Re: peak-to-peak voltage to watt

Charles G. Fry (fry@bert.chem.wisc.edu)
Wed, 10 Jun 1998 18:03:19 -0500

Woody has a typo; the last equation is easy to remember:

P = V(RMS)^2 / R

V(RMS) = V(p-p)/[2*sqrt(2)] ...peak-to-peak is always bigger than RMS!

then if R=50 ohms

P = V(p-p)^2 / 400

Cheers,

At 03:43 PM 6/10/98 -0700, you wrote:
>Power = V(RMS)^2 / R
>
>1 volt(RMS) = 2.828 volt(Peak to Peak)
>
>woody@acornnmr.com
>
>> -----Original Message-----
>> From: Qing Ning [mailto:q-ning@nwu.edu]
>> Sent: Wednesday, June 10, 1998 3:18 PM
>> To: ammrl@wwitch.unl.edu
>> Subject: peak-to-peak voltage to watt
>>
>>
>> Hi! there,
>>
>> I am measuring output power of transmitter and decouplerof my VXR300 using
>> Tektornix scope. The reading I got is the peak-to-peak voltage. I am
>> wondering if anyone has a equation to convert it to watt. Thanks
>> in advance!
>>
>> Qing Ning
>>
>> ----------
>> Qing Ning, Ph.D
>>
>> NMR Specialist Analytical Service Lab
>> q-ning@nwu.edu Department of Chemistry
>> tel. 847-491-7080 Northwestern University
>> Fax 847-491-7713 2145 Sheridan Road
>> Evanston, IL 60208

----------------------------------------------------------
Charlie Fry Tel: (608)262-3182
Director, MR Facility Fax: (608)262-0381
Chem. Dept., Univ. Wisconsin
Madison, WI 53706 USA email: fry@chem.wisc.edu
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