From: sweir (firstname.lastname@example.org)
Date: Mon Mar 26 2001 - 22:11:58 PST
Yes, the derivation is pretty simple. For a series terminated driver, a
new edge passes through a divider network that consists of:
The driver, assumed zero ohms.
The series termination resistor, R,
The transmission line, assumed R matches the real component.
This is just a 2:1 voltage divider between the new value at the driver
output, ( slew rate assumed infinite ), and the prior state of the line,
assumed to be the opposing level, ie +/-Vswing. So, this is where the
Vswing/2 term comes from.
The power transferred to the resistor is the power formula V^2/R, which
This condition persists until the wave propagates to the far-end, where
it reflects 2:1, ( completely open end ), and then propagates back to
the driver. When the return wave reaches the driver, the line is at
equilibrium, and the voltage across Rterm collapses to zero. So, this
yields an energy for a change from one symbol to the other of:
(( Vswing/2 )^2)/Rterm * 2 Tflight.
I actually made a mistake in the original representation, as this
applies to all symbol transitions.
So, power is just the energy per symbol multiplied by the symbol
(( Vswing/2 ) ^2)/Rterm * 2 Tflight * Symbol_rate = Vswing^2 * Tflight *
Symbol_rate / ( 2 * Rterm ).
The actual power is just a tiny bit more, as loading at the far end
prevents a perfect 2:1 reflection. However, the residual power is
Could you please suggest how to derive the equation, you show below?
Series R power = ( ( Vswing / 2 ) ^ 2 / Rterm ) * Tflight * Symbol_rate.
( 2 * Tflight cancels Symbol_rate / 2 )
Does somebody have any clues about the derivation of the equation?
Thank you for your comments.
Senior SI Engineer, Server Team, ARD4
Quanta Computer Inc.,Taiwan, R.O.C.
Tel: 886+3+3979000 ext. 5183
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