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Re: [Amps] load capacitor ratings?

To: <amps@contesting.com>
Subject: Re: [Amps] load capacitor ratings?
From: Karl-Arne Markström <sm0aom@telia.com>
Reply-to: Karl-Arne Markström <sm0aom@telia.com>
Date: Tue, 31 Aug 2004 08:03:51 +0200
List-post: <mailto:amps@contesting.com>
Actually, no.

A peak reading wattmeter will show PEP, which by definition is the power
delivered to the load during one or more RF cycles at the peak of the 
modulation envelope.

If the peak voltage for calculation of component stress safety factors is 
needed,
you will have to multiply the voltage derived from SQR(PEP*R) by SQR(2).

A worked example:

1500 W PEP into 50 ohms load:        U(rms) = SQR(50*1500) = SQR(75000) = 273 
V. 
                                                         U(peak) = SQR(2)*273 = 
387 V

73/

Karl-Arne
SM0AOM
----- Original Message ----- 
From: "Will Matney" <craxd1@ezwv.com>
To: "R. Measures" <r@somis.org>; <amps@contesting.com>
Sent: Tuesday, August 31, 2004 7:42 AM
Subject: Re: [Amps] load capacitor ratings?


> 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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> >> http://lists.contesting.com/mailman/listinfo/amps
> >>
> >>
> > Richard L. Measures, AG6K, 805.386.3734.  www.somis.org
> >
> >
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> >
> >
> 
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