David wrote:
Can anyone explain why commercial instruments use DC, despite that small DC
voltages will be developed by unwanted thermocouples? I would have thought
that using AC was a no-brainer no very low resistance measurements, but
commercial instruments don't use to use AC.
Difficulty
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
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
bandwidt
On 17 September 2017 at 21:58, Mitch Van Ochten <
mi...@vincentelectronics.com> wrote:
> The Keithley 2002 uses DC but automatically takes a reading of any offset
> voltage and subtracts it (offset compensation). Rated accuracy on the 20
> ohm range (2 years) is +/- 26 ppm, and with 10 averages i
On 17 September 2017 at 20:12, wrote:
> The question is what accuracy you need.
>
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
The Keithley 2002 uses DC but automatically takes a reading of any offset
voltage and subtracts it (offset compensation). Rated accuracy on the 20 ohm
range (2 years) is +/- 26 ppm, and with 10 averages it has a resolution of 0.1
microohms.
Best regards,
mitch
-Original Message-
Fro
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 c
In message
, "Dr.
David Kirkby (Kirkby Microwave Lt
d)" writes:
>I want to measure the resistance between two bits of aluminum. Each are 40
>x 30 mm across. One is 250 mm long, the other is 8 mm long.
The dominant factor in the resistance you want to measure is the pressure
on the cont
I want to measure the resistance between two bits of aluminum. Each are 40
x 30 mm across. One is 250 mm long, the other is 8 mm long. I'm wondering
is surface oxides are on the faces, so despite being held together with
bolts, the resistance is perhaps not as long as I would expect. There's
also a