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UPDATE: This question triggered what might fairly be called a research obsession for me. I've gotten pretty close to the bottom of it I think, I've posted my findings as an answer below.


There was a similar question here but it did not ask for nor did it receive a general account in its answers.


Noise gain turns out to be an infrequently mentioned and apparently ill-understood concept that is redeemed by the fact that it provides the power to flexibly adjust the stability of your op amp circuit if you know how to use it.

Just when you thought there was one equation you could absolutely count on, the well-known gain equation for op amps turns out to be situation dependent.

$$G = \frac{A_o}{1 + A_o\beta}$$

It turns out, it depends on which definition of \$\beta\$ you use.

The unsuprising part (background)

I'll start with a brief accounting of what I know and can demonstrate to be true, just so you can tell I've done my homework and discourage hasty answers:

\$\beta\$ in known as the feedback fraction, (sometimes feedback factor), and is the proportion of the output voltage fed back to the inverting input.

Considering the non-inverting amplifier below, the fraction of \$V_{out}\$ that reaches the inverting input is readily determined to be \$1/10\$ by inspection of the voltage divider:

enter image description here

$$V_- = V_{out} \frac{R_g}{R_f + R_g}$$

$$\beta = \frac{V_-}{V_{out}} = \frac{R_g}{R_f + R_g} = \frac{10\mathrm{k}}{90\mathrm{k} + 10\mathrm{k}} = \frac{1}{10}$$

Returning to the formula we started with, \$A_o\$ stands for open-loop gain, about 100,000 in this case. Substituting into the formula, the gain is:

$$G = \frac{A_o}{1 + A_o\beta} = \frac{100,000}{1 + (100,000\cdot \frac{1}{10})} = \frac{100,000}{10,001} = 9.999$$

Which is awfully darned close to \$10\$, which is why we usually drop the \$1 +\$ bit and just say \$G = 1/\beta\$. This is what a simulation predicts and is very close to what is observed on the bench. So far, so good.

\$\beta\$ also plays a role in the frequency response.

enter image description here

The yellow trace is the open loop gain (\$V_{out}/(V_+ - V_-)\$, the purple one is the closed-loop (CL) signal gain (\$V_{out}/V_{sig}\$).

It's tricky to see without expanding the image, but the open-loop gain crosses 0dB at 4.51 MHz; the 3dB down point on the closed loop gain is 479 kHz, so right about a decade below. The closed-loop gain "consumes" open-loop gain to boost the signal. When the open-loop gain isn't enough to do that, the closed-loop gain drops and hits its 3dB down point, in this case where the open-loop gain is 10 (20dB). Since \$A_o\$ drops at 20dB/decade, that's a decade below \$A_o\$'s 0dB point.

So in this case:

$${BW}_{CL} = \beta \cdot {BW}_{OL} = 0.1 \cdot 4.51 \mathrm{MHz} \approx 479 \mathrm{kHz}$$

The surprising part

Ok, so maybe I was wrong? This all seems to work just fine. Hmm, what if we make a little tweak to the circuit. Let's pop in this innocent-looking resistor \$R_n\$:

enter image description here

And take a look at the gain over frequency again:

enter image description here

Whoa! What's up with that?

  • The closed loop signal gain (purple trace) is still 10 (20dB)
  • but it's bandwidth is reduced by a further decade, down to 43.6 kHz!
  • There's a cyan trace that bumps into \$A_o\$ in the right way, but it's up at 40dB

What I've worked out so far

Over the weekend I was studying Walter Jung's excellent book Op Amp Applications. In the first chapter he introduces the notion of noise gain, to be carefully distinguished from signal gain. This seemed simple enough at the time as he defined noise gain as simply \$1/\beta\$ and suggested the notation \$NG\$.

For the first non-inverting amplifier above, the noise gain is equal to the signal gain \$(G)\$, which is perhaps why one so seldom encounters the distinction.

However, I've collected a variety of factoids from various sources:

  • The cyan trace above is the noise gain (actually, it's only where it would be if I was able to plot it with SPICE). I was able to find a handful of references after extensive online searching, but no description of how to determine it when it's not the same as the signal gain. In the second circuit above, it value is:

    $$\frac{R_f}{R_g\parallel R_n}$$

  • Noise gain is what actually determines the frequency response, not signal gain. Noise gain is what SPICE (and your circuit) use to determine frequency response on an AC analysis.

  • Loop gain is (\$A_o\beta\$) and determines amplifier stability. But the \$\beta\$ in that expression is the noise beta (1/noise-gain), not the signal beta. Note that I've never seen either the term noise beta or signal beta in print, I've just invented (or perhaps reinvented) them here to distinguish the two.
  • As demonstrated above, noise gain can be manipulated without changing the signal gain. This turns out to be a very powerful way to tune in the bandwidth of an amplifier to get just the phase margin you want without monkeying around with the signal gain your circuit needs.
  • The terminology is a bit vexed, but this app note from AD seems clearest to me by saying there is open-loop gain and closed-loop gain, but there are two types of closed-loop gain, signal gain and noise gain.

A few things I've tentatively inferred

Note: this hypothesis turns out to be false. An op amp is a DC amplifier, and so it's essential circuit characteristics (including noise gain) can be measured at DC, at which it turns out to be the same as for low frequencies.

  • Hypothesis: Signal gain is determined by DC analysis. Noise gain is determined by AC analysis. I suspect this is not the whole story and is one of my main questions below. But it seems to produce the right value for noise gain in the cases I've tried so far if you short independent voltage sources and then work out the voltage gain transfer function of the feedback network. This would imply that:

$$\beta_{noise} = \frac{\Delta v_-}{\Delta v_{out}}$$

Why this is really handy

Let's take a look at the loop gain, where the stability of the circuit is determined. I'll substitute in \$R_n\$ values of 1k (as just above), 2k, 5k, and 100Meg (like no resistor at all). I added a 5 nF capacitor across the output to reduce the uncompensated circuit to 45 degrees phase margin:

enter image description here

I'll just jump to the punch line here. By adjusting \$R_n\$, I can manipulate the phase margin between wherever it is (46° in this case) and 90° and anywhere I want in-between. This comes at the cost of bandwidth, so it's not an entirely free lunch, but it allows me to optimize that trade-off wherever I want. This translates into the ability to tune my step response between the yellow and purple traces below:

enter image description here

Questions that a full and general account would answer

I'm not looking for individual answers to the following questions. What I'm looking for is the explanation of noise gain that would allow me to readily answer these questions for myself. Think of these as the "test suite" for the answer :)

  • How can the op amp have two distinct feedback fractions? Since signal gain can be calculated at DC and noise gain seems to be at AC, maybe we could consider one of them the DC feedback fraction and the second the AC feedback fraction?

  • If noise beta is the AC feedback fraction, why does the DC feedback fraction determine the signal gain? The signal is AC, so I don't see how it would be treated differently.

So my actual question is:

  • What is noise gain really?
  • How and why is it different from signal gain, in the sense of "why are there two and not one"?, and
  • How does one determine the noise gain via circuit analysis in the general case? (i.e. What equivalent model is used.)
  • Bonus points if you happen to know how to plot it in SPICE :)
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  • 3
    \$\begingroup\$ Interesting question. Can't wait to see what the really knowledgeable folks have to say. \$\endgroup\$ – JRE Aug 13 '16 at 9:02
  • \$\begingroup\$ The cyan trace = 10* Vout so is irrelevant. This question is far too long winded and you are missing the point. Noise gain is nothing to do with what you are showing. \$\endgroup\$ – Andy aka Aug 13 '16 at 10:05
  • \$\begingroup\$ @Andyaka - On the contrary, the cyan trace is the point; it is the noise gain of that second circuit, $$1 + \frac{R_f}{R_g \parallel R_n} = 1 + \frac{90k}{10k \parallel 1k} \approx 100 = 40dB $$ If you don't believe me, believe Walter Jung: analog.com/library/analogDialogue/archives/31-2/Graphics/…. The second circuit above is the same as the one on the left in Walter's image. I couldn't plot the noise gain directly, so I approximated it with 10*V_out, which is quite a good approximation in this case, at least to the 0dB crossover. \$\endgroup\$ – scanny Aug 14 '16 at 1:30
  • \$\begingroup\$ but that is my point. Drawing it as ten times Vout is a wholly preposterous thing to do. It lowered tour question towards the gutter. Redemption is needed! \$\endgroup\$ – Andy aka Aug 14 '16 at 9:15
  • \$\begingroup\$ You may find this useful: analog.com/library/analogDialogue/archives/43-09/… \$\endgroup\$ – Peter Smith Aug 14 '16 at 15:28
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Okay, after a lot more research, I think I've gotten to the bottom of this. Actually I'm certain it's only approaching the bottom, as I've found this topic area quite deep, but I think I've gotten close enough to shed some light.

A basic misconception

A turning point in my understanding was when I realized that the equation I led off with in the OP:

$$ G = \frac{A_o}{1 + A_o\beta} $$

is a block diagram equation, not a circuit equation. Those are two different things and the translation between one and the other is often not trivial. The fact that the translation is trivial for the simple non-inverting op amp case is perhaps a trap for the unwary, certainly one I fell into head first :)

We'll see why that matters shortly.

What is noise gain, really?

Noise gain (in an op amp circuit) is the gain experienced by a small signal applied at the non-inverting (+) input.

It is so called because noise is frequently stated as "referred to the input", meaning the noise signal that would need to be present at the input to produce a specified noise output. This allows noise that originates in various parts of the op amp to be "lumped" into a single equivalent value, simplifying any analysis that doesn't really care where inside the black box the noise originates.

In a simple non-inverting amplifier, noise gain is the same as the signal gain:

enter image description here

That makes sense when you consider that the signal is applied directly to the non-inverting input, and a small differential voltage applied at that node would experience precisely the same gain as the signal does.

I think a block diagram view shows this most clearly. It's probably not strictly necessary for understanding this simple non-inverting case, but I found it crucial for understanding the general case. Also, \$\beta\$ is a block diagram variable, so we can avoid repeated mental block-to-circuit translations if we stay in the block diagram domain when we use it for reasoning.

enter image description here

The \$+\$ node on the summing block corresponds to the non-inverting input of the op amp (in this case, but not in general as we'll see below). It's easy to see there's no difference between a noise signal and a "real" signal applied there and the noise gain in this case is:

$$ NG = \frac{A_o}{1 + A_o\beta} $$

Now, in his book, Walter Jung defines noise gain as \$1/\beta\$. And, I expect we all know that the equation just above is approximately \$1/\beta\$ when \$A_o\beta \gg 1\$. In fact, this approximation is great for getting the DC noise gain, the Y-axis position of the long flat part at the beginning of its magnitude curve on the Bode chart. But if you want to see its frequency dependent behavior (for example, to plot it in SPICE), you need to use the long form.

Ok, so we're well on our way to how to calculate noise gain in the general case, but one challenge remains: How do we determine the value of beta (\$\beta\$)? It might not be obvious at first, but this is a challenge because the components that contribute to beta may also contribute to other blocks. There's no guarantee the feedback network has them all to itself; in fact, we need look no further than the inverting amplifier configuration for an example of them being "shared" (perhaps more precisely, interdependent).

Consider the inverting amplifier circuit below:

enter image description here

The block diagram for this circuit turns out to be this:

enter image description here

I won't go through the details of how you get here from the circuit diagram, but that might make an interesting follow-on question if you wanted to post it. Basically you create a Thevenin equivalent looking into \$R_f\$ from the inverting terminal and then use superposition to get the two contributions to the summing node. Note that here, \$V_e\$ represents \$V_- - V_+\$ at the op amp inputs, which is why \$A_o\$ and \$\beta\$ have minus signs in their expressions.

There are a couple interesting things we can see:

  1. The input signal \$v_{in}\$ does not appear directly at the summing node. It is first attenuated by \$T_i\$ (\$T_i\$ here stands for input transmittance). This explains why the noise gain is not equal to the signal gain for the inverting topology. Noise gain is an attribute of the core amplifier loop, not the overall circuit.

  2. \$\beta\$ is the same as it is for the non-inverting case (once you get the signs sorted out). This explains why the noise gain is the same for the inverting and non-inverting topologies.

  3. \$R_f\$ and \$R_{in}\$ appear in both the \$\beta\$ and \$T_i\$ block expressions. This reflects the interdependence between the feedback network and the input attenuation network. Changing one of the impedances therefore changes both the signal and the noise gain. So it's not possible to modify them separately by changing values of the existing feedback network components.

So what is "forcing the noise gain" and why does it work?

I got into this question of noise gain pursuing an interest in op amp stability/compensation, not noise. I found a couple references that claimed (paraphrased) "... forcing the noise gain is a powerful compensation technique that many analog engineers don't know about ...". My reaction was: "Hmm, sounds interesting! I love the analog black arts! What's noise gain? And how do I force it to do something it doesn't want to?"

Well, after this recent research, I'm inclined to think "forcing the loop gain" (downward) is a more apt expression, since that is what enhances the stability. The loop gain is \$A_o \beta\$; changing \$\beta\$ isn't the only way to change that product. This will become more clear in a minute.

As a reminder, this is what the "forced noise gain" circuit from above looks like, as applied to a non-inverting amplifier:

enter image description here

If we do the same Thevenin equivalent analysis to isolate the feedback and input blocks, we end up with a block diagram that looks like this:

enter image description here

We can observe a few interesting points:

  • The feedback path is attenuated by \$T_f\$. This effectively reduces the feedback fraction, increasing the closed loop gain of the core amplifier loop, otherwise known as the noise gain.

  • The input is attenuated by \$T_i\$, which is exactly the same as \$T_f\$. This normally would have the effect of decreasing the overall circuit signal gain. However, in this case, that decrease is exactly offset by the increase in noise gain and overall signal gain is unaffected.

  • Because \$T_i\$ and \$T_f\$ are the same and because they both appear immediately prior to a summing block, block diagram algebra allows us to move that block to the other side of the summer as in the figure below. Just a caution though, while manipulations of the block diagram like this still give you the right answer for the overall transfer function \$V_{out}/V_{in}\$, the correspondence of any given signal (connecting line) to a physical point on the circuit can be disrupted.

enter image description here

Embracing the equivalent diagram this gives us, we see that the desired reduction in loop gain can be achieved by attenuating the gain of the main amplifier, without producing a change in the overall signal gain (at low frequencies).

There is a really excellent video development of this by the late Professor James Roberge of MIT (starting about 35:17). I ended up watching the whole 20 lecture series (most of it twice :) and highly recommend it :)

I also worked out how to directly plot the noise gain in LTspice, I've posted that as a follow-up question if you'd like to take a look: How do I plot noise gain of an op amp circuit in SPICE?.

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  • \$\begingroup\$ Scanny,I think you have provided a rather comprehensive, exact and illustrative derivation. With this comment I like to mention that providing a resistor Rn - or a series connection of a suitable Cn and Rn - between both opamp input terminals is one of the classical methods for external frequency compensation (improving the stability margin). This works because the loop gain is reduced. More than that the signal gain will be not influenced because - as you also have shown - the "forward damping" is affected by the same factor. However, the signal bandwidth is also reduced correspondingly. \$\endgroup\$ – LvW Aug 24 '16 at 7:44
  • \$\begingroup\$ Another solid question and another solid answer. Fantastic. Do you have a link to "... forcing the noise gain is a powerful compensation technique that many analog engineers don't know about ..." ? Seems like it might be worth a good read. \$\endgroup\$ – efox29 Sep 14 '16 at 23:48
  • \$\begingroup\$ @efox29: Here are a couple of the ones I was referring to :) link 1, link 2. \$\endgroup\$ – scanny Sep 15 '16 at 5:21
  • \$\begingroup\$ Follow up question: What would then be the noise gain of a simple follower? Simply 1? And how is noise treated for a follower? \$\endgroup\$ – Irenaius Oct 29 '18 at 11:26
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The noise gain of an op-amp is always given by \$G_N\$ = \$ 1\ +\ \frac {R_F} {R_{IN}}\$ assuming that the open loop gain \$AV_{OL}\$ is >> \$A_{CL}\$ (the closed loop gain) where for your circuit, \$R_{IN}\$ is given by (as you note) \$R_G\$ || \$R_N\$. This is the non-inverting gain of the amplifier and is true for both inverting and non-inverting configurations.

The noise gain is used for stability criteria, not the signal gain.

Here is a handy little graphic:

Gain definitions

If the amplifier has very high open loop gain, then the closed loop gain is the noise gain.

Signal gain and noise gain for various topologies

Your circuit above is the same as circuit C.

As you have found, by varying \$R_{IN}\$, you can change the stability margin at the expense of more noise and offset.

Definition of the closed loop gain of the amplifier:

Closed loop gain

[Update]

In response to the comments:

The noise gain of the amplifier is not a special case; it is always the non-inverting gain of the amplifier and ultimately sets the closed loop gain of the amplifier.

The noise gain is \$1\ + \frac {R_F} {R_{IN}}\$ and the signal gain is 1 + \$ \frac {R_F} {R_G}\$.

Note that \$R_{IN} \$ is always the input impedance as seen from the inverting input at AC (so in this case it is shorted inputs).

Your ac source has zero impedance and therefore connects (for ac purposes) \$R_{IN}\$ to ground for the purpose of analysis; try adding a source impedance to see why this might change things.

Source material.

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  • \$\begingroup\$ I think you're hot on the trail here Peter :) A couple things: (1) I think my circuit is not pictured in Figure 1.9 as C is an inverting amp. That doesn't change the validity of the answer, but might mislead a future reader. (2) I believe the \$1\ +\ \frac {R_F} {R_{IN}}\$ equation is a special case (e.g. doesn't hold for Fig 1.9c) unless you say \$R_{IN}\$ is defined as the resistance seen from the inverting terminal with sources shorted. \$\endgroup\$ – scanny Aug 14 '16 at 17:44
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Noise gain is how the noise (internal to an op-amp's input) is amplified by the feedback resistors IN CONJUNCTION WITH (very importantly) the "invisible" capacitance from the inverting input to ground i.e. the inputs parasitic capacitance. Consider the standard non-inverting amplifer: -

enter image description here

We normally assume that the output voltage equals \$V_{IN}\times 1 + \dfrac{R2}{R1}\$ until the frequency reaches the limit where the falling open loop gain causes the closed loop gain to fall accordingly. I'm going to add two things to the above circuit that make things more relevant in terms of how noise gain is analysed: -

enter image description here

The two components added are the leakage capacitance of the inverting input and the internal noise source inside every op-amp input.

From the perspective of noise (and signal), the gain is increased by the added capacitor across R1. R1 is shunted (at high frequencies) by the reactance of the capacitor. This means that both signal gain and (shall we say) noise amplification is increased.

So, the final part of this story is a bode plot: -

enter image description here

From DC upwards, amplification is determined by the conventional gain i.e. 1 + R2/R1 then, at some point, C1 starts to progressively shunt R1 and gain rises with frequency. This rising gain contines until it meets the open-loop response then, naturally falls as the open loop gain falls.

This is what noise gain is all about when applied to a non-inverting op-amp circuit.

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I've been pretty confused with all the instructions I've read, too, since they only apply to certain types of circuits.

I think this is the easiest way to understand it, and works in all scenarios:

  1. Replace your sources with shorts or open circuits, following the Superposition theorem
  2. Disconnect the non-inverting input of the op-amp and insert a noise voltage source in series with it.
  3. Noise gain is the gain from that noise voltage source to the output.

So for this circuit:

Inverting amplifier schematic with resistor between inverting and non-inverting input

  • The signal gain is 10/2 = 5× ≈ +14 dB
  • The Req = 1 kΩ || 2 kΩ || 10 kΩ = 625 Ω

Change it to this circuit:

Inverting amplifier schematic with noise voltage source in series with non-inverting input

  • The noise gain is 10/(2 || 1) = 15× ≈ +24 dB

Examples:

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The term "noise gain" comes from the convention of referring the equivalent noise of the internals of an op amp to the non-inverting terminal. So for example the voltage noise in the op-amp is transformed to an equivalent voltage source in series with the non-inverting terminal, in volts per root-hertz. This allows you to calculate the output noise by multiplying by the non-inverting gain, figuring in the bandwidth.

When figuring out the bandwidth of an amplifier with a dominant pole, you also have to use the "noise gain" or the gain seen from the non-inverting input. That way, the bandwidth is simply the GBW product over the noise gain.

That's basically it- Noise gain is the gain from the non inverting terminal. In an inverting amplifier the signal gain is different, but the bandwidth and noise would be calculated with the non-inverting gain from the + terminal to the output.

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  • \$\begingroup\$ How would this account for the difference in noise gain and signal gain in the second circuit? The signal is being applied to the non-inverting terminal and is seeing a gain of 10 (signal gain), not 20 (noise gain). \$\endgroup\$ – scanny Aug 13 '16 at 4:38
  • \$\begingroup\$ I don't see a difference- Why do you think the noise gain is 20? Signal gain is 10, noise gain is 10, right? If it were an inverting amplifier then signal and noise gain would be different. \$\endgroup\$ – John D Aug 13 '16 at 5:07
  • \$\begingroup\$ Noise gain in that circuit is 40dB (100), (sorry, not 20, got my dBs mixed up :) But it's definitely not 10. That's why the bandwidth is reduced by 2 decades instead of 1. It's the fact that signal and noise gain are not the same in that circuit that originally gave rise to my question :) (It's also what makes it interesting design-wise.) \$\endgroup\$ – scanny Aug 13 '16 at 5:10
  • \$\begingroup\$ This is interesting- For an ideal op-amp your Rn does nothing and noise gain is the same as signal gain, right? (zero volts between + and - inputs.) For a real op-amp there will be some effect due to adding a resistor between the + and - terminals, granted, but it doesn't intuitively seem like it could change the gain from the non-inverting terminal by an order of magnitude. How did you get the cyan trace showing the noise gain = 100? \$\endgroup\$ – John D Aug 13 '16 at 5:21
  • \$\begingroup\$ It is interesting, isn't it? It kept me puzzling all weekend :) I don't think it has to do with real vs. ideal. It does seem to have to do with AC analysis vs. DC analysis though. If you do AC analysis of the feedback fraction (short independent V source V_sig), then it produces exactly the right result, with is 90k/.909k = 100 (40dB). \$\endgroup\$ – scanny Aug 13 '16 at 5:25
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Regarding the inverting configuration, it´s said: "Rf and Rin appear in both the β and Ti block expressions. This reflects the interdependence between the feedback network and the input attenuation network. Changing one of the impedances therefore changes both the signal and the noise gain. So it's not possible to modify them separately by changing values of the existing feedback network components"

But I think is posible:

Inverter with Rn compensation

Inverter with Rn compensation

Inverter block diagram

Inverter block diagram

The close loop gain is

The close loop gain is

The input is attenuated by The input is attenuated by

β is: β is:

The gain is the same that without compensation

The gain is the same that without compensation

Noise gain now is:

Noise gain now is:

Instead of:

Instead of:

Conclusion: We modified the noise gain without affecting the signal gain in the inverting configuration.

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