curiosity; then get on with more important
stuff.

niftydog

Doggy,

You're on the wrong trail. The best way to learn complex digital circuits is
a good understanding of the basic blocks. In this particular case the OP
understands the inner working of the circuit well enough. He was only
confused as it did not function like he expected. Expectations based on
general statements about logic circuits that seemed to be contraditionary to
his experience. (Somehow he's right too. The terminology of logic people is
not always as logic as it should be.) I'm not a fortune-teller but I expect
him to become a great logic designer. And as I love the profession I'll try
to answer any fair question, especially newbees questions. This group is
supposed to exist for it.

petrus
 
S

Steve

Jan 1, 1970
0
I wasn't referring to regeneration or memory in my analogies, I was
referring to the amount of effort going into understanding the
circuit.

Besides, I thought it was funny... get off your high horse and loosen
up!
A worse waste of time is your complaining

I wasn't complaining.
about something which, obviously, has nothing to do with you.

WTF? This is Usenet. Of course it has nothing to do with me, just as
it technically has nothing to do with you. You chose to post your
opinion, as did I.
This is sci.electronics.basics

Really? Sheesh, I thought I was in the Bahamas!
and a subject which may seem mundane to you

You're reading things into my post. I never implied that it was
mundane.

CHILL OUT!

niftydog
 
S

Steve

Jan 1, 1970
0
The best way to learn complex digital circuits is a good
understanding of the basic blocks.

I repair digital video equipment, and in four years of my current job
such a thorough understanding of an SR flip flop built from NAND gates
has NEVER been needed.

I appreciate that it's good to know, but it seemed like the OP was
delving into it in immense detail. From a quick look around the
website given, this effort was disproportionate to the amount of
attention paid to the device in the course.

I wasn't saying "don't bother learning this", I was suggesting that
sometimes you can try too hard to understand that which is relatively
unimportant. You can end up chasing your tail for days over a trivial
detail.
And as I love the profession I'll try to answer any fair question, especially newbees questions. This group is supposed to exist for it.

I was trying to express that perhaps the OP may be spending precious
time and effort chasing their tail when there are far more exciting
and important things about logic circuits to learn.

In the end, the OP said:
"...S=1 and R=1 is something designers try to avoid as it doesn't make
sense to reset and set the latch simultaneously. This clears
everything up..."

This is the easy way out I'm talking about. Forget trying to analyse
what happens when you DO set and reset at the same time, just accept
that it's unpredictable and should be avoided.

There's a lot of reading between the lines going on in this thread...

niftydog
 
R

R. Steve Walz

Jan 1, 1970
0
Steve said:
understanding of the basic blocks.

I repair digital video equipment, and in four years of my current job
such a thorough understanding of an SR flip flop built from NAND gates
has NEVER been needed.
 
G

George

Jan 1, 1970
0
[email protected] (David Taylor) wrote in message news: said:
More importantly, if I feed in previous values of Q and QN, I don't
get the right result, I should find that
when R=1 S=0 then Q should equal 0
also
when R=0 S=1 then Q should equal 1
And I'd have thought the previous values of Q and QN are irrelevant.

The past values are why this circuit holds memory, the main function
of a latch.
I'll describe the results i'm getting-
suppose R=0 S=1 and the previous outputs now going in are Q=0 QN=1
I expect a result of Q=1 as R means reset, and S means set.

In the top NOR gate, (R NOR QN)=(0 NOR 1)= 0 so new Q=0
This is not right, Q should be Set, Q should = 1.

Im the bottom NOR gate, new QN=0.

You need to give the circuit time to stabilize. You do not just accept
the initial outputs from the latch, you give it time to stabilize.
Take your R=0, S=1, and your new outputs Q and QN =0. Put those
through the circuit a second time, you get Q=1 and QN=0. This is will
stay that way until the inputs change. This is the correct outputs.
This circuit has no clock so the NOR calculations will keep happening
until they stabilize. The reason you never input S=1 R=1 is that there
is no way of predicting what the circuit will stabilize at in most
sequences, or that it will stabilize at an unacceptable combination.
If I had not fed in previous inputs and just said that as R=0 NOR
can't be evaluated yet. S=1 so QN=0 Feeding in R with the new QN, R
NOR QN gives 1 so Q=1. If I do that, then I get the right result.

Another thing i don't understand is that I keep reading that when S=0
and R=0 then Q and QN should retain their original values. I find
this only to be true when Q=0 and QN=1 or Q=1 and QN=0. Clearly,
when Q=0 and QN=0, you feed in
R NOR QN that's 0 NOR 0 =1 so Q=1
you feed in that old Q
S NOR Q so QN=1
Clearly Q and QN have flipped, and not retained their original values.
Similarly, with R=0 S=0 and Q=1 QN=1, the NOR gates see the 1s coming
in, evaluate to 0, and so Q=0 QN=0, so they've flipped too.
Definitely not retainig their values as they should be.\

As other posts have said, you should never have Q and QN equal each
other so that is not a problem that matters.
An additional problem I have, is that the definition of sequential
logic circuit "one whose output depends not only on its inputs but
also on its past sequence of inputs" Surely this NOR SR latch is
feeding in past outputs rather than past inputs. I don't see any old
Rs and Ss going in, only an old Q and QN.

The outputs are based on past inputs. Thus, the fact that you include
past outputs in the circuit automatically means you include past
inputs. Hence a latch is a sequential.
 
J

John Fields

Jan 1, 1970
0
I wasn't referring to regeneration or memory in my analogies, I was
referring to the amount of effort going into understanding the
circuit.
---
I'm quite aware of what you _weren't_ referring to.

If the OP wanted to spend the rest of his life, and a fortune, trying to
understand the circuit then that's his business and good for him. When
you come along with your destructive criticisn and, instead of offering
an explanation which might clarify things for him, tell him that he's
wasting his life, that's _not_ a good thing.
---
Besides, I thought it was funny... get off your high horse and loosen
up!

---
What about it did you think was funny?

On the one hand you present a scenario where a herculean task is being
attempted with trivial resources (the OP is too stupid to be able to
understand the magnitude of the task at hand) and on the other you
present a scenario where a trivial task is being performed with
herculean resources. (The OP is too stupid to understand the magnitude
of the task at hand)

Thank you, but I'd rather ride than crawl along down there with the
likes of you.
---
I wasn't complaining.

---
Webster disagrees with you, but don't bother to look it up. Just take
my word for it.
---
WTF? This is Usenet. Of course it has nothing to do with me, just as
it technically has nothing to do with you. You chose to post your
opinion, as did I.

---
By 'not having anything to do with you' I was referring to your
failure/inability to address the technical content of the OP's post. My
post, while admittedly off-topic, was in response to the content of your
post (which I found objectionable) and therefore became interesting to
me and therefore had spmething to do with me.
---

Really? Sheesh, I thought I was in the Bahamas!
---
Trolling?
---



You're reading things into my post. I never implied that it was
mundane.

---
Yet, in another post, you said:

"I was trying to express that perhaps the OP may be spending precious
time and effort chasing their tail when there are far more exciting
and important things about logic circuits to learn."
---
 
D

David Taylor

Jan 1, 1970
0
This is a very basic logic building block treated in every book a ever saw
on logic systems. You call it a latch. It's often called "trigger" or
"flipflop" as well but it has no use to start a discussion about that.

but all good books state that latches don't have clocks, and flip
flops do. Good books and good lecture slides state that bad books and
bad slides call latches flip flops. So I don't call a latch a flip
flop. I won't cite examples as it's not relevant to the thread.

You're mixing logic with levels. The outputs are inverted with respect to
the inputs.

Isn't the logic the meaning of the levels, it's what the levels
represent. HIGH represents 1. LOW represents 0. So If I say 0 and 1,
then i'm looking at the meaning. If I say HIGH and LOW, then I'm
looking at the implementation. I don't see how it's wrong to say
R=1 S=0 e.t.c. Truth tables do it, books do it when they show truth
tables, what's the problem with it?

As I said the outputs are inverted. You will have the "correct" results by
inverting them again using inverters between the outputs of the gates and
the external signals. In a practical situation you will simply exchange Q
and QN.

From what i've read - inverters are used in the NAND latch for correct
results, not in the NOR latch.
Q and QN need not be exchanged in the NOR latch.
My problem, is that if I put Q and QN into the gates first, I get
one result, and if I put R and S in first, I get another result (the
right one). Do R and S have to reach the gates before Q and QN in
order for the circuit to work? That's the problem I'm having.

The older problem of R&S=1 and oscillations is now resolved to a level
i'm comfortable with.

thanks.
 
D

David Taylor

Jan 1, 1970
0
I'll describe the results i'm getting-
You need to give the circuit time to stabilize. You do not just accept
the initial outputs from the latch, you give it time to stabilize.
Take your R=0, S=1, and your new outputs Q and QN =0. Put those
through the circuit a second time, you get Q=1 and QN=0. This is will
stay that way until the inputs change. This is the correct outputs.
This circuit has no clock so the NOR calculations will keep happening
until they stabilize. The reason you never input S=1 R=1 is that there
is no way of predicting what the circuit will stabilize at in most
sequences, or that it will stabilize at an unacceptable combination.


I had no idea that R=0 S=1 could ever be unstable initially, as none
of my books treat latches in such detail.
It all makes sense now.
You've sorted it out.
many thanks.
 
P

petrus bitbyter

Jan 1, 1970
0
David Taylor said:
but all good books state that latches don't have clocks, and flip
flops do. Good books and good lecture slides state that bad books and
bad slides call latches flip flops. So I don't call a latch a flip
flop. I won't cite examples as it's not relevant to the thread.

I know. Just wrote it down for your information.
Isn't the logic the meaning of the levels, it's what the levels
represent. HIGH represents 1. LOW represents 0. So If I say 0 and 1,
then i'm looking at the meaning. If I say HIGH and LOW, then I'm
looking at the implementation. I don't see how it's wrong to say
R=1 S=0 e.t.c. Truth tables do it, books do it when they show truth
tables, what's the problem with it?

It's a matter of theory. This circuit is supposed to remember the last
action. If the last action was set (a positive pulse on S) it is remembered
by keeping Q low. So the output is inverted with respect to the last action.
You can say that Q remembers the last reset correctly. But by convention the
Q is related to the set, not to the reset.

Theoretical logic knows "true" and "false" also named "1" and "0"
respectively. Practical circuits knows "high" and "low" also named "on" or
"off". Most of the times "true" and "high" are considered to be the same.
This is called positive logic. But there's no reason to call "low" "true".
No surprise that's called negative logic. Then there is mixed logic as well.
In practical circuits you often see the term "active low" for inputs as well
as for outputs. That's where the positive logic is not so logical as the
active or "true" level is "low". In alphabetic terms there's often an "N"
placed behind it or a "/" for it. There are other notations as well. There's
no problem with it as long as you talk about simple gates. But you can run
into a lot of confusion when you have to analyze or design more complicated
digital circuits. You'll find it out for yourself soon enough.
From what i've read - inverters are used in the NAND latch for correct
results, not in the NOR latch.

NAND latches can be considered to have active low inputs, where NOR latches
can be considered to have active low outputs. They both need inverters if
you want to keep positive logic only. As I said nobody cares in practical
situations.
Q and QN need not be exchanged in the NOR latch.
My problem, is that if I put Q and QN into the gates first, I get
one result, and if I put R and S in first, I get another result (the
right one). Do R and S have to reach the gates before Q and QN in
order for the circuit to work? That's the problem I'm having.

Q and QN are outputs. They are in a kind of a feedback loop but nevertheless
you cannot separate them from the inputs of the gates. You also cannot
expect a digital circuit that has undifined inputs to have well defined
outputs. Theoretically the ouput of a gate follows the input immediately.
Practical gates have a delay time, that's the time required to stabilize the
new output as a reaction on an input change. But in both situations you
cannot put Q and QN into the gates but by controlling set and/or reset. The
only exception is at power on, but even then it gets some value somehow and
you can't be sure whether it will be high or low. That's why the state of
this circuit is not defined after power on. You have to initialize it
somehow. Which requires extra hardware.

petrus
 
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