In message CANX10hA5BeZYCXJORQ5r4h=CbbV2JpZedsWKFswg0hU6HNP-QA@mail.gmail.com, "Dr. David Kirkby (Kirkby Microwave Lt
d)" writes:

The dominant factor in the resistance you want to measure is the pressure
on the contacting surfaces.

If the contact pressure is high enough, making the joint "gas tight",
you can treat it as one contiguous piece of aluminium and just do
the math (= differential equations for complex geometries)

If the contact pressure is too low, trying to measure the joint
resistance is a fools errand, because it will not be stable over
any parameter (voltage, current, humidity, vibration, air composition,
...)

The only known way to measure ohmic resistance in the micro-ohm
domain with any precision is caloriometric methods, and they are
experimentally troublesome to no end.

--
Poul-Henning Kamp      | UNIX since Zilog Zeus 3.20
phk@FreeBSD.ORG        | TCP/IP since RFC 956
FreeBSD committer      | BSD since 4.3-tahoe
Never attribute to malice what can adequately be explained by incompetence.

A DMM with good low resistance capability will have an "offset compensated
ohms" feature. A large current is used and then a small current is used.
The slope of the line formed by V-I points is the true resistance.

Measuring tiny DC voltages is easier than measuring tiny AC voltages. More
bandwidth is more cost for an amplifier.

Pass an amp or more through the device and use a KE 2182A to measure the
microvolts.

Since I do not own a 2182A, I use a KE 147. You will need a DMM on the
analog output of the 147 if you want digits of resolution.

When I do low resistance, I am just looking to bracket the resistance, not
measure to 6 1/2 digits.

The thermal offsets can require a lot of time to stabilize.

I built a Tetrajunction demo device that is in the tens of nano-ohms. It
takes all day to set up the measurement and have the offsets stabilize and
take a few measurements.

-Brian Smith

On Sun, Sep 17, 2017 at 1:23 PM, Dr. David Kirkby (Kirkby Microwave Ltd) <
drkirkby@kirkbymicrowave.co.uk> wrote:

The question is what accuracy you need.
The classical way to do that (achieving high accuracy) is to apply a known accurate current (say 10A) and measure the voltage drop accross the rod with a nanovoltmeter.
As the piece of aluminum is isothermal you should not expect a big thermovoltage. You could also compensate for this by reversing the current and take the average, also by nulling the voltage reading prior to applying any current. Generating precisely known AC currents (low uncertainty) is difficult (i.e. measuring it precisely), therefore DC currents are ususaly used also in metrology for this.
If you do some internet search you will find metrology reports about this. If you do not have a nanovoltmeter you could build a measurement amplifier with not that much of an effort (based on chopper amp or low drif precision opamp)

On 17 September 2017 at 20:12, acbern@gmx.de wrote:

No a lot. I just want to find out if there's any voltage drops that are
significantly higher than I would expect. The unit makes an RF transmission
line, and the loss at RF is significantly higher than predicted by a
computer model, which takes into account the skin depth of the materials.
I'm wondering if there's something odd going on. I suspect the problem is
the current in the aluminum is not being computed properly due to the oxide
on the surface. But I just wanted to make sure there was no unexpected DC
resistance. I don't think there will be, but I want to climate that
possibility.

The only nV meter I have is the lock-in amplifier, which has a full-scale
sensitivity of 2 nV to 1 V in a 1-2-5 sequence.

The only instrument I have able to measure > 3 A of current is a handheld
multimeter. One of my power supplies can supply 35 A, and has an ammeter in
it. I don't have any particularly accurate means of measuring DC current
outside the limited of the 3457A.

In terms of simplicity, getting a $10 audio amplifier from China and using
the lock-in amplifier is the way to go, but I accept a metrologist would
not like that idea!

Dave

On 17 September 2017 at 21:58, Mitch Van Ochten <
Mitch@vincentelectronics.com> wrote:

Also very pricy!

I've made an offer on a Keithley 580 micro ohm meter, and ordered a couple
of $3 audio amplifier boards. I will try one with the lock-in amplifier. It
it does not work, or works poorly, it will not have broken the bank.

Dave

David wrote:

Difficulty of measuring AC, compared to DC, is one reason, as has been
mentioned.  But the main reason is that skin effect (and usually to a
much lesser extent, inductance) is a significant factor at surprisingly
low frequencies, particularly when the expected value is in the micro-
to deci-ohm range.

Also, since you said the waveguide is aluminum (and didn't say anything
about plating), be aware that aluminum exposed to air is covered by a
thin aluminum oxide layer (Al2O3), which forms within seconds after a
new surface is exposed.  This layer is thin -- generally about 4 nm --
but the bulk resistivity of Al2O3 is very high, so there is a finite and
variable resistance across the interface between two joined pieces of
aluminum (depending on the area of the joint, the joining pressure, and
the extent to which the joining method produces a clean [oxide-free],
gas-tight interface between the joined surfaces).

Best regards,

Charles

I use a Keithley 2182 and 6221 nano-ohm setup at work. It is a combination of a reversing precision current source and a nanovoltmeter with embedded software to manage the process. I can reliably measure into the 50 nano-ohm regime.The surface chemistry of the metal joint is very important. Both nickel and aluminum have thin tenacious oxides. It takes a compliance setting of >65 V to punch through nickel oxide films on pressure contacts.If a bolted contact has sufficient contact pressure to crack the film yo can obtain dry circuit conduction. My experience is that a silver plated contact surface will have a decade lower contact resistance than an otherwise identical nickel plated one.Keithley has an excellent Low Level Measurement Handbook and appnotes for download. 
Howard Davidson

Sent from Yahoo Mail for iPad

On Sunday, September 17, 2017, 4:41 PM, Charles Steinmetz csteinmetz@yandex.com wrote:

David wrote:

Difficulty of measuring AC, compared to DC, is one reason, as has been
mentioned.  But the main reason is that skin effect (and usually to a
much lesser extent, inductance) is a significant factor at surprisingly
low frequencies, particularly when the expected value is in the micro-
to deci-ohm range.

Also, since you said the waveguide is aluminum (and didn't say anything
about plating), be aware that aluminum exposed to air is covered by a
thin aluminum oxide layer (Al2O3), which forms within seconds after a
new surface is exposed.  This layer is thin -- generally about 4 nm --
but the bulk resistivity of Al2O3 is very high, so there is a finite and
variable resistance across the interface between two joined pieces of
aluminum (depending on the area of the joint, the joining pressure, and
the extent to which the joining method produces a clean [oxide-free],
gas-tight interface between the joined surfaces).

Best regards,

Charles

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On 18 Sep 2017 00:43, "Charles Steinmetz" csteinmetz@yandex.com wrote:

about plating), be aware that aluminum exposed to air is covered by a thin
aluminum oxide layer (Al2O3), which forms within seconds after a new
surface is exposed.  This layer is thin -- generally about 4 nm -- but the
bulk resistivity of Al2O3 is very high, so there is a finite and variable
resistance across the interface between two joined pieces of aluminum
(depending on the area of the joint, the joining pressure, and the extent
to which the joining method produces a clean [oxide-free], gas-tight
interface between the joined surfaces).

Thanks.  You have confirmed what I was thinking - it is probably the
oxide causing the problem.

It's not a waveguide in the normal sense of the word, transmitting a TE or
TM wave down a hollow tube,  but more like a coaxial line transmitting
something close(ish) to a TEM wave. The outer conductor is uncoated
aluminum and rectangular in cross section.  The inner conductor is brass.
See pictures attached (I made them small, so quality his not great, but it
should not too use much bandwidth)

Attached are a couple of pictures, and also S11 measured on a VNA, with one
connector shorted Since this is a reflection measurement, the EM wave
travellels along this twice, so about half the loss would be in each
direction. It is only a rough measurement, but a transmission measurement
showed similar results, but half as much attenuation, as it is only being
attenuated one way.

Maybe I need to use brass, or silver plate the aluminum.

The purpose of this was to measure the loss of a very low loss liquid
dielectric, but from discussions I had with someone at NPL, such a
structure is not suitable if the loss is very low.

Anyway, I have put it an offer on a Keithley microohm meter. I notice there
are a lot of Chinese ones at quite low priced. I've no idea how good/bad
they are. But they are much more modern and cheaper than an affordable
Kiethley meter. A Keithley 2002 is well outside my budget.

Dave