All right. Good morning,
all.
You have two handouts, lecture notes and an article on
mixed signal chips. A mixed signal stands for
circuits that have both analog and digital components to them.
The reason I am giving you the handout is that Lab 4 and also
your last homework involve designing and building a mixed
signal circuit. It's a real fun exercise.
And I just wanted to tell you that from past experience people
who have taken 6.002 often view the last lab as the single most
fun thing they did in all of
So, as you go into Lab 4, you should be telling yourself
I should be having fun, I should be having,
I should be having fun. You have to positively psych
yourself. Otherwise, it's going to go by.
And then you're going to say boy, that was fun,
I wish I had savored the moment as I was doing it.
All right. Let's see.
What do we do today? Today's lecture is actually
going to be a fair amount of fun.
We are going to blast through a bunch of fun things.
And some things that you will be quite unprepared for.
Until now, in the last two lectures with op amps we talked
about negative feedback. That is applying some portion
of the output voltage to the negative input so that I could
control this high strung device, my op amp.
Today, what we are going to do is try to get a handle on what
happens if we use positive feedback.
It's the usual curious child. You tell them to do this,
and of course they're going to try to do this as well.
And we are going to try to do that and see what happens and
look to see if we can build some useful circuits.
Today --
As motivation, let me do a quick review of a
circuit that should now become affixed in your brains in a
standard pattern. This is a circuit that gives
you negative feedback.
R1 and R2.
And I apply a vIN. By now you should be able to
look at this pattern. And this is your inverting
amplifier pattern. So, you should be able to write
down by inspection this is simply vIN or the minus vIN
times R2 divided by R1. This is an amplifier whose gain
is controlled by the ratio of R2 and R1.
This is a negative feedback circuit because it is always fun
to do the intuition thing and say that look,
if this voltage tends to go more positive than I care then
this negative input goes more positive than I care.
If that goes more positive then the negative input v minus
becomes more positive in the plus input which yanks the
output down. So, there is a nice
counteracting force that keeps the output stable.
Let's look at this circuit. Being curious engineers,
let's look at the opposite here where I give myself some
positive feedback in this op amp.
And it is going to be interesting to analyze this
because what we find out on the face of it is not quite actually
how it behaves. We are going to spend most of
the lecture today on understanding the dynamics of
circuits that look like this and to see if we can build some fun
and interesting circuits and systems based on this kind of
positive feedback. It is positive feedback because
I am feeding back a portion of the output to the positive
input. And you should be able to stare
at this and already begin to intuit what should happen to
this. Let's think about it.
This is zero. Remember, with positive
feedback, the famous v plus is equal to v minus method doesn't
apply anymore. Let's apply very simple
analyses. If this is zero,
let's say for example that this output tends to go a little bit
more positive. This output,
due to some noise or perturbation,
tends to go up a little bit. If that goes up a little bit
then because of feedback this node tends to go up a little
bit. If this node tends to go up a
little bit this exacerbates the positive input here and this one
goes cachunk, whacks into the positive rail.
Let's take the other point of view and look at it intuitively.
What if this one tries to droop a little bit?
If it droops a little bit then the input at the plus terminal
droops a little bit. If that tends to go down a
little bit, that makes the output droop further and it goes
and hits into the negative rail. I can see that this circuit
wants to hammer into the positive rail or hammer into the
negative rail because of the positive feedback.
It is like if you give incredibly positive feedback all
the time, and by positive feedback I mean feedback
encouraging the child to do whatever the child is doing.
It could be if he does bad stuff you give a lot of positive
feedback or good stuff you give a lot of positive feedback then
you are guaranteed to have a very good child or a very bad
child. You are not going to have
anybody in the middle. Same way here.
By giving positive feedback you're going to drive this into
the positive rail or drive this into the negative rail.
Now, I am going to analyze this in two steps.
First I am going to analyze this using a method you've seen
before which is replace the op amp with its equivalent circuit
and analyze it statically. And by analyzing it statically
we are going to show that the simple static analysis will
yield the following expression. I put this in quotes,
well, for a reason you will see shortly.
When I apply a plain and simple static analysis here is what I
find. Let's go ahead with the
analysis and see what is basically different about these
two. And, first of all,
I will confirm for you that our naive analysis we have seen so
far will give rise to that expression.
So, let's go ahead and analyze that circuit.
And to analyze that circuit what I will do is replace the op
amp with its equivalent circuit. If you remember the op amp is
characterized by the following circuit, A times v+ minus v-,
vOUT. This is the equivalent circuit
of my op amp. And let me just impose that
external circuit on this op amp. I have grounded my v- terminal.
My v+ terminal goes through a resistor and a supply,
the v into ground, it's the resistance R1.
This terminal goes to the output through a resistor R2.
So, this is the equivalent circuit.
And I can apply the same good-old techniques I have
learned about all through this course to this circuit and see
what vOUT looks like. Very simply,
vOUT is this expression here A times v+ minus v-.
And because of my ground connection v- is zero.
Then let me go ahead and replace v+ with the voltage that
relates vOUT and vIN. What is v+?
v+ is simply the current through this part of the
circuit, the current flowing here times the resistance R1.
That gives me the drop across R1.
And to that I add vIN and that will give me V+.
And then of course I multiply this by the gain here.
So, let me write down that expression.
The current through this is simply vOUT minus vIN.
That is the voltage drop between these two points.
I divide that by the resistance R1 plus R2.
That gives me the current flowing through here.
That times R1 is the drop across resistor R1.
And to that I add vIN and that gives me the voltage v+.
So, this is v+. That is simply vIN plus the
drop across the resistance R1. Let me shuffle things around
and put all the vOUT terms on this side here.
I get a 1+ for that vOUT and let me move AR1 divided by R1
plus R2 to the left-hand side. And I pick up a minus sign.
So, I get AR1 divide by R1 plus R2.
I pick up that. And on the left-hand sign I end
up with vIN, and my vIN here is a function of the vIN that I
have here. I have an A multiplying both
the vINs. And then I get a one for this
vIN here and there is a minus sign, so I get a minus R1
divided by R1+R2. That is the expression that I
have. Let me go ahead and simplify
that a little further and move this whole thing down here.
That gives me my expression as a function of vIN.
What I will do is, let me continue here.
vOUT=vIN A(1-R1/(R1+R2)). By the way, you may be
wondering why I am going through so laboriously what is seemingly
a very simple exercise. The reason I want to do is it I
want to very carefully show you that the result produced by this
exercise is exactly that. No magic here.
No cheating. We are going to get exactly
that. And then stare at it and say
huh, how did that happen? And then we are going to try to
figure out how it actually behaves following that.
I divide this by 1-AR1/(R1+R2). And by now you should be
familiar with the technique of ignoring small numbers when I
have a big number next to it. So, AR1/(R1+R2) can be very
much larger than one because A is very large.
So, I can ignore my one there. And then what I am going to do
is multiply the numerator and denominator by R1+R2.
Oh, this A and this A is going to cancel out.
This A and this A will then cancel out.
And then I multiply the numerator and denominator by
R1+R2, so this R1+R2 vanishes. I get R1+R2 here.
R1+R2 minus R1 is simply R2. And then down here I get a R1
and then I have a minus sign out there.
Notice that vOUT we have found to be equal to vIN R2 divided by
R1. That is not wrong.
That is correct. Technically that is correct.
But you will see in a few seconds that in practice that
that's rarely what you are going to see happen.
And we will try to understand why that is so.
What we have done so far, if you stare at these two
panels here, first of all, we know that the inverting
amplifier has the expression for vOUT up there.
And through this laborious exercise we have also shown that
even with positive feedback, if I take a static view of the
circuit -- If I take a snapshot of the
circuit and simply analyze it as a static circuit,
I get the same expression vOUT. But what we are going to do is
when I explain to you that look, a small perturbation in vOUT is
going to drive the op amp to the positive and negative rail,
that is where the insight begins to show.
That if everything were magical and I could somehow exactly keep
things just so that will be true.
I will be able to build that positive feedback circuit where
the output is equal to R2/R1 vIN.
But remember even the slightly amount of perturbation is going
to send the op amp scurrying off to the positive rail or the
negative rail. How do we analyze that?
How do we analyze the behavior of a circuit that based on a
small perturbation begins to move one place or another?
We want to analyze the dynamics of the op amp.
And to analyze the dynamics what I need to do is give you a
slightly more detailed view of the operational amplifier.
If the operational amplifier is not moving instantaneously
between the plus and minus rail, I need to give you a more
detailed model that encapsulates the behavior of the op amp.
And so let me do that. If you want to study the
dynamics of an op amp -- By dynamics I mean how an op
amp moves as I perturb the input or the output and so on.
To capture the dynamics of the op amp we build a slightly more
involved circuit, so v+ and v-.
This is what we've seen before, two terminals and dependent
source that amplifies the difference input here by a large
amount. Instead what we are going to do
here is something slightly different and interpose the
following circuit in the middle here.
This is a model of the dynamics of an op amp.
We are going to impose a small RC circuit in here.
This is R. This is C.
And I am going to call the voltage across the capacitor v*.
Notice what I have done is rather than say this is Av+
minus v- I am breaking it apart in two dependent sources,
the first dependent source, which is simply v+ minus v-,
and there is a RC time constant surrounding it and then here I
simply add on my gain Av*. Notice that if it turned out
that the resistance here, for example,
was zero then v+ minus v- would appear across v* and this would
be A(v+ - v-), what you have seen before.
It is always good to take a look at circuits and look at
what happens when some component goes to an extreme value.
This would give you your basic op amp circuit.
What I would like to do next is analyze the following circuit to
understand how positive and negative feedback work together.
And by understanding that then be able to explain how a
positive feedback circuit works or a negative feedback circuit
works. Here is what I will do.
This part simply corresponds to my positive feedback circuit,
R2, R1. So, that is my positive
feedback circuit. And I will do the same thing on
this side.
All I am doing is applying both a positive feedback through R2
and R1 and negative feedback through R4 and R3 and
representing the dynamics of the op amp and then standing back
and ee, all right, let's see what happens to you.
So, I am sticking positive feedback, negative feedback,
the dynamics of the op amp here and let's see what happens.
What I would like to do is impose this circuit on top of
this op amp model. To save myself some effort,
let me just go ahead and modify this circuit directly.
I get an R2 here, an R1 here, and then up here I
get an R4, R3 here. The math is going to be just a
little bit grubby but the result is actually pretty spectacular.
So, all I have done is replace the op amp with its internal
circuit out here. And now we are going to take a
look at what happens to op amp dynamics when there is a small
perturbation. Let's develop an equation of
this circuit containing a capacitor using techniques that
we already know. Just to give you some insight
into what you're going to see, notice that if I make a small
perturbation in the voltage across the capacitor,
let's say I make a small perturbation to the capacitor
voltage let's say by applying some initial condition kind of
thing onto the capacitor. Then let's say that the output
changes to some value K. So, the change on the capacitor
must have been K divided by A. And what you are going to see
is what happens to the op amp when the initial condition on
the capacitor is such that this output gets perturbed to the
value K. Let's write an equation for
this little circuit and see what happens.
Recall our goal was to understand what happens when I
perturbed the output a little bit.
Here I perturbed the output such that its value goes to K.
And I can perturb the output by changing what happens at the
capacitor. Let me write the equation for
this circuit now and then to understand what happens to this
capacitor circuit if I let go after giving it a small
perturbation. What I am going to do is let me
start by writing the good old equation for this little circuit
here. And that equation is simply the
voltage here v+ minus v- equals the voltage across the RC.
So, v+ minus v- will be equal to the voltage drop across the
resistor plus that across the capacitor.
The voltage across the capacitor is v*.
The voltage across the resistor is the current through the
capacitor C dv*/dt times R. So, v* plus RC dv/dt is equal
to v+ minus v-. RC dv*/dt plus v* is v+ minus
v-. You have done this millions of
times before, but yet again.
This voltage here is equal to the drop across these two,
and the drop across these two is v*, the drop across C,
plus the current through the capacitor C dv/dt times the
resistance R. Or you can apply the node
method as well and get the same expression.
Now, we also know here that vO divided by A is v*.
I can go ahead and replace this guy here, v* by vO divided by A.
RC/A dvO/dt. Recall, I want the dynamics of
vO so let me just get an expression in vO.
So, I get vO divided by A plus v+ minus v- equals.
Now, I want an expression in vO, an equation in vO,
so I need to express v+ and v- in terms of vO.
What are these expressions? The expression for v- is vO and
this voltage divider, so it's vOR3/(R3+R4).
And just for simplicity, let me call this some constant
gamma minus. This is some fraction
R3/(R3+R4). And let me call that fraction
gamma minus. Similarly, v+ is vO R1/(R1+R2).
And let me call that gamma plus.
All I am doing is replacing v+ and v- in terms of vO.
So, effectively, what I have here is v+ is some
fraction of vO. That's the best intuitive way
of thinking about it, some fraction of vO.
And v- is some fraction of vO as well.
And I just stick these. I now have an expression in vO.
Don't get psyched by gamma plus and gamma minus.
Simply read this as if it is an F1 and F2 if you would like.
So, vO times some fraction minus vO times some other
fraction. I am feeding back some fraction
of the output to the positive and to the negative terminals.
Then, just moving things around a little bit,
dividing throughout by A divided by RC.
So, I divided by A divided by RC.
Plus vO divided by RC. And what I am going to do here
in a second, vO gamma plus minus gamma minus.
And I have multiplied by A divided by RC throughout.
Finally, collecting all the vO terms I get vO times one divided
by RC plus A divided by RC. I got a plus sign here so I
will just reverse these two guys in there, gamma minus minus
gamma plus equals zero. All I have done here is simply
grunged through some math to express this equation in terms
of vO. And just to make it even
simpler, I will just replace this thing by one divided by T,
much as we did for first order equations.
What I end up with is dvO/dt+vO/T=0.
Despite all the grubbiness, I end up with something that is
very, very familiar to all of us.
I went through a bunch of gyrations to substitute for v+,
v- and v*, but at the end of the day I got the simple
expression which was dvO/dt+vO/T=0.
Where capital T is the time constant of the circuit,
and the time constant of the circuit relates to the
expression in there 1/RC+A/RC(gamma minus - gamma
plus). The gamma minus and gamma plus
are the respective portions of the output fed back to the
negative input and the positive input.
Now, as we all know, based on very simple intuition
that we can completely predict the behavior of a first order of
an RC circuit once we know what the initial condition of the
capacitor is and once you know the time constant.
That's it. We know, we are masters at the
fact that the capacitor is going to behave like this.
It is going to be exponential. And I do know that the time
constant capital T. What's here?
It is simply the initial condition.
There is no drive input. I am not driving this with any
input here. There is no input drive
anywhere here. This is simply the natural
dynamics of the system. And, recall,
I start off with bumping the capacitor voltage such that the
output starts off being K. That is it.
You should be able to write down this expression and the
form of the response simply based on this.
So, this is what I bumped up the output to be by perturbing
the capacitor voltage. My output response based on
this equation is going to look like that.
Let's try to understand what that means.
It is actually quite a lot of fun.
How do we plot that response? You all learned that the way to
plot the response is plot the initial value,
plot the final value, and go cachoock,
right? It's pretty simple.
I am going to start at K. I know that.
I am going to start at K and I am going to go and find out what
the steady state value is. Here is where the interesting
stuff comes in. The final value on the
capacitor depends a lot on whether T is positive or
negative. In my RC circuits that I looked
at what was T? In the very simple RC circuit
we looked at what was capital T? What was the time constant?
RC. This was RC.
This was a positive quantity. When capital T is positive my
output is going to look like this.
When T is positive. And T is positive when this
expression is positive. And if A is so large that I can
ignore the 1/RC term, if A is very,
very large and I can ignore the left-hand term here then T is
positive when gamma minus is greater than gamma plus.
So, when gamma minus is greater than gamma plus,
I have a stable circuit, this is the good-old stuff we
have seen before. Now things begin to make sense.
Intuitively, what am I saying here?
All the gammas and other pieces of crapola aside,
what am I really saying here in English?
What I am saying here is that if the portion of the output fed
to the negative input is greater than that fed to the positive
input then I have net negative feedback.
I have net negative feedback. I am feeding the output back to
both the positive and negative inputs.
And if my negative input has a stronger effect then I am going
to see the op amp output decay down to a value that I expect
which is going to be zero. Notice that since I am not
applying any input here, I expect the stable point for
this to be output going to zero. I don't have any input there.
Let's take a look at another situation.
What happens when the opposite is true?
What happens when gamma minus is less than gamma plus?
When I feedback more, what happens when I do this,
when gamma plus is greater than gamma minus?
The opposite is true. This means that I am feeding
back more to the positive input. A bigger proportion goes to the
positive than the negative. What happens then?
Then what happens is capital T becomes negative.
We cannot see this happening on the RC circuit because capital T
is equal to RC, but here we have a more
complicated circuit and capital T can go negative.
If capital T goes negative then this whole thing in the exponent
there goes positive. If that goes positive what
should the output look like? It should take off into
never-never land. There we go.
I start off at zero and a make a small perturbation,
and the output should go as t divided by capital T.
The dynamics of this it goes berserk, so it is net positive
feedback. This is called a stable
situation. This is unstable.
What happens when capital T goes to infinity?
When capital T goes to infinity, spend five seconds
thinking about what it means physically.
What does it mean for the time constant of an RC circuit to go
to infinity? That means that your R and C
are very, very, very large.
That means that circuit is going to be very,
very sluggish. Think elephant.
A big time constant. I want to move a leg.
It takes a while to do that. Think big.
Big time constant. So, everything is going to
happen really slowly. It's like moving in molasses.
Big time constant. Everything is going to happen
really, really slowly. If gamma minus is greater than
gamma plus with a huge time constant it is going to look
like this. And the output is going to look
like this. I make T even larger.
All right. It is going to like this.
I make these so large that T tends to zero,
T tends to infinity in which case I get this situation.
The output goes dah. OK?
Very slow. Very lethargic.
Big time constant. T tends to infinity.
And so if this is stable, this is unstable,
this is called corresponding neutral.
And there is a mechanical analog to all of this.
You can show that this situation is akin to let's say I
had a physical well of the sort and I had a ball in there.
I let the ball go. Then the ball will come down
here and settle down in a stable state.
Any small perturbation of the ball will get it to come down
and settle down here. The unstable situation is this
situation where I have a ball sitting up here where any small
perturbation will get it to zip down to a positive rail or to a
negative rail. So, this is an unstable
equilibrium situation. And exactly the reason we got
this analysis in the static situation is that this can
happen. If I do this circuit here and
don't perturb it then I could get the output sitting at zero,
but the slightest perturbation, boom, it is going to fall down
or go up. What about the neutral
equilibrium state? That can be modeled like a
table top and the ball is here. It doesn't matter where you go.
There you are. How many people saw the
Buckaroo Bonzi thing? Possibly well before your time.
OK. I have this table here.
No matter what I do to it, it just goes and settles down
where it is, and that is neutral equilibrium.
But what this gives you is a fun view of the dynamics of the
operational amplifier as I make small perturbations to it.
And the even more interesting thing here is you have the tools
based on your first order RC analysis to analyze the dynamics
of a simple op amp circuit. OK, so much for theory.
Now let's get to some action here.
All right. Fine.
That is really pretty, good and so on,
but what can you do for me? What good does this property do
for me? What can I build?
What we will do is look at the op amp circuit and focus on the
situation where I have net positive feedback.
In particular just look at this circuit with R1 and R2 and send
both to infinity. So, I have no negative feedback
and I ground this terminal here and take a look at what happens
to a circuit with positive feedback and see if I can build
some interesting circuits. What you are going to do is
build on a circuit called the basic comparator.
What is that? If I have an op amp that looks
like this, and remember a VS rail and minus VS supply there,
this is v+, this is v-, I can build a very basic
comparator by doing the following.
All the circuits I am going to show you are going to build on
this basic little circuit. What I am going to do is
consider applying an input to the v- terminal,
applying some sort of an input and taking a look at how the
output behaves. So, I apply some input vIN.
And if I just do that, if this is v+ minus v- here
then I am going to get something that goes like this.
That is when this is positive here then this guy is going to
go to the VS rail and this guy is going to go to the minus VS
rail. In terms of the,
if I plot the same thing, in terms of vIN,
and this is vOUT, if I plot the thing in terms of
vIN then notice that as vIN increases this guy should go to
a negative rail. So, in terms of vIN it looks
like this. What this says is that as the
input becomes more and more positive applied to v- then the
output goes to minus VS, and if the input becomes more
and more negative then the output goes to VS.
This is what is called a very basic comparator circuit.
It compares the two inputs and goes up if the input is in one
direction and goes to the other rail if the input is in the
opposite direction. So supposing I feed this- I can
plot this is a function of time. Let's say I plot vIN.
Let's say I feed some vIN here. Let me just call this.
I feed some vIN to this circuit here, then what do you expect
the output to look like, the output wave form?
For all positive vINs the output is negative.
So, my output vO is going to be negative as long as vIN is
positive. And when vIN becomes negative
this one shoots up and behaves like this.
This is minus VS. That is plus VS.
This is my input vIN. Then this guy is going to be my
output. As vIN is positive output slams
to the negative rail. When vIN becomes negative the
output slams to the positive rail.
So, that is quite nice. And so such a circuit is pretty
useful to me. Let's say, for example,
I want to build a little digital circuit that is fed ones
and zeros. I can use a comparator to turn
my vIN voltage into a sequence of ones and zeros.
When vIN is positive I produce a zero and when vIN is negative
I produce a one. I can get this one,
zero, one, zero sequence coming out corresponding to the values
of vIN being greater or less than zero.
Now, one problem with something like this is that this circuit
can be quite messy in the following situation.
Suppose I superimpose a small amount of noise in vIN.
In particular, let's say that I have some
amount of noise on vIN. I get a bunch of noise sitting
around here. What happens is that at this
point where the value goes negative, I do bump up.
But when for a second I have my input going above zero again --
-- this output comes down again and out here it goes up again.
I get this nasty behavior at the point where the input is
around zero. When the input is around zero,
the input is meandering around zero because of noise,
I get a huge amount of up and down glitches on the output.
That's not very nice. And we will do a little circuit
that attempts to fix that little problem.
What we are going to do is use positive feedback.
And I am going to build you a circuit that shows that we can
eliminate this for small noise on the input.
So, let's build the following circuit.
So I still feed vi to the negative input,
but this time around I give it some positive feedback.
So, I give it some positive feedback.
And what I am going to do is feedback a portion of vO to the
positive input. This is positive feedback.
And, in particular, let's assume that VS equals 12
volts. And to the negative one I
connect -VS. This guy is going to go between
12 and -12. And correspondingly because
these two are equal this one is going to go between 6 and -6.
This is going to be a 12 or -12.
Remember, the top rail and the bottom rail.
And this one is going to be a +6 or -6.
And let's understand how this circuit works when I apply an
input vIN. Let's start by saying that
assume my input is zero for a moment.
And let's say my output starts off being 12 volts.
The output is 12 volts then the input here is going to be 6
volts. In this case v+ is going to be
6 volts. The output is 12,
v+ is going to be 6 volts. And my circuit is sitting out
there doing nothing. Now, this started off being
zero. Let's say vIN increases.
As vIN begins to increase what happens?
Well, nothing until vIN reaches 6 volts.
Since this is 6, vIN has to go up to 6 volts,
has to equal this voltage before I can flip the circuit.
What happens when vIN is greater than 6 volts,
if vIN goes above 6 then I have more voltage on a negative
terminal than the positive so the op amp flips its state.
And vO gets to -12 volts. When vi goes above 6,
vO gets to 12 volts. And what does v+ go to?
In this state v+ goes to half of -12 which is -6 volts.
Now, this guy is sitting at -6 and this guy is sitting at -12.
If this one keeps rising nothing happens,
so output can stay at -12. So I am pretty safe.
Then let's say v begins to come down.
As v begins to come down, does anything happen when v
gets to 6 again? If v is equal to 6 what
happens? Nothing because this is at -6
now. So, there is still a huge net
negative voltage here from v+ to v-.
And so therefore I sit at -12. Oh, well, I keep coming down
until I reach -6. When I reach -6 here these two
become equal. And what happens when this
becomes less than -6? v- becomes less than -6.
If this one goes below this voltage, this is -6 and this is
-7. There is a net positive voltage
between v+ and v-, so this output swings to the
positive rail like so. We will spend a lot more time
on this in the next few minutes to really hammer the point home.
What is interesting about this is that even though the moment
vi became more than 6, I swung to the positive rail,
and then I had to go all the way back down to -6 before I
could change state. I had to go way down before it
could flip again. How can we make use of that?
Well, let me draw you a little vi versus vO diagram and then
talk about how that can be useful to us.
This is vi, this is vO, this is zero.
Let's say this is 12, -12, -6, +6.
Let's plot that on the screen and see what it looks like.
As I told you, the output was at 12 volts to
begin with and my input was at zero.
So, my input kept increasing. When the input hit +6 what
happened to my output? My output swung down to -12.
As the input kept increasing nothing happened.
This was step one, this was step two,
step three. My input kept increasing and
output stayed at -12 volts. Then what I said was well,
let's bring the input down. So, my input began to go down,
step four, became more and more negative.
Nothing happened until I reached -6.
When I reached -6 I swung positive, step five.
Again, one, two, three, four,
five. I am going up here.
It came up here. And nothing happens until I
reach -6, but at -6 boom, I switch to the positive rail.
And as I get more and more negative I stay there.
Then again, as I start increasing again,
nothing happens until I reach +6.
Think of that as your seventh step.
What is spectacular about this is that I seem to have a circuit
that now has some knowledge of where it came.
If it is coming from here it switches at +6,
but if it is coming from here it switches at -6.
So, there seems to be sort of a lag in the behavior of the
circuit or some memory property in the circuit.
This kind of behavior is called hysteresis.
The word comes from magnetic circuits where,
or rather elements that you're trying to magnetize.
Where if you take a magnet and move it over a piece of metal it
may leave some residual magnetism in it.
And, in the same way, that is called hysteresis.
Same way here. As the voltage increases it
seems to leave some residual in the circuit so that it effects
when it shifts. The good news with this is that
now, if I take the same kind of noisy wave form that I had
before and do this -- If this is vi then what is
going to happen is for vO I am going to be negative at this
point. Nothing happens here because I
have to get to -6 or +6 before something happens.
Out here I get to -6 and I switch state and go up to +12.
And then this one comes up above -6 very slightly out
there. Nothing happens because the
next change will happen only when the input goes to +6.
So, if eventually the input gets to +6 and then I am going
to change state again. It is actually a really cool
property and something that is completely non-obvious.
In the last 30 seconds let me show you a quick demo.
And, based on this property of hysteresis, I have actually
built a little circuit. Let me do that first.
Notice here that I am showing you the input on the X axis vi
and vO on the Y axis. Notice how the output switches
at +6 volts and switches at a -6 volts to +12 or -12.
That's the hysteresis property. And we can actually use this
property to build a clock circuit, which is on page 9,
build an oscillator that sits there and oscillates by itself.
And you will see details of that in recitation tomorrow.