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In Reply to: RE: Other Methods IT posted by cpotl on March 05, 2017 at 09:42:22
"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?
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"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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"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.
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