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In Reply to: RE: other circuit mods... posted by PakProtector on January 18, 2017 at 04:23:27
Doug
You have becareful with that circuit because it operates on a 100% FB principle in the output stage but you could mod the bootstrapped driver I suppose and that would allow the 12BH7 voltage amp to run cooler.Now the Luxman MB3045 is kind of interesting and I did pick one up but now I need a mate for it.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken
Follow Ups:
nah...it is not 100% FB, if that were true the gain would be 1, instead of the obvious ~2( voltage-wise that is ).
Besides, getting rid of positive FB increases stability, and getting rid of the global loop of NFB will also increase stability.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Doug
I don't how to convince you but there is no voltage gain in that output stage. The cathode windings are in parallel and you cannot add the voltages measured in parallel. This was confirmed by Doc Hoyer and another MAc winder as well.There is 325vac at the grids of Mc60s when you run the amp up to 60 watts.A 2 to 1 ratio would take that to 650vac.I get the output looks like a concertina but cathode windings would have to be in series in order add them and they are not. Being Hoyer winds the transformers daily,I'm pretty sure he knows what he is talking about.
Here is another way to figure. The step ratio of an Mc60 output trafo is 14.9 to 1 at the 8 ohm tap. There is 325vac coming in between the G1 at the 6550s. 325 divided 14.9 equals 21.8. Square the 21.8 and divide by 8 ohms and you come out with 60 watts.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken
Edits: 01/18/17 01/19/17
Well Mike, Hoyer is wrong. It is a split load, where the output voltage is ~2x the input voltage. Measure grid to ground input voltage, and measure voltage cathode-anode on the finals. I'll predict with certainty you will get a ratio of 1.8:1, give or take just a wee bit.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
If you measure g1-cathode and cathode-anode voltage, I think you'll discover the gain to be just like a plain anode load of the same load value.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
ff you measure g1-cathode and cathode-anode voltage, I think you'll discover the gain to be just like a plain anode load of the same load value.
That I can do.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken
the most difficult bit is measuring the anode-cathode voltage. Well...it is just as hard as measuring the g1-cathode voltage since they are both not AC grounded.
Tell you what, measure anode-cathode, and g1-ground and tell me what you find.
cheers,
Douglas
you ought to be able to conduct this experiment on paper. You move the g1 up a volt, the cathode will go up around .9 volts. Since the plate winding has the same number of turns it will move the same amount, but in the opposite direction. So, anode-cathode voltage change of 1.8V for a 1V change at the input.
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
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