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Old 18th June 2004, 03:51 PM   #1
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Default calculating NFB dB

Hi, and thanks for reading my first post!

I'm just wondering how to calculate the amount of NFB applied in a circuit, from the schematic.

Say, for instance, there's a 100k feedback resistor, carrying the signal back from point 'B' (after the O.T., possibly) to point 'A' (phase splitter, or some earlier pre-amp stage cathode). How would I arrive at the amount of NFB in dB's?

(I know how to calculate dB's, BTW, just need to know how to apply it in this instance).

Thanks in advance!

Steve
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Old 20th June 2004, 11:33 AM   #2
benny is offline benny  Australia
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ok,

first, you need to break down your output stage into a couple of basic sections.

Ao, which is the open loop (with no feedback) gain factor of all the stages encompased by your feedback loop.

Ro, which is the internal output resistance of the amplifier, or the transformer primary impedance.

Rf, which is the series feedback resistor, in this case it's the 100k one you mentioned.

Ri, which is the shunt feedback resistor, or the one the resistance between your feed back loop and ground.

now, once you have these

Acl = Ao*(Ri+Rf) / (Ri + Rf + Ro + Ri*Ao)

Acl is the closed loop gain.

For gain reduction (Ar) you use

Ar = Acl/Ao

in dB terms, this is

Ar(dB) = 20*log(Ar)


hope this helps.


cheers
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Old 20th June 2004, 06:51 PM   #3
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*Blank stare*

What I do is figure the gain of the encompassed stages (multiply together or measure), find the impedance of the driven point (usually a cathode... remember cathode follower effect: Z will be lower than just Rk) and figure how much current it will take to drive it. Now, the voltage you need to drive is as follows:
First, you know you need 10mV to reach 5V output. Open-loop gain is thus 5/.01 = 500. If you want to increase this to 1V, gain will drop to 5, the amount of NFB being 500/5 = 100 times, or 20 log(100) = 20dBV. To increase drive voltage to 1V, you need to add .99V to the signal (since .01V of that actually drives the amp). Thus, the current input to the driving point = V/Z. Now, you know these figures are for 5V output, so you can run from .99V on the cathode to 5V at the output = 4V across that NFB resistor.

Your case really is too general to say anything about it, but you can get an idea about how to do it from these steps. For instance, Z can be a grid leak resistor on a LTP (where signal input is on one side, NFB the other).

Tim
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Old 22nd June 2004, 03:49 PM   #4
benny is offline benny  Australia
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simplest thing you can do to obtain simply just the amount of neg f/back is just to calculate the voltage of the signal where you're taking the feeback from... so for example, say you use the 16 ohm tap on an OT at 100W, you are getting v = sqrt(100*16) = 40V... now, just think of the feedback resistor and the resistor to ground as making a voltage divider... voltage divider formula is

Vout = [R2/(R1+R2)] * Vin

so, say you have 100K f/back resistor, and 5k resistor from here to ground, you get

Vout = [5k/105k] * 40

Vout = 1.9V

so you get 38.1V across that f/back resistor... i don't know if this helps.
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Old 23rd June 2004, 02:26 AM   #5
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I've found the articles at Aiken Amps to be very helpful in understanding how to calculate such things as gain, impedance, feedback, etc.

Go to URL: http://www.aikenamps.com/ and select 'TECH INFO', then 'Advanced (Aiken)' and you'll see a menu of articles.
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