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I have an AQZ204 AC/DC type solid state relay AQZA204 datasheet

Basically I want to calculate the minimum supply voltage required to turn the relay on

and the minimum supply voltage required to turn the relay off. These are similar to Pick/Drop voltages of mechanical relays or must be on voltage, must be off voltage..

I'm using 31.5 volt DC supply, with a series 4.7k Ohm Resistor to the input LED of the relay.

The data sheet is a little confusing to me, it saids the LED max dropout voltage is 1.5V so I calculated around 6.38mA forward current for the LED:

31.5VDC - 1.5Leddrop = 30VDC/4700 = 6.38mA

I'm confused by the IFon current being 3mA max?

So I can test the relays to see if they are in spec.

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The IfOn Max current is the current that the least sensitive relay should require to turn in - most will turn on at lower currents, but you should supply at least 3 mA to ensure that any relay will be on. The Ablosute Maximum LED current is 50 mA, so it is safe to supply more than 3 mA.

For turn-off, there is IfOff Minimum of 0.4 mA - most units will turn off at a higher current, but the most sensitive may stay on with currents as low as 0.4 mA, so you should ensure that the LED current is below 0.4 mA when you want the thing to be off.

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  • \$\begingroup\$ So since I have a 4700 Ohm resistor, to get 3mA to turn the relay on I would need (4700* 3mA) = 14.1VDC to turn any AQZ304 relay on \$\endgroup\$ – zacharoni16 Jun 11 '13 at 19:28
  • \$\begingroup\$ do I need to add the 1.5V LED dropout voltage to the 14.1VDC? for 15.6VDC? \$\endgroup\$ – zacharoni16 Jun 11 '13 at 19:30
  • \$\begingroup\$ @zacharoni16 - to be absolutely correct, yes - but the great majority of relays will operate at less than 3 mA (or 15.6V with your 4K7 resistor) \$\endgroup\$ – Peter Bennett Jun 12 '13 at 1:18
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It's usually only a small control signal which controls a larger load current or voltage on a relay. If you take a look at Mouser you'll see the control voltage ranges from 1.5 - 5V for this with 400V being max load output at 500mA. If you have 31.5V going into the control input it seems like a bit much for the input LED.

You can see the same on Page 3 where it says AC/DC absolute maximum ratings. Input 5V, and 1A.

while the output can max out at 400V for this AQZ204.

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  • \$\begingroup\$ Well I have a series 4.7K 1/2 watt resistor in series of the relay LED. From my understanding, the LED will drop 1.5V leaving 30 volts to be dropped by the resistor? In the datasheet they do give a typical Turn-on current of 1mA which I'm getting a minimum input voltage of 4.7 Volts to turn on the relay Then a 0.4mA turn off current which would be 1.88V to turn off the relay. Not sure if this is correct or not. \$\endgroup\$ – zacharoni16 Jun 11 '13 at 19:24
  • \$\begingroup\$ You could just set up a voltage divider and connect the LED to different points in the divider to see what the suits your situation the best if you are still working with the 30V's. Just make sure you have the proper current as mentioned above to swing it on and off. \$\endgroup\$ – crackhaus Jun 11 '13 at 20:03

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