I have a specification says the power level at the transmitter's antenna should be at +15 dBm. If I assume the antenna's impedance is 50Ohm, can I calculate the peak to peak voltage to be
Vpp = { 2 * 50 * [10^(15/20)] }^0.5 = 16.8mV?
Question - How do I convert dBm into miniVolts?
Jun 16, 2005
3 Replies
"Nomad MP3" ...
** Errrr - no.1 mW at 50 ohms = sq rt 0.05 = 0. 224 volts rms.
- 15 dB = 5.62 times 0.224 = 1.26 volts rms
1.26 volts rms = 3.56 volts p-p.
............ Phil
**Sounds suspiciously like an assignment question to me.
Trevor Wilson
www.rageaudio.com.au
Phil's answer is correct. you are on the right track, but have made several errors:
1) its power, so dBm = 10log(P/1mW) - you used 20log... which is correct for voltage or current.so you should have used 10^(15/10)*1mW = 31.6 *mW*
2) you have ignored the mW scalar, 1/1000so you should have used 10^(15/10)*1/1000 = 0.0316 W
3) your 2* is wrong, *and* shouldnt be inside the square root.{50*10^(15/10)*1/1000} = Phils 1.257V, which is the RMS voltage required across 50 Ohms to generate 31.6mW = 15dBm
multiply by sqrt(2) to convert to 1.778 Volts peak
double it to get 3.556Vpp
Cheers Terry
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