0
\$\begingroup\$

I am trying to calculate the current required to energize and hold a 24 VAC relay coil while switching no load. I did the same for a 24 VDC relay coil using ohms law which is very simple and matched my measured current. But my calculation for the AC coil did not match my measured current value from my ammeter.

Measured values of AC relay coil:

  • DC coil resistance: 13.5 ohms

  • Coil inductance (while manually latched/energized): 168 mH

  • Operating voltage: 26.2 VAC RMS

  • Operating current: 262 mA

I created a series RL equivalent circuit of the relay coil by first working out the inductive reactance of the coil at 50 Hz to be 52.78j ohms, giving an overall impedance of 54.48 < 75.65 degrees (in polar form). Then simply V/Z, 26.2/54.48 = 480 mA.

This is considerably more than what I physically measured!

I think the reason for this is because I have measured the DC coil resistance instead of the AC resistance. If I work my calculation back from my known measured values, my coil resistance should be 85.94 ohms (this is assuming my inductance measurement is accurate as I used an inductance meter).

If anyone thinks there is something else I'm doing wrong, please let me know. If you agree I need to measure the AC resistance, can I do this accurately without an oscilloscope?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ DC coil resistance is what's required for your calculations which look OK. See keisan.casio.com/exec/system/1258032600. How did you measure inductance? \$\endgroup\$
    – Transistor
    Mar 23, 2018 at 20:36
  • \$\begingroup\$ Hi, thanks for the reply. I just measured directly across the coil terminals with an inductance meter while I manually ‘energised’ or latched the relay to get the inductance while in that state. It is a cheap meter, maybe it is not very accurate. \$\endgroup\$
    – David777
    Mar 23, 2018 at 21:52
  • \$\begingroup\$ the Coil inductance in armed state is greater the core is closed the 168 mH is rise by two or more \$\endgroup\$
    – manuel
    Jul 5, 2020 at 5:09

2 Answers 2

Reset to default
1
\$\begingroup\$

An AC relay is not a simple inductor, there's a shading coil, and that's going to mess up your inductance measurement unless you did it at AC frequency. eddy currents will skew the result too,

What are the differences between an AC and DC coil relay?

\$\endgroup\$
2
  • \$\begingroup\$ Thanks Jasen. So this is not the best component to do this experiment with. Could I repeat the same procedure with a 3 phase ac induction motor under no load and create a per phase equivalent circuit? Is the dc resistance and coil inductance of one phase all I require or do I need ac resistance? \$\endgroup\$
    – David777
    Mar 24, 2018 at 10:27
  • \$\begingroup\$ I can't think of a good way to measure it without actually running it.and measuring voltage and current. \$\endgroup\$
    – Jasen Слава Україні
    Nov 17, 2018 at 23:50
2
\$\begingroup\$

The key is what R and X are at frequency. R is actually a reflection of energy adsorbed, as noted above will reflect the shading coil and hysteresis losses, eddy current losses, etc. Again the inductance has all sorts of influences that will not be reflected by an inductance meter. the shading coil puts a time variable into the measurement. The overall impedance can best be determined by energizing the relay coil and measuring voltage and current, as suggested above. How you get the phase angle probably is going to require a scope. A power meter would get you the actual "R" value, but on a relay such a power meter would be pretty small.

\$\endgroup\$

Your Answer

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

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