Sometimes I read things like "This amplifier has 30dB of feedback", what figure is that refering to? the \$\beta\$ of the amplifier? or the loop gain \$1+A\beta \$? or the negative feedback factor \$\frac{1}{1+A\beta}\$?

I know that the general closed loop equation for gain is \$A=\frac{A}{1+A\beta}\$

But if the figure quoted refers to \$\beta\$, then a \$\beta=1\$ is maximum feedback, but \$20\log(\beta)\$ with a \$\beta=1\$ would mean 0 dB of feedback which doesnt make sense, because I believe that a feedback factor of 1 should give an infinite dB of feedback. So which one is it?

  • 2
    \$\begingroup\$ Feedback is attenuation so -dB \$\endgroup\$ Dec 15, 2018 at 0:28
  • \$\begingroup\$ Fortunately, we can read such a wording only sometimes. I must admit, it happened only once that I could read such an information (in dB). Normally, the feedback factor is given as an absolute figure or in %.. \$\endgroup\$
    – LvW
    Dec 15, 2018 at 9:32
  • \$\begingroup\$ I believe I found what I was looking for, the amount of feedback in dB seems to be the difference between the open loop gain and the close loop gain at a specific frequency. \$\endgroup\$
    – S.s.
    Dec 20, 2018 at 3:18
  • \$\begingroup\$ That's true, the amount of feedback is the difference between the open loop gain and the closed loop gain when both are specified in dBs. If both the open loop gain and closed loop gain are specified as a ratio (instead of in dBs) then the amount of feedback, known as the feedback factor (1+BAol for the non-inverting amp), is the ratio of the open loop gain to the closed loop gain (Ao/Acl). See my equations below. \$\endgroup\$
    – user173271
    Aug 8, 2022 at 10:45

2 Answers 2


When considering the non-inverting amp, feedback factor = Loop Gain +1 = BA+1 = (Voltage at amplifier's input)/(difference Voltage between inputs) quoted at certain frequency. Say 30 or 30dB at 20kHz for an audio power amplifier. Note that, strictly speaking, Feedback factor = Return Ratio +1 because Loop Gain = -BA where as Return Ratio = BA. Or, you could say that Feedback factor = 1 - Loop Gain.

Feedback factor can be quoted as a ratio or in dB. Feedback factor = BA+1 or in dB = 20log10(BA+1).

Note that feedback factor is different to 'feedback fraction' which = B = R1/(R1+R2).

Feedback factor equations

  • \$\begingroup\$ The feedback factor is \$1+A\beta\$ whereas the loop gain is \$A\beta\$. They are not the same thing. \$\endgroup\$
    – edmz
    Aug 25, 2019 at 11:10
  • \$\begingroup\$ I've edited my answer to take into account your comment. I'd appreciate any further comments you can make regarding the correctness of my equations, particularly equation 4. \$\endgroup\$
    – user173271
    Sep 8, 2019 at 15:13
  • \$\begingroup\$ It's now July 2022 and I feel much more confident about the correctness of all four of those equations. \$\endgroup\$
    – user173271
    Jul 6, 2022 at 4:54

Take this all with a grain of salt, I don't work in dB on a daily basis. An RF guy could probably answer this blindfolded with their hands tied behind their back. But I'll take a crack with what I remember from university many years ago.

When you hear things described in decibels, that's almost always expressing a ratio. When it comes to a ratio of power, the formula you want to remember is: dB = 10 * log10(P_o / P_i). When it comes to a ratio of voltages the formula is dB = 20 * log10( P_out / P_in). This is due to the fact that Power is proportional to the square of Voltage, and exponents become multipliers in logarithmic space (because of the algebraic laws of exponents and logs).

Then it comes down to understanding what the ratio being described is. So when I see something that says "This amplifier has 30dB of feedback", what I take that to mean is the ratio of the power fed back to the total power output of the circuit, which is to say: 30 = 10 * log10( Power_feedback / Power_totalout). I could be wrong about that, but it's a decent guess. It matches a rule of thumb I remember, which is 3dB equates to 50%.

If 10 watts is fed back, and 10 watts is delivered to the load, the dB according to this logic would be 10 * log10 (10 / (10 + 10)) = 10 * log10( 0.5 ) = -0.3 * 10 = 3dB. Again according to this logic if its 30dB, that means the fraction of the output power which is fed back is:

-30 = 10 * log10(P_feedback / P_total)

So P_feedback / P_total = 10^(-30/10) = 0.001 = 0.1%

Hopefully someone else will either ratify this logic or refute it (in which case I will readily withdraw the answer).

  • 1
    \$\begingroup\$ I'm afraid you have it backwards. By definition dB is defined as a power ratio, specifically as 10logpower ratio). Power is proportional to the square of voltage. If the 2 powers involved in the ratio are associated with equal resistances, then the power ratio is given by the square of the voltage ratio. In this case, the ratio in dB is given by 20log(voltage ratio) because of the squaring of the voltage. \$\endgroup\$
    – Barry
    Dec 15, 2018 at 4:35
  • \$\begingroup\$ ah that's right, duh P ~ V^2. so log(P) ~ 2*log(V), I'll update my answer and maths. Thanks for the review @Barry \$\endgroup\$
    – vicatcu
    Dec 15, 2018 at 18:28

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