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When designing a closed loop amplifier for audio the poles of the open-loop transfer function are usually only vaguely known.

\$\frac{1}{(s-p_1)(s-p_2)...}\$

These poles can be modeled as \$RC\$ low-passes which attenuate the signal from their break frequency and, more importantly, add phase shift to the signal.

To avoid to much phase shift before gain reaches the 0-dB limit usually a pole is deliberately added at a very low frequency so the open loop gain falls below 0 dB before the other poles kick in.

So far so good. But how can the second, third etc pole be estimated to get a clue where the first pole has to be?

What does the notion of "so much dB feedback is ok to be applied without threatening the stability" mean? How is the applicable amount of dB feedback defined?

An example: I want an amplifier to amplify a signal 30 times and have an open loop gain of say 10000. What does it mean when I say "I apply ...dB of feedback"? Usually I would make a voltage divider of 29/1 and therefore get a gain of 30 (factor, not dB). I don't know how to put this any simpler, but doesn't the applicable amount of feedback depend on how my closed loop gain of the amplifier should be? It's often said that the more feedback the better but when I make a unity gain buffer It is useless since I want to amplify my music, right?

Long story short:

What is meant by the applicable amount of feedback?

How do I estimate the other poles?

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  • \$\begingroup\$ It seems you are missing the distinction between the open-loop and closed-loop transfer functions. When we talk about the gain bandwidth product (GPB) we are talking about the product of the DC gain the -3dB frequency of the closed-loop transfer function. When I talk about the unity gain frequency of the open-loop transfer function, that is a different thing. As for how you specify the feedback, dB is always a relative measure. 0dB is the same as 1. You can specify the feedback gain (actually an attenuation) in dB. 0dB would be unity feedback, a 1:10 attenuation would be -20dB, a 1:100 attenua \$\endgroup\$ – user49826 Jul 22 '14 at 2:10
  • \$\begingroup\$ So, am I getting you right? 0dB feedback = unity gain buffer, -20dB feedback = 1:10 feedback = amplification of 10 and so on? But I read some recurring pattern that the amount of feedback is measured positively and it was said that the amount of fb shall not exceed a certain value. E.g. 20 dB feedback is ok, 40dB will cause oscillation. How does that fit into your explanation? \$\endgroup\$ – jjstcool Jul 22 '14 at 10:10
  • \$\begingroup\$ If the feedback network is a passive divider, then 0dB feedback is the most you could get. It is possible, of course, to have a feedback network with gain > 1 (> 0dB), but this is not common in audio amplifier design. I don't know that I've often seen feedback discussed in that way, however, so I'm not sure I can really answer you. \$\endgroup\$ – user49826 Jul 22 '14 at 15:05
  • \$\begingroup\$ @user49628 yes, seems there are various ways to define it. But thanks for thinking about it with me:-) \$\endgroup\$ – jjstcool Jul 23 '14 at 11:30
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BODE diagram

I think the keyword is "loop gain" - and the most important point (as far as stability is concerned) is the frequency where the loop gain Alo is unity (0 dB).

Because the loop gain is the product of the amplifier´s open-loop gain Alo and the feedback factor Hr (Alo=Hr*Aol) you can find the BODE diagram for the loop gain response very easily:

Draw Aol=f(w) and 1/Hr(w) as a BODE plot and the difference between both curves gives you the loop gain Alo (in dB). The stability criterion requires that at the frequency where both curves meet (Alo=0 dB) the phase shift of the loop gain must not equal to (or even more than) 180 deg.

That means: The "rate of closure" (slope) of the loop gain at this crossing frequency must not be -40dB/dec. (Rule of thumb: If the second pole is identical to the frequency with Alo=0 dB we have app. 45 deg phase margin).

Based on this requirement you can derive the necessary/desired location of the second pole.

Remark ("It's often said that the more feedback the better"): You strictly must distinguish between DC and dynamic stability. Heavy feedback improves the stability of the operating point but - at the same time - degrades the dynamic stability (against oscillations).

EDIT: Sorry for the large picture. I don´t know if/how the size could be reduced.

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  • \$\begingroup\$ The attached diagram shows how the BODE plot for the loop gain Alo is produced. The frequency response for the gain of the amplifier (Aol) has a second pole which gives a phase margin of app. 45 deg for 100% feedback (unity closed-loop gain). \$\endgroup\$ – LvW Jul 22 '14 at 9:26
  • \$\begingroup\$ Ok, I think I'm with you. To get on the same page: assume 100dB open loop gain (olg). For simplicity we assume there are no poles. I want to amplify my signal voltage by say 10. I make a voltage divider (vd) 1:10 and feed it back. Is then my feedback amount 90dB? If this is right a vd of 1:10 applies different amounts of feedback, dependent on olg, right? Means: Hr for a given vd is a function of Aol, right? Could you define loop gain more specifically, is that the gain of the feedback part or the gain of the overall amp with closed loop? \$\endgroup\$ – jjstcool Jul 22 '14 at 10:21
  • \$\begingroup\$ To your remark, yes I'm aware that this is the tradeoff. If it was different, there would be no problem and nothing to talk about:-) \$\endgroup\$ – jjstcool Jul 22 '14 at 10:23
  • \$\begingroup\$ @jjstcool, to answer your question: A voltage divider 1:10 in the feedback path means: Feedback resistor Rf=9k and input resistor Ro=1k (Ro/(Ro+Rf)=1/10). Thus, your feedback factor is -20dB and for the BODE plot 1/Hr=+20dB. Regarding the loop gain: There is only one commonly agreed definition: Loop gain Alo=Aol*Hr. For negative feedback, this results, of course, in a negative loop gain. However, this cannot be seen in the BODE diagram because it shows magnitudes only. Please note that the feedback factor of 0.01 gives an inverting gain of -10 only in case the opamps gain Aol is very large. \$\endgroup\$ – LvW Jul 22 '14 at 10:58
  • \$\begingroup\$ @jjstcool, regarding your question "Hr for a given vd is a function of Aol, right?" I can answer as follows: The opamp´s open-loop gain Aol influences the closed-loop gain only if it is "rather small". Better: As long as the loop gain (Aol*Hr) is much larger than unity (roughly: larger than 40 dB) the closed-loop gain will be determined primarily by the feedback factor only. By the way: That is the main reason for Aol values as large as 100...120 dB (at dc). \$\endgroup\$ – LvW Jul 22 '14 at 12:04
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Estimating the location of the other poles would require a thorough analysis of the specific circuit you are using. Hard to even crack into that without more information.

Regarding the effect of feedback gain on closed-loop stability: consider that when you put a feedback divider in the loop (1/29 in your case) you reduce the open-loop gain. In other words, the gain of the loop will fall to 0dB earlier. For a typical single-pole dominant loop, this will benefit stability because the unity gain frequency will be further from the detrimental phase shifts introduced by your high-frequency (parasitic) poles. In this sense, unity gain configuration is the most stressful for a high-frequency amplifier. You see some op-amp datasheets that list "unity gain stable" as a feature.

When you see a specification for the amount of feedback you can apply, it is referring to this phenomena. 0dB feedback would be unity (closed loop) gain configuration, while in your case the feedback is -30dB. Note that you introduce an attenuation in the feedback transfer function in order to achieve an amplification in your closed loop transfer function. I.e. your divider is -30dB or 1/30 and your closed-loop gain will be 30dB or 30x.


It seems you are missing the distinction between the open-loop and closed-loop transfer functions. When we talk about the gain bandwidth product (GPB) we are talking about the product of the DC gain the -3dB frequency of the closed-loop transfer function. When I talk about the unity gain frequency of the open-loop transfer function, that is a different thing.

As for how you specify the feedback, dB is always a relative measure. 0dB is the same as 1. You can specify the feedback gain (actually an attenuation) in dB. 0dB would be unity feedback, a 1:10 attenuation would be -20dB, a 1:100 attenuation would be -40dB and so on. So it would make sense to say that an amplifier is stable with no more than -20dB feedback...

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  • \$\begingroup\$ As far as I understand it doesn't. My open loop gain solely depends on the first pole (provided we only have one). When introducing negative feedback the GBP stays the same, only at a higher frequency it will fall below my desired amplification. Of course for a gain of 0dB it stays the longest this gain. When making a unity gain stage, the feedback is in my opinion the biggest since I feed back the most (namely all, as opposed to open loop, where I do no feedback). So I conclude that there must be a different measurement of the negative feedback than you propose. Sorry, I think youre wrong \$\endgroup\$ – jjstcool Jul 22 '14 at 1:01
  • \$\begingroup\$ Or I don't get it right. Had to use two comments... \$\endgroup\$ – jjstcool Jul 22 '14 at 1:02
  • \$\begingroup\$ Your open loop UGF (for a single-pole system) depends on both the DC gain and the pole location. When you change your feedback network, you change the former... You are right, however, that the unity gain is the "biggest" feedback. \$\endgroup\$ – user49628 Jul 22 '14 at 1:09
  • \$\begingroup\$ Also, note the distinction that your closed loop GBP stays the same when you change the feedback gain, but not your open loop UGF... \$\endgroup\$ – user49628 Jul 22 '14 at 1:13
  • \$\begingroup\$ I don't think so. Assume a open loop gain of 100dB and a pole at 1Hz. then the gain goes down with 20db/dec. If you lower the closed loop gain to say 30dB you have more bandwith which sustains this gain. The UGB will stay the same. See here en.wikipedia.org/wiki/Gain%E2%80%93bandwidth_product this again leads to the question how is feedback measured in dB? I'm pretty sure that unity gain is not 0dB feedback, but rather open loop could be considered 0dB feedback. So how is it? \$\endgroup\$ – jjstcool Jul 22 '14 at 1:25
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I found the answer to my question by accident:Leach explains it

Just for those who are still interested in what I asked. Now everything makes sense and gets way clearer. Thanks for the help anyways!

Edit : I made up an example to clarify it and make it more tangible: Assume a system which has 3 Amplifiers (x100, x100, x10) and 3 Low Passes which have poles at 1Hz (all for mathematical simplicity). We obtain \$\frac{1e5}{(s+1)^3}\$ and bode plotting it gives a phase shift of 180° at about 1.7 Hz. When looking into the output, the gain has fallen to about 80dB. To stay mathematically stable we need to attenuate the signal by more than 80dB (1/10000) to have the phase-shifted signal which is put into the negative input lower than one (0dB). That makes for \$1+Ab\$ (omitting the 1) \$1e5/1e4\$ which is equal to 100dB (A)- 80dB(b) = 20 dB. There we are: the highest applicable feedback factor is 20dB. Adding more poles makes this even lower (critical phase shift is added up earlier; imagine 4 poles, all a 1Hz, then 180° is reached while gain is still 87dB, so we have to attenuate the outgoing signal by 87dB to stay stable, hence we have 100-87 = 13dB left for feedback.) This concludes to: b = A minus maximum-applicable-feedback. Note that the amount of applied feedback cannot be determined solely from b, you have to know A! Done:-)

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  • \$\begingroup\$ I cannot follow your explanation. At first, we have a phase shift of -180 deg at w=1.73 rad/s. Hence, the frequency is f=1.73/6.28=0.275 Hz. More than that, this transfer function with three equal poles (decoupled) is known from the oscillator theory. From this, we know that the circuit oscillates for a loop gain of app. "6". That means: The circuit is stable for loop gains below 6 (15.5 dB). \$\endgroup\$ – LvW Aug 8 '14 at 7:18
  • \$\begingroup\$ Ok, I did not match the formula correctly,lost the minus sign, sorry, nevertheless the point remains the same: The amount of feedback is defined as : \$1+bA\$. So if you have an amplifier with a total gain of 100dB you cannot just close the feedback loop since the signal will be fed back at a level of 80dB and thus oscillate. Your feedback path has therefore to attenuate the signal at least by 80dB to stay under 0dB at -180 deg. Putting it to the formula given above this makes 1+Ab equals (omitting the one) 100dB-80dB, so the amount of feedback applicable for stable operation has to be less \$\endgroup\$ – jjstcool Aug 8 '14 at 19:57
  • \$\begingroup\$ than 20dB. This is not the overall gain, nor is it the factor \$b\$ but the expression \$1+Ab\$. Hope this explains it better. \$\endgroup\$ – jjstcool Aug 8 '14 at 19:58
  • \$\begingroup\$ In principle, you are right. I have mentioned a loop gain of "6" (oscillation case). That was wrong - instead it is "8", eqivalent to 18dB (my error). Applying your method, we can see that the gain at 180 deg phase shift will be 82 dB (instead of 80 dB). Thus, the difference is 100-82=18 dB. Hence, we arrive at the same value for the allowed amount of feedback. \$\endgroup\$ – LvW Aug 9 '14 at 8:55
  • \$\begingroup\$ good, so we're on the same page. The reason why this "weird" measure is used is the different angle of looking at the topic. In osc. theory you want to amplify the output signal back to the negative input, so it makes sense to to say the feedback path has to have a specific gain. A different angle would be to specify \$b\$ in terms of how much overall gain you need. But when looking from the angle of stability, it makes sense find out how much the signal has to be attenuated by the fb-path in order to stay stable. This also depends on the o/l gain of the amp so this measure is used. \$\endgroup\$ – jjstcool Aug 9 '14 at 21:02

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