# Op-amp application

Discussion in 'General Electronics Chat' started by Dritech, Nov 11, 2013.

1. ### Dritech Thread Starter Well-Known Member

Sep 21, 2011
756
5
Hi,

Can someone please explain what is the function of the op-amp circuit in the attached diagram? It is like a non-inverting amplifier but resistor R1 is connected to the input instead of the ground.

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2. ### #12 Expert

Nov 30, 2010
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It looks like a voltage follower to me. The only thing I can come up with besides that is a frequency dependent change due to an RC node at the inverting input, but that doesn't seem like it is predictable enough to use as a design parameter. I wonder if it would suddenly become obvious if this was shown in the context of a larger circuit.

Nov 23, 2012
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Attenuator?

4. ### ScottWang Moderator

Aug 23, 2012
4,933
777
Where did you get this circuit?
Any explanation from that source?
I saw and checked much more op amp circuits, but I can't find anyone similar like this.

5. ### GopherT AAC Fanatic!

Nov 23, 2012
6,321
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It's just a buffer when R1 is large (above 100 ohms). Out perfectly matches input when simulated. R1 and R2 can be anything but no change.

As R1 becomes small (e.g 0.1 ohm) it becomes an attenuator - at R2 = 10k and R1 is 0.1ohms, attenuation is 50%. About 90% when R2 is increased to 100k.

Last edited: Nov 11, 2013
ScottWang likes this.
6. ### LvW Active Member

Jun 13, 2013
674
100
Dritech, it is a rather tricky circuit.
It is a unity gain amplifier with one big advantage if compared with the classical buffer having 100% feedback:
It is a circuit with adjustable feedback - independent on gain (that is always unity).
Thus, you can use uncompensated opamps (normally not unity gain stable).
The factor R1/(R1+R2) - and, thus, the loop gain - can be made as small as necessary to allow stable operation - without influencing the closed-loop gain.
Such a circuit is also used for compensated opamps (unity gain stable) to improve the phase margin (step response with less overshoot).

Explanation for operation:
*The voltage at the non-inv. input always is Vin.
*Therefore, the voltage at the inv. input also is identical to Vin (idealized gain of infinite assumed).
* Thus, no current through R1.
* Therefore, no current through R2.
* Result: Vout=Vin

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7. ### #12 Expert

Nov 30, 2010
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"Very interesting". (Arte Johnson)

All these things I know about op-amps and, even after you show it to me, I am stretching to visualize it. Changing the closed loop gain of an amplifier that doesn't have any gain (in the usual sense). That's brilliant, and you'll never see it in an op-amp tutorial. (I think I read all of them.)

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
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You never heard about noise gain vs signal gain ?
http://www.analog.com/library/analogdialogue/archives/47-03/raq_91.html
http://www.analog.com/library/analogdialogue/archives/31-2/appleng.html

9. ### LvW Active Member

Jun 13, 2013
674
100
Another example for an amplfier having a noise gain that is independent on signal gain is an opamp with negative as well as positive feedback (for stability reasons, the negative feedback must dominate, of course). This option also allows decompensated amplifiers to operate for low signal gain valus.
However, one should be aware that the signal bandwidth is determined by the noise gain bandwidth (unity loop gain).

10. ### GopherT AAC Fanatic!

Nov 23, 2012
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If I remember correctly, the whole Arte Johnson quote was, "very interesting, but stupid". Maybe that is what #12 was implying?

I have seen better attenuator circuits. I'm still trying to get my head around the rest of the story. The Jury is still out, Arte may be right.

11. ### #12 Expert

Nov 30, 2010
16,704
7,354
ROFL!

Still learning something almost every day on this site.

12. ### LvW Active Member

Jun 13, 2013
674
100
"attenuator circuits" ?