Rich,
1500 watts PEP on a 300 ohm load. Sqrt of 1500 X 300 = 670.8 Volts. 1500
v x 300 ohms = 450,000 and the sqrt of 450,000 = 670.8. On a 50 ohm
load, 1500 watts PEP x 50 = 75,000 and the sqrt of 75,000 = 273.86 V.
That would be in peak volts though as I would figure the power measured
with a true peak reading wattmeter. The peak voltage would then have to
yield a peak voltage in this circumstance wouldn't it? An RMS reading
watt meter would be 1.414 times the voltage calculated yielding peak
voltage?
Will
R. Measures wrote:
On Aug 30, 2004, at 6:20 PM, Will Matney wrote:
I failed to mention that the values I were mentioning earlier, in the
designs I was using, were for a Pi tank circuit and not a Pi-L. Thus
the load C would have seen a different load resistance of about 300
ohms or so. For 1500 watts PEP on a 300 ohm load, a voltage of 671
volts would be incurred. At the 50 ohm load for a Pi tank, load C
would see about 274 volts. So a capacitor with a rating of at least
2.5 : 1 difference would be needed. Where V = sqrt of W x R.
- Peak-V or RMS-V?
Will Matney
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Richard L. Measures, AG6K, 805.386.3734. www.somis.org
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