a question about local feedback

[hacknet]

i was pondering bout my next amp project and it hit me that i didn`t have an idea what local feedback did to the output impedance of the stage. and if implemented in an output stage, what kind change in output impedance would i see...? is it very much like global feedback?

thanks!

 

[Eli Duttman]

Local or global, voltage NFB is voltage NFB. It raises I/P impedance and lowers O/P impedance. The math stays the same.

 

[hacknet]

with 3db of feedback, how much change in the o/p Z am i expecting to see change?

thanks!

 

[Sch3mat1c]

3 decibels (power) = x 2 = 6 decibel-volts.

dBV are double (duodecibels?) because doubling voltage doubles current into a resistance, so power actually goes up by a factor of four (i.e., +6dB).

At any rate, linear negative feedback reduces Zo, distortion, gain and, if connected properly, increase Zin all by the same factor. This factor is the amount of feedback.

Tim

 

[tubetvr]

3 decibels (power) = x 2 = 6 decibel-volts.

Not quite.... 3dB increase for voltage means that power also increase 3dB

dB for voltage = 20 * LOG(V2/V1), so if V2/V1 is 2, (double voltage) it gives 6dB

dB for Power = 10 * LOG(P2/P1), so if V2/V1 is 2, (double power) it gives 3dB

However 3dB for voltage and 3dB for power is the same thing

V2/V1 = 1.4142 gives 3dB and an increase of 1.4142 in voltage gives an increase of 2 times in power, (P ~ V^2) and 2 times in power is also 3dB.

Back to the original question, 3dB feedback is 1.4142 times in voltage so the decrease in output impedance will be 1.4142 times, (I assume that what is meant with amount feedback here is how much the gain decreases)

Regards Hans

 

[hacknet]

thanks a million, now i`ve got an idea how my amp will turn out.

 

[Sch3mat1c]

Not quite.... 3dB increase for voltage means that power also increase 3dB

Ok, I got that a bit ambiguous. Lemme try again...

By numbers alone, 3dB (power) = factor of two = 6dBV.

I noted lower that, into a resistance, 3dB = 6dBV.

Tim

 

[tubetvr]

By numbers alone, 3dB (power) = factor of two = 6dBV.

Tim I understand what you write but it is not correct, as I wrote:

"dB for voltage = 20 * LOG(V2/V1), so if V2/V1 is 2, (double voltage) it gives 6dB

dB for Power = 10 * LOG(P2/P1), so if V2/V1 is 2, (double power) it gives 3dB"

This is the definition of dB.

What it means is that 3dB is 2 times for Power but only SQRT(2) for voltage.

An example: an amplifier with 40dB gain has power gain of 1000 times but a voltage gain of only 100 times, (assuming same input and output impedance)

In this case, (what Hacknet asked about) we had a feedback of 3dB which means that the power AND voltage gain decrease by 3dB i.e. SQRT(2) ~= 1.4142 times but a power gain decrease of 2 times.

Regards Hans

 

[ray_moth]

This thread, which I strarted last January, discusses the effectiveness of partial/local feedback for improving damping factor. There were many helpful contributions from people with a lot of experience.