I'm doing a DIY project of bulding a 1kW Induction heater. in the process I have built a poormans PSU out of a MOT ( and a 50V 25A brigde rectifier. with not even nearly enough of a smoothing cap (2,500 uF from a broken switched psu for my old comp).. but anyway:


I am going to build a Volt/Amp-meter with a Aurdino tiny and as I am testing my prototype something is a bit of..

I'm just wondering what I'm doing wrong..

According to the internet copper has a resistivity coefficient (q) of \$1.72\cdot 10^{-8} = q\$. Resistance (R) in a conductor is given by \$(q L)/A = R\$ where L is length in meter and A is Surface area in \$m^2\$.

The resistance I'm using for a shunt is a 2.5mm^2 copper cable (European way. I don't know gauge)

now, I decided I want 2mV/A in drop and that, witch all other variables now given gives me a Length (L) of 0.727 m for a


simulate this circuit – Schematic created using CircuitLab

Issue When I hook it up to a known circuit that handles 20A+ and go to calibrate it in reference to my multimeter I end up at 5.7mV/A drop instead of 2.0mV/A witch was calculated.

It's quite a difference and since this stuff is pretty straightforward \$U=RI\$ I wonder what I'm missing?

  • \$\begingroup\$ Can you describe in more detail exactly how you connect to the shunt? The circuit diagram above shows the voltmeter connected on the circuit side of the resistor terminals, so you're measuring voltage drop on the connections and also the shunt. It would be easy to have a few mOhm there. Better to connect the shunt to the circuit, and then a short way from the ends of the wire, clamp on your sense terminals. Also describe how the Arduino is powered and if its ground is common with anything else? \$\endgroup\$
    – tomnexus
    Jul 5, 2015 at 16:23
  • \$\begingroup\$ Yes thanks, forgot about that, i shopuld connect it to the return path as close to the end as possible.. And i havent connected it to my tiny yet, i just tried it on low amp with my multimeter and osc as reference \$\endgroup\$ Jul 5, 2015 at 17:17
  • \$\begingroup\$ Here is a recent answer with a nice diagram of the way to connect to a shunt. It shows all the parasitic resistances, you can easily see how they will affect your results if they're included in the measurement. \$\endgroup\$
    – tomnexus
    Jul 7, 2015 at 6:25

1 Answer 1


A is not surface area it is cross-sectional area.

So L = \$\frac {A R}{\rho}\$ = 0.291m

A = 2.5E-6 m^2

R = 0.002\$\Omega\$

\$\rho\$ = 1.72E-8 \$\Omega\$ m (your number)

You're right, it's a lot easier (from first principles) using European units rather than gauges, but usually we just look it up in a table.

Copper has a quite high temperature coefficient (about +4000 ppm/K) so it's only useful as a crude shunt. Even brass is better, but nichrome constantan or manganin would best (in order of possible difficulty to source).

  • 1
    \$\begingroup\$ hrmm.. my actual misstake i see now was that i took A=.0025^2=6.25E^-6 i meant cross-sectional area. \$\endgroup\$ Jul 5, 2015 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.