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Re: [Amps] load capacitor ratings?

To: "R. Measures" <r@somis.org>,"amps@contesting.com" <amps@contesting.com>
Subject: Re: [Amps] load capacitor ratings?
From: Will Matney <craxd1@ezwv.com>
Date: Tue, 31 Aug 2004 02:07:46 -0400
List-post: <mailto:amps@contesting.com>
Rich,
Yep, your correct, I wasn't thinking about the differences in the readings. So in turn it would double the voltage values. Don't know why I was thinking that, must be getting senile?


Will

R. Measures wrote:


On Aug 30, 2004, at 10:42 PM, Will Matney wrote:


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?


For a sinewave, RMS V is 0.707 of peak V. When talking about AC, the type of V must be specified as either peak, peak-to-peak, or RMS. Also, PEP is not peak power, it is based on the RMS value of the observed peak. Thus, 1500W-PEP = 3000W-peak.


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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Richard L. Measures, AG6K, 805.386.3734. www.somis.org


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