Optocoupler question

Hi guys, i have a quick question about implementing an optocoupler into my project.

I have load that runs on 12V AC or DC.

I want to connect the load to the input pads of an optocoupler and then aurduino on the other side to generate an output.

My question is:

Can all optocouplers run off AC and DC or am I looking for a specific type of component?
I was looking at a 4N33,

http://www.vishay.com/docs/81865/4n32.pdf

in the datasheet under INPUT it shows FORWARD CURRENT 60mA max - is this the max current I can run through the inputs?

I am trying to understand that happens if my load is 2 AMP, will I burn the optocoupler?

From the little that I know if seems that I would probably need a current limiting resistor after my load to ensure that I do not exceed 60mA ?

Thank you guys in advance.

So you see it can take a maximum forward current of 60mA (0.06A) and want to know if it can do 1A?

No.

It is there when you want to electrically isolate 2 circuits. What you would then do is use the output of the opto coupler to drive a transistor which would then drive a relay.

Oh and happy new year!

Can all optocouplers run off AC

No opto coupler can run off AC.

If you want to use AC then rectify and smooth the AC first into DC.

I am trying to understand that happens if my load is 2 AMP,

Absolutely nothing to do with anything. You do not connect the opto in series with the load you connect it in parallel.

Well Mike, I have seen optocouplers with two antiparallell LEDs made for AC input.
I have also seen optocouplers with TRIAC outputs made for AC output. (MOC3020)

Pelle

Yes I have used triac output LEDs, link please to the anti parallel ones.

Grumpy_Mike:
Yes I have used triac output LEDs, link please to the anti parallel ones.

With the greatest of pleasure!

I have some from eBay - not actually used them yet but ...

You need a resistor in seriese with the LED, so you arrange the value to be such that you do not exceed the maximum churrent with the maximum voltage you will get. Remember that AC voltage is normally expressed as an RMS value multiply this by 1.414 to get the peak voltage.

@Paul__B - thanks for the link.

Grumpy_Mike:
You need a resistor in seriese with the LED, so you arrange the value to be such that you do not exceed the maximum churrent with the maximum voltage you will get. Remember that AC voltage is normally expressed as an RMS value multiply this by 1.414 to get the peak voltage.

@Paul__B - thanks for the link.

Thank you mike, so I understood correctly. I just need a resistor in series.

"Thank you for the first link. I think the H11AA1 would work for what I want."
No thanks though for the second link?

"A resistor in series".
That's what I showed at my link (the second link.)
I didn't declare the value, so into the pillory I goeth.
Use 22K, 2W (1W would be OK, but 2W is better.)

My assumption is that you too reside in the "Lands of 110 vac".
If not, re-calculate accordingly.

^ Of course, thank you for both links - heck thank you for even taking the time to look at my newb post.

I had to do some reading on what a triac is and how it works and why it works with AC current before I could even try to read your blog post.

Yes I live the land of 110vac.

Hmmm... The words newb and "land of 110V" tend not to go well together.

I think lots of studying are in your future before you start doing practicals. I'm including finding the perfect part(s) is in there somewhere...

"Hi guys, i have a quick question about implementing an optocoupler into my project.
"I have load that runs on 12V AC or DC. "

OK, your deal, still, is 12vac/12VDC.
So, your series resistor value, for the input of that H11AA1 is 2.2 kΩ, for 5 ma/mA or 1kΩ for 10 ma/mA. (1/2W resistor - Go Big!)

How did you arrive at the 2.2Kohms?

I used an online calculator and even did it by hand and I got:
http://led.linear1.org/1led.wiz?VS=12;VF=1.5;ID=60
180
or (if I do the math myself)
12v-1.5vf/0.06mA = 175 Ohms?

What am I doing wrong or different?

crullier:
12v-1.5vf/0.06mA = 175 Ohms?

What am I doing wrong or different?

Everything?
Determining R for pre-determined (desired) current:
Ω = (12 V - 1.5 V) / 0.005A
Ω = 10.5 V / 0.005 A
Ω = 2100
*Back-calculating, *using nearest "standard value":
A = (12 V - 1.5V) / 2.2kΩ
A = 10.5 V / 2.2kΩ
A = 0.00477
[ 4.77 mA ]

Can you tell me where the 0.005A come from?

here is the datasheet, I thought we were concerned with the forward current on the input side which according to the sheet is +-60mA.

Are you looking at the "collector emitter leakage current"?

crullier:
Can you tell me where the 0.005A come from?

here is the datasheet, I thought we were concerned with the forward current on the input side which according to the sheet is +-60mA.

No; common "newbie" blunder - you completely misread the datasheet. Look at the heading above the "± 60". Now, what exactly does it say?


Ah! "ABSOLUTE MAXIMUM RATINGS". Do you know what that means? It means you do not feed it with 60 mA. At all. Ever.

RP nominated a sensible current of 5 mA. Note by the way, that its Current Transfer Ratio is rated at a maximum of 20%. This means that feeding it 5 mA, you may only expect it to conduct 1 mA, so you use a minimum pull-up value of 5k. In fact, you would use the internal pull-up of the Arduino device, which is approximately 10k.

thank you very much. I am such a newb, but thank you for showing me the ropes !!

Needless to say, don't touch any components running more than 30V unless your wearing gloves.

see pages 2 and 3 of this.

Also, you could use a relay or SSR to drive the load with an opto coupler. You can tie the +Vdc to the + Vdc
input of the SSR and the - Vdc input to the collector of the opto output and use the opto to sink the led
current to turn on the SSR. A similar method can be used to turn on a 5V relay (tie the +5V to the input (and Vcc of the relay) and tie the relay input ground to the opto collector. When the opto turns on , the
relay input will have 5V across it because the opto collector will connect the relay GND to the circuit ground
turning on the relay. There are probably many more variations of this all involving the use of the opto
collector to complete an input circuit to some device.

Can you tell me where the 0.005A come from?

OK, your deal, still, is 12vac/12VDC.
So, your series resistor value, for the input of that H11AA1 is 2.2 kΩ, for 5 ma/mA or 1kΩ for 10 ma/mA. (1/2W resistor - Go Big!)

I missed your post above. Let me read though it.
You said to wear gloves? (stupid question ahead) why / how, what do you mean.