[SI-LIST] Re: TDR

Yue,
Each equipment manufacturer approaches TDR stimulation differently, and how
they generate the TDR stimulous can have a significant impact on apparent
impedance shifts like you desribe.

For example, one equipment manufacturer states that their
signal generator is a -10 mA to 0 mA current source in parallel
with a 50 ohm shunt resistor, which feeds a 50 Ohm transmission
line to the coaxial connector.

Under open condition, the -10 mA is shunted through the internal
50 Ohm resistor, and the voltage at the output is -500 mV.  When
the current source transitions to 0 mA, the signal will
propagate/reflect/etc and eventually settle out at 0 V.

Under a 50 Ohm load condition (i.e., ideally terminated), the
-10 mA is split between the internal 50 Ohm load and the external
50 Ohm termination.  As such, the output voltage from the -10 mA
source will be -250 mV.
Again, when the current source transitions to 0 mA, the 
output voltage will eventually settle out at the same 0 V.

Under a short condition (i.e., 0 Ohms), the -10 mA is shorted
to ground, and the output voltage is 0 V.  After the current
source transitions to 0 mA, the signal will propogate/reflect/etc
and again eventually settle out at 0 V.

Looking at these three conditions, you'll notice that the starting
voltage is very dependent upon the load.  That is, the higher the
load resistance (shunted to ground), the lower the starting
voltage.  Also, regardless of the load or starting voltage,
the voltage eventually settles down to 0 V.

Oh yeah, the above example was for a current source in parallel
with an internal 50 Ohm shunt termination.  However, per Thevenin
equivalence, the same exact scenario holds true for voltage sources
with series termination, so this is a basic issue with all TDRs.

The trick is for the TDR to compensate for this load-dependency.
Each manufacturer has it's own method of compensation, and some do
it better than others.

We don't have the exact TDR head you're working with (per
separate e-mail), so I apologize for not having a specific
answer for your situation.  That said, I'll provide a few
ideas to consider.

An important point with all this is that the impedance of
your 'high quality' cables does not change with load; it remains
constant.  If the cables are 50 Ohms under ideal conditions, then the
impedance will remain at 50 Ohms under non-ideal conditions (barring
bending/melting/destroying the cables...).  You can use this
fact by using your cables are references.

One option is to adjust your TDR time scale such that a small
portion of your 'high quality' cables is always on the screen.
Then, place a marker on that portion of the TDR response and
continually record its value.

Then, as you change loads, note the apparent impedance at the
load and at the DUT (device under test).  From this, you can
post-calibrate the impedance.

We've used this approach where we use NIST-traceable 50.00 Ohm
impedance coaxial lines as references (in place of your 
"high-quality" cables).  In doing so, we're fairly confident
in our reference impedance.

Note that there are two potential ways of performing the post-calibration
that are dependent upon how your TDR is trying to
compensate for the load shift.  In some TDRs, you can do a simple
impedance subtraction.  In others, you'll need to convert
impedance to reflection coefficient (rho), subtract rho values,
then convert back to impedance.  (some TDRs offer this as a
feature called baseline correction or zero-rho)

Hope this helps, and good luck,
Pat Zabinski
Principal Engineer                
Mayo Clinic               Ph: 507-538-5499
200 First St SW           Fx: 507-284-9171
Rochester, MN 55905       Em: zabinski.patrick@xxxxxxxx




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