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Re: [Amps] Measuring RF Power

To: "R.Measures" <r@somis.org>
Subject: Re: [Amps] Measuring RF Power
From: Gary Schafer <garyschafer@comcast.net>
Reply-to: garyschafer@comcast.net
Date: Tue, 22 Mar 2005 11:16:57 -0500
List-post: <mailto:amps@contesting.com>

R.Measures wrote:
> On Mar 21, 2005, at 10:06 AM, Gary Schafer wrote:
> 
> 
>>
>>R. Measures wrote:
>>
>>>On Mar 21, 2005, at 9:08 AM, Gary Schafer wrote:
>>>
>>>>
>>>>Tony King - W4ZT wrote:
>>>>
>>>>
>>>>>Harold B. Mandel wrote:
>>>>>
>>>>>
>>>>>
>>>>>><snip> PEP is different
>>>>>>than RMS, and RMS is different than Average. <snip>
>>>>>>
>>>>>>
>>>>>It may be worth mentioning that RMS is the usual method of defining 
>>>>>a DC
>>>>>equivalent which, given a pure waveform, will be about the same as 
>>>>>the
>>>>>average. With distorted waveforms RMS will not define the average
>>>>>properly.
>>>>
>>>>
>>>>It may also be worth mentioning that there really is no such thing as
>>>>"RMS power". The proper term is average power. RMS is only valid in
>>>>terms of voltage or current.
>>>>
>>>>You can find an RMS value of power but it is not useful for anything.
>>>
>>>It's fairly useful for heating water.
>>
>>Only average power is useful.
> 
> 
> Average power is not a measure of heating ability.  RMS power is a 
> measure of heating ability.
> cr

That's a common misconception.
Only RMS voltage or RMS current through a resistor produces the 
equivalent heating that the same value DC voltage or current provide 
into that same resistor. When calculated for power they are all average 
power.

2 amps DC into 1 ohm = 4 watts average power.
2 amps RMS into 1 ohm = 4 watts average power.

You can not find RMS power by multiplying RMS voltage by RMS current. 
That gives average power.

If you square RMS voltage and divide by resistance that gives average power.

If you square RMS current and multiply by resistance that gives average 
power.

73
Gary  K4FMX


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