I am using a (rather cheap-ish) chinese solid state relay (ssr) with 5V DC on the driving, 230V AV on the driven side (there's an ATtiny85 randomly letting devices run for 10-30 minutes, then cuts the power for 10 seconds, then all over again for stress-testing my embedded audio player).

After the ssr was "up", it only very slowly drops the driven side down to around 10 V AC after the primary side goes down to 0VDC if I don't have a very notable load on the secondary side.

As I'm only turning off a handfull of USB power adaptors, it takes the ssr over 20 seconds to get down to around 10V AC (which seems to be it's off state).

Any idea what I could be doing wrong? I tried "switching" the hot and cold AC line with the ssr, same effect. For now I also switch a lamp which lets the voltage drop almost instantaneously, but in the ling run that can't be the best solution...

As always, thanks for any tips and pointers in the right direction!

  • \$\begingroup\$ Trigger Current: 7.5mA / 12V ... Have you measured the actual input current on the ATTiny side of the device? Could it be that the MCU isn't able to pull/discharge the input resistance/capacitance quick enough? \$\endgroup\$ – Dzarda Mar 21 '14 at 10:48
  • \$\begingroup\$ No I haven't measured that (could I use some sort of add'l dropdown resistor?). But why would it drop quickly if there's a "real" load on the secondary side and not drop if there's virtually none? \$\endgroup\$ – Christian Mar 21 '14 at 11:09

Your SSR includes a snubber on the AC side that can pass up to 3.0 mA of leakage current in the "off" state. Your USB power adapter only draws something like 20-25 mA even at full load. (I'm assuming it's a fairly efficient switching converter.)

When you switch off the drive to the SSR, the leakage current allows the AC voltmeter to read essentially full voltage, clipped only by the remaining voltage on the USB adapter's input capacitor. As this capacitor discharges into the load, the waveform gets clipped at lower and lower values, until it reaches an equilibrium where the 3 mA matches the residual current draw of the adapter.

Looking at the AC waveform with an oscilloscope would make this process very obvious — but be very careful about how you hook it up. Do NOT connect the scope's ground lead drectly to either side of the power line; instead, use an isolation transformer or use a 2-channel scope in differential mode.

Connecting the lamp as a parallel load "soaks up" the leakage current, preventing it from driving the voltage up to the clipping level imposed by the USB adapter. The primary of a small power transformer (leave the secondary unconnected, the magnetizing current alone will be enough) would work just as well, without the excessive dissipation when "on".

  • \$\begingroup\$ Thanks for the technical background! Even if I had an oscilloscope I wouldn't directly hook it up to mains voltage (or, any high(er) voltage at all). That much my ham radio license has taught me ;) I'll give the idea with the transformer coil a go once I've located one in my junk-drawers. \$\endgroup\$ – Christian Mar 21 '14 at 11:53

Might be leakage current that causes the problem. Check Omrons appnote on SCR http://www.omron.com/ecb/products/pdf/precautions_ssr.pdf Page 2

  • \$\begingroup\$ Thanks for the link, a valuable read. I think this is pretty much what @dave-tweed said above. \$\endgroup\$ – Christian Mar 21 '14 at 12:03

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