Drive H11L1 Optocoupler issue

Thread Starter

daryooosh

Joined Jun 15, 2019
35
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.
 

Thread Starter

daryooosh

Joined Jun 15, 2019
35
See Fairchild AN-3006 fig3 for a recommended way.
Max.
Here you go.
Thanks a lot

1.What dose "Emitter" in Absolute Maximum table at datasheet mean?
2.What does "Detector" in Absolute Maximum table at datasheet mean?

I think Emitter means IR LED and Detector means BJT Opto transistor.Am i right?
If i am right then what does mean "Reverse Voltage" in "Emitter" section?
 

JohnInTX

Joined Jun 26, 2012
3,911
Correct, Emitter is the LED.
Reverse voltage is spec’d because the H11L1 just has the LED input. You can’t drive it with AC without a series diode like a 1N4004 to block the negative 1/2 cycle or a bridge rectifier to to convert ac to full wave rectified DC. You also can get optos that have two inverse parallel LEDs to handle AC inputs.
 

JohnInTX

Joined Jun 26, 2012
3,911
The LED is a diode that emits light when it conducts current. Diodes conduct in one direction and block current in the reverse direction -UP TO A MAXIMUM VOLTAGE. More than max reverse volts destroys the LED. Therefore you can’t use that LED (6v max Vr) on a 220vac input. You either have to use an additional diode with a high reverse voltage in series to block the other 1/2 cycle, use a bridge to make both 1/2 cycles the same polarity or use an opto with AC input.
 

Thread Starter

daryooosh

Joined Jun 15, 2019
35
I am a magician;).I created a prototype two days ago with out a Diode and until now it has been working correctly.What does happen?
 

Thread Starter

daryooosh

Joined Jun 15, 2019
35
The LED is a diode that emits light when it conducts current. Diodes conduct in one direction and block current in the reverse direction -UP TO A MAXIMUM VOLTAGE. More than max reverse volts destroys the LED. Therefore you can’t use that LED (6v max Vr) on a 220vac input. You either have to use an additional diode with a high reverse voltage in series to block the other 1/2 cycle, use a bridge to make both 1/2 cycles the same polarity or use an opto with AC input.
I am a magician;).I created a prototype two days ago with out a Diode and until now it has been working correctly.What does happen?
 

JohnInTX

Joined Jun 26, 2012
3,911
Your series resistor is limiting the power dissipation so it doesn’t blow up immediately but it is not happy and it will eventually punish you by failing.
 

Thread Starter

daryooosh

Joined Jun 15, 2019
35
Your series resistor is limiting the power dissipation so it doesn’t blow up immediately but it is not happy and it will eventually punish you by failing.
If i use any Diode, it has a very small reverse current(in u amp range). When use diode in series with the LED, in negative cycle acts as a Voltage divider and may causes a Voltage greater than 6v across the LED.
 
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JohnInTX

Joined Jun 26, 2012
3,911
Unlikely, but a fair point. In that case, you can put the 1N4004 in inverse parallel with the LED. It will forward bias on the negative 1/2 cycle and clamp the reverse LED voltage to its Vf.

Edit: you beat me to it. Good call. That will increase the power dissipation of the series R but worth it.
 

JohnInTX

Joined Jun 26, 2012
3,911
Actually, I usually use a small resistor across the LED when driving from a solid state device like an SSR or diode to swamp leakage currents. A side benefit is that that resistor lets you sat a current threshold below which the LED won’t light. Vf/Racross sets the threshold current. Set it to whatever current you have at 500us before zero crossing
 

Thread Starter

daryooosh

Joined Jun 15, 2019
35
Actually, I usually use a small resistor across the LED when driving from a solid state device like an SSR or diode to swamp leakage currents. A side benefit is that that resistor lets you sat a current threshold below which the LED won’t light. Vf/Racross sets the threshold current. Set it to whatever current you have at 500us before zero crossing
So one other question :
1.What does "Input Current" in "Safety and Insulation Ratings" table mean?
2.What does "Continuous Forward Current" under "Emitter" section on "Absolute Maximum Ratings" table mean?
 

JohnInTX

Joined Jun 26, 2012
3,911
So one other question :
1.What does "Input Current" in "Safety and Insulation Ratings" table mean?
2.What does "Continuous Forward Current" under "Emitter" section on "Absolute Maximum Ratings" table mean?
1-not sure, it’s not on the datasheet I have.
2- theoretically the most current DC you can safely run through the LED. Good design practice says to run it less than that.
This turns on at 1.6ma. You need to ensure that at Vpeak, you stay safely below the max rating 1.2A Ipeak.
 
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Thread Starter

daryooosh

Joined Jun 15, 2019
35
1-not sure, it’s not on the datasheet I have.
2- theoretically the most current DC you can safely run through the LED. Good design practice says to run it less than that.
It is datasheet that i have.
"Input Current" in "Safety and Insulation Ratings" table is 350 mA but "Continuous Forward Current" under "Emitter" section on "Absolute Maximum Ratings" table is 30 mA.This datasheet drive me crazy!
 

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JohnInTX

Joined Jun 26, 2012
3,911
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.
 
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