Question - How do I convert dBm into miniVolts?

[Nomad MP3]

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?

 

[Phil Allison]

"Nomad MP3" ...

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?

** 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

 

[Trevor Wilson]

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:3hciipFg9u16U1@individual.net...

"Nomad MP3" ...
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?



** 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.

**Sounds suspiciously like an assignment question to me.


--
Trevor Wilson
www.rageaudio.com.au

 

[Terry Given]

Nomad MP3 wrote:

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?

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