# 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,