Finding the output current in an op-amp.

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The discussion focuses on calculating the output current of an ideal op-amp given specific conditions. The initial approach involved applying Kirchhoff's Current Law (KCL) at the Vn node, leading to an incorrect output voltage of -20 volts, which exceeds the op-amp's range. After realizing the output voltage must be capped at -15 volts, the user recalculated Vn and subsequently determined the output current. The final calculation yielded an output current of -3.167 mA, aligning with the expected answer. The conversation highlights the importance of considering the op-amp's limitations in voltage output during calculations.
mnvaughn
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Homework Statement



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

Homework Equations



vp=vn
in=ip=0

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
But the answer to the question is -3.167mA. Please help!
 
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Re-consider your assumption "Since vp=0 then vn must equal zero".
 
The Electrician said:
Re-consider your assumption "Since vp=0 then vn must equal zero".

Nice!
 
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?
 
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
 

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