Ah. So the safety ratings are for when you are complying with the DIN specification cited to provide a primary electrical isolation function. I’m not familiar with that one but I would guess that they’re saying that your external circuitry has to limit the input current to 350ma worst case, limits output power etc if you’re going for that certification.

Note that the datasheet you linked is for the H11L1M. That is a different device spec than the H11L1 you originally posted. The M is fully specified for those DIN specs (and derated). The other one is for more general use.

Late here, goodnight and good luck.

Thankful
good night

 

Hello
I am reading the H11L1M datasheet. There is two number for IR emitter LED forward current on datasheet one of them on "Safety and Insulation Ratings" Table (is 350 mA) and other is on "Absolute Maximum Ratings" Table (is 30 mA). I don't the difference between them. Can anyone help me?

Thanks

 

Absolute Maximum Ratings indicate max current. higher will kill the opto-coupler.
safety and isolation >> no current mentioned in data sheet as far as I cab See
' UL1577, 4,170 VACRMS for 1 Minute– DIN-EN/IEC60747-5-5, 850 V Peak Working Insulation Voltage '


Picbuster

 

Absolute Maximum Ratings indicate max current. higher will kill the opto-coupler.
safety and isolation >> no current mentioned in data sheet as far as I cab See
' UL1577, 4,170 VACRMS for 1 Minute– DIN-EN/IEC60747-5-5, 850 V Peak Working Insulation Voltage '


Picbuster

Thank
Could you explain more clearly?

 

you should look at
UL1577
DIN-EN/IEC60747-5-5 https://www.vishay.com/docs/83707/83707.pdf
Both will explain.

Picbuster

 

Hello
I want to drive a H11L1 Opto.My line voltage is 220v 50Hz,I want to calculate series resistor.The opto must be turned off 500 us before zero so i calculate the voltage at that time [ V1 = 48v],The minimum current to drive emitter is 1 mA so calculate the resisto [R1 = 47 KΩ].
now is turn to check the Absolute Maximum values :

1.Maximum forward current is (311-1) / 47k = 6.5 mA(ok).
2.LED power dissipation is less than 10 mW(ok).

But i don't know how to calculate the reverse voltage. Please help me to calculate it.
Thanks.

Hi

First.....Please understand that you are working with voltage levels that can cause injury or even be FATAL if handled improperly.

If the input is AC, use an AC optocoupler like an H11AA1 or IL250. The output current will depend on the current transfer ratio.

eT​

 

PCB layout topic for this circuit is now here. Comments invited there, particularly for spacings, safety concerns etc.
https://forum.allaboutcircuits.com/threads/designing-a-isolate-pcb-for-h11l1-opto.161342/