# OPAMP question

Discussion in 'Homework Help' started by Schoppy, Jul 9, 2014.

1. ### Schoppy Thread Starter New Member

Jul 7, 2014
26
0
Ok so not directly an OPAMP question but rather a very simple one and I just can't seem to understand it. For V1=-40E3(i2) where does the -40E3 come from? It's been two semesters since I took my first Networks class and I'm embarrassed to say I can't recall how to arrive there, Voltage Division maybe? That doesn't seem to work though. Thanks guys.

File size:
3.5 MB
Views:
32
2. ### Schoppy Thread Starter New Member

Jul 7, 2014
26
0
Got it, it's because i2 = -v1/R because of it being an inverting amp, makes sense now

Apr 5, 2008
19,902
4,138
Hello,

I have turned the picture over and changed the perspective.
This way the others can have a better look at the picture.

Bertus

• ###### schoppy_20140709_150603.jpg
File size:
45.1 KB
Views:
56
Last edited: Jul 9, 2014
4. ### Schoppy Thread Starter New Member

Jul 7, 2014
26
0
Thank you, and I was wrong with my original assumption, it looks like it equals that because let's call the note next to the 2k resistor Vs. Then i2 comes from (Vs-Vi)/40k.
But why does Vs = 0? That's the only thing that seems to make sense but I am having trouble seeing why.

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,667
745
Because of Virtual Ground, note the use of capital letters.

The none inverting input of op amp is connected to ground. There is 0 volts at none inverting input.

The inverting input of the op amp MUST be equal to the none inverting input. So, V inverting = V none inverting, therefore V inverting = 0 volts because V none inverting is connected to ground which means it is 0 volts.

If you want more background, google: op amp virtual ground.

That is also how they found i3.
V none inverting = 0 volts.
V inverting = V none inverting = 0 volts.
i3=(V source - V inverting)/2k=(150 mV - 0)/2k = 75 microA