Thevenin Equivalent with dependent sources

[Marshillboy]

Homework Statement

See attached image

Homework Equations

Thevenin's theorem
Node voltage method?

The Attempt at a Solution

So if I want the current Io, I believe I need to treat the 1K resistor as an open circuit in finding the Thevenin equivalent. I think setting ground to be the bottom node is a good idea.

One trick I'm familiar with to find the Thevenin equivalent resistance is to replace the current sources with open circuits and the voltage sources with short circuits and find the resistance between the two points. I think this gives me just 1kOhm between "a" and "b"(ground) for my Thevenin resistance. Does this only work with independent sources?

I'm not sure how to find the Thevenin voltage. All these dependent sources are psyching me out. I've tried to use the node voltage method, treating the top right dependent source as a supernode, but I'm pretty sure I got a wrong answer from that, since my system of equations tells me there is an independent variable (which shouldn't happen when solving these equations simultaneously).

 

[gneill]

Homework Statement

See attached image

Homework Equations

Thevenin's theorem
Node voltage method?

The Attempt at a Solution

So if I want the current Io, I believe I need to treat the 1K resistor as an open circuit in finding the Thevenin equivalent. I think setting ground to be the bottom node is a good idea.

One trick I'm familiar with to find the Thevenin equivalent resistance is to replace the current sources with open circuits and the voltage sources with short circuits and find the resistance between the two points. I think this gives me just 1kOhm between "a" and "b"(ground) for my Thevenin resistance. Does this only work with independent sources?

I'm not sure how to find the Thevenin voltage. All these dependent sources are psyching me out. I've tried to use the node voltage method, treating the top right dependent source as a supernode, but I'm pretty sure I got a wrong answer from that, since my system of equations tells me there is an independent variable (which shouldn't happen when solving these equations simultaneously).

You're certainly thinking along the right lines. If you pull out the 1k resistor where you suggest and find the Thevenin equivalent 'looking into' where it was, you're golden. As you say, the controlled sources do make things 'interesting'; They are going to mess with both the Thevenin voltage and Thevenin resistance. So stick your own source there to set the 'measurement' conditions.

Your supernode suggestion is good. Note also that the node where the 6 V source connects with the 4Ix source is a supernode of your chosen ground node. So in the end you're left with a 2-node circuit! So node voltage equations would seem to be a natural for this one.

Start by writing equivalent expressions for Vx and Ix in terms of the node voltages. You can substitute these into your node equations as you go.

 

[Marshillboy]

Thanks for the reply!

So, I solved the circuit you provided using the node voltage method and substituted in the equivalent equations for Vx and Ix (Vx = 6-a, g' = 6, Ix = b-a/1k, etc). However, when I got to the point that I needed to solve my equations, the I from the imaginary source was a third unknown between two equations and I just dropped it, and I think I still got the proper number for the open circuit voltage between a and ground.

From there, I used my trick to zero out the sources to find the thevenin equivalent resistance between a and g, which I think is still 1kOhm. Plugging everything into the final thevenin equivalent circuit, it looks like I have my Vth which I just found, in series with Rth (1K) and the load 1K resistor.

I plugged these numbers into a SPICE program and verifies my answer. Thanks for the help!

 

[gneill]

Thanks for the reply!

So, I solved the circuit you provided using the node voltage method and substituted in the equivalent equations for Vx and Ix (Vx = 6-a, g' = 6, Ix = b-a/1k, etc). However, when I got to the point that I needed to solve my equations, the I from the imaginary source was a third unknown between two equations and I just dropped it, and I think I still got the proper number for the open circuit voltage between a and ground.

From there, I used my trick to zero out the sources to find the thevenin equivalent resistance between a and g, which I think is still 1kOhm. Plugging everything into the final thevenin equivalent circuit, it looks like I have my Vth which I just found, in series with Rth (1K) and the load 1K resistor.

I plugged these numbers into a SPICE program and verifies my answer. Thanks for the help!

Yes, dropping the I term would lead you to the Thevenin voltage. But the controlled sources are still going to futz with the Thevenin resistance. If you'd left it in place you would end up with something like:

$Va = \frac{32}{5} - 200\;I$

From that you can pick out both the Thevenin voltage of 32/5 V (setting I = 0) and the Thevenin resistance since ΔV/ΔI = 200

 

[Marshillboy]

Yes, dropping the I term would lead you to the Thevenin voltage. But the controlled sources are still going to futz with the Thevenin resistance. If you'd left it in place you would end up with something like:

$Va = \frac{32}{5} - 200\;I$

From that you can pick out both the Thevenin voltage of 32/5 V (setting I = 0) and the Thevenin resistance since ΔV/ΔI = 200

Ouch, you're right. That's what I get for being lazy. I did happen to get the correct value for the Thevenin voltage at least.

I understand how you got the Thevenin voltage when I=0, but how did the expression give the Thevenin resistance?

 

[gneill]

Ouch, you're right. That's what I get for being lazy. I did happen to get the correct value for the Thevenin voltage at least.

I understand how you got the Thevenin voltage when I=0, but how did the expression give the Thevenin resistance?

You want to know how the "output" voltage varies with the "output" current, since by Ohm's law R = V/I. The expression for the node-a voltage, if plotted, would be a straight line with slope 200. That slope is the resistance: ΔV/ΔI.

 

[Marshillboy]

You want to know how the "output" voltage varies with the "output" current, since by Ohm's law R = V/I. The expression for the node-a voltage, if plotted, would be a straight line with slope 200. That slope is the resistance: ΔV/ΔI.

Oh I see, that's pretty clever. You're basically creating a linear equation for the potential across the desired terminals and analyzing its slope and intercept for determining the Thevenin equivalent voltage and resistance. Interesting.

 

[gneill]

Oh I see, that's pretty clever.

That's why they pay me absolutely nothing to be here