I'm trying to use solid state relay (Panasonic's AQY277 / AQV215) to switch a light bulb (100 V AC). To control the input, I'm using a signal generator in DC mode with 5 V as input. I have not yet succeeded in making it work. Anyone having any idea about wiring details of these relays?
closed as not a real question by Olin Lathrop, Brian Carlton, W5VO♦ Nov 15 '12 at 4:46
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You should have PIN 2 connected to the
pin 1 to the driver among a 1.5K resistence ( that guarntee a correct current driving the internal LED ) of about
(5-1.2)/1.5K = 2.53 mA that according to the datasheet is enought to switch the device on.