# Looking for and SMD optocoupled TRIAC

I'm looking for a SMD optocoupled TRIAC part for driving EL Wire (110VAC, 1000Hz, low current non-resistive load) with an ATTiny85 pin (TTL, 20mA max) running at 3.3V.

The previous design used the Z0103 non-optocoupled TRIAC, but it seems there is a lot of problems related to noise and the AC ground floating in relation to the DC ground. I've searched for some parts on Mouser but all of them have a very low maximun input voltage (~1.5V, and I need 3.3V) or extremely high (> 64V).

Any advice or recommendation is welcome, but please suggest something I can buy from Mouser (the only store that ships to my country). I not totally sure what I'm doing so any correction is welcome too.

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The 1.5V you mention is the voltage drop across the opto-coupler's LED, and this is fairly constant. You can't just apply 3.3V because a way too high current will flow, which may destroy the LED. In this case the current will be limited by the Arduino, but you have to limit it to a lower level. The MOC3023 only needs 5mA maximum to operate the triac, and has an AMR (Absolute Maximum Rating) of 60mA. (Note: you're not supposed to operate a device continuously at AMR!).
Let's pick a safe value: 10mA. $V_F$ is typically 1.15V, then we can calculate the series resistor we'll need to set the current to 10mA:

$R = \dfrac{\Delta V}{I} = \dfrac{3.3V - 1.15V}{10mA} = 215\Omega$

We pick the closest E12 value, 220$\Omega$. But wait, that's for the typical $V_F$, the datasheet says it can be as high as 1.5V. What will the current then be?

$I = \dfrac{\Delta V}{R} = \dfrac{3.3V - 1.5V}{220\Omega} = 8.2mA$

So that's still enough to trigger the triac, and well within the limits of what the Arduino can deliver.

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Thank you very much, that's make a lot of sense now. –  Havok Oct 12 '11 at 8:25
I think I'll go with the S2S3LA0F, what do you think? It's SMD, cheap, good ratings. The Maximum Continuous Output Current is 50mA, greater than the one from Z0103, so, can I use this one directly to switch the EL Wire or do I always need to use another triac as the switch and this one just as the trigger? –  Havok Oct 14 '11 at 15:27
@Havoc - looks OK to me. I don't know how much current the EL-wire uses, but most likely it will be far less than the 50mA (that would be 5.5W). In that case you can drive it directly, no power triac needed. Anyway remember what I said about AMR. (BTW, the MOC3023 is cheaper) –  stevenvh Oct 14 '11 at 15:38

NB!!! - An "opto"couple is so called because it uses optical means to convey data. The optical data sender is an LED and the input you provide turns on the LED. So you need to understand how to drive LEDs to understand how to drive optocouplers of this type.

LEDs need to be current driven, not voltage driven, So, Vf for an optocoupler is NOT the voltage you drive it at - it is the voltage "that happens" when you provide the correct current.

The optocoupler input current , which sets the Vf accordingly, can be accommodated by the use of one resistor with suitable design.

Your optocouple does NOT expect a voltage input!!! It expects a current input.
It should NEVER be driven from a voltage source.
Always from a current source or approximately so.

Consider first Mouser part MOC3023.
MOC3023 Datasheet

If abs max = 60 mA. page 2 If operating = 5 to 30 mA (varies with version) page 3 So lots of leeway.

Vf = 1.15 typ, 1.5V max.

Design for 30 mA, 3V drive, 1.5V Vf. R = V/I = (3-1.5) / 0.03 = 50 R

At other extreme, 30 mA, 3.3 V drive, 1.15 V Vf - what current with 50R? I = V/R = (3.3-1.15)/50 = 43 mA so still in spec.

Depending on version you may not need 30 mA - change (increase) R as required.

Note that Atduino may not be able to source 30 mA!
Optocoupler chose may want very low If.

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It might help conceptually to note that the input side of an optocoupler is just an LED. When the LED is turned on, it illuminates and activates the junction of a triac/SCR/transistor on the output side of the package, across an air gap that keeps the two sides isolated. –  mng Oct 12 '11 at 3:10
@mng - Thanks. Done. –  Russell McMahon Oct 12 '11 at 8:17