From: Larry Miller ([email protected])
Date: Wed Jan 24 2001 - 12:09:35 PST
The loss is proportional to I, not V. So with a higher Zo it takes less I to
get a given V at the receiver. The kind of driver makes no difference-- you
can always substitute a Thevenin voltage source for a Norton current source
or vice-versa. That's why they use high voltage transmission lines for mains
power distribution (when you can get power >:^( ). The same equations we use
for GHz in Inches-world work at 60 Hz in Miles-world.
From: [email protected]
[mailto:[email protected]]On Behalf Of Farrokh Mottahedin
Sent: Wednesday, January 24, 2001 11:44 AM
To: [email protected]
Subject: [SI-LIST] : Skin Effect
There seems to be a phenomenon that on a differential transmission line,
an increase in the characteristic impedance (Zo) will help to reduce IR
losses due to skin effect.
Now, we know that Zo = sqrt(L/C). Likewise the IR losses due to skin
effect can be summarized generally as 4.34 (R/Zo+GZo) in dB/meter. R and G
are the load resistance and admittance.
Conceptually, it also makes sense that if a transmitter sees a larger Zo
(the transmitter does not see the load directly, but only sees the line
ahead), less current will flow, and since the load doesn't change, there
will be less power loss. But if the drivers are current sources, then the
current should be constant, and a larger Zo serves only to cause more IR
loss. Here I am looking for some math to clear all this up rather than to
rely on intuition.
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