dbm to vrms

Given input power in dBm, we can convert to mW by 10^(dBm/10). However we can't use 10^(dBm/20)*1E-3 to convert to Vrms, right? But if we are doing a calculation that involves a ratio of two voltages then its ok to use the above expression without taking load into account, right? The formula I want to use is LO leakage = 20log(2*Vdc/Vm) and I want to compute Vm from input power given.

Thanks

Hi,
dB is always used for power ratio. To convert it any other form use the formula for the power
P = VI = I*I*R = V*V/R
and use it. It is straigh forward.

B R
M

CATV dBm, dBmV, and dBμV Conversions
http://www.maxim-ic.com/appnotes.cfm/appnote_number/808
http://pdfserv.maxim-ic.com/en/an/AN808.pdf

http://www.ctjc.com/units.htm
http://www.tonercable.com/Toner%20We...ableofconv.pdf
http://www.scte.org.uk/members/techn...nversions.html
http://store.noblepub.com/pdfs/KinleyChapter_1.pdf

PdBm=10log(Pwatt/1mW)

Pwatt= vrms2/R