# Thevenin and Norton Conundrum

1. Sep 23, 2014

### amenhotep

Hi,
I'm trying to find the output voltage of a load attached to the output of a common emitter amplifier. The collector is represented as a current controlled source. Attached to current source are two resistors, $R_{C}$ and $R_{L}$.
The circuit is already in Norton's equivalent circuit form. The two resistors $R_{C}$ and $R_{L}$ are in parallel with the current source.
Now transform the circuit into its Thevenin equivalent circuit. From the perspective of $R_{L}$, $R_{th}=R_{C}$, meaning the same resistors that were in parallel now appear in series with the voltage source.
Now the conundrum :
1. Norton Circuit : If $R_{C}=\infty$, the current source gives all its current to $R_{L}$.
2. Thevinin Circuit : If $R_{C}=\infty$, No current flows in the circuit.
So what's going on ??
Amplifiers that want to pass on most of their voltage have low output impedance. In a Common Emitter amplifier, the output impedance is approximately $R_{C}$. The thing is when you look at it from the perspective of the Norton equivalent circuit, it seems that with a very large $R_{C}$, almost all of the current is through $R_{L}$, meaning $R_{L}$ voltage drop is high. So here it seems that large $R_{C}$ is good. From the perspective of the Thevenin equivalent circuit, a very large $R_{C}$ will take all of the voltage by virtue of its size since its in series with a smaller $R_{L}$. So here, it seems a small $R_{C}$ is good.

I was reading a book on amplifier design when I was surprised when the author converted a Norton Circuit to a Thevenin circuit for him to declare that a small output impedance is desirable to pass on the voltage. Something weird seems to be going on !!

Please tell me where I'm wrong.

Thank You.

2. Sep 23, 2014

### amenhotep

A short version of the question :
In analyzing amplifiers, why is the output impedance described from the Thevenin equivalent circuit and not the Norton equivalent circuit because even though they are equal, their interpretation is different because one is in series and the other is in parallel with the load ?

3. Sep 23, 2014

### analogdesign

Amplifiers usually have a voltage mode output. Even transconductance amplifiers virtually always dump their output current into some high impedance to generate an output voltage. Thevenin eqiv. circuits just make it a lot easier to analyze a chain of amplifiers.

The author is correct. Just look at the simple voltage divider circuit to see a small output impedance is desirable to pass a voltage. The author could have also made the equivalent statement that a high output impedance is desirable to pass a current.

The key point you're missing is that in the Norton version of an amplifier all the current is passing through its own output impedance, not the load. So you see, in an ideal voltage amplifier the series output impedance is zero (so it drives no current into the load).

4. Sep 23, 2014

### amenhotep

I don't really understand what you're saying. Here's the circuit as it looks from in Norton's form. $R_{C}$ is the output impedance. Clearly in this form, a large $R_{C}$ means that most of the current is flowing through the load meaning that a large output impedance is fine. But I know that is wrong. The question is why ?

5. Sep 23, 2014

### analogdesign

Maybe you'll understand it if you actually do some calculations.

Suppose Rc = 100 Ohm and RL = 50 Ohm. Assume for the time being that the Norton current In = 1 mA. What is the current in the load? Well, by the principal of current division, IL = In * (Rc / (RL + Rc) ) = 1 mA * ( 50/150) = 333 uA. This will generate an output voltage of iL * (Rl || Rc) = 1 mA * (33.3 Ohms) = 33.3 mV.

Summary of Norton: Load current = 333 uA. Output voltage = 33.3 mV.

Ok, now let's take the equivalent thevenin circuit:

Vth = In*Rc = 1 mA * 100 = 100 mV. Rth = Rc = 100 Ohms.

What's the output voltage? We can use the voltage divider rule. Vo = Vth*(Rl / (RL + Rth)) = 100 mV * (50 / (150) = 33.3 mV.

So far so good. What's the current in the load?

IL = Vo/RL = 33.3 mV / 100 Ohms = 333 uA.

Summary of Thevenin: Load current = 333 uA. Output voltage = 33.3 mV.

Isn't the something? You get the same value for output voltage AND current delivered to the load regardless of a Norton or Thevenin circuit.

One way to think of it is like that: A voltage amplifier (like a Thevenin eq.) WANTS a low output impedance. A current amplifier (like a Norton) WANTS a high output impedance.

6. Sep 23, 2014

### amenhotep

I actually know that because I've already done the calculations myself. I would like you to address my problem which is how $R_{C}$ appears in the circuit below. Looking at this circuit, if $R_{C}=0$, then no current flows through the load meaning that the voltage across the load is zero. But this contradicts the fact a small output impedance is desirable. Answer why my thinking is wrong and you would have helped me.

7. Sep 23, 2014

### analogdesign

I don't usually quote myself, but...

So, where did you get the idea that a small output impedance is desirable for a current amplifier? The thevenin voltage goes to 0 as Rc goes to 0 so that should show you that Rc can't equal zero.

8. Sep 23, 2014

### amenhotep

Why do you consider a Norton Circuit a current amplifier and a Thevenin circuit a voltage amplifer. Why can't you represent an amplifier by any equivalent circuit you like ?

9. Sep 23, 2014

### analogdesign

You're really close. WHen you define the output as a current (norton) you want large output impedance. When you define it as a voltage (thevenin) you want small output impedance. You're mixing concepts a bit, as Norton and Thevenin are equiv. ways to look at a circuit, that's all.

The key is that if you change from Norton to Thevenin the Thevenin voltage takes Rc into account so there is no conflict there.

10. Sep 23, 2014

### amenhotep

You know what, thanks for your time. I don't think I understand what's going on. For a CE amplifier, the Norton Circuit was converted to a Thevenin circuit just to point out that the output impedance should be small. For the emitter follower, a Thevenin circuit was converted to a Norton Circuit just to show that a high impedance is great for a current source. I don't understand why, if both circuits are equivalent, why bother transform them when you can readily use one in the form that its given which only includes a few resistors. So I am lost and will just have to pretend that I know what I'm doing and continue reading. All what you're doing is pointing out what I am stating which doesn't alleviate my confusion. My question is why transform ?? What can't I analyze a circuit in Norton's form to see whether a large impedance is a good thing or not. I don't want to convert to Thevenin !!

11. Sep 23, 2014

### analogdesign

Why transform? It makes the algebra easier. Similar as to why use nodal or mesh analysis. They give the same answers, but one gives easier algebra in some instances than the other. That's it. If you prefer one, stick with it. I find thevenin's equiv to me more intuitive so I use that almost always when I think about a circuit.

12. Sep 24, 2014

### psparky

Thevenin and Norton are directly related. They both based off of V=IR. One has a current source in parallel to a resistor (Norton)....the other has a Voltage source in series with a resistor. (Thevenin) They are the same thing.

Once I learned them well, I often used thev and norton to solve bigger circuits. Whether I used thev or norton was a simple matter of which one is more convenient to use. Other than that, same, same.

13. Sep 24, 2014

### amenhotep

I think no one really understands my question. My question is with Norton's theorem a large output impedance seems better whereas if you transform the same circuit to Thevenin, a small output impedance seems better. That is my question. Why the same impedance is better in one case and bad in the other case when the two circuits are the same circuits.
Please don't be telling me again what Thevenin or Norton means or which one is easier to use. I have solved enough problems to know which one I prefer.

14. Sep 24, 2014

### Jony130

But this is not about Thevenin/Norton. But about ideal voltage source and ideal current source.
Ideal voltage should have 0 ohm internal resistance. If not the output voltage is not equal to source voltage (see voltage divider).
In case of a ideal current source, the internal resistance of an ideal current source should be equal to infinite. If not the output current is not equal to source current (current divider).

15. Sep 25, 2014

### Staff: Mentor

It does not. Though I can understand your confusion.
Yes, but it's the wrong question, and based on a false premise. :

Let's consider a voltage source of 10V having an impedance of 0.1 Ω
Ideally, when we connect a load, or vary the load, the voltage across the load won't change much. So we'd like the Thévenin resistance to be small in order that changes in load cause only small changes in load voltage.

Let's study the Norton equivalent, viz., a 100 ampere source having a parallel resistance of 0.1 ohms. Is that low resistance desirable? Most definitely! The low parallel resistance means that when you cannect a load, or vary the load, there is not much change in the current through the 0.1 ohms. No change in the current through the 0.1 ohms means no change in the voltage across it and therefore no change in the output voltage, and isn't that what we want in a good voltage source: no change in load voltage even though there may be changes in the external load!

So when you model a good voltage source using its Norton equivalent, it's desirable that it have a low Norton resistance.

Last edited: Sep 25, 2014
16. Sep 25, 2014

### psparky

Yatzee!!!! I think we have a winner ladies and gentleman!!!