I work at company producing household appliances, and I'm looking into a way to test compliance with part 27.5 of the NEN-EN-IEC 60335-1 standard (which is derived from IEC 60335-1).

It states (emphasis theirs):

The connection between the earthing terminal or earthing contact and earthed metal parts shall have low resistance.

If the clearances of basic insulation in a protective extra-low voltage circuit are based on the rated voltage of the appliance, this requirement does not apply to connections providing earthing continuity in the protective extra-low voltage circuit.

Compliance is checked by the following test.

A current derived from a source having a no-load voltage not exceeding 12 V (a.c. or d.c.) and equal to 1,5 times rated current of the appliance or 25 A, whichever is higher, is passed between the earthing terminal or earthing contact and each of the accessible metal parts in turn.

The voltage drop between the earthing terminal of the appliance or the earthing contact of the appliance inlet and the accessible metal part is measured. The resistance calculated from the current and this voltage drop shall not exceed 0,1 Ω.

NOTE 1 In case of doubt, the test is carried out until steady conditions have been established.
NOTE 2 The resistance of the supply cord is not included in the measurement.
NOTE 3 Care is to be taken to ensure that the contact resistance between the tip of the measuring probe and the metal part under test does not influence the test results.

I thought of the following way to do this:


simulate this circuit – Schematic created using CircuitLab

By switching lamps on or off, the current can be chosen, in increments of about 8.3 amps. The diode is there to ensure the voltage doesn't go above 12 volts. Test point B will be connected to the device's earthing contact/terminal, and test point A will be used to test the various earthed metal parts of the device's housing.

My questions are these:

  • Is this a good way of testing compliance to this part of the standard?
  • I'm thinking of using a computer PSU (ATX) to supply the 12 volts. Is there a reason I shouldn't?
  • The lamps are halogen ones. Good idea? (I am aware of their PTC tendencies, the presence of the ampmeter fixes that.)
  • I figure 15AWG (or 1.5 mm²) will be sufficient for wiring all of this?
  • Anything else I'm not thinking of?


  • Power resistors will be used in stead of lamps (due to initial resistance being to high, and PTC behavior)
  • 4mm² (or 11-ish AWG) was suggested, to prevent measurement error problems
  • 1
    \$\begingroup\$ (1) THe diode doesn't do what you say it does. You can leave it out. (2) Aluminium clad wirewound resistors on a heatsink are made for this sort of job, use them instead of bulbs. (3) I wouldn't use 1.5mm^2. Since the minimum acceptable test current is 25A I'd use at least 4mm^2 to reduce measurement error problems. (4) Get an automotive fuseholder and a bag of fuses(30A and up) in case of little accidents. \$\endgroup\$ Jan 27 '17 at 10:16
  • \$\begingroup\$ @BrianDrummond Could you elaborate on the why of (1)? Thanks for the great suggestions btw! \$\endgroup\$
    – KJdev
    Jan 27 '17 at 10:25
  • \$\begingroup\$ Because a forward biased diode will just increase a relatively small voltage drop (about 1V) whether the battery is 3V or 48V. You may have been aiming for a "crowbar" circuit (search term...) and missed ... a zener diode (reverse biassed and connected ACROSS the battery AFTER the fuse) is another option, but not at 25A. \$\endgroup\$ Jan 27 '17 at 10:30
  • \$\begingroup\$ Maybe I'm reading your comment wrong, but I don't understand how using a diode to drop the voltage from 12 to 11-ish volts is wrong. Also, what battery are you talking about? Or was that to illustrate your point? \$\endgroup\$
    – KJdev
    Jan 27 '17 at 10:38
  • \$\begingroup\$ It isn't wrong if that's what you want to do - but that's not what you said it was for in the question. You can simply measure the PSU voltage (and some allow adjustment) if you need a specific voltage. But for compliance with "not exceeding 12V" from an ATX PSU it's fine. I read the question as implying it was some sort of protection e.g. in case of power supply failure, and the diode won't do that. (by "battery" I meant "supply" and it's too late to edit now) \$\endgroup\$ Jan 27 '17 at 10:46

No, using lamps is not a great idea for one simple reason; when cold, the lamp filament resistance is significantly lower than when it is running hot and producing light so, the instantaneous current when a lamp is intitially switched in will be two or possibly three times higher than when operating a few tens of milli seconds later.

  • \$\begingroup\$ That's why I'm switching them on one by one. Would that still not be good enough? What if made it so they were warmed up before the test? If this is still not good enough, do you have an alternative load that's equally cheap and easy to come by? \$\endgroup\$
    – KJdev
    Jan 27 '17 at 9:39
  • \$\begingroup\$ How does switching them on one by one alter what I'm saying? Use power resistors. 12 volts and 8.3 amps is 1.445 ohms. How long the test takes largely determines the power rating of the resistor. \$\endgroup\$
    – Andy aka
    Jan 27 '17 at 9:51
  • \$\begingroup\$ If the lamps were on (warmed up) before the test, wouldn't the problem be eliminated? And 2 to 3 times the current doesn't seem like a problem in this case, or am I judging that wrong? Whatever the case, I think I'm gonna go with your suggestion of power resistors. \$\endgroup\$
    – KJdev
    Jan 27 '17 at 10:23
  • \$\begingroup\$ Power resistors will ALSO have significant drift with temperature, and then you also need to find a way to keep them from vaporizing. So that means heat sinks and fans as well. I don't see the issue if the testing is done after the bulbs had time to warm up. Bulbs are just power resistors with integrated heat sinks. \$\endgroup\$ Feb 24 '20 at 7:10

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