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.
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)
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.
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.
"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!)
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 ]
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.
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!)