Op-Amp related question


by Inept
Tags: circuits, ece, non-inverting, op-amp, voltage
Inept
Inept is offline
#1
Apr4-11, 09:18 PM
P: 5
1. The problem statement, all variables and given/known data




2. Relevant equations

vo = G*vs

where G = ((R1+R2)/(R2)

therefore,
vo = vs * ((R1+R2)/(R2)

would vo be independent of Rs ?

3. The attempt at a solution

I assumed it was a Non Inverting Op Amp...but I haven't come across one with (two?) input resistances? I'm not sure where to go with it. I began to use the equation I have above, but I don't think it's right. I think (or I keep telling myself) that I have to use Rs or Rx in part a somehow.

For part b) I believe I have to figure out the linear range of Rx, which should be doable. So I'm not worried about that, assuming what I'm thinking is correct...

If anyone can help thanks! If not, thanks anyways!
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phyzguy
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#2
Apr4-11, 09:23 PM
P: 2,068
One question that may help - What is the voltage at the + input of the Op Amp. Is it 750 mV, or something else?
Redbelly98
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#3
Apr4-11, 09:24 PM
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Quote Quote by Inept View Post
1. The problem statement, all variables and given/known data




2. Relevant equations

vo = G*vs

where G = ((R1+R2)/(R2)

therefore,
vo = vs * ((R1+R2)/(R2)

would vo be independent of Rs ?

3. The attempt at a solution

I assumed it was a Non Inverting Op Amp...but I haven't come across one with (two?) input resistances? I'm not sure where to go with it. I began to use the equation I have above, but I don't think it's right. I think (or I keep telling myself) that I have to use Rs or Rx in part a somehow.

For part b) I believe I have to figure out the linear range of Rx, which should be doable. So I'm not worried about that, assuming what I'm thinking is correct...

If anyone can help thanks! If not, thanks anyways!
For starters, what is the voltage at the + input (in terms of Vs)?

EDIT: phyzguy had the same thought.

Inept
Inept is offline
#4
Apr4-11, 09:26 PM
P: 5

Op-Amp related question


I solved it, I had to solve for a non-inverting (positive) input voltage (vp) first and then created a "new" circuit. Using that circuit I found the output voltage using that equation I listed above while replacing vs with vp .


I got vp = [(80k/(80k+20k)]*750 mV = 600 mV

then vo = [(3.3k+46.2k)/3.3k]*600 mV = 9000 mV


It feels good to figure my own answer out...lol


EDIT: I didnt refresh, so I didn't see your guys replies. Thanks all! :)


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