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Help on correct hook-up of an optocoupler/optoisolator

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pocketlink

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I am a new poster and would greatly appreciate assistance on the correct hook-up of an optocoupler/optoisolator. I am working with a NTE3098 (spec sheet link below).

https://www.nteinc.com/specs/3000to3099/pdf/nte3098.pdf

My input source is 3.6v from DC Supply (or equivalent battery) to 100ohm, 1/4w resistor on the (+) side to the Anode and the (-) side to the Cathode. The output is to 100ohm, 1/4w resistor on the (+) side to the Collector and the (-) side to the Emitter terminating to the leads of a Multimeter set to read DC output. I’ve confirmed a good connection on all working components but the consistent DC output is “0” on the Multimeter.

My understanding is that a function (among others) of the optocoupler/optoisolator is to transfer 100% of the voltage and current from the LED side to the photo-sensor side with ground loop isolation in between.

My application is a simple DC circuit that feeds continuous low voltage and current to a second circuit with ground isolation between them so, I thought, this would be the ideal solution. I’ve now tried several NTE3045, NTE3041, NTE3098 in combinations of 100ohm, 360ohm and 1k resistors respectively with similar results. Only with the NTE3041 (6-pin) did I get a trace .48v output – but that was with the (+) side to the Base pin and the (-) side to the Collector pin.

I am a not a professional so I am obviously not hooking-up correctly and could use any help or advice that would solve this problem. Or, if you have a better suggestion for me to achieve my purpose it would be greatly appreciated.
 

First of all, in opto-couplers there is NO transfer of energy from one side to the other side ..
It is not a transformer, it is an optical barrier, that’s all ..

If you know how a transistor works just imagine this: the diode emits light and that light controls the Base of the output transistor ..
If diode emits light – the Collector-Emitter channel is “open”, if there is no light – the Collector-Emitter channel is “closed” ..

If you want to conduct current through the output transistor it has to be fed from an external power supply .. see attached picture ..

Rgds,
IanP
 

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  • opto_circuit_2083.jpg
    opto_circuit_2083.jpg
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Thank you for your answer and the image. So, if no “energy” is being transferred from one side to the other and the light emitted serves only to declare an “open” or “closed” state on the other side, then it will not serve my purpose, if I must power both sides independently. My need is a simple DC circuit that feeds continuous low voltage and current to a second circuit with ground isolation between them. Do you know of an alternate way I can achieve this?
Thanks again for your reply.
 

My need is a simple DC circuit that feeds continuous low voltage and current to a second circuit with ground isolation between them. Do you know of an alternate way I can achieve this?

There are small dc-dc modules that can transfer power from one side to the other side, for example, the NMA0505S transfers up to 100mA@5V from 5V source to the other side with ≈1kV isolation ..
http://www.datasheetcatalog.org/datasheet/candd/NMA0512S.pdf

Rgds,
IanP
 

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