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

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    \$\begingroup\$ do you have a wiring diagram ? \$\endgroup\$ – Felice Pollano Nov 14 '12 at 14:28
  • 1
    \$\begingroup\$ This 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. \$\endgroup\$ – Olin Lathrop Nov 14 '12 at 14:38
  • \$\begingroup\$ While I feel the question is vague, closing it just makes this forum discouraging for people trying to learn how to use it. \$\endgroup\$ – speciousfool Nov 16 '12 at 8:25

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.

  • \$\begingroup\$ 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 ? \$\endgroup\$ – user16119 Nov 14 '12 at 17:32