# Op-Amp Question

Discussion in 'Homework Help' started by dli286, Mar 9, 2009.

1. ### dli286 Thread Starter New Member

Mar 9, 2009
6
0
For the op-amp below, I am solving V_o in as a function of V_i.

Here are the equations (using KCL) I have so far.

V_i - V+ = I_1 * R_1
V + = I_1 * R_2
V_i - V- = I_2 * 10k
V_o - V- = I * 10k

I am not sure how to apply KCL at the node between the two 10k resistors. Thanks for your help.

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Assume that the op amp is ideal and thus no current flows in its inputs.

Thus,

Current through right 10K:

Io=[Vo-(V-)]/10K

Current through left 10K:

Ii=[(V-)-Vi]/10K

Then V- equals V+ which equal:

(V-)=(V+)=Vi*R2/(R1+R2)

Make the substitutions and hopefully you will find it.

Mar 9, 2009
6
0

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Of course, that is the reason you assume no input current.

5. ### dli286 Thread Starter New Member

Mar 9, 2009
6
0
Thank you very much for your help.

Feb 4, 2008
4,846
63

7. ### Ratch New Member

Mar 20, 2007
1,068
3
dli286,

By now you should have found that Vo/Vi = (R2-R1)/(R1+R2) . If you did not, then you should ask for more help.

Ratch

8. ### dli286 Thread Starter New Member

Mar 9, 2009
6
0
My answer is V_o = V_i ((2kR_2 - R_1 - R_2) / (R_1 + R_2)), which is the same thing as V_o = V_i (R_2 - R_1) / (R_2 + R_1).