[volt-nuts] Best way to measure micro Ohms

[Todd Micallef]

Dave,

Another meter is the Cambridge LOM-510A. I am not sure if it is in your
budget but there has been a review made on EEVBlog. There is one currently
on eBay with a current amplifier that I have never seen before today. It
may be worth reviewing if it meets your needs.

Todd

On Mon, Sep 18, 2017 at 12:31 PM, Dr. David Kirkby (Kirkby Microwave Ltd) <
drkirkby at kirkbymicrowave.co.uk> wrote:

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