Re: [BP] thermal noise question for signaling conference call today
john,
Yes, -140dBm is 1e-17W.
P = V^2 / R
V = sqrt( R * P ) = sqrt(1e-17 *100) = sqrt(1e-15) =sqrt(1e-16 *10)
= sqrt(10) * 1e-8 = 31.6nV
charles
> I am confused what the -140dBm number refers to in the discussion today (I
> do not recall who put it forth).
>
> -140dbm/Hz is equivalent to 3.2nV/sqrt(Hz).
>
> [-140dBm/Hz = -170dBw/Hz = 1e-17W/Hz = 3.2nV/sqrt(Hz)].
>
> You only get to 32nv/sqrt(Hz) (the number that was said to result) if you
> multiply the original number by 100 (1e-17*100 W/Hz = 32nV/sqrt(Hz)) The
> factor of 100 was Mike's resistor value. If this is why this was done then I
> am assuming that the person who mentioned this is using the standard spot
> noise equation for noise from a resistor:
>
> Pspot = 4kTR = 1e-15 (derived from 32nV/sqrt(Hz))
>
> 4(1.38e-23)T(100) = 1e-15W/Hz
>
> T = 181e3
>
> This seems bizarre. Can someone explain why we multiplied -140dBm = 1e-17W
> by 100? If we do not do this and use 3.2nV/sqrt(Hz) then we end up with
> 224uV of noise. Mike's value would have led to about 100uV of noise.
>
> I am not sure what the proper number to include for thermal noise, but I
> think that 2.2mV is a bit high.
>
> jts
>
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> John T. Stonick, Ph.D.
> Scientist
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| Charles Moore
| Agilent Technologies
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