NFB for 20 watt/8 Ohm tap?

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Normster
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NFB for 20 watt/8 Ohm tap?

Post by Normster »

I did a little research through old threads and found this comment from Dogears:
The 8 ohm tap has about 40% more voltage than the 4 ohm tap. Multiply the 4.7k fb resistor by 1.4 to get the equivelent.
I'd try about a 6.5k on the 8 ohm tap. Leave the rest of the circuit unchanged.
And this comment from Gil:
1. 100W Dumbles use a 4.7K feedback resistor, 390 shunt resistor, so the feedback voltage transfer function is 390/4700 = 8.3%.
2. 50W Dumbles used a 8.2K feedback resistor and 1K shunt resistor, so the feedback voltage transfer function is 1/8.2 = 12.2%....
With a little more math, their ratio is: 12.2/8.3 = 1.47
So, for a 20 watt 6V6 amp with a single 8 Ohm tap, what should the series and shunt resistors be for NFB? If I follow the logic here, my series resistor should be about 9k2 with a 390R shunt. Somehow that doesn't sound right.
llemtt
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NFB

Post by llemtt »

The logic behind NFB is how much negative feedback you put in place, it is measured in dB so you usually hear statements like "this amp has 20dB of negative feedback".

How to measure it? Take the open loop gain in dB (the gain of the amp without the loop in place) and subtract the closed loop gain in dB (the gain of the amp with the loop in place), this gives you how many dBs of negative feedback has your amp.

Amp distortion (below clipping) is reduced by the same amount of negative feedback, although the harmonic spectrum become worst, while the damping factor is increases by the same amount. (i.e. 3dB of negative feedback means half distortion and double damping factor)

If the amp has an "almost" infinite open loop gain like usual opamps, the amount of negative feedback can be easly calculated from the ratio of feedback and shunt resistor. Tube power amps have a relatively low open loop gain so calculation isn't easy but you can find all the formulas on the Aiken site.

I usually prefer the PC way, so I just simulate the amps via Spice to take open and closed loop measures, you can do it directly on the real amp by inputting a 1volt (or less) sine signal and measuring the output voltage at the speaker with and without the loop in place. Then tweak the resistors until you get the amount of negative feedback you desire.

The 100w amp should be between 5dB and 6dB -> http://www.m-elli.it/dumble_50w-100w_feedback.pdf

teo
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Darkbluemurder
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Re: NFB for 20 watt/8 Ohm tap?

Post by Darkbluemurder »

I have a 6k8 series and a 470R shunt resistor (didn't have a 390R) in my 2 6V6 amp. The feedback runs from the 8 Ohms tap. The amp sounds fine (most recent improvement: added a 1MA global master volume pot with a 47pf bright cap before the PI).

You could temporarily wire in a 25k pot in place of the series resistor, turn it to where you like the sound most, measure the value and put the next closest standard value in.
Normster
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Re: NFB for 20 watt/8 Ohm tap?

Post by Normster »

Thanks very much guys! I hope to get this build started tomorrow and maybe finished this weekend. The NFB was my last little bit of confusion. I guess there's no harm in substituting a trimmer for the resistor and just tweaking until I get the tone I like. :wink:
Pete
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Re: NFB for 20 watt/8 Ohm tap?

Post by Pete »

1. 100W Dumbles use a 4.7K feedback resistor, 390 shunt resistor, so the feedback voltage transfer function is 390/4700 = 8.3%.
2. 50W Dumbles used a 8.2K feedback resistor and 1K shunt resistor, so the feedback voltage transfer function is 1/8.2 = 12.2%....
So would it be correct to assume that the schematics which show the 4.7K/390R values with 2 output tubes are incorrect? Just hoping to clear this up.
dogears
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Re: NFB for 20 watt/8 Ohm tap?

Post by dogears »

Could be...... I have found that if your amps output is really beefy and can handle the lows, the 100watt feedback arrangement sounds great. However, if you hear some thuddy saturating flub on the lows, try something with more NFB. I had to do this on one of my amps.
Pete wrote:
1. 100W Dumbles use a 4.7K feedback resistor, 390 shunt resistor, so the feedback voltage transfer function is 390/4700 = 8.3%.
2. 50W Dumbles used a 8.2K feedback resistor and 1K shunt resistor, so the feedback voltage transfer function is 1/8.2 = 12.2%....
So would it be correct to assume that the schematics which show the 4.7K/390R values with 2 output tubes are incorrect? Just hoping to clear this up.
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