I'm not sure I understand you entirely... I see what you mean about writing separate equations for the pos. and neg. V2 nodes... but not about my v3 equation.... If I write two equations for V2, then that will bring me to four equations (including my v3 one already written) and I will proceed and solve... correct?

Also, what is your reasoning that I have to write two equations for them... by definition in an ideal Op-Amp they are the same voltage... but that doesn't necessarily mean they have the same current sums entering and exiting them? If this is the case, I think someone who suggested that this is the way to do it to me is wrong

 

Assuming you are dealing with an ideal opamp, you will not need an explicit equation for v3.

You should be able to form two independent equations, one for the voltage at the negative input and one for the voltage at the positive input. Then you set them equal to each other and solve. V3 will fall at naturally from this process since it appears in the expression for the voltage at the negative terminal.

You may want to take a more simple opamp and get your confidence up before you finish up the one you are working on. Take a classical non-inverting opamp and use the technique I am advocating. You will then see what I am talking about.

hgmjr

 

I redid the problem... and I'd like to think my algebra is decent... but I don't see a way to solve this now... v3 is dependent on v2-.... I don't see a way to solve for v3...

http://img97.imageshack.us/img97/9113/img0003ez.jpg

 

I redid the problem... and I'd like to think my algebra is decent... but I don't see a way to solve this now... v3 is dependent on v2-.... I don't see a way to solve for v3...

http://img97.imageshack.us/img97/9113/img0003ez.jpg

Here are your equations as I interpret them:

Equation for V1

\(\frac{(V_1+j0.004)}{10000}+\frac{(V_1-V_{2+})}{-j5000}+\frac{V_1}{-j10000}+\frac{(V_1-V_3)}{20000}=0\)

Equation for V2 at the positive terminal

\(\frac{(V_{2+}-V_1)}{-j5000}+\frac{V_{2+}}{10000}=0\)

Equation V2 at the negative terminal

\(\frac{(V_{2-}-V_3)}{4000}+\frac{V_2}{20000}=0\)

Have I transcribed them correctly?

hgmjr

 

Yes, that's correct.

 

Step 1: Solve the first equation for V1
Step 2: Plug the expression for V1 from step 1 into the equation for V2 at the positive terminal and solve for V2.
Step 3: Solve the third equation for V2 (not V3)
Step 4: Set the equation from step 2 equal to the equation in step 3 and solve for V3/Vin.

hgmjr

EDIT: Changed V3/V1 to V3/Vin

 

In other words, don't forget that V2+ and V2- are the same thing.... that messed me up.

If anyone has done the numbers and wanted to check:

I got v3=vo= -1.57*10^-3 + j9.23*10^-3 = 9.37<99.7 mV

 

In other words, don't forget that V2+ and V2- are the same thing.... that messed me up.

Does it make sense now?

hgmjr

 

Which is not what Matlab gives.... v1, v2, and v3 are

-0.0010 - 0.0016i
-0.0002 - 0.0016i
-0.0002 - 0.0020i

 

What you need to do is show your work so that we can see how you go to your final answer. Then we can figure out where things went off the rail.

hgmjr

 

First off, I am pretty sure there are some sign inconsistencies in your original equations.

For Example:

Equation V2 at the negative terminal

\(\frac{(V_{2-}-V_3)}{40000}+\frac{V_{2-}}{20000}=0\)

I would have written the equation as:

\(\frac{(V_3-V_{2-})}{40000}-\frac{V_{2-}}{20000}=0\)

Opppsss!!! Just notices the denominator in the first term is incorrect.

hgmjr

 

http://img24.imageshack.us/img24/264/img0004nf.jpg

http://img44.imageshack.us/img44/5817/img0005wf.jpg

 

My originals are exactly the same as what you posted earlier.

Caught a mistake on the second page... didn't multiply by 5 to 0.008 term... gonna see if that fixes answer.

 

I think all of your equations could use a revisit to make sure that the signs of the currents are consistent.

hgmjr

 

I think all of your equations could use a revisit to make sure that the signs of the currents are consistent.

hgmjr

But why? the equations I start off with are exactly the ones that you posted in the earlier thread. By the way fixing the error I spotted does not give the right answer still.

Here is the matlab code:

A =
6.0000 - 3.0000i -4.0000 0 + 1.0000i
-2.0000 2.0000 - 1.0000i 0
0 6.0000 -5.0000
>> b=[-.008;0;0]
b =
-0.0080
0
0
>> V=sym('[v1;v2;v3]')

V =

[ v1]
[ v2]
[ v3]

>> X=A\b
X =
-0.0010 - 0.0016i
-0.0002 - 0.0016i
-0.0002 - 0.0020i

 

Earlier, I just parroted your equations to see that I had read them correctly. I did not intend to infer that they were correct in every detail.

hgmjr

 

I just looked again, and they look correct to me.

 

I will look over your equations and see if I can find where they may be off.

hgmjr

 

Found the error... equation 2 needs a denominator of 40,000... not 4000

thanks hgmjr

 

What final result did you get for the voltages?