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Question - How do I convert dBm into miniVolts?

Discussion in 'Hobby Electronics' started by Nomad MP3, Jun 16, 2005.

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  1. Nomad MP3

    Nomad MP3 Guest

    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?
  2. Phil Allison

    Phil Allison Guest

    "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
  3. **Sounds suspiciously like an assignment question to me.
  4. Terry Given

    Terry Given Guest

    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

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