Blair,
Yes the lead resistance was taken into account.

Phil.

-----Original Message-----
From: TechNet [mailto:[log in to unmask]]On Behalf Of Blair K. Hogg
Sent: Thursday, 16 December 2004 12:04 AM
To: [log in to unmask]
Subject: Re: [TN] Resistive internal layers


Phil,
 
I'm curious, when you measured these did you use a 4-wire meter or some
other method to eliminate the meter lead resistance?
 
Blair

>>> [log in to unmask] 12/14/04 07:26PM >>>
Just some further findings on this particular problem;

There are four through hole connector pads (A,B,C,D) in a row,  each pad
0.8 inches from the next pad.
There is a 6 thou, half ounce track (same internal layer) running in
parallel with the four pins, about 100 thou from the pads.
The four pins are connected to the pads with stubs. IE: the traces do
not go through the pads

The resistance measurements are as follows:
A to B =  1.2 ohms
B to C =  1.2 ohms
C to D =  1.2 ohms

A to C =  2.5 ohms
B to D =  2.5 ohms

A to D =  3.7 ohms

Which makes the trace stub and pad/via resistance approx 0.05 ohms
And the parallel trace resistance between any two adjacent pins approx
1.2 ohms

The other internal layer is the same, but the A to D resistance is 2.4
ohms

Some other PCBs show approx identical resistances
While others have different resistance but the same ratios
Some are OK

None of the external traces show this effect. Even those going through
vias from the top to the bottom of the PCB.

regards,

Phil.

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