Finding the output current in an op-amp.

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Discussion Overview

The discussion revolves around finding the output current in an ideal operational amplifier (op-amp) circuit. Participants explore the application of Kirchhoff's Current Law (KCL) and the characteristics of op-amps, particularly focusing on the output voltage and its implications on the output current.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant initially assumes that since the non-inverting input (vp) is 0, the inverting input (vn) must also equal zero.
  • Another participant challenges this assumption, suggesting a reconsideration of the relationship between vp and vn.
  • A different approach is proposed where the output voltage (Vout) is calculated using the formula Vout=(-Rf/Rs)*Vs, leading to an initial output of -20 volts.
  • Participants discuss the implications of Vout being outside the linear range of the op-amp, with one concluding that Vout should be capped at -15 volts instead.
  • After adjusting Vout, a participant recalculates vn and finds it to be 5/3, leading to a new calculation for the output current (Iout) of -3.167 mA.

Areas of Agreement / Disagreement

There is no consensus on the initial assumptions regarding vn and vp, as participants challenge and refine these assumptions. The discussion includes competing views on how to handle the output voltage in relation to the op-amp's linear range.

Contextual Notes

Participants express uncertainty regarding the assumptions made about the input voltages and the output voltage limits of the op-amp. The calculations depend on these assumptions, which have not been fully resolved.

Who May Find This Useful

Students and practitioners working on circuit analysis involving operational amplifiers, particularly those interested in understanding KCL applications and output characteristics of op-amps.

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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