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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?

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    \$\begingroup\$ Feedback is attenuation so -dB \$\endgroup\$ – Sunnyskyguy EE75 Dec 15 '18 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 '18 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 '18 at 3:18
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Feedback factor = Loop Gain = BA = (Voltage at -ve input)/(difference Voltage between inputs) quoted at certain frequency. Say 30 or 30dB at 20kHz for an audio power amplifier.

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

Feedback factor (Which is equal to loop Gain) must be less than '1' by the time the loop phase reaches -180 degrees for a stable amp.

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

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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).

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    \$\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 '18 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 '18 at 18:28

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