Thevenin equivalent in a nonlinear circuit

Question

Is it possible to use the Thevenin equivalent of a voltage divider driving an emitter follower?

^{simulate this circuit – Schematic created using CircuitLab}

In general, when is it safe to use the Thevenin equivalent of a source if it is connected to nonlinear components? Common exercises always imply a resistor as the load.

Answer 1

Yes, it is possible to use the Thevenin equivalent of a voltage divider driving an emitter follower.

The subcircuit you are replacing by a Thevenin Equivalent is a linear circuit (just the part consisting of 15V constant voltage source and the voltage divider; you are not replacing any of the non-linear part). Therfore using the Thevenin Equivalent for the subcircuit is not only an approximation, it is completely equivalent and perfectly ok. It doesn't matter that that subcircuit is connected to a non-linear subcircuit.

If you would put both implentations of the linear subcircuit each in a black box, there'd be no way to distingiush them from outside; not even by connecting an external non-linear circuit like your emitter follower. That's why it is called Thevenin Equivalent and not Thevenin Approximation.

What you can not do (in general) is to replace the non-linear subcircuit (right part with transistor) by a Thevenin Equivalent. There isn't a definition of a Thevenin Equivalent for a non-linear circuit.

Answer 2

To answer this question you need to consider two things:

• will the transistor have the same biasing ?
• will the small-signal transfer change ?

Let's make things easy on ourselves and assume the beta of the transistor is very high so we can ignore the base current. Then the base voltage would be the same in both circuits. The rest of the circuit (V2,Q1, R2) is identical so YES, the transistor will have the same biasing.

This means that the small signal parameters of the transistors will also be the same !

But will the overall small signal equivalent change ? We only need to consider how the signal comes from V1 into the base of Q1 compared to V3 into the base of Q2. Now I do see a difference ! Do you see it ?

R1 // R3 is indeed 333.3 ohms (= R4). BUT while the signal from V1 is attenuated by R1 and R3, the signal from V3 is NOT !

So the circuit with Q2 will have a higher voltage gain ! Even though from a DC point of view, the transistor cannot tell the difference if it was in either circuit.

This is assuming the input signal has the same amplitude on V1 and V3. But V3 is 10 V DC instead of 15 V DC. If you would also lower the signal amplitude of V3 with the same ratio then the signal at the output would be the same ! Then the circuits would indeed be equivalent !!!

Do you also see that it is not so much the non-linear component that makes the difference ? I mean, the transistor is biased in the same way so it behaves identical in both circuits. But the transfer to the Thevenin equivalent did change the way the signal is attenuated through the circuit.

I do not think you can state that in general it is safe to use the Thevenin equivalent. You have to look at it case by case. And use common sense !

Answer 3

Generally - no.

It's because the Thevenin's theorem is defined for linear circuits. For other (nonlinear) circuits one can't say about Thevenin equivalent.

However, you want something that is similar in its concept but for non-linear circuits. Normally, there are two ways to approach this:

1. you solve circuit equations and check if for some values it is linear or similar to linear, and then you try to approximate,
2. you transform the set of equations of circuits (eg. with Laplace's theorem) and find the transfer function for this circuit. It's generally solving algebra equations.