[sdiy] Equation for converting V PP to dBm?

[Magnus Danielson]

On 12/29/2011 06:01 PM, Justin Owen wrote:
> Hello,
>
> I've been researching how to convert a Peak to Peak Voltage into dB and I've come up with 4 slightly different equations which all give different answers. I've also had a few different people tell me that different ones were the 'right' one - so I thought I'd ask here.
>
> 1) dB = 20*log_10(Vpp/0.707)
> 2) dB = 20*LOG10(Volts_peak_to_peak/SQRT(0.008*Z))
> 3) db = (20log10) (RMS )&  RMS= 0.707*P-P
> 4) dB = 20 log10 (Vout / Vin)
>
> Recognise any of them?
>
> What I'm trying to do is plot the response curve of the 2164 based on the data sheet spec of -33mV/dB. I figure if I can convert V PP to dB I'll also have mV to V PP.
>
> Takers? Thanks.

So you want dBm? OK, lets assume sine waveform and 600 Ohm impedance, we 
then have

Z = 600
Vpp = 2*Vp
Vp = sqrt(2)*Vrms
W = Vrms^2 / Z
LdBm = 10*log10(W/0.001)

Reversing for Vp

Vp = Vpp/2

Reversing for Vrms

Vrms = VP/sqrt(2)

Inserting that results in

LdBm = 10*log10(Vrms^2 / 0.6) = 20*log10(Vrms/sqrt(0.6))
      = 20*log10(Vp/sqrt(1.2)) = 20*log10(Vpp/sqrt(4.8))

So, an approximation would be

LdBm = 20*log10(Vpp/2.191)

The 0 dBm voltage of 2.191 V is 2*sqrt(2) higher than the 0.775 for dBu 
comming from the crestfactor (peak/RMS) of 1.414 and peak-to-peak/peak 
factor of 2.

Recall that dBm is power and not voltage, and a system impedance must be 
assumed. For dBu this is 600 ohm, and 0.775Vrms is an approximation.

Cheers,
Magnus