anonymous
2 December 2004
I was trying recently to figure out why 2Vrms is 6.02 dBV instead of 3.01 dBv, and I learned that it is of course because the power in constant-load circuit is proportional to the square of the voltage. So,
dBV = 20 * log (Vmeas / 1V)
Here's my question: Does the same thing apply to dBm? If 0dBm is 1 milliwatt into a 600 ohm load, then is 2 milliwatts into the same load 3.01 dBm or 6.02 dBm?
Here's a good explanation of dB in general (tho it doesn't quite get to my question):
http://en.wikipedia.org/wiki/Decibel
Thanks,
RR
Comments
post your q here if you still need help... http://www.dutchforc
post your q here if you still need help...
http://www.dutchforce.com/~eforum/index.php?s=c31d7133ad3160992b71144f5e00b8a1&showforum=8
I'm pretty sure i've answered my own question. dBm = 10 * log (
I'm pretty sure i've answered my own question.
dBm = 10 * log (Pmeas / 1mW)
Not that I think I'll ever actually use this stuff, but I get curious...
RR