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In Reply to: RE: These are the right adapters if you want to run 6CG7s in the MAC amps in place of the 12au7s. posted by rosendds@frontiernet.net on January 15, 2017 at 12:56:46
What Jim said below.Also,te tube has almost twice the dissipation so distortion is also lower at that point and in the final product.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken
Follow Ups:
My Hagerman Clarinet line stage uses two 2au7's.I'm thinking about buying a couple of those tube adapters.
David
If the Coronet uses 12.6vdc,you can't use 6SN7s.You can use 12SN7s tho.Get the ones for the MAC tho.Will love the improvement.
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken
beware! you will have doubled the heater current requirement by running decent tubes.
Also, plate dissipation is not an input for distortion...:) the 6CG7 is just a better tube than the low-cost 12AU7 any day of the week, and twice on Tuesday...and especially at night.
So check on the heater voltage, and crank 'em up next time you play Fizbin.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
beware! you will have doubled the heater current requirement by running decent tubes"
While this is true,keep in mind that it is no problem for the MAC amps because the Mac amps have the potential to run two C8 preamps which have three 12AX7s each,that you will not be running.So ask yourself the question.Is the difference in current that the 6CG7s draw vs the 12AU7s going to even come close to what six 12AX7s draw,running at 6.3vac each?
"For every complex problem there is an answer that is clear, simple, and wrong" H. L. Mencken
I'd probably focus on other circuit modifications. The output stage has *PLENTY* of NFB. Get rid of the PFB applied to the 12BH7 stage, without getting rid of the headroom delivered by the current PFB/Bootstrapping topology first...then get rid of the NFB loop. Gain should be about the same, but with stability markedly increased...LOL
Then move onto the finer details. CCS in the PI cathode circuit, axe the input 12AX7 in favour of a 12AY7, 6CG7 for the PI tube itself...and so forth. And of course the PS mods you have developed...:)
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
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
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.
LOL...I did not consider that question.
Six more 12AX7 per 240 amp is an extra 1.8A of 6.3...:) 12AU7 swap to 6CG7 will eat up .6 A of that leaving an extra .6A per channel so there is some room there...:P Depends on what you want to do next, focus on the cathode followers or the last signal amp tubes.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
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