From: [email protected]
Date: Wed Oct 27 1999 - 18:51:11 PDT
Larry,
Perhaps my past message is not clear.
While I agree with rms current and average power ideas, as stated below,
what I was saying in that special case is, rather than getting into
averaging the instantaneous power, you can get the answer
of average power by the average current. And that's why I always
used average current for my past discussions.
Nothing is trying to be offending. Sorry for the confusion.
Raymond
---------------------- Forwarded by Raymond Leung/QSA/AU on 28/10/99 11:48
---------------------------
Raymond Leung
27/10/99 08:47
To: [email protected]
cc:
Subject: RE: [SI-LIST] : Heating resistor: RMS or average current (was Re:
Effect of low Zo for unterminated lines)
I would standby Andy's idea.
In the case of the power dissipation in a driver + a capacitive load,
the average power delivered from the power supply to that whole
thing is V*I(average), and most of the power is anticipated
consumed in the driver.
Raymond
---------------------- Forwarded by Raymond Leung/QSA/AU on 27/10/99 08:40
---------------------------
"Ingraham, Andrew" <[email protected]> on 27/10/99 04:40:59
Please respond to [email protected]
To: "'[email protected]'" <[email protected]>
cc: (bcc: Raymond Leung/QSA/AU)
Subject: RE: [SI-LIST] : Heating resistor: RMS or average current (was Re:
Effect of low Zo for unterminated lines)
>RMS IS the heating effect in terms of watts; good RMS meters use bolometers
>to measure it with. "Average" is the wrong answer.
Be careful here. The correct thing is RMS current, and Average power.
I'm just trying to steer everyone away from thinking along the lines of "RMS
power" (taking the square root of the mean average of the square of the
instantaneous power) or RMS watts, which would be the wrong thing.
Heating watts is a simple time-average of the instantaneous watts, or can be
found using an RMS averaging of current and/or voltage.
Andy
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