PROFESSOR: Last time we talked about op-amps and the fact
that they allowed us to abstract away certain
components of a circuit and sample particular voltages
from a circuit and then modify those voltages as we would
like in an LTI fashion.
Today I'd like to talk to you about some other interesting
things that fall out of the fact that we're only dealing
with LTI systems.
In particular, Thevenin/Norton equivalence and superposition.
Thevenin/Norton equivalence is the idea that if you have a
very complex circuit, but it's still a LTI circuit, you can
express it using a linear curve.
And superposition is the idea that because you're also
dealing with LTI components, if you're trying to solve a
circuit, you can take the individual contributions of
independent sources to that circuit and sum them in order
to find out either the total current flowing through a
particular component or the total voltage drop across a
particular component.
At this point, I'll walk you through both.
Thevenin/Norton equivalence is an important concept in that
you may have a very complicated circuit.
And you don't really want to talk about the entire
complicated circuit.
You just want to sample the voltage drop or current in a
very particular location.
Because we're dealing with LTI systems we can actually
express that particular sample as its relationship between I
and V and possibly whatever resistive component is
associated with the voltage drop across that sample.
We can solve for this curve by looking at the location that
were sampling and finding the open circuit voltage
associated with that position.
Or if we were to leave the two terminals from where we are
sampling open, what is the voltage drop across that
section of our circuit?
That's the point right here, where the current flowing
through the system is 0.
And there's a given voltage associated with it.
Likewise if we want to find the other intercept, we can
close those two terminals by running a wire across them and
then look at the current that flows across that wire.
That's the close circuit current.
The slope is going to tell us about our resistance, if any.
If we're only dealing with a voltage source or a current
source, then we're only going to be dealing
with a straight line.
Once we've solved for these values, we tend to express
them either as a Thevenin equivalent circuit or a Norton
equivalent circuit.
And you can convert between the two.
Let's walk through an example.
Here I've got a simple circuit.
And I would like to find the Thevenin equivalent of that
circuit, given that I'm sampling from above this
resistor and below this resistor.
The first thing that I'm going to do is find the
open circuit voltage.
Or if I were sampling from this point to this point, what
is the voltage drop across this section of the circuit?
Well, I've got 36 Volts.
There's 12 Ohms of resistance in the
section that I'm sampling.
And there's 6 Ohms of resistance outside the section
that I'm sampling.
So 2/3 of my total voltage drop across this voltage
divider is going to be accounted for inside my
Thevenin equivalent circuit.
So my Thevenin equivalent voltage is 24 Volts.
Now I want to go after the short circuit current.
If I were to draw a wire from this terminal to this
terminal, I'm curious what this value would be.
As a consequence of connecting these two terminals, these
resistors are going to be completely bypassed.
We don't actually have to drop any voltage or put any current
through these resistors because there's actually an
infinitely easier way for the current to go.
In this case, we're just going to follow Ohm's law.
36 divided by 6 is 6 Amperes.
But because of the directionality on this short
circuit current, current in this system is going to flow
in this direction.
So are short circuit current is
actually negative 6 Amperes.
That just leaves Rth.
Once again we're going to rely on Ohm's law.
If we divide our voltage by our current--
you can't have a negative resistance.
We're actually going to divide by the negative of this.
We'll get out 4 Ohms.
Let's do it again, this time with a slightly different
sample point and find the Norton
equivalent of our circuit.
So if I'm only sampling across the 9 Ohm resistor--
I personally still solve for the Thevenin equivalent
voltage first because I think it's easiest.
I'll just make note of it over here.
The voltage drop across this resistor is determined by the
fact that this is a simple voltage divider.
The other resistors in this circuit--
7, 2, 9 Ohms.
So the voltage drop across this resistor is going to
represent half of the total voltage in our circuit.
That's 18 Volts.
The short circuit current associated with connecting
these two terminals is 36 Volts divided by the sum of
these two resistors.
Once again we're going to end up bypassing this resistor
entirely because, think about it this way, if there was a
resistor across here with 0 resistance, all the current
would sink through it, or it would represent the location
for which infinite current would flow.
36 divided by 9 is 4.
And because we're following in the opposite direction we're
going to say this is negative 4 Amperes.
In this orientation, the Norton equivalent amperage is
actually the negative of this value.
You sometimes see the arrow pointing downwards.
This preserves our property of short circuit current.
If this was short circuited, Isc would still
be negative 4 Amperes.
And our Vth--
if our Vth was the voltage drop across these two points--
ha, I haven't solved for Rth yet.
There are a few ways to do this.
But I'm going to use Vth and In.
18 over 4, we can reduce it to 9 over 2.
And if I were to solve back out for the voltage drop
across this resistor just as a consequence of I Norton, I
would get 18 Volts, which is also the Thevenin equivalent
voltage for this circuit.
That covers Thevenin/Norton equivalence.
And we're going to talk about superposition really quickly.
Superposition is yet another circuit-solving strategy.
It falls out as a consequence of only working with LTI
components.
And it means that in order to solve for a given component
current or component voltage, you can actually find it by
taking the linear combination of every contribution as a
consequence of independent sources.
What do I mean when I say that?
Well, if you have a circuit like this--
and you might recognize it from the
NVCC lecture earlier--
you can actually express it as a linear combination of just
the voltage source and just the current source.
Note what when I remove the current source, I end up with
no connection because if you have 0 current flowing through
a system, it's the same as if the
connection has been removed.
Note that when I remove the voltage source, I no longer
have a voltage drop across that connection, but I
maintain the connection.
If in this example I was just looking for I1.
I1 would be the linear combination of the
contribution of I1 in this particular circuit plus I1 in
this particular circuit.
So this is the expression that I'm looking for.
In the first circuit, I'm just going to use V over IR --
or V equals IR --
and I'm going to have 3 Amperes in this direction.
In this circuit I'm actually going to have to
use a current divider.
Identify that I have five parts total to distribute
among these two branches.
Identify that this side is going to get the inverse of
this side's contribution to the total parts, or this side
is going to get the proportion of parts equivalent to this
side's contribution.
And vice versa.
That this side is the side that I'm interested in.
And so 2/5 of the total current flowing through this
current divider is going to end up in this branch.
2/5 of 10 is 4.
It's in the opposite direction of this current flow.
So I'm actually going to contribute negative 4 Amperes
from this subcircuit.
This should match the results from the NVCC lecture.
That's my super fast coverage of superposition and also
Thevenin/Norton equivalence.
Next time we're going to start a whole new module.