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.

 

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.

 

Thanks a lot for your fast reply. Can I let Io=Ii?

 

Thanks a lot for your fast reply. Can I let Io=Ii?

Of course, that is the reason you assume no input current.

 

Thank you very much for your help.

 

What is your answer?

 

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

 

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).