Finding Thevenin Equivalent Circuit

1. Jan 21, 2008

opticaltempest

Hello,

I am working on the following problem:

I am stuck on this seemingly simple circuit problem. I've worked many Thevenin circuit problems in my introductory circuit analysis. This problem is from a new class and seems different. I realize that I must first find the open-circuit voltage at the terminals. After that I need to the Thevenin equivalent resistance by using

$$\LARGE R_{Th}=V_{Th}/I_{sc}$$.

First let me find the open-circuit voltage $$v_{oc}$$. Is this correct?

We know the circuit current is $$g_mv$$. Thus the open circuit voltage which is the voltage across the voltage-controlled current source is

Using KVL:

$$\LARGE -v_s+v-v_{oc}=0 \implies v_{oc}=v-v_s$$

and

$$\LARGE v=(g_mv)R_1$$.

So we have

$$\LARGE v_{oc} = (g_mv)R_1-v_s \implies \LARGE v_{oc} = (0.002v)(50k\Omega)-v_s \implies \LARGE v_{oc} = 100v-v_s$$

Does this look correct for the open circuit voltage?

Last edited: Jan 21, 2008
2. Jan 21, 2008

mjsd

V_oc = voltage across controlled current source, and assuming +polarity on the arrow-head end, you then have V_oc = Vs - V = Vs - IR where I is given by -gm V (passive ref scheme)

3. Jan 22, 2008

opticaltempest

Does this solution look correct? I see where I was wrong while finding the open-circuit voltage.

4. Jan 22, 2008

mjsd

judging by the final diagram with + on top and - on bottom at output, your short circuit current may have been defined in the wrong direction, hence leading to an overall -ve sign error.

5. Jan 23, 2008

CEL

The current $$g_mv$$ is entering the minus terminal of $$v$$, so you should have $$v = -g_mvR_1$$ or $$v\left[1+g_mR_1]=0$$ so, $$v = 0$$ and $$v_{oc}=v_s$$

6. Jan 23, 2008

opticaltempest

How did you get this? Why isn't it

$$v_{oc}=v_s-v$$ ?

Thanks

7. Jan 23, 2008

CEL

It is! But since $$v=0$$, $$v_{oc}=v_s-0=v_s$$.

8. Jan 23, 2008

mjsd

your expression is pretty much correct, all you need is to look at the limit as v->0. basically R_th has to be infinite..... such equivalent circuit does not exist in real life, only in the realm of an analysis tool.

9. Jan 23, 2008

opticaltempest

I'm still confused. Why is $$v=0$$ and not left as $$v=-g_mvR_1$$?

Why do we need to look at the limit of $$v$$ as $$v \rightarrow 0$$ ?

Correction: limit of $$R_{Th}$$ as $$v \rightarrow 0$$

Last edited: Jan 23, 2008
10. Jan 24, 2008

CEL

You have $$v$$ in both members of the equation $$v=-g_mvR_1$$. So, unless $$g_mR_1=1$$, which it is not, you must have $$v=0$$.
Short circuiting the output you have $$i_{sc}=\frac{v_s}{R_1}+g_mv$$.
Since $$v=v_s$$, $$i_{sc}=\frac{v_s}{R_1}+g_mv_s=v_s\left[\frac{1}{R_1}+g_m\right]$$
So
$$R_{th}=\frac{v_{oc}}{i_{sc}}=\frac{1}{\frac{1}{R_1}+g_m}=\frac{R_1}{1+g_mR_1}$$.

11. Jan 24, 2008

opticaltempest

Thanks CEL and mjsd for all of the help on this problem. It now makes sense. I found out the answer is R=495 ohms and vth=vs. This is definately the correct answer when I substitute the given values into the problem. Thanks again!

12. Jan 24, 2008

mjsd

ok... i shouldn't have said R_th is infinite.. I evaluated the limit incorrectly.. since vs = v,