# Thevenin Resistance

Discussion in 'Homework Help' started by ihaveaquestion, Oct 4, 2009.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
http://img269.imageshack.us/img269/8452/img0001nw.jpg

I'm finding Rth and applied the 1V source.

My nodal equations give me two different values of V1...

I have a feeling it has something to do with my defining vo in the circuit as V1... isn't it?

2. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Figured this one out

3. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
http://img223.imageshack.us/img223/6213/imgoj.jpg

http://img14.imageshack.us/img14/9519/img0001ge.jpg

This is probably the third time I've encountered this problem... I set up my node equations, solve for the unknowns manually, and end up getting an equation such as 2v3 - 6v3 + j8v3 = 0.... I simply can't get a value for v3 out of this... I'm not sure what's going on, but again I'm fairly sure my nodal equations are correct.. I plug the matrix into MATLAB and get values for v1,v2 and v3.... specifically
v1 = -10.1484 -11.8731i
v2 = -1.1498 -11.9856i
v3 = -7.1489 -11.9106i

if someone could help me out and pinpoint why this continues to occur I'd appreciate it.

4. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
http://img257.imageshack.us/img257/4830/imgha.jpg

http://img525.imageshack.us/img525/5749/img0001gb.jpg

This has me stumped... I set up my node equations (which I am fairly sure are correct), and while solving for the unknowns by hand, I end up getting something along the lines of 2v3 - j7v3 - 14v3 = 0 or something... which will not give me a value for v3.... this is like the third time it's happened and I'm not sure why.

When I plug the equations into a matrix in matlab I get the values
v1= -10.1484 -11.8731i
v2 = -1.1498 -11.9856i
v3 = -7.1489 -11.9106i

Any ideas? Thanks.

5. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
The example with the number I made up above... I was simply trying to explain how my equation has every term with a v3 in it... thus resulting in me not being able to solve for v3.... you can see where this happens on the second link at the end

6. ### syed_husain Active Member

Aug 24, 2009
61
5
hi ihaveaquestion,

the op-amp circuit that u provided, there is a voltage source in timedomain signal 4 sin(400t) mV in the frequency domain it is 40. which one is right voltage source? i hope that was the problem.

cheers.

7. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Thanks for spotting that numerical error, but that still does not solve the problem that I mentioned above.... symbolically if you follow my work, in the end of my work you see I don't have anything to solve v3 for..

8. ### hgmjr Moderator

Jan 28, 2005
9,030
214
One of the terms you are missing.

$\frac{V_{2(-term)}}{20000}=\frac{V_3-V_{2(-term)}}{40000}$

The above equation can be solved for V2 at the the -terminal of the opamp. This can then be used to solve for V3 using the assumption that the voltage at the negative terminal and the voltage at the positive terminal are equal in an ideal opamp.

The expression you have for V3 is incorrect.

Your other equations are suspect as well.

hgmjr

9. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
hgmjr,

I included the equation you wrote on the left-hand side of my equation '@ V2.' I'm assuming it's incorrect to do this? I should instead write two equations for V2 for the + and - terminals instead of all including it in one equation?

Also, why is v3 equation wrong? It looks right to me. We don't know what the current leaving the op-amp is, so I only have the two terms for v1 and v2.

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I mistakened your colon for an equal sign in all of your equations. I could not figure out why you placed the equation for V3 in brackets and then multiplied by 40000. I guess I am unfamiliar with your notation method.

hgmjr

11. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
I do that to make it easier without denominators.

So do I have to write two equations for v2? And do you agree with my v3 equation now?

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Your equation for V3 is not correct. V1 is affected by V3 and so it appears in the expression for V1. V3 is not affected by V1 since V3 is a voltage source.

hgmjr

13. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Yes, that is the method you should use.

I would recommend that you not use the same node identifier for two different nodes. It has the strong possibility of creating confusion.

hgmjr

14. ### syed_husain Active Member

Aug 24, 2009
61
5
hi ihaveaquestion,

in the 2nd page when u found out v1 in terms of v3 u forgot the term 80. so, v1=(0.834+j*0.107)v3+80. put this equation in the next equation u will definitely get a answer for v3 other than zero. hope this helps.

cheers.

15. ### syed_husain Active Member

Aug 24, 2009
61
5
it should be v1=(0.834+j*0.107)v3-80. my mistake

16. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
I think I need help understanding this then.... do you never apply KCL to the output node? How do you derive the equation?

17. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If you are dealing with an ideal opamp then it is not necessary to create an expression for V3.

If you are not dealing with an ideal opamp then you would need to form an equation similar to (in your case) something like V3 = A{V(positive) - V(negative)}. Where A would represent the open loop gain of the opamp.

hgmjr

May 1, 2009
314
0
19. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
hgmjr,

It is not necessary, but is what I did wrong?

20. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If you are referring to your equation for V3, I'm affraid the answer would be yes. You need to write your third equation as the expression for V2 (at the negative terminal of the opamp per your labeling convention). That would be the equation that I provided in my earlier reply (#8 in this thread).

hgmjr