Re: [BP] thermal noise question for signaling conference call today
John,
Since I brought it up, I'll take a shot at answering your question.
The 100x multiplier comes in when you convert W to V^2, assuming that
the load is a 100ohm resistor. This is valid for a 100ohm differential
impedance system, which I was assuming the backplane channel to be.
But like I said at the end of the call, I was proposing the -140dBm/Hz
number as the background noise level, not the thermal noise level. I
didn't realize that Mike had broken out those two parameters.
Vivek
Vivek Telang
Broadcom Corp.
-----Original Message-----
From: owner-stds-802-3-blade@listserv.ieee.org
[mailto:owner-stds-802-3-blade@listserv.ieee.org] On Behalf Of John
Stonick
Sent: Friday, October 29, 2004 3:29 PM
To: STDS-802-3-BLADE@listserv.ieee.org
Subject: [BP] thermal noise question for signaling conference call today
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.
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