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?
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2do you have a wiring diagram ? – Felice Pollano Nov 14 '12 at 14:28
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1This is impossible to answer without knowing exactly what you did. You should be able to see for yourself we can't know what exactly you connected to what. – Olin Lathrop Nov 14 '12 at 14:38
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While I feel the question is vague, closing it just makes this forum discouraging for people trying to learn how to use it. – speciousfool Nov 16 '12 at 08:25
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You should have PIN 2 connected to the GND and 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.
Felice Pollano
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Thanks for your reply. I did this before as you suggested. I used a potentiometer (2K) but it didn't work. Is it possible that the signal generator is not able to sink enough current ? – Nov 14 '12 at 17:32