Operational amplifier

hgmjr

Joined Jan 28, 2005
9,027
Hmm, suit your self, but there are valid ways to work this out so that you do not end up with negaitve ohms as your answer.
It would be interesting to see your work where you derive the transfer function of this particular opamp circuit and end up with a final answer that is positive rather than negative.

Do you mind posting it here?

hgmjr
 
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hgmjr

Joined Jan 28, 2005
9,027
At the risk of being accused of trolling, I think that your second equation should be:

\(\frac{u_0R_p}{R_p+R_3}-(-R_1i_s)=0\)

Based on the direction of current indicated in the current source, the voltage at the end of R1 into which current Is enters is considered more positive than voltage at the end of R1 out of which current Is exits. Since the point at which it enters is contrained to be 0V, then the voltage at the other end of R1 must be negative.

hgmjr
 

BillO

Joined Nov 24, 2008
999
Trolling? Not at all.

If we are going with convention, then Uo is clearly marked as positive, which puts us in a odd spot.

I'm simply choosing, because of the contradictory nature of the drawing, not to assume anything about the nature of Is or Uo, which is perfectly valid.

Remember, this is not a 'normal' transfer function either. The answer here has units of resistance. Not a unit-less number given by the normal Vo/Vi calculation.

To me, a result of negative resistance, especially in ambiguous case like this, is aesthetically unpleasing.

If your Is is my –Is, or your Uo is my –Uo, then so be it. I see either way as valid in this case. I just like my way better.
 

BillO

Joined Nov 24, 2008
999
No Ron, it's not. The diagram in unambiguous and the result is in accordance with it. Even if it is a bit of an odd way to express this situation.
 

Ron H

Joined Apr 14, 2005
7,063
No Ron, it's not. The diagram in unambiguous and the result is in accordance with it.
I didn't see anything ambiguous about the original schematic.:confused:
I saw your comments on current polarity and output polarity. I thought they were defined unambiguously. Takes all kinds, I guess.
 

BillO

Joined Nov 24, 2008
999
The output voltage is bound to be negative.

Well, the output is clearly marked as positive. That is where the ambiguity is Ron.

Yes, it takes all kinds. If you will feel better in believing I am an idiot, foolish or some sort of weirdo, then please, go ahead. I can assure you it will be no skin off my nose.

So, you are right, I am wrong. Ok?
 

Ron H

Joined Apr 14, 2005
7,063
BillO, I don't think this is about you, or me. It's about the OP and other readers who are reading this thread and don't know what the answer is.
Regarding the output polarity, those + and - signs don't preclude the output going negative. They just mean that the transfer function is defined when the positive probe of the measuring instrument is on the output, and the negative probe is on GND.
 

BillO

Joined Nov 24, 2008
999
BillO, I don't think this is about you, or me. It's about the OP and other readers who are reading this thread and don't know what the answer is.
Regarding the output polarity, those + and - signs don't preclude the output going negative. They just mean that the transfer function is defined when the positive probe of the measuring instrument is on the output, and the negative probe is on GND.
Agreed Ron.

I've been away from electronics for some years and am not familliar with the notation you're referring to. Given that then, the ambiguity goes away and the case is more specific.

Sorry about that.
 

Ron H

Joined Apr 14, 2005
7,063
Agreed Ron.

I've been away from electronics for some years and am not familliar with the notation you're referring to. Given that then, the ambiguity goes away and the case is more specific.

Sorry about that.
No problem. I'm just glad we got it resolved.:)
 

Thread Starter

regexp

Joined Nov 20, 2010
24
err...

So i have \(vb=-R_{1}\)

Which gives:

\(I_{s}=\frac{R_{1}}{R_{2}} + \frac{-R_{1} - V_{out}}{R_{3}} +1\)

Doesn't feel correct =/
 

Ron H

Joined Apr 14, 2005
7,063
err...

So i have \(vb=-R_{1}\)

Which gives:

\(I_{s}=\frac{R_{1}}{R_{2}} + \frac{-R_{1} - V_{out}}{R_{3}} +1\)

Doesn't feel correct =/
No, Vb=-R1*Is.
You are not trying to solve for Is. It is a given.
Even if you were solving for current, your units don't match. Every term in the equation should have, in this case, units of Amps. On the right side, the first term is unitless (a resistor ratio). The second term mixes ohms and volts. Unit analysis (or whatever the correct term is) is a very simple and powerful tool for finding mistakes when solving equations.
 
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