-1
\$\begingroup\$

I stumbled across some unexpected data points when measuring the AC resistance of magnet wire. I varied the frequency of the AC voltage and measured the corresponding impedance of half a meter of 32 gage magnet wire. As expected, due to eddy currents the AC resistance of the wire is greater than the DC resistance for 1 kHZ - 10 MHz. However, the AC resistance of the wire drops below the DC resistance from 12 MHz to 20 MHz.

This is a hard trend to justify, since increasing the frequency of the voltage would increase the magnitudes of the eddy currents in the wire. Has anyone experienced a sudden drop in the resistance of a wire above a certain frequency?

Thanks guys!

\$\endgroup\$
  • \$\begingroup\$ You mean above the self-resonant frequency of the wire? \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 15 '15 at 3:17
  • 3
    \$\begingroup\$ How was your measurement setup? \$\endgroup\$ – Lorenzo Donati Jul 15 '15 at 3:33
  • 2
    \$\begingroup\$ It seems more likely you're getting a transmission line effect which you're measuring as a decreased impedance. \$\endgroup\$ – Samuel Jul 15 '15 at 3:43
  • 2
    \$\begingroup\$ What instruments are you using to measure resistance at 20 MHz? And how are you compensating for the inductance of the wire under test? \$\endgroup\$ – tomnexus Jul 15 '15 at 4:56
  • \$\begingroup\$ Bifilar-wind the length of wire (fold it in half along itself) and re-run the test. That should all but nullify it's self-inductance. \$\endgroup\$ – rdtsc Jul 15 '15 at 5:18
1
\$\begingroup\$

It not hard to justify and is a very logical problem during AC operation the impedance(no resistance) is dependent on the capacitance and inductance of the wire. Where as in DC operation these elements play no role.

So to say Your capacitive impedance Xc ~ 1/f ie,inversely proportional to the frequency,hence it decreases as the frequency rises. Similarly your inductive resistance is proportional to frequency X~ f hence your inductive reactance increase with the frequency.

Are we Done?

Thats what most people would have thought.

But what you forgot was very important phenomena of skin Depth Skin depth tends to decrease as frequency increases hence conductor penetration decrease and current flow on surface,rather than like Dc where current flow through whole wire.

For more reference see this page
New resistance after skin effect calculation is given as R=Rdc * k* sqrt(f). Also how much your material is affected by skin effect depends on what you are using.

\$\endgroup\$
  • \$\begingroup\$ Skin effect (and proximity effect, which is actually what most people forget) both lead to increased resistance. I'm not seeing where the second half or your answer helps explain a decrease in resistance at higher frequency. \$\endgroup\$ – Samuel Jul 15 '15 at 3:46
  • \$\begingroup\$ @Samuel I think the OP measured ac resistance using Y matrix,or some other method or his conductor is smaller that the skin depth,I still need more reference hence i left that for later addtion. \$\endgroup\$ – MaMba Jul 15 '15 at 3:58
  • \$\begingroup\$ I measured the resistance using a 2 wire setup. The resistance is on the order of 0.5 Ohms, so a 4 wire setup would definitely be more accurate. \$\endgroup\$ – fermi Jul 16 '15 at 5:15
  • \$\begingroup\$ And I simply measured the AC resistance using a digital multimeter. The diameter of the 32 gage wire is 0.2 mm and the skin depth of copper at 10 MHz is 0.02 mm. The current is being transported through a much smaller region at 10 MHz than at 10 KHz, so why is more current being transported at 10 MHz? \$\endgroup\$ – fermi Jul 16 '15 at 5:35
  • \$\begingroup\$ yes the Dia of 32 wire guage is 0.2mm,the skin depth at 1khz(copper) is 2061.6648 um and skin depth (copper) at 10 MHz is 20.6166 um, As Data tells the skin depth reduced significantly your resistance should go up,Physics just dont fail,either your multimeter is not designed to measure at 20 MHz or you missing something,becuase values never lie. \$\endgroup\$ – MaMba Jul 16 '15 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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