I have a zero-cross solid state relay (Crydom CKRD2420) that regulates the power output of a 2kW@240V heater using 10Hz PWM (e.g. here to illustrate). I need the power output of the heater to be under ~2W when OFF. However I measured the leakage current is 25mA RMS (2.5 times the 10mA garanteed in the datasheet, by the way...), which means that the power leakage is about 13W (the mains supply voltage reads 264V).

My first thought was a current divider, but the bleed resistor would reduce by a factor of 10 the available power, and reduce significantly the efficiency. I can't use a normal relay because of the lifetime required (I need at least 50-100M cycles of operation), and power MOSFETs would be tricky to implement. Are there any simple-ish fixes I have overlooked?

  • \$\begingroup\$ To me, the leakage current being out of spec says the SSR is damaged. \$\endgroup\$
    – Matt Young
    Dec 16, 2015 at 14:35
  • \$\begingroup\$ To me as well, but I've got 4 of them and they all show the same leakage current - I don't see how they could all be damaged in the same way, especially given that they are used correctly... \$\endgroup\$ Dec 16, 2015 at 14:35
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    \$\begingroup\$ Perhaps they are not? \$\endgroup\$
    – Eugene Sh.
    Dec 16, 2015 at 14:36
  • \$\begingroup\$ BTW, Are you sure an AC SSR is the right thing for PWM? I would say you need a DC relay here... Also I am not quite sure how do you modulate AC with PWM. Can you provide some kind of schematic? \$\endgroup\$
    – Eugene Sh.
    Dec 16, 2015 at 14:41
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    \$\begingroup\$ @MisterMystère: use some relay in series that you PWM with <1Hz frequency when in the lower enegy requirements range? \$\endgroup\$
    – PlasmaHH
    Dec 16, 2015 at 15:00

1 Answer 1


The 'leakage' is no doubt coming from the internal snubber network. Perhaps they increased the capacitor value from when the datasheet was written.

The load resistance is R = 240^2/2000 = 29\$\Omega\$, so with 25mA leakage you'll have a power in the heater of 18mW (I^2R), which is not a problem. The '13W' you calculated is actually 13VA, almost all leading in phase and does not represent power either in the relay or the load.

So, I don't think you have a real problem. Your cycle time is a bit short- you may run into beating issues with the mains, but that's another issue.

Anyway, to answer your question, the easiest way to avoid the snubber would be to make your own SSR with a opto-triac and a snubberless triac (aka alternistor). But it's not necessary.

  • \$\begingroup\$ Of course! I stupidly multiplied the leakage current by the mains voltage, but if the leakage current is due to a snubber network and therefore not in phase, then it makes no sense to do so. If I had used RI² instead of UI in the first place I wouldn't have run into that trouble, but it's a good refresher of reactive power. \$\endgroup\$ Dec 16, 2015 at 15:21
  • \$\begingroup\$ So it means the leakage current out of the SSR is leading in phase compared to the mains feed, and since a purely resistive load is connected, the voltage out of the SSR is also leading in phase, of amplitude Rx25mA=0.725VRMS, right? \$\endgroup\$ Dec 16, 2015 at 15:23
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    \$\begingroup\$ Yes, that's right. 0.7V only across the heater. You should be able to measure it directly. \$\endgroup\$ Dec 16, 2015 at 15:57

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