# Finding the output current in an op-amp.

1. Nov 9, 2011

### mnvaughn

1. The problem statement, all variables and given/known data

The problem is gives an ideal op-amp and wants me to find the output current of it. Here is a picture of the problem.
http://i.imgur.com/NfO6q.png

2. Relevant equations

vp=vn
in=ip=0

3. The attempt at a solution
So what i did is take KCL at the Vn node....

(Vn-10)/5K + (Vn-Vout)/10K = 0

Since vp=0 then vn must equal zero

-10/5k-Vout/10k=0

So Vout=-20

But because it's out of the amplifiers range which is from -15 to 15 volts, then Vout is 15 volts. So I then do KCL at the final node to solve for iout....
(15-0)/10K + 15/10K = -iout
-3mA=iout
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 9, 2011

### The Electrician

Re-consider your assumption "Since vp=0 then vn must equal zero".

3. Nov 9, 2011

Nice!

4. Nov 9, 2011

### mnvaughn

Ok so I instead I did Vout=(-Rf/Rs)*Vs
So (-10K/5K)*10
Vout=-20
But since Vout is still not with the linear range of the power rails, isn't Vout=15?

5. Nov 9, 2011

### mnvaughn

NVM I got it. So my mistake was that since Vout is -20 and outside the linear range of the op-amp then Vout is -15. From there I did KCL at the Vn node with the new Vout.
(Vn-10)/5K + (Vn+15)/10K = 0
Vn=5/3
From there I did KCL at the output Vout:
(-15-5/3)/10K - 15/10K = Iout
Iout=-3.167 mA