# Solid State Relay Voltage When Off

Please excuse my ignorance, but I am relatively new to electronics. Sorry, also, for using the LED schematic. I could not find one for a CFL lamp.

I am using a solid state relay (SSR40DA http://www.fotek.com.hk/solid/SSR-1.htm http://www.datasheetspdf.com/datasheet/download.php?id=789332) to control a CFL lamp. I have measured voltages across 3 points in the circuit (AB, AC, and BC). I did the measurements in various states (VDC applied/not applied, lamp on/off), and I got a surprising (to me) result. When the lamp switch (D) is on and the relay is off, the voltage across it goes to 28V. Can someone please explain why that would be?

Thank you very much in advance,

Fed

• What wattage is your lamp? The snubber network internal to the SSR can conduct a small amount of current while off. Commented Sep 27, 2017 at 17:08
• @sstobbe the datasheet shows a 3-mA off-state leakage. To give a 28V drop, the lamp should have a cold-state resistance which is 9.3kOhm. When the lamp's filament becomes hot, the resistance will increase many times. But let's assume that 9.3kOhm is the HOT resistance. That lamp's wattage would be 1.5W... Commented Sep 27, 2017 at 17:21
• @next-hack Sure but that's assuming the leakage is only 3mA... Either way measuring 28Vrms across a lamp means current is flowing, what we don't know is how much. Commented Sep 27, 2017 at 17:52
• I used a clamp meter, and it showed no current to the lamp when off. 150mA when on. It isn't a very good meter, however. The lamp is 13W 190mA. It is a CFL, not Tungsten.
– Fed
Commented Sep 27, 2017 at 18:59
• I think it would be helpful for future readers with a similar problem to edit your diagram in paint to read CFL not Tungsten. Or use the site schematic tool. A CFL is a non-linear load versus a tungsten lamp Commented Sep 27, 2017 at 19:07

Had it been really tungsten lamp, the second column would have been 124V, 124V, 0V.

It turned out in the comment section that you had a 13-W compact fluorescent lamp.

You are seeing a small but non-zero voltage because the impedance of the compact fluorescent lamp is very high (in your case, about 1/4 of the OFF-state resistance of your SSR). You're just creating an AC voltage divider.

In fact, an OFF-state (i.e. when the voltage is very small) compact fluorescent lamp has a very high impedance, because its switching circuitry is OFF, so there is not power delivered to the tube. Therefore you can only measure the leakage of the brige diodes and startup circuitry consumption (if any is present.). Since it's small, the impedance is high. As soon as the voltage reaches a threshold level, the switching circuitry turns on to power the tube, and the impedance suddenly decreases.

Many low power LED lamps (especially "filed" lamps) have a similar behavior, as they consist of a capacitor dropper, a bridge and some tens LEDs in series. If the applied voltage is smaller than the total drop of all the LEDs, the current consumption will be extremely small and the impedance will be high.

• Thank your for your explanation. I changed the initial post from tunsgten to CFL.
– Fed
Commented Sep 27, 2017 at 23:58
• One more question, is there much wasted energy?
– Fed
Commented Sep 28, 2017 at 0:00
• @Fed The problem is that we do not know the current (just the 3-mA maximum leakage value, which is frankly, quite high). Still, if it is really 3-mA, then you are wasting 0.372 VA (not W!) of APPARENT power. The amount of active power depends on where the impedance actually comes from. Is it resistive? Is it capacitive? If the (small) current flow is just due to capacitive effects, then no active power is dissipated (only reactive, which is not counted on your bill). Commented Sep 28, 2017 at 5:50