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

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?

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

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


1 Answer 1


As I read it, the IFon Max rating of 3 mA means that even the least sensistive unit will switch on with 3 mA, while most will switch on at 1 mA (or even less).

IFoff Min of 0.4 mA and Typical of 0.9 mA says most wiil turn off if the LED current falls below 0.9 mA, but some may stay on until the current drops below 0.4 mA

  • \$\begingroup\$ ah ok, the maximum rating scares me when it saids 3mA maximum for IFon it sounds like if you exceed 3mA it will blow the LED, but it clearly can operate 5 to 10mA up to 50mA max... so 3mA max means that if you apply at least 3mA it is guaranteed to be on? \$\endgroup\$ Commented Jun 6, 2013 at 17:37

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