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
Best Answer
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.
The answer now is simple.
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.