In this video, we're going to extend our knowledge of Equivalent Circuits to the
most general case. And last time we asked, what does the
source see? And we figured out what we mean by sources seeing things, but in
this video we're going to talk about what the circuit elements see.
This is going to open a door to, the generalization of what we've already
talked about namely Thevenin and the Mayer-Norton equivalent circuits.
These are very important to understand and appreciate the knowledge they give
us. So let's Talk about a little bit of an
extension of what we've already talked about.
You recall this was our original circuit. We had a source, and our system was a
series resistance and we now, we know That, for that case V-out was R2/1+R2 *
VN, we use voltage divider to find it. Well, if you're going to actually use
that voltage for something it has to be attached to the sync.
Remember our whole system thinking that we have to send our message to the sink.
And the input to most circuits that we care about look like a simple resistor.
And so the question is, if I attach my circuit to.
The receiving circuit which is going to look like a resistor, does that change
the output voltage V out. You may say well the voltage v out is
across both the RL and across our original R2.
But in reality it's really across the parallel combination.
Well that changes the resistance, so you should expect there might be at least the
possibility of some change in V out, so I want to explore that.
By the way, RL, this is standard terminology, this L here stands for load,
and that's a term that comes from the power in the circuit.
Well, how do we think about this? And here's the way I want to do this.
I want to try to figure out what does the load resistor see? What is the circuit
that it sees looking back in that way? That's going to be a key for this, so
that's what I meant by what are the circuit elements C.
Well, it's not just a simple resistor of course, there's a voltage source in here,
so the series and parallel rules don't quite apply.
However, we can use the Principle of Superposition to figure this out.
If you think about this for a little bit If I drew a box around here, this just
looks like a regular old generic circuit with a current define going in the
positive side, of the voltage defined for it.
And I could figure out the, voltage if you just gave me the current.
But because of the presence of the source There's going to be an additional term,
which is the part of v that's due to the source.
So the principal of superposition says that voltage V will be equal to the sum
of the voltage you get when you set the voltage source that's inside, V in to 0.
Plus the part you get when you set the input current to 0.
And that will give you the, the part of V, this is to the V in.
Okay? So, let's do the first one. When we set the voltage source to 0,
that's the same as a short circuit. So remember our voltage source supplies
some voltage for any current. Well, if you set that source to zero now
the voltage is zero for any current. That's the definition of an ideal wire,
or a short circuit. So now, what do we see looking into The
circuit, we see R1 in parallel with R2. And so we know that at least that
component of the output voltage is given by a simple resistive like term.
And now, if you go on to setting I to 0. So I'm going to turn the source back on.
Our source is alive and well. In a setting i=0, means there's nothing
attached to the terminals, and what we have is the same circuit that we had
before that we already analyzed. You simply use voltage divider to figure
out what vout is and this is what we got. Last time.
So, v, is going to be the sum of 2 terms. If you look, think about, that, as a
system, and you want to know what the v i relation is, you can use this super
position principle. And we see that, at least for this case,
that's what we get. We get that v is equal to some resistance
linked term times the current plus a source linked term.
Well, that turns out, opens the door to a really interesting result.
So, let's define something. I'm going to call that r e q, equivalent
resistance. I'm going to call the second term v e q,
equivalent voltage. And I want to point out, that something,
that of this form. Which can be written like that, that
looks, that is the same via our relationship that applies to this
circuit. And of course the subscripts EQ here mean
equivalent. So these two circuits cannot, you cannot
tell them apart Just from the terminals. They look identical because their VI
relationships are exactly the same. So I can express in particular as a very
simple circuit, a very complicated circuit, containing voltage sources.
And resistors. We're going to show you in a second that
you can in addition to alter sources, you can have current sources too.
So this is called the Thevenin equivalent circuit, and it's named for a late 19th
century enigneer Who developed the equivalent circuit idea, while he was
learning to teach a course at the eco polytectic in Paris.
So, that's quite interesting. But there's even more.
So, here's our simple little circuit that we've been talking about.
And here's the, Thevenin equivalent. relationship for it.
I'm going to turn that equation around. I'm going to write this as i equals
something, and I'm going to call that last term, IEQ.
And I want to point out that that equisite of that equation Describes
behavior of this circuit. Where there is a current source, in
parallel with REQ, it's the same REQ that we had, when we talked about the Thevenin
equivalent, there's no difference. This is called the Mayer-Norton.
Equivalent circuit. Mayer-Norton is named for two engineers,
Edward Norton who is from the US and Hans Ferdinand Mayer who is from Germany, they
developed this idea, at the, almost exactly same time, but independently.
So. We have two, different ways, of thinking
about, a circuit. So.
We think about these, terminals, this pair of terminals out here, as a place
where you're about to attach something, some other circuit.
And you want to know. What does that circuit look like? In
order to simplify your calculations, you can describe it either as a voltage
source in series with a resistor, or you can describe it as a current source in
parallel with the same resistor. Whichever seems to be the easiest
approach for your application. They both apply.
They, you can do either one. So, this really simplifies things.
In fact, this is a profound result. This says that no matter how complicated
A circuit is in side consisting of sources, current sources, voltage sources
and resistors. It all can be summarized as a simple
voltage source and series with equivalent resistance.
This equivalent resistance is sometimes cal, called the output resistance.
And we'll see that this is an important quantity in just a second.
Now, and these, this equation ties these two equivalent circuits together, so that
VEQ is REQ times Ieq so if you find one. You, you've got all you need to find the
other, no problem. So, one issue with finding equivalent
circuits is, can you really, do it. Find the, equivalent voltage, equivalent
resistance. Without looking inside the box can you
make measurements just at the terminals to figure out what those quantities are?
And the answer is yes. The first thing we do is to find what we
call the open-circuit voltage and that means we set this current to zero.
And we look at the voltage we get. Well, if you look at this equivalent
circuit. If you set that current to 0, that means
there's no voltage drop across that resistor, so v is just equal to v(eq).
So if you do that for a real circuit You just open-circuit the terminals, don't
attach anything, and you just measure the voltage.
That voltage that you measure is the equivalent, the Thevenin equivalent
voltage. Similarly, you can short-circuit the
current, so what that means Is that you short circuit these terminals, you put a
short across there. That sets the voltage to, zero of course
and then you measure the current. And, when you do that, you look at this
equivalent, circuit, that short circuit current, is just going to be minus VEQ,
over REQ, and that's the negative. Of the Myer-Norton equivalent current
source so, I'm making measurements at the terminal, the open circuit voltage and
the short circuit current. You do not have to look inside the box,
to figure out what the equivalent circuits are, so let's
go on, and to a simple example. And so, to show these ideas, I want to
find the equivalent circuits for this example circuit.
And I could find, explicitly, the. Thevenin equivalent.
I could find the minor Norton equivalent, but I'm just going to find, the
equivalent, quantities, the Veq, Req, and Ieq.
so, the first thing we can do is set all the sources to zero.
If you set the source to zero in the Thevenin equivalent, that's your short
circuit, and what you see from the terminals is just that resistor.
If you set the current source to zero, that's the same as opening it up, and
then again, you also see Req. So, here if you look inside the box we
see we have a current source, a zero valued current source.
It's the same as an open circuit, no current flows through any voltage.
That's the definition of an open circuit. And what I see when I look this way.
Into there, I see R3 is in parallel with this series combination.
So we can use our series and parallel formulas and we get that for Req.
So, that's the source resistance of the Rs, of our circuit.
Now, suppose we want to find VEQ. I'm just trying to use our terminal only
results, I'm going to open circuit voltage so there's nothing across here,
this is 0. And now we measure the voltage.
Well this voltage. Is, going to be given by some dependence
on this current source, and since it's a current source, it's a whole lot easier
to find the current through here, and then I'll multiply that by R3 to get the
voltage. Well.
Because we have this in parallel with this now, from the point of view of the
volt of the current source. I'm going to use current divider which is
going to be the other resistor divided by the sum of that one plus this series
combination. And when you all get said and done,
that's what you get. So, here's the, current divider part, and
then I just multiply by R3 to get our voltage.
So that is the Thevinin equivalent, voltage, for this circuit.
It's a little more complicated than you might think, but it's pretty easy to
find. And then finally I'm going to find IEQ,
which I'm going to do by finding the short circuit current.
And I just put a wire across here, and, so what that does by setting the voltage
to zero, that shorts out this, so that resistor isn't there, doesn't matter if
it's there or not. And so, again, I'm going to use current
divider, because I have a current source here.
Well, the current i is going to be, i is going to be the negative of that current,
and I used current divider, so current divider says that current is R1 divided
by the sum of the resistances times I in. And I stick a negative sign in front
because of the directionality of the current and I get this for isc, the short
circuit current which, I know, is the negative of Ieq.
So, we could have divided The equivalent voltage, by the equivalent, current,
defined req. We u-, in finding req up here, what we
did was look inside the box. And we don't really have to do that.
But it's nice to know we have a way of checking your answers.
So if the source is to zero to find Req. Find the open circuit voltage to find an
evident equivalent voltage. find the short circuit current.
Find the Meyer-Norton equivalent current source, and then your ratios should work
out to be the other. Everything should check.
Okay. So, what did we discover? Well, we
discovered that all circuits consisting of sources and resistors are equivalent
to one source, be it a voltage source or a current source, that's in series or in
parallel with a single resistor. No matter how complicated that circuit
is. From the point of view of the connection
points, the terminals. It is a very, very simple circuit.
So, that means that, we can use either of these equivalent, circuits.
The Thévenin or Mayer-Norton form. The voltage source or the current source
form. Whichever seems to be appropriate.
It makes solving the problem easy. What I mean by that.
Is you're now going to attach something to the terminals, and, and des, determine
things. And, depending on what you're interested
in finding, how complicated the circuit is you attach, you would use one or the
other. Now.
Remember how we got started on this. Remember we were attaching our, Sync to
our original circuit That we've been talking about.
And the issue was what's called Loading. And, Loading is a term that means, that
if I attach to the wrong kind of load resistor for the given application.
This V out will change. That's called loading down the original
circuit. Well, I want to figure out what values of
RL will not contribute to loading. What kind of resistor values can the sink
have so the The alk doesn't change from what it is if nothing is attached.
Well, the, we now know that our system has, looks like this.
This, which is, I'm going to use the Thevenin equivalent.
Because I have voltage that's the output. And the voltage I want is across RL.
Well, what is v out for this really simple circuit? Well, it's the voltage
divider. It's very easy to find.
And, what values of RL. We'll allow V out to be = VEQ.
Because when RL is infinity, that's the same as attaching nothing.
and that's what we started. So I know that the output that I want is
VEQ, so I think it's pretty clear that what we want is RL to be big.
But big relative to what? And the answer is it has to be big
relative to Req, which is R1 in parallel with r 2.
So as long as the load is big compared to the Whatever this parallel, resistance
turns out to be, then the, you can attach the sync to the circuit and it won't
change V out. However, if RL is smaller than that
quantity or comparable to it, that outputs going to change and that may not
be what you want It depends on the application.
This little, simple example points out the power of using equivalent circuits.