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In Reply to: RE: Other Methods IT posted by Tre' on March 03, 2017 at 09:48:35
"Even in A1 the driver tube must be able to deliver power, not just voltage, to satisfy the Miller capacitance"
The need for a low-Z driver regarding Miller capacitance has to do with dissipation in the driver, not the need to deliver power to the load. Like any pure reactance, Miller capacitance doesn't consume power. Rather, the circuit's dissipation results from the resistance of the driver in series with the Miller capacitance. Current into the Miller capacitance causes power to be dissipated in the driver's series resistance. Reducing that resistance serves to reduce the dissipation (and the frequency-sensitive voltage drop). No power is delivered to the load in any case.
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Buy Chinese. Bury freedom.
Follow Ups:
You said "Current into the Miller capacitance" but then you said "No power is delivered to the load"
If the driver stage has to deliver current to the Miller capacitance then isn't the Miller capacitance a load for the driver tube?
Maybe I'm not understanding what you mean by "the load" or maybe I don't understand what you mean by "consume power".
by Thorsten Loesch
"In addition, a directly heated triode (and any similar triode) has a quite substantial Input Capacitance comprising mostly the miller amplified Anode to Grid Capacitance.
The input Capacitance of a given Valve is (Cga + Cstray) * Mu + Cgk. This capacitance must be driven from somewhere, not only with respect to the frequency response, but also concerning the current draw at higher frequencies.
For a 300B Valve the input capacitance is usually around 70pF, a little more in reality due to stray capacitance's from the wiring or pcb. If we take 80pF Input Capacitance and a Bias of 70V for our above mentioned operating point we must be able to supply the current drawn by this capacitance up to at least 100kHz (usually a much higher frequency is strongly advisable) at full voltage swing.
The impedance of the 80pF Capacitance at 100kHz is around 20kOhm, the peak current drawn by this capacitance at full signal is 3.5mA."
70v times 3.5ma of current is .245 watts of power that needs to be sourced from the driver stage to satisfy the Miller.
I refer to that, rightly or wrongly, as the driver stage delivering power.
and my point was, regardless of the type of coupling the Miller capacitance current requirement still needs to be satisfied.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
"You said "Current into the Miller capacitance" but then you said "No power is delivered to the load"
If the driver stage has to deliver current to the Miller capacitance then isn't the Miller capacitance a load for the driver tube?
Maybe I'm not understanding what you mean by "the load" or maybe I don't understand what you mean by "consume power"."
Perhaps he was making the point that no power is delivered into a purely capacitive load, since the voltage and the current are 90 degrees out of phase. In your example of 70v times 3.5ma of current, if that were into a resistive load it would indeed represent a power dissipation of 0.245W, but if the load is pure capacitive, the power dissipated into it would be zero.
Whether that is a relevant issue as far as demands on the driver is concerned is another question.
Chris
"if that were into a resistive load it would indeed represent a power dissipation of 0.245W, but if the load is pure capacitive, the power dissipated into it would be zero."
That's the part I don't understand.
"We know that in Pure capacitive circuit, current is leading by 90 degree from voltage ( in other words, Voltage is lagging 90 Degree from current) i.e the phase difference between current and voltage is 90 degree.
So If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero."
So "power" is the wrong word for me to use. It seem like a matter of semantics. Current still needs to be delivered from the driver stage even if it causes no heat in the tube (load).
am I on the right track?
Thanks.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Yes, that's what I meant, but it's not just a matter of semantics. The fact that the load doesn't consume power, and that there's a significant power factor involved, might affect the topology and tube type selected for the driver. I admit that I haven't needed to deal with this apparent contradiction as a mathematical problem, but that's not to say the principles can simply be ignored. In any event, I just wanted to clarify the operation. My comments weren't intended to nitpick your post. :)
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Buy Chinese. Bury freedom.
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Have Fun and Enjoy the Music
"Still Working the Problem"
"So "power" is the wrong word for me to use. It seem like a matter of semantics. Current still needs to be delivered from the driver stage even if it causes no heat in the tube (load)."
Yes, I would agree that in the context of discussing the requirements for the driver to be able to supply the necessary voltage and current signal to the output stage, the fact that the power dissipated into a capacitative load is zero is probably hardly relevant.
I haven't thought much about it, but on the face of it I would suppose that the design criteria for a driver that can supply the voltage and supply the current are not really that much affected by whether the voltage and the current are 90 degrees out of phase or not. What matters should be that the driver can provide the necessary current, at the required voltage, without significant voltage drop at the higher frequencies where the capacitance matters more.
Chris
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