Finding Resistances for EE Fall Homework Assignment

[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.

Homework Statement

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

Homework Equations

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?

 

[gneill]

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

Why would R₃ be parallel to R'_eq?

 

[NHLspl09]

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

Why would R₃ be parallel to R'_eq?

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

 

[gneill]

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

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

g_mv_x is what's known as a dependent current source. The current it produces is a function of the voltage across R1 (V_x) 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.

 

[NHLspl09]

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

g_mv_x is what's known as a dependent current source. The current it produces is a function of the voltage across R1 (V_x) 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.

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 I_R1 = Vx/R1. I am on the correct track so far right?

 

[gneill]

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 I_R1 = Vx/R1. I am on the correct track so far right?

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.

 

[NHLspl09]

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.

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)

 

[gneill]

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)

Yes

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

 

[NHLspl09]

Yes

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

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

 

[gneill]

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?

 

[NHLspl09]

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?

I believe:

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

 

[gneill]

Yup again

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

 

[NHLspl09]

Yup again

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

Nice

I attached what I got for R'_eq:

 

[gneill]

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.

 

[NHLspl09]

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.

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:

 

[gneill]

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:

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 R_eq.

 

[NHLspl09]

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 R_eq.

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

 

[gneill]

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

 

[NHLspl09]

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

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 = R_eq

 

[NHLspl09]

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

 

[gneill]

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

Looks fine.

It sure looks easy once its cleaned up!

 

[NHLspl09]

Looks fine.

It sure looks easy once its cleaned up!

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