# Optocoupler Project

Discussion in 'Homework Help' started by ghebaur, Jun 17, 2014.

1. ### ghebaur Thread Starter Member

Jun 17, 2014
35
2
Hello, i have followed some tutorials here but i still need some help. I want to hook up this optocoupler to a 5V DC source that can give a maximum of 450mA. I don't know if i used the correct values from the data sheet and if my algorithm for calculating the resistor is correct.

In the schematics i represented only the diode part of the optocoupler for simplicity.

I used the VF=1.25V from and IF=20mA under TEST CONDITION from ELECTRIC-OPTICAL CHARACTERISTICS table as diode characteristics. I first calculated the remaining voltage drop on the resistor and then the value of the needed resistance.

The resulted resistance R1=187.5ohms is between this two available values that i could use: 180ohms and 200 ohms. Should i use one of them, and which one if so, or should i group resistor to get exactly 187.5ohms?

One more question. Do i need to lower to the current that comes from the source to 20mA or the LED draws only what it needs? I didn't find anywhere any explanation for this kind of situation.

PS: Sorry, i forgot the datasheet.

Last edited: Jun 17, 2014
2. ### Bernard AAC Fanatic!

Aug 7, 2008
4,550
477
I would use the 180Ω, either one would work.

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3. ### ghebaur Thread Starter Member

Jun 17, 2014
35
2
So, everything i said is ok? Even that LED draw what current it needs?

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,173
1,188
Diode is a "current driven" device and this is why you need to use a resistor to set the current.
The phrase "current driven" refers to a drive scheme where the current is held constant to a reasonable degree, and the voltage is determined by the I-V curve of the diode.

5. ### ghebaur Thread Starter Member

Jun 17, 2014
35
2

Then the resistor would be R1 = 5V / 20mA = 250ohms? Is that right? Then then the voltage drop on the diode will be 0V.

It's getting confusing, i don't know if i'm getting this right.

PS: If the voltage on diode will be 0V, doesn't it means that it will emit no light?

Last edited: Jun 17, 2014
6. ### crutschow Expert

Mar 14, 2008
16,509
4,443
No, you had it right in the beginning, you need to include the diode forward drop in your resistor calculations. The exact diode current is not critical as it will operate over a wide current range, so the calculated resistor value is approximate. Either 180Ω or 200Ω should be fine.

Certainly if the voltage across the diode is zero then it is off and will emit no light.

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7. ### ghebaur Thread Starter Member

Jun 17, 2014
35
2
Yey! That's what i wanted to hear. Thank you very much. I'll keep working and come back when i have other questions.