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Do low power amps lack bass? Or do they clip?

33AndAThird

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Hi ASR,

I was listening to Darko.audio podcast where the Enlium AMP-23R was discussed - yes I know Darko is a subjectivist, but he is listenable and I enjoy trying to figure out where the holes are, which brings me to my question. The amp is 25w per channel into 8 ohms, 45w into 4 ohms. Darko and Srajan rightly say that for many speakers in many rooms, 25w per channel ought to be enough. But then they say that while the amp sounded beautiful, it was 'cut off at the waist', effectively implying that on the low sensistivity standmounts they were using, the amp didn't have the power to drive the lower frequencies. This didn't quite make sense to me, so wanted to check my knowledge here.

To my understanding, when an amp runs out of power it clips, and the top and bottom of the audio signal are effectively cut off. My questions are:
  • Even if we grant that the musical signal contains higher amplictues at lower frequencies, given the additive nature of the complex waves that make up music would clipping not affect all frequencies equally, rather than just the bass?
  • If this is the case, would this not result in distortion across the frequency range rather than appearing as 'missing bass'?
An alternate explanation is posited by Srajan; he references another similarly powered amplifier - the Pass Labs 30.8 - which he says has 20 output devices per channel where as the Enlium has one output device per channel, and hence can provide more current at the same wattage (I guess implying lower voltage in the signal). This sounds bogus to me, surely greater current delivery just allows you to drive a lower impedance speaker, and given the enlium is rated at 45w into 4 ohms, it can already do a reasonable job of that given 25w into 8. They mention impedance swings up to 50 ohms at responance that they claim requires a higher power amp to tame, but I don't understand why super high impedance requires more power either.
Or is it something about current swings? I've seen curent swings referenced in amplifier revews before (where amps of similar power outputs have different current swings eg. 25 amps compared to 9 peak-to-peak) but not sure how this manifests or how this can be independent to wattage at a given impedance.

Is everything they are saying nonsense? Ie. their sighted listening of a 25w amp leads them to believe its 'cut off at the waist', or is there some possible explanation for this that wouldn't manifest in clipping / distortion?

Thanks!
 

Blumlein 88

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Do yourself a favor and even more than their subjective ideas, don't pay attention to them on technical concepts.

I could maybe decode something like what they mean even if they know, which often they are kind of foggy in what they think is happening.

Now most power is needed for low frequencies. So with a small power amp low frequencies will be why it runs out of power. Maybe they just limit the playback level to prevent that and yes at lower sound levels we perceive less bass.

The comparison with the Pass amp and number of output devices is all down to specifics and the impedance of the speaker in use when it runs out of power. If it is a low impedance ported speaker then maybe something to the extra current (which btw means at that load it has more power) or if a sealed box or other high impedance speaker at low frequencies their idea has no merit. If could even be a speaker with very high impedance at low frequencies and that small amp doesn't have enough voltage to develop its full 25 watts. It is all in the details.

As for clipping etc. When you clip something you get odd order harmonics up quite a ways. If it clips in the bass, it would sound different than if it clips elsewhere. Also clipping isn't usually the complete black and white idea of clipping and one watt less no clipping. Clipped really hard yes everything will sound awful. Moral of the story is don't clip your amps.
 
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oleg87

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Your intuition is correct, this seemingly anthropomorphic view of amp "power" is not how it works. An amp hitting its limits manifests as distortion and clipping, there is no mechanism by which insufficient power would make it politely decline to reproduce low end while remaining well-behaved elsewhere.
 

wwenze

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  • Even if we grant that the musical signal contains higher amplictues at lower frequencies, given the additive nature of the complex waves that make up music would clipping not affect all frequencies equally, rather than just the bass?
  • If this is the case, would this not result in distortion across the frequency range rather than appearing as 'missing bass'?

This is a track with 100Hz and 2000Hz at the same amplitude

1712966265090.png


This is the same track amplified 14dB so that it is +6dB clipping

1712966397270.png


So yes, clipping affects all frequencies equally, in the mathematical sense all other things equal.

The subtle difference: This is the waveform from a music when the bass drum was just kicked:
1712966735859.png


So essentially you'd have HF superimposed on the LF waveform. You can think of the HF's "mid point" as tracking the LF waveform. Clipping cuts away the extreme values, and since your HF signal is already closer to the extreme values, a bigger % of the wave gets cut at that moment.
But at the same time, this also means that there are times where the HF would be away from the extreme values and hence does not get cut, while the LF is cut all the time. So this is a source of the myth. However this myth forgets the part where we hear things for more than 1 cycle, and when the whole thing is averaged over time, it's the same.
And mathematically speaking, at the point of clipping, the "cut" energy comes from both HF and LF, and not HF only, since the original waveform consists of energy from both.

Now let's discuss the "LF gets affected more part"

This is a 1kHz sine wave recorded from an amp with defective power capacitor
1712967291453.png


So the capacitor is running out of juice at the end of that 100Hz charge cycle. (I live in 50Hz country).

Bad capacitor = Worse LF? Certainly doesn't look like that here.

Neither me nor the owner noticed less bass. Rather, the complaint is about cracking sound when volume is turned up. That kind of cracking sound characteristic of clipping. Heck, clipping happens so often in our everyday lives (microphones, phone calls) you'd know it when you hear it.

Moral of story is,

They can throw all the theories they want, but except for the scientifically motivated, nobody will go and find out how much each of these effects with opposite outcome contribute to the final result, and the scientifically motivated will tell you to look at the final result.

They mention impedance swings up to 50 ohms at responance that they claim requires a higher power amp to tame, but I don't understand why super high impedance requires more power either.

No it doesn't. High impedance swings requiring more power to tame is a misconception, one that shows up even on this forum from time to time, but usually by usernames with low likes-to-message ratio.

This sounds bogus to me, surely greater current delivery just allows you to drive a lower impedance speaker

Exactly. Amps have max current and max voltage limits. Amps are designed to have these limits appropriate for the load type, as having more max current will not benefit something that does not need more current (e.g. 1000W PSU on a 30W computer), and having more max voltage does not benefit something that does not need that much voltage to push that amount of current/power.
The max current/voltage combination is chosen to optimize the performance you get for the cost. Let's say we start with a standard amp. I can double the power supply and output transistors so it has double max current. But it does not increase the power I can push in a voltage-limited situation, so all that max current capacity is wasted.

The rabbit hole goes deeper when we consider what those additional max capacity do to help linearity or lower distortion, and this is indeed something always being done in super low distortion amps (even in audio, look at some of the >110dB SINAD amps measured here), to put it using a simple example it's like using transistors rated for 40A for an amp rated at 20A. But this applies to both voltage *and* current: Having high spare voltage or a super tightly regulated* power supply are also known to increase output linearity since the olden days because frankly speaking each individual components' linearly is horrible so we only use the best part of their operating range in the first place. And of course, this low-hanging fruit was already plucked and designers have moved towards tackling other sources which have become the new bottleneck for SINAD. But lately it seems that allocating excessive capacity has become cost-effective again, a logical outcome considering cost is decreasing and engineering difficulty is increasing, especially since we can easily find cheap 50W class-AB and 500W class-D running around nowadays.

But all this above is not what the influencers are talking about, while those in the actual business are too tired to talk about, so just read up to the best of your ability so that you know influencer bs when you see them.
 
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Doodski

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Srajan; he references another similarly powered amplifier - the Pass Labs 30.8 - which he says has 20 output devices per channel where as the Enlium has one output device per channel, and hence can provide more current at the same wattage (I guess implying lower voltage in the signal).
The extra unity gain stage (Current amp section.) transistors lower the output impedance so that increased magnitude happens at the musical peaks, (Not the bottom and top because there is no bottom and top because they are called peaks.) A lower output impedance with a suitable power supply will result in a more linear amplifier stage output.
 

wwenze

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Output impedance is difficult to quantify... We can use a simulator to get the value if we know the entire schematic and model, else we would have to guess-calculate based on the device's open-loop impedance divide by how much gain you're willing to sacrifice to give to negative feedback. Or just measure it, which is what the manufacturers list as damping factor.

The Enleum doesn't say how much it does, but Pass Labs 30.8 value of 150 can't be called good by today's standards either.

*And i forgot to explain the asterisk in previous post so I might as well do it here - In the same vein, regulating a power supply via negative feedback also reduces the PSU's output impedance. If your voltage stays the same regardless of load current changes, mathematically you'd have an output impedance of zero. Of course every regulator will shift it's output voltage depending on load, but a cheap LM317 can do 0.1% load regulation. At 5V, a 5mV voltage change from a 1.5A current change implies a output impedance of 3.33mohm.
 
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Doodski

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how much gain you're willing to sacrifice to give to negative feedback.
Interesting question arises from that comment. How much energy is lost at the output to negative feedback? Can it be quantified?
 

wwenze

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Interesting question arises from that comment. How much energy is lost at the output to negative feedback? Can it be quantified?
The transfer function of the circuit is changed. Consider the typical op-amp circuit:
1712971778353.png

I just grabbed it off Google and it's an inverting amp, oh well.

Case 1) In the above image, the right leg of R2, and consequently Vout, will be mathematically -1 * V1 according to op-amp formula. Easy calculation.
Case 2) But if you connect the right leg of R2 to the left side of R1, then your Vout becomes the output from OA1 minus whatever voltage drop across R1, which is dependent on load current. The output impedance becomes pretty much the value of R1.

In both cases, output current will still have to flow through R1.
Without negative feedback to reduce the effect of R1, a reduced voltage will reach the load.
With negative feedback to reduce the effect of R1, the op-amp will just increase the output voltage at its output pin so Vout remains the same i.e. Vout will always follow the formula. Even if you increase the value of R1, the op-amp will just increase its output voltage but Vout remains the same.

So you will use more power but that's because you're trying to make Vout the correct voltage. Conversely, without negative feedback, you will use less power but that's because Vout is lower. In either case, the ratio of the power dissipation between Zout and load is the same.


Add: I think I misread the question. If you mean how much % of the energy is being lost to the negative feedback current, then it's just the inverse of their respective impedance values. In the above picture, 1A of current into load will have 0.1A flowing through R2. In more realistic circuits the power will be negligible... else we'd need heatsinked resistors for a 500W amp.
 

Doodski

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So you will use more power but that's because you're trying to make Vout the correct voltage. Conversely, without negative feedback, you will use less power but that's because Vout is lower. In either case, the ratio of the power dissipation between Zout and load is the same.
IC... Makes sense now that you've laid it out so well.
Another Q.
If for example a class AB amp with multiple output current amp transistors are used with their mating emitter resisters the impedance should go down if the power supply has the current reserve and that will still depend on the negative feedback determining the total gain of the voltage amp section and not the unity current gain stage? How does the negative feedback in a class AB usually get injected back into the circuitry. Is it usually into the voltage amp section or before it or is it at the current amplification stage?
 

Ramon Cota

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I have wondered this myself as i read many times statements such as " this bass/subwoofer driver takes 300 Watts so you need a powerful 400 Watt amp to control it etc " but what if i always listen at 70 db and peaks are 85 db and my puny 70 W amp is able to satisfy it without clipping ? Is 70 W from 70 W amp worse than 70 W from 400 W amp ?
 

Blumlein 88

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I owned a McIntosh 752 for a while. It had a Powerguard circuit that triggered optically to prevent clipping. If you pushed it hard enough, it was a compressor, but it didn't clip. It had lights to let you know when this circuit was being activated. Activated hard enough it turned down peaks so no clipping occurred, but sound would get compressed. It was interesting and instructive in regards to clipping amps. That amp was 75 wpc @ 8 ohms and I think 100 wpc at 4 ohms.

At the time I was using Acoustat Two electrostats. They really could use more power and were not efficient at all. It was enough most of the time, most being like 95% of the time. When I really cranked something up now and again the lights would activate some. I tried it more and you could get them to flash almost continually. It was a very compressed sound. No clipping though. I kept it around as a spare for a few years and let friends use it, tried it on several speakers. It was pretty hard to get those lights to come on with Maggies or Thiels or Meadowlarks. You could get them to come on with Hales only because those seem to handle the power enough you might try it. Most other speakers started to fall apart before you clipped anything (or actually I guess activated the no clipping circuitry). 75 watts is a lot of power really if it is clean. I also owned a Spectral DMA-50 which was an 80 wpc @ 8 ohm amp and maybe something close to double that into 4 ohms. It could be clipped, but there wasn't much you really needed more power on. I mean 160 wpc is only 3 db more and 320 wpc only 6 db more. Enough to notice, but pretty rarely needed. Sure there are speakers that can take that much and if in a large room you might hit that on peaks at times. I do think it would be nice if more amps had clipping indicators at least.

Now with good quality clean power as cheap as it is now, why would you use a 25 watt amp on anything unless it was plenty of power, and wonder about it? Pretty foolish thing to do. And this is a $6k+ 25 watt amp. Foolish product if you ask me. I notice in specs for it they quote 4 wpc into 60 ohms. Strong for a headphone amp for a speaker power amp not so much. Indicates to me it runs on maybe 22 volt rails or something like that. So not a good amp on an ESL because it will have no power in the lower frequencies where an ESL will have 30 ohms or more impedance. It also won't have much current output more than likely. I know it is a high end audiophile thing, but this fashion creates lots of goofy products.
 
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oleg87

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How does the negative feedback in a class AB usually get injected back into the circuitry. Is it usually into the voltage amp section or before it or is it at the current amplification stage?
In the conventional power amp topology the feedback goes to one side of the input differential stage, which has the role of "calculating" the error between input and output.

More exotic topologies may use more complicated feedback schemes to eke out more loop gain.
 

wwenze

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IC... Makes sense now that you've laid it out so well.
Another Q.
If for example a class AB amp with multiple output current amp transistors are used with their mating emitter resisters the impedance should go down if the power supply has the current reserve and that will still depend on the negative feedback determining the total gain of the voltage amp section and not the unity current gain stage? How does the negative feedback in a class AB usually get injected back into the circuitry. Is it usually into the voltage amp section or before it or is it at the current amplification stage?
So the power supply has its own output impedance, and the transistors have their own output impedance. If power supply impedance is negligible (or, if it has the current reserve to not drop voltage upon load), then the emitter resistors become the main output impedance, then yes, double the transistors+emitter resistors will half the open-loop output impedance, which is then divided further by NFB.

Negative feedback can be fed in a lot of places. A practical circuit would have both local (within one amp stage) and global NFB (final output fed back to input pins).

Unity-voltage-gain current buffer stages are less straightforward, not for technical reasons but for marketing reasons. Most designs if they want to actually reproduce the voltage signal accurately, it will have feedback, and the components used to build it will typically have some form of gain and feedback too i.e. Basically they will use transistor with high transconductance, and then use feedback to set gain to unity or whatever they want. In normal amplifiers, this last big transistor stage also provides voltage gain anyway so I prefer to call them output transistors. For products that advertise the inclusion of a current-gain final stage stage... it's like... *brptbsthbraack* I can't predict what is actually in it: Transistors fundamentally work on tranconductance anyway (input voltage changes output current) and we wire the circuit / we arrange the circuit's formula so it produces a voltage gain. But what I can say is, watch out for designs that have very low open-loop gain so they can run the product with no feedback, because those won't give a SINAD over 60dB.

For perspective these are some different products and their applications. Note how the global NFB is connected in all cases.
1712979203345.png

1712979262216.png

1712979315468.png


And also a simplified circuit diagram of the inside of LME49600
1712979773734.png
 
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Doodski

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Here is a rough example of something I would have been presented with in the past. Differential pair at the input, FET outputs and voltage gain stages in between. So the negative feedback is through R13 @ 33k Ohms resister to the base of Q2. How does all that other peripheral stuff fit into the feedback loop. For example through R11 @ 1k Ohm and through R10 @ 10k Ohms to the base of Q3 and then back further through C2 at the base of Q1. That is some complex math going on there. At what point is the feedback not considered feedback in this sort of more complex discreet circuit? Is it considered feedback at the base of Q2 but not at the base of Q1?
mosfet.png
 

wwenze

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Q3 is the current source for Q1 and Q2, together forming the long-tail-pair. The current source provides common-mode rejection and high gain for the LTP.

The node between R11 and R10 is just ground, nothing much else to say regarding ground. So there's no positive feedback (no feedback signal reaching base of Q1).

R11 and C3 allows base of Q2 to move like it's the top node of R2.
 

solderdude

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Here is a rough example of something I would have been presented with in the past. Differential pair at the input, FET outputs and voltage gain stages in between. So the negative feedback is through R13 @ 33k Ohms resister to the base of Q2. How does all that other peripheral stuff fit into the feedback loop. For example through R11 @ 1k Ohm and through R10 @ 10k Ohms to the base of Q3 and then back further through C2 at the base of Q1. That is some complex math going on there. At what point is the feedback not considered feedback in this sort of more complex discreet circuit? Is it considered feedback at the base of Q2 but not at the base of Q1?
View attachment 363400
The negative feedback = R13 and R11 (voltage division determines the gain for frequencies > 10Hz) for DC the feedback is 1x due to C3.
There is a small amount of current going into the base of Q2 but the effective input impedance of Q2 is much, much higher than what flows through Q2.
So the feedback path is R13, R11+C3 only.
The idea is to make the input voltage of Q2 (feedback) the exact same value as the input voltage of Q1. So feedback path is Q2 only it does not make it into Q1.


  • Even if we grant that the musical signal contains higher amplitudes at lower frequencies, given the additive nature of the complex waves that make up music would clipping not affect all frequencies equally, rather than just the bass?
When clipping it will cut off all of the signal above the clipping level so also the rest of the signal that is superimposed on that bass note. This creates the 'roughness' in the sound.
But here's the thing. When clipping is reached (input level exceeds the maximum value) the out clips at the highest peaks. Those peaks are short. This means that when the peak is limited by clipping the bass levels won't increase in loudness anymore (well.. the harmonics will) but as soon as the 'clip' is gone again the rest of the signal won't clip and thus become 'louder' compared to the initial peak of the clipped bass note (at least the first bit that clips and is used by the brain to determine loudness).
So only the bass note will sound sound lower in perceived loudness (clipping) while the rest of the time the amplitude just sounds louder.
Effectively the bass will not increase in 'perceived power' while the rest will so the sound will be less 'bassy'.
Of course, above a certain level the sound will just turn nasty as too much clipping occurs.

  • If this is the case, would this not result in distortion across the frequency range rather than appearing as 'missing bass'?
Yep it will. Fortunately the hearing is a funny thing and is not really bothered by certain types of 'distortion' of peaks only. This is how loudness wars 'work'. Above a certain point it becomes sound degrading.

Do low power amps lack bass? Or do they clip?​


They could have both, lack bass (but not yet clip) or clip.
Here's the thing though... the effect one has between an (8ohm) 15W rated tube amp and 15W rated SS amp can sound quite different where the tube amp could sound more powerful. This has to do with the circuit design combined with the speaker impedance as well as the clipping behavior.

Lacking bass could be because of clipping levels being reached or a circuit having a high pass filter (incorrectly rated transformer and/or input/output capacitors in the signal path)

An alternate explanation is posited by Srajan; he references another similarly powered amplifier - the Pass Labs 30.8 - which he says has 20 output devices per channel where as the Enlium has one output device per channel, and hence can provide more current at the same wattage (I guess implying lower voltage in the signal).

Nonsense, that amp may well be able to provide more current into lower impedances but in the end the clipping voltage is the same if both are rated 25W in 8ohm.
The only difference could be (depends on power supply) that in 2 ohm the Pass could deliver more current and the clipping voltage will be the same as for 8ohm loads where the 1 transistor amp might have a lower output voltage in 2 ohm than it would have in 8ohm.

Speaker impedances are not flat which is the difficulty here. In 8 ohm they would supply the same clipping voltage but one amp may still be able to supply the same output voltage in lower impedances while the other one may not. In this case a speaker that dips at 2ohm might go louder one one amp.

Then there is clipping behavior. Almost never measured but can make a difference between how amps 'sound' once clipping is happening. It could be the difference between 'harsh unpleasant' and just 'limited in power' (up to a certain level of clipping).

So there is some truth but it is more complex than just 'current'.
 
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33AndAThird

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Just checking in to say thanks to everyone for the detailed replies. My questions have been answered, my understanding of clipping enhanced and my scepticism of some of the subjectivist explanations justified. As a bonus the conversation seems to have extended into deeper topics on feedback that are a bit beyond my understanding, but happy that the conversation has helped others.

As an aside I read an interesting essay by Bruno Putzeys on negative feedback a while ago which I was able to follow a little more closely - different questions answered but I found it very interesting.
 

kemmler3D

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Just wanted to mention that with most poweramps you'll ever see, the clipping isn't "hard clipping" that's instantly obvious, like with a digital signal. It's "soft clipping" where the THD increases rapidly, but not instantly, depending on the signal level.

As @solderdude mentioned, this can effectively lower the bass relative to the rest of the signal, because bass almost always draws the most power.

Also, we're accustomed to a lot of harmonic distortion in the bass region anyway, and supposedly people are also less sensitive to THD in this range. Most speakers that don't have large woofers have a lot more THD in the <100hz range than amps do even if they're "clipping". An amp can be considered to be clipping at 1% or 10% THD, but some speakers will reach 100% THD or more when you push them hard to produce low bass.

So, the subjective impression of being bass-shy could easily happen because you're clipping the amp, but it won't sound like "clipping" per se. This doesn't mean the amp has a non-flat frequency response, though.
 

JW001

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Interesting question arises from that comment. How much energy is lost at the output to negative feedback? Can it be quantified?
No energy is lost due to negative feedback, at least not any meaningful amount. A very low current is fed back to a high-impedance input stage via high-value resistors. For example, look at Figure 1 here: https://sound-au.com/project3a.htm. The AC feedback is fed back to transistor Q2 via a voltage divider R4/(R4 + R5), fixing the amplifier gain at 27dB.
 

JW001

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Here is a rough example of something I would have been presented with in the past. Differential pair at the input, FET outputs and voltage gain stages in between. So the negative feedback is through R13 @ 33k Ohms resister to the base of Q2. How does all that other peripheral stuff fit into the feedback loop. For example through R11 @ 1k Ohm and through R10 @ 10k Ohms to the base of Q3 and then back further through C2 at the base of Q1. That is some complex math going on there. At what point is the feedback not considered feedback in this sort of more complex discreet circuit? Is it considered feedback at the base of Q2 but not at the base of Q1?
View attachment 363400
For global feedback analysis, your example amplifier can be represented by this block diagram:
https://upload.wikimedia.org/wikipedia/commons/6/6e/Block_Diagram_for_Feedback.svg

From that diagram, it would be trivial to calculate the gain of the whole amplifier. It is:
G = 1 / (1/Aol + Beta)
where: Aol is the gain of the amplifier with the feedback loop removed (open loop). Normally its value is very high (in thousands) Beta is the gain of the feedback loop itself. For the amplifier above, Beta = R11 / (R11 + R13) = 0.029
Now, since Aol is very large: G = 1 / Beta = 34.5 = 30dB
 
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