I have to confess that I hadn't given this any thought until I heard from you, caninus auricularis. This is what I think: the negative feedback loop takes a sample of the output signal as seen at the OT secondary, scales it down and feeds it back to the phase inverter. I need to make a couple of assumptions here, which I believe are reasonable:dogears wrote:Hi Teo,
Assuming that the taps are rated correctly, I can't see how power scaling would change anything. The 50 watt amp has less gain than the 100 and therefore the db drop would scale as well... no??
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The actual feedback ratio is what counts and where it is tapped from. Not amp total output.
On the other hand, I am just a player and I could be really wrong. I just don't think so.![]()
But maybe I am.... Sheesh this is confusing..... LOL
1. For all practical purporses, the signal generated at the phase inverter remains unaltered -- if no feedback is applied to it -- regarldess of the power being dissipated by the power section.
2. Assume that the OT has perfectly sized taps such that going from using one pair of tubes to two pairs of tubes can be compensated for by using another tap of the OT, such that the power transfer to the speaker will be optimum for a given load rating (call it 4 ohms, since that's where Dumble taps the NFB loop from).
Now, if you call V1 the voltage seen at the 4 ohm tap when two tubes (50W) are being used, the power dissipated is (V1)^2/R. And we said that R is the load impedance.
If we now turn on two more tubes and change OT taps, the new voltage appearing at the OT secondary is V2 (100W case). The power now dissipated by the load is (V2)^2/R, and R is still the same load impedance.
It's pretty easy to see that V1 and V2 are not the same since:
(V1)^2/R = (1/2) (V2) ^2/R
It follows that V2 = 1.4 V1...
The above, coupled with the assumption that the PI is not loaded by the power section (#1 above), means that if I need to feed back the same "amplitude" voltage in the 100W case as we did in the 50W case, the scaling factor will have to be 1.4 times smaller, since the signal that appears at the secondary of the OT is 1.4 times bigger than in the previous case. Conversly, the scaling factor for a 50W amp needs to be 1.4 times bigger than in the 100W case.
Since the NFB loop is a voltage divider with the FB resistor being much larger than the shunt resistor (390 ohms in Dumbles), then it follows that the feedback resistor for a 100W amp should be 1.4 times higher than that of a 50W amp. We know that value is 4.7K for the 100W amps, so for the 50W amps it should be 4.7/1.4 = 3.4K.
I believe this result correlates nicely with Teo's simulation findings.
Cheers,
Gil