[SOLVED] thevenin equivalent for source follower

Hello,

I have a question concerning the derivation of the thevenin equivalent circuit shown on page 16 (upper part) in
https://www-soc.lip6.fr/~hassan/lec3_single_stage.pdf

I know that the small signal equivalent of a diode connected transistor corresponds to a 1/gmb resitor.
It is also clear how to retrieve the gain Av in the left circuit directly (by KVL/KCL).

But I'm missing an systematic way of how to get from the left circuit to its Thevenin equivalent.
I would proceed as follows:
- the Thevenin equivalent is calculated by considering only Vin and the gm-current source (the circuit part with 1/gmb is not considered in the calculation since it is left the same in the Thevenin equivalent circuit)
- for the Thevenin equivalent resistance it is then clear: short-cut Vin and get a diode connected transistor 1/gm.

But how to retrieve the thevenin equivalent voltage source value Vin?
- If I measure the voltage of the open-circuited Vout in the left circuit (ignoring the 1/gmb part of the circuit as mentioned above) I cannot retrieve any voltage "Vin" since the gm-current source is ideal.
- If I add the 1/gmb part to the calculation then it's not a "real" Thevenin equivalent" anymore (consisting of only a voltage source and a "single" resistor)

Does someone know what is missing in my approach?

Thanks in advance!

wasserkasten,

You have an dependent current source that is defeating your attempt at making a Thevenin equivalent. What you need is the General Immittance Theorem (GIT). This theorem states that if you know the transfer function of a circuit, you can get the input and output impedance from the denominator of the transfer function by eliminating the source impedance from the denominator. Let's illustrate. You are given Av, the transfer function. Convert the conductances to resistances and we get Rmg/(Rg + Rmg). Now look at the denominator carefully. If we eliminate Rmg, we get the impedance looking into the circuit from where Rmg was connected. That impedance is the Thevenin impedance Rg. Converted to conductance it is Rg = 1/gm in series with Vin. This is shown on the right side schemat. If you need help finding Av, I can show you how to do that also.

Ratch

Hello Ratch,

thank you for your quick response.
I'm interested in finding out more about this "General Immittance Theorem". Do you know where I can find its proof? (I could'nt find any in the web - maybe it's name is different)

Nevertheless for this case I'd like to use Thevenin's theorem since it was also used in Razavi's book and I'd like to understand how he managed it.
So how to retrieve the "value Vin" of the equivalent voltage source (retrieving 1/gm or Av was not a problem)?

Attached I depicted my approach of finding the equivalent circuit.

wasserkasten,

I'm interested in finding out more about this "General Immittance Theorem".

Click to expand...

I cannot find out anything about it in textbooks or on the web either. I found this method in Electronic Design, Feb, 1965. I consider this "lost" knowledge. If you find something about it, let me know.

Nevertheless for this case I'd like to use Thevenin's theorem since it was also used in Razavi's book and I'd like to understand how he managed it.
So how to retrieve the "value Vin" of the equivalent voltage source (retrieving 1/gm or Av was not a problem)?

Click to expand...

He found the Thevenin equivalent, but who says he used Thevenin's method? Especially for dependent sources. I think the applied source method was used. It consists of applying a voltage, calculating the current, and finding the impedance by the resistance formula.

In the example you submitted, removing Rmg and applying a voltage Va will result in Vi - V1 = Va, -V1/Rm = Ia . So Va/Ia = [-(Vi-V1)*Rm]/V1 . Setting Vi to 0 gives Rm, which is the Thevenin impedance. That converts to 1/gm.

Ratch

The knot in my head was solved by a lecturer:
gm*Vin equals 0 (since the output impedance Rds (approximation) is infinity and there is no way for the current to flow)
Hence the last equation in the picture leads to Vin=Vx