[10GBASE-T] clarification on voltage level
In response to some questions at last weeks meeting (and to stimulate some reflector activity), I wanted to clarify the launch voltage level
For 10 dBm, ideal PAM10 (random data and PSD ~sinc^2) and 100 ohm load
Vrms = sqrt(P*R) = 1 V
Vpeak = Vrms*sqrt(3*(M-1)/(M+1)) where M is the number of PAM levels
= 1.57 V
Vpp = 2*Vpeak = 3.13 V
For 9.5 dBm
Vrms = .944 V
Vpeak = 1.48 V
Vpp = 2.96 V
Note: this is based on an ideal PAM signal, any filtering in the transmit path will change the PAR and peak voltages.
William W. Jones, Ph.D.
Director of Systems Engineering
SolarFlare Communications, Inc.
949-581-6830, ext. 2550