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ads dbmtov expression to convert dBm to V

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ads dbmtov expression

Hi all,

This is the expression used by ads to convert dbm to V, can anyone shed some light on where the 8 comes from.

Example:
y = dbmtov(10 (power) , 50(Zref))
returns 2.0

Expression:
value = ( 8 * (10^((v(power)-30)/10) ) * R(Zref) )^1/2

from my hand calculation, I always get value = 1 based on input power of 10dBm and reference of 50 ohm.

Thanks a lot
 

Re: ads dbmtov expression

It is because dbmtov(power_dBm, Zref) converts the dBm measure into open circuit voltage, Vemf_peak(peak value not rms value), given the reference impedance.

power_Watt=(Vemf_peak/(2*sqrt(2)))^2/real(Zref)=Vemf_peak^2/(8*real(Zref))
power_dBm=10*log10(power_Watt)+30
power_Watt=10^((power_dBm-30)/10)

Vemf_peak=sqrt( 8*real(Zref)*power_Watt )=sqrt( 8*real(Zref)*10^((power_dBm-30)/10) )
 

Hi Panchoo,

Thanks, but this is the part which I still eludes me, the 2 as shown:


power_Watt=(Vemf_peak/(2*sqrt(2)))^2/real(Zref)=Vemf_peak^2/(8*real(Zref))
^
|
|

I understand that,
Vrms = Vpeak/sqrt(2), but why is the 2 there?

Thanks

Added after 6 minutes:

Hi Pancho,

I wanted to highlight the 2 but the arrow spacing is off

power_Watt=(Vemf_peak/( 2 *sqrt(2)))^2/real(Zref)=Vemf_peak^2/(8*real(Zref))

I understand that,
Vrms = Vpeak/sqrt(2), but why is the 2 there?

Thanks
 

Again, dbmtov(power_dBm, Zref) converts the dBm measure into open circuit voltage, Vemf_peak(peak value not rms value), given the reference impedance.

power_Watt=(Vemf_peak/(2*sqrt(2)))^2/real(Zref)
Click to expand...
Red is because Vemf_peak is open circuit voltage.
Blue is due to conversion of Vemf_peak/2 to RMS value.
 

Hi pancho,

Yes I think I got it.... Thanks a lot!!!!!
 

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