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I want to boost the output voltage of op amp to a higher value. I found the conceptual circuit below online. I understand that Q1 and Q2 are used to provide the necessary supply voltage for the op-amp.

link : https://www.analog.com/media/en/technical-documentation/application-notes/28080533AN106.pdf figure 14

But how do the other parts work ? In particular, the voltage gain is apparently Av=3 (written in the text). How did they arrive at that? Should the (voltage gain)Av not be -33 ?

What is the function of R4 and R3?

EDIT: In the answer by Photon, it is stated that the gain of the whole circuit is 33 while the gain of the output is 3 . So you can have 2 gains in an op-amp ? So one is the gain of the input to the Vout? and the other is the gain of the op-amp output to the actual Vout?(the second one is bit confusing)

enter image description here

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    \$\begingroup\$ Hint, when the op-amp output is sourcing current, the current in to its positive supply pin is very nearly equal to the output current. When it is sinking current, the current out of its negative supply pin is very nearly equal to the current flowing in to the output pin. \$\endgroup\$ – The Photon Nov 1 '18 at 0:57
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    \$\begingroup\$ With no signal, it looks like the quiescent current in the op-amp will be insufficient to turn on Q3 or Q4 so the output will be driven via R4 and C1. As signal is applied, the opamp will drive current into R3 which, as @ThePhoton suggests, will also flow in the opamp's power pins. As the output voltage increases, R4 will also drive current into R3 so perhaps providing negative feedback in the inner loop. C1 could be for stability of the inner loop. Time domain simulation might give you some insight. Agreed about the -33 gain. I can also see a 3:1 ratio from R4 and R3. A link to the text? \$\endgroup\$ – Steve Hubbard Nov 1 '18 at 1:09
  • \$\begingroup\$ analog.com/media/en/technical-documentation/application-notes/… figure 14 @SteveHubbard \$\endgroup\$ – Navaro Nov 1 '18 at 1:13
  • \$\begingroup\$ Look here electronics.stackexchange.com/questions/338809/… \$\endgroup\$ – G36 Nov 1 '18 at 16:40
  • \$\begingroup\$ Not the "gain of the output"; The gain of the "stage" formed by R3 and R4. \$\endgroup\$ – The Photon Nov 1 '18 at 17:07
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About the gain being stated as \$A_V=3\$, the complete relevant text is

R3 and R4 form an output voltage gain stage whose gain, \$A_V=3\$, is reduced to unity at high frequencies by C1 to maintain stability.

What this is saying is that R3 and R4 form a voltage divider so that

$$v_o = \frac{v_{out}}{3}$$

where \$v_{o}\$ is the voltage at the output of the op-amp IC.

Or, turned around,

$$v_{out} = 3 v_o.$$

This works because the negative feedback around the op-amp will cause it to push or pull current from its output pin to make it work.

The overall gain of the circuit is 33, as you calculated.

, it is stated that the gain of the whole circuit is 33 while the gain of the output is 3 . So you can have 2 gains in an op-amp ?

No, the "stage" formed by R3 and R4, with gain 3 doesn't really involve the op-amp.

But even within the op-amp integrated circuit itself, of course every stage in the design can have a different gain value.

So one is the gain of the input to the Vout? and the other is the gain of the op-amp output to the actual Vout?

33 is the gain from \$v_{in}\$ to \$v_{out}\$

3 is the gain from \$v_o\$ to \$v_{out}\$. I think Tim's comment does a better job explaining it than I could:

If the voltage on the op-amp output is not equal to 1/3 of the output voltage (ignoring C1), then current will flow in it's positive or negative power pin. That forms a negative feedback loop composed of the four external capacitors and the op-amps (internal) output stage.

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  • \$\begingroup\$ This clears things up ...can you explain a bit further what you mean by "This works because the negative feedback around the op-amp will cause it to push or pull current from its output pin to make it work." . do you mean to say that because of the negative feedback the op-amp will do anything (push or pull) to maintain its output node to vout/3 ? And can you also give some explanation what the difference is with the total gain ? the total gain does not affect the voltage gain? \$\endgroup\$ – Navaro Nov 1 '18 at 1:48
  • \$\begingroup\$ 1. Sorry, I'm in the middle of something and can't think of a great way to explain it. 2. The gain of the R3/R4 stage is only part of the whole circuit, which has an overall gain of 33. \$\endgroup\$ – The Photon Nov 1 '18 at 1:59
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    \$\begingroup\$ @Navaro: yes. If the voltage on the op-amp output is not equal to 1/3 of the output voltage (ignoring C1), then current will flow in it's positive or negative power pin. That forms a negative feedback loop composed of the four external capacitors and the op-amps (internal) output stage. Because C1 boosts the gain at high frequencies it tends to stabilize that loop. \$\endgroup\$ – TimWescott Nov 1 '18 at 2:55
  • \$\begingroup\$ which 4 external capacitors ? \$\endgroup\$ – Navaro Nov 1 '18 at 9:54
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I'm not sure where the Av = 3 part comes from -- it may be the gain of the funky booster stage.

That looks like an old method for doing HV op-amp circuits, before you could just buy HV op amps. The basic idea is that Q1 and Q2 are emitter followers (as you figured out) powering the OP-37. As the OP-37 tries to raise it's output voltage, it draws current from Q1. This lowers the base voltage of Q3, it starts conducting, and voltage (and power) boost happens. Ditto with lowering the output voltage, Q2 and Q4.

I've never tried to design one of these and, frankly, I think if one came up in a design review the first question I'd ask, gently but persistently, would be "are you sure there are no alternatives?" I suspect that -- particularly without careful choice of the op-amp, and some deep knowledge of its innards -- this is just a good way of turning electronics into smoke.

I'd love to hear from someone who does have some experience with a stage like this.

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  • \$\begingroup\$ A topology quite similar to this one was personally recommended to me by one of the authors of The Art of Electronics. All opamps require reading their datasheet, of course, to do a design. But the topology itself isn't as "funky" as suggested. I've already posted here at EESE two different circuits based on the idea, as well. It's possible you even saw Dr. Hill's post in the sci.electronics newsgroup. Do you recall? \$\endgroup\$ – jonk Nov 1 '18 at 9:41
  • \$\begingroup\$ @jonk so, how'd it work in production? I haven't been on s.e.d for years; I may well have seen such a post years ago and forgotten it. \$\endgroup\$ – TimWescott Nov 1 '18 at 14:47
  • \$\begingroup\$ I think one of the reasons I don't like this approach is that -- however well it may work for a given op-amp -- it basically depends on a side effect. In general, counting on side effects to make your circuit work leaves you vulnerable to design changes in the part, and makes it harder to swap parts out for other, newer parts. It's a matter of opinion, which I know is discouraged here -- but even though this is probably the most trustworthy side effect of op-amps in general, I'm going to let the "funky" stand. \$\endgroup\$ – TimWescott Nov 1 '18 at 14:48
  • \$\begingroup\$ @jonk can you add a link with the circuits you refer to? \$\endgroup\$ – Navaro Nov 1 '18 at 16:14
  • \$\begingroup\$ @Navaro Here is one on one of two, or none, LEDs lit up where some of the ideas were getting creative and I'd added one of my own to throw people off a bit. Another is about a circuit similar (not exact) to one I have used for a voltage-controlled two -quadrant sink-source. \$\endgroup\$ – jonk Nov 1 '18 at 20:17

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