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?

- posted
17 years ago

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

Loading thread data ...

- posted
17 years ago

"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

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

- posted
17 years ago

-- Trevor Wilson www.rageaudio.com.au

- posted
17 years ago

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, 1/1000

so 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

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)

2) you have ignored the mW scalar, 1/1000

so you should have used 10^(15/10)*1/1000 = 0.0316 W

3) your 2

{50*

multiply by sqrt(2) to convert to 1.778 Volts peak

double it to get 3.556Vpp

Cheers Terry

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