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In Reply to: RE: What tubes to use with this Interstage Transformers? posted by Tre' on September 04, 2016 at 11:35:43
R is the fixed R between the plate of the Driver and the Grid of the Output tube in the Direct Coupling Circuit.
Or if it's eliminated, it's the Rp of the Driver / Source.
And you are saying 160Kohm impedance of the IT as the R in this case.
I don't think Z(IT) is in series with the Shunt C at all in the IT circuits. It is effectively Rp.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
Follow Ups:
And with this, we are now back at Square 1.
Reducing Rp will have more current available to the C. I don't think I disagree with that. It just meant a different driver tube other than 5K Rp in the case of 801A.
But really, how much current a Grid needs? Elimate the IT and the fixed resistor R in the direct coupling circuit and substitute with the short so the Vsource is the same as the low Rp tube like 500ohms. Fc will shift due to the lower Rp (or R in the RC) value but passing that Fc point, the attenuation will occur the same. Rp values are fixed too. In a way, it is more current avilable till Imax is reached. But beyond Xc point, it'll all be just the same all over. Fc is shifted due to the R being smaller.
So, you can't make more current available without changing Rp (the tube itself) or the IT. I think it's all you have been saying that 1:4IT won't supply enough current due to it's high Z value and the Fc point will occur within AF band.
I also think I have been saying that Triode Roll Off will happen. But when you say it won't work, I started misunderstanding why won't it work.
But I think we are all saying the same thing in a different way.
Essentially, your choice is not to use the 1:4IT which I understand and respect that.
But my choice is to use the IT I have and to work around it and find a suitable tubes for it. I think you have mentioned that it would be very difficult if not next to impossible?
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
"Fc is shifted due to the R being smaller."
Correct but with a low enough Rp (and it not being multiplied by a step up IT) Fc is high enough so it no longer matters.
"Essentially, your choice is not to use the 1:4IT which I understand and respect that.
But my choice is to use the IT I have and to work around it and find a suitable tubes for it. I think you have mentioned that it would be very difficult if not next to impossible? "
Yes, and I will say it again.
If you had a driver stage with (let's say) a 500 ohm output impedance, that 500 ohms would look like 8k ohms to the output tube's grid and Miller C. As long as the output does not have too much Miller C then the 8k ohms would drive it just fine. That is to say the Fc would be high enough as to not matter.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Yes! I agree to that.
That's truly make it difficult to design a circuit for Fc out of 200KHz not to be effected by the RC filter slope.
So, let's say your Fc moves out to the 40KHz instead of 20KHz, the low pass attenuation starts at around 4KHz?
Now, here is the advantage of the 1:4 IT I have. It slightly adds more Voltage gain at the rising frequency so it'll negate that effect of the Low Pass filter loss and puts an overall FR of the final output a little flatter. You normally don't have that advantage with a constant voltage sources.
Now back to the transformer impedance, it was specified by the manufacturer at a certain (unknown to me) frequency as the starting point. But it is the Audio IT, I can safely assume that is the the optimum load it present to the Source during the AF Band.
Since I can't measure and don't know the exact value of it, I must use it as an optimal loading point to the Driver Tube and find an appropriate driver tube. Lower Rp driver tube does not pose a problem but Higher Rp driver more than 10K Rp will be loaded by the Transformer Primary.
It's the same reason you can't put 600Ohm Primary IT on the 5K Rp tube since it'll just load the Voltage in the primary and drop most of it on the primary side and less current (or AC Flux) is available to the core and to the secondary side. It's the opposite of Saturation (current starving core) which looks like a saturated core. I think Dave and I had quite a lengthy discussion of that.
Anyway, I appreciate you taking seriously to make me see through this.
Sometimes, things got lost in translation and my apologies for that. But I think we are essentially on the same page with a different approach to the solution.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
"And you are saying 160Kohm impedance of the IT as the R in this case."I am not saying that. I never said that.
"I don't think Z(IT) is in series with the Shunt C at all in the IT circuits. It is effectively Rp."
Correct, if the IT is 1:1 then Rp (not really Rp but the output impedance of the driver stage) is reflected to the secondary and that is the source impedance and the "R" used to calculate the effect of the Miller Capacitance.
But with a 1:4 winding ratio IT we have a 1:16 impedance ratio IT so "R" is the output impedance of the driver stage times 16. That's is what I have been saying.
BTW Transformers do not have impedances. They only have winding ratios.Transformers are made to work with certain impedances. The inductance and winding capacitance dictate that and that is what the "impedance rating" of a transformer is all about.
But that is not to say that they have impedances of their own. They don't.
Transformers only "reflect" impedance according to their impedance ratio.
The "R" presented to the primary will reflect to the secondary and the "R" presented to the secondary will reflect to the primary.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 09/04/16
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