# Homework Help: Finding Resistances

1. Jul 18, 2011

### NHLspl09

Starting the last problem on my EE summer homework assignment for the Fall semester, I wanted to post my thoughts on how to find the resistances before I went ahead and did all the work. I am given what seems to be a fairly simple circuit and asked to find two resistances.

1. The problem statement, all variables and given/known data

(Attachment 1 - EE P2)
Derive an expression for the resistances, R'eq and Req, for the small-signal circuit shown below.

2. Relevant equations

3. The attempt at a solution

(Attachment 2 - EE P2.1)
My thought process is exactly what I have written:
1. Solve for R'eq - First R1 and R2 are in series, and then the result is in parallel with GmVx
2. Req = R'eq in parallel with R3

Any thoughts or input/am I wrong in doing this?

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• ###### EE P2.1.JPG
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2. Jul 18, 2011

### Staff: Mentor

vx is an unknown, so it shouldn't appear in your equivalent resistance. Also, gmvx should be a current value, so it doesn't 'play well' as a parallel resistance.

Why would R3 be parallel to R'eq?

3. Jul 19, 2011

### NHLspl09

I see, I've never actually seen current sources in this form until this homework so I keep forgetting until I'm reminded by you. Do you have any suggestions for that? I firgured R3 would be in parallel with R'eq (once R'eq was found) because they would be sharing the node when I set it up and sketched the circuit

4. Jul 19, 2011

### Staff: Mentor

I see. Let's concentrate on R'eq first. Start by having a look at what depends upon what in the circuit.

gmvx is what's known as a dependent current source. The current it produces is a function of the voltage across R1 (Vx) The voltage across R1, in turn, depends upon the current flowing through R1 (and R2 since the two are in series). This current is set by the voltage that appears at the right hand end of R2 where it meets the top of the dependent source. All right so far?

Now, one way to determine the resistance of such a network is to imagine a fixed voltage source, let's call it Vo, attached across the network at its output where you want to find the equivalent resistance. This voltage source will drive a current, call it Io, into the network. The resistance of the network is then given by Vo/Io.

You should be able to solve for Io in this circuit. Start by determining the current through the resistors and finding Vx.

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5. Jul 19, 2011

### NHLspl09

Alright I understand that, and I know that once the current Io is driven into the network an equivilent amount of current must be driven back out (KCL - $\sum$currents in = $\sum$currents out). With this being said Io = Current over R1 + Current over R2. The current over R1 is IR1 = Vx/R1. I am on the correct track so far right?

6. Jul 19, 2011

### Staff: Mentor

The current through R1 and R2 is the same; they are in series. So don't add the current in each! And Io is going to split between the resistor path and the controlled current source path; Those two currents will sum to Io. At this point in time Io is the unknown that you want to solve for.

Vo is fixing the voltage across the series connected R1+R2. So find the current through the resistors using Ohm's law. Then determine Vx, again using Ohm's law. This Vx will then depend only on Vo and the resistor values.

7. Jul 19, 2011

### NHLspl09

I feel like I'm making this more difficult than it needs to be and end up confusing myself, so I apologize in advance for the dumb statements/questions so just to be sure there's no confusion and to answer your questions/statements - the current through the resistors is Vo/(R1+R2)

8. Jul 19, 2011

### Staff: Mentor

Yes

So if that's the current through R1, Vx must be ______ ?

9. Jul 19, 2011

### NHLspl09

Vx = ($\frac{Vo}{R1+R2}$)R1

10. Jul 19, 2011

### Staff: Mentor

Yup.

So now you can use this expression for Vx to get rid of the Vx in the controlled current source. Voila! No more unknown Vx.

Time to find Io. You have the current through the resistors, and your newly minted expression for the controlled current source current. What's Io?

11. Jul 19, 2011

### NHLspl09

I believe:
Io = ($\frac{Vo}{R1+R2}$) + gm($\frac{VoR1}{R1+R2}$)​

12. Jul 19, 2011

### Staff: Mentor

Yup again

So now you can form Vo/Io, yielding the equivalent resistance R'eq.

13. Jul 19, 2011

### NHLspl09

Nice

I attached what I got for R'eq:

#### Attached Files:

• ###### EE P2.pdf
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14. Jul 19, 2011

### Staff: Mentor

Alright. I see that you've put in I(R1 + R2) for Vo. Why not leave Vo as is and then it'll cancel with the Vo's in the denominator. All the unknowns and 'working variables' will then have 'vanished', leaving just component values. You can then simplify the expression a bit more.

Last edited: Jul 19, 2011
15. Jul 19, 2011

### NHLspl09

Yeah when I initially did that it seemed a bit cloudy, I gave what you suggested a shot and think I have found it through algebra:

#### Attached Files:

• ###### EE P2.1.pdf
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16. Jul 19, 2011

### Staff: Mentor

Bravo A nice, simple expression for R'eq.

To summarize the plot so far: The circuit was "broken" to remove R3, and the equivalent resistance of the remaining network, R'eq was found by applying a fixed voltage source Vo to the network at that point and determining the current it would produce. The math went (essentially) as follows,

Now you can tack R3 back onto the circuit and add its effect, yielding Req.

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• ###### Fig1.gif
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17. Jul 19, 2011

### NHLspl09

Awesome question: what exactly do you mean yielding Req? My first instinct when looking at this was that R'eq and R3 would be in parallel since they 'shared' the same node

18. Jul 19, 2011

### Staff: Mentor

If you look again I think you'll find that they are in series.

#### Attached Files:

• ###### Fig1.gif
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19. Jul 20, 2011

### NHLspl09

Awesome!! I didn't think of it like that - I figured the node to the left of R'eq meant it was in parallel (stupid mistake I know). So all I would do is have R'eq + R3 = Req

20. Jul 21, 2011

### NHLspl09

Just rewrote everything out and cleaned it up a bit! Just checking if Req is correct or not, sorry it's so difficult to read :grumpy:

#### Attached Files:

• ###### EE P2.2.pdf
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21. Jul 21, 2011

### Staff: Mentor

Looks fine.

It sure looks easy once its cleaned up!

22. Jul 21, 2011

### NHLspl09

Haha you can say that again!! Thanks for all the help and input gneill!! Much appreciated