I'd like to add a correction to otherwise pertinent comments of Peter:
Ten amps is the PEAK value of current and therefore it's not correct to
calculate dissipation based upon this value. It can be done either by
calculating RMS current and then P=Irms^2*R or by integral calculus of the same
formula with instantaneous values of current. Anyway, lost power is
significantly lower than 175W in Peter's example.
I guess that what Peter really says is that for a tuned circuit, losses are
proportional to QL/Q0, where QL is loaded Q and Q0 is unloaded Q. And yes, more
Q0 is better.
73 de Mike, YO3CTK
-----Original Message-----
From: Peter Chadwick [mailto:Peter_Chadwick@mitel.com]
Sent: marti 5 ianuarie 1999 15:07
To: 'amps'
Subject: RE: [AMPS] Plate load charts
As Rich says
? In any event, reducing RF resistance in L1 improves tank efficiency.
Get this in perspective. If you have a 4CX1000 with 2000 volts plate swing
(Peak, not peak to peak) and a 2000ohm plate load with a Q of 10, and Ip
peak of 1 amp, the circulating current in the tank is 10 amps. With a coil
Q of 100, on 20m the reactance is about 175 ohms and the series resistance
1.75 ohms. This gives 175 watts loss in the tank coil. No wonder they get
hot!
If coil Q is 200, the series resistance is .875, and the loss 87.5 watts.
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