Hello, this is Doctor Cynthia, first at the University of Utah, and today we're going to talk about
Thevenin and Norton equivalent circuits. Thevenin and Norton are used when
we analyze systems to allow us to simplify a complex system into simpler block diagrams that we can
analyze or design individually. The concept of input and output resistance
becomes very important in this case and can or not lead to voltage loading.
We always consider designing our circuit as simple feminine equivalent
to a Norton equivalent shown here, and they can be converted back and forth between each other using
source transformation. We'll also talk about maximum power transfer and the updated ECE toolbox
using these two powerful methods. We actually do this all the time we if
we want to design a simple lamp or a fan, we don't actually consider the entire
Power Distribution shown in this box. We convert it instead to a feminine equivalent circuit.
Here is the 110 Volt source in your house and an equivalent source resistance, say that's 10 ohms.
What we're saying is that the voltage and the current seen at these two
points are the same as the voltage and current that they're seen at these two points.
So we've done an equivalent circuit. We can then analyze this circuit
as a set of blocks, as shown here. Let's consider the total resistance and find the current,
and let's see what voltage we're measuring across our fan in our lamp.
We can see that these two are in parallel, so 100 in parallel with 100 is 50.
Plus the 10 in series means that our total resistance is 60 ohms. If we consider the current here,
that's 110 volts divided by the total resistance, or 1.83 lamps, 3 amps.
Now let's consider what's the voltage across the fan in the lamp using a voltage divider.
That's going to be the 110 volts divided by their fifty ohms divided by sorry times their fifty ohms
divided by 10 + 50 or 60 ohms, which gives us 92 volts. Now wait a minute.
That is significantly less than the voltage that we started with. That's called voltage loading.
Where did the voltage go? It went across our S right here. What could we have done to reduce
this voltage loading? We could have reduced our source resistance to a much smaller value
so that we got a higher value here. Take a minute, stop the video and see if you
can calculate what the voltage is across the fan in the lamp.
Is it closer to 110 volts? Now let's talk about the input and output resistance.
The input resistance is the resistance looking into the circuit, and the output resistance is the is
the resistance that's looking into the output part of the circuit.
So here is an output resistance and here is another input resistance. Let's number these.
This is R out for the first block, R in and R out for
the second block, and R in for the third block.
Here's how you can decide if you need to consider voltage loading or not.
If R out from our first block is much less than, not less than,
or equal to much less than R in for our second block, then no voltage loading occurs and the
circuits can be considered decoupled. So in this case,
is 10 ohms much less than 100 ohms? No, In fact, probably not.
How about one ohms? Yes. So in this case they would still be considered coupled.
In this case they would be considered decoupled. Let's consider the next set.
Let's say R out two.
Is that much less than R in 2R in three? Absolutely not.
In this case we have 100 ohms. Is that much less than the 100 ohms input to R3? No.
So these circuits must always be considered coupled because they are not much less than.
This tells us that we need to analyze these two together, and in fact we did because we
considered them in parallel. But if we were using our one ohm source,
if this was a simple one ohm source, we could at least consider the source
block independent from the load blocks. OK, now let's go back to look at
our Thevenin equivalent circuit. What we have done is considered two points,
A and B in our circuit, and we have analyzed those as a very We've taken our actual circuit and
converted it into a much simpler Febinine equivalent circuit. In order to do this,
we need to find two things, the V Thevenin and the R Thevenin. Let's see how to do that.
In order to find the V thevan and the first thing we do is what's called the open circuit method.
We remove the loads and we open circuit the actual circuit and we measure, calculate or simulate the
voltage between points A and B. That's called the open circuit voltage. So similarly,
as soon as we measure that open circuit voltage, we can see that Vth is equal to Voc.
Again, we can measure, calculate, or simulate this. There are three ways to
find the Fevodin resistance, and I'm going to show you all three. The first one can only be used if
there are no dependent sources, and it's probably the easiest of the methods.
What you do is you deactivate all the independent sources, you convert a voltage to a short circuit,
a current to an open circuit, and of course you can't have dependent sources.
Then you simply look into points A and B and you see what the equivalent resistance is there.
And that's equal to the Fevodin resistance. Let's do this. Here's an example.
We want to do a feminine equivalent circuit of this more complicated circuit between points A and B.
The first thing that we do is deactivate the voltage source by shorting it,
and deactivate the current source by opening it
up as we have here. Then we look to see what does the
resistance look like when we look inside. Well, we can see that we have 250
ohm resistors in parallel, so that gives us a 25 ohm resistance. That block of two would be
in series with this 35 ohm. So combine the 25 and 35 and that's going to give you 60 ohms,
which when you put it in parallel with this 30 ohm is going to be the Rth is equal to 20 ohms.
We could have measured, simulated or calculated this. In this case we calculated it.
Now here's the 2nd way to do this. Let's use the open circuit short circuit method.
We have already opened the circuit in order to find Voc. We can and so that's our. That's the Voc.
Then we can short circuit the load and measure the current. In that case,
R Thevenin is equal to Voc divided by ISC. Again, we can measure,
calculate or simulate this. Let's calculate it. Here's another example.
Let's find the Thevenin equivalent between points A and B. Now, the first thing that we're going to
do is measure the Voc across here. So let's do that using node voltage.
In this case, we're going to have VC
-24 / 6 plus VC -0 / 12 plus A7 amp outgoing.
And then we don't need to consider this because any resistor that is connected to
an open circuit does not have a current going through it and can be neglected.
So we can use this to calculate that V thevenant is -12 volts. That shows up right here.
And notice we simply turned the V thevenant upside down. OK, now let's find the R thevenin.
We need the short circuit current, so let's take VC prime because this is a different circuit.
VC prime -24 / 6 plus VC
prime -0 / 12 + 7 amps. And this time the two ohms does show up.
So it's VC prime -0 / 2 ohms that I'll
write here says VC prime is -4 volts. Then we consider the I short circuit current.
We know what VC prime is, so we can find ISC. That's -2 amps.
In this case R Thevenin is V Thevenin
which was -12 / I short circuit -2. That's six ohms. A common mistake.
Please don't make this is do not take the VC prime and put it in here.
It's supposed to be the V Thevenin divided by the I short circuit current.
So that gives us V Thevenin is 6 ohms.