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I was looking into designing circuits around photo-couples/opto-isolators instead of transistors, but I have found that I need an inverter of sorts. I need something that stops current flow through the outputs when current flows through the inputs. So the LED opens the circuit using the collector/emitter rather than closing it.

My understanding of photocouplers was derived from the description of a very simple four pin photocoupler I got from a computer power supply. I was very intrigued by the way it isolated inputs and outputs, so I wanted to see if I could build logical devices out of them instead of transistors. I quickly discovered that I needed something that turned the output off rather than on when the current flowed through the input.

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You could simply attach a PNP transistor to the opto. – Polynomial Apr 11 '12 at 0:12
Yes, I know, but I like optos because they have two inputs and two outputs, so I am trying to use those. – Feynman Apr 11 '12 at 1:51
You could flip it so that the collector is tied to +ve and your load sits between the emitter and ground. Makes sense, but I've never tried it before - YMMV! – Polynomial Apr 11 '12 at 2:04
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You should formulate your needs better. The answer you accepted doesn't agree with what I read as your requirements. – stevenvh Apr 12 '12 at 7:16
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I think the problem here is that OP has an over-simpliefied model of how a optocoupler works: he imagines it like a relay. Most optocouplers have phototransistors, though, which makes it easy (actually it seems to be the default) to invert the signal. – noah1989 Apr 12 '12 at 9:15
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3 Answers

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See these Application Circuits for the Phototransistor Switch:

The common-emitter circuit will generate an output signal which transitions from high to low high when the phototransistor detects light or infrared radiation. This is commonly referred to as an “inverting logic condition”.

common-emmiter circuit

Is that what you're looking for?

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This circuit doesn't do anything: the output will always be low. – Federico Russo Apr 11 '12 at 6:40
@FedericoRusso obviously, in a real application, you don't connect R1 directly to VCC but to an input signal... – noah1989 Apr 11 '12 at 8:55
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It's fundamental to the way photodetectors work that applied light generates carriers (electrons or holes). This is the photovoltaic effect, closely related to the photoelectric effect that Einstein got his Nobel prize for explaining.

Whether you add it as an extra circuit or its built in to a device you buy, using light to block an electrical current would require some kind of additional inverting device.

As an alternative, if you have some current running through some load, you could use a photo-activated circuit to divert that current so its not available to your load. But giving more detail on how to do that would depend on knowing a lot more about what you're trying to do.

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"From the horse's mouth" :-) - assuming your eponym is meaningful :-). – Russell McMahon Apr 11 '12 at 6:26
Chose the name when I was editing Wikipedia a few years ago... – The Photon Apr 11 '12 at 16:30
I think it would still be possible to make something like this. You could, for example, make a device that uses the electric field generated from the photoelectric effect to deflect the path of the electrons, reducing current. – noah1989 Apr 12 '12 at 7:30
@noah1989, That would basically be the same as a photoconductive device plus a FET. A perfectly good idea, but not what the OP is (unrealistically) asking for. Also, the circuit in your answer is a good example of the general principle mentioned in my 3rd paragraph. – The Photon Apr 12 '12 at 15:22
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edit

I need something that stops current flow through the outputs when current flows through the inputs.

Whatever the optocoupler's configuration the LED will always cause a current in the phototransistor when it's on.

So the LED opens the circuit using the collector/emitter rather than closing it.

It's not clear what you mean by opening or closing the circuit. In its standard use, emitter to ground, the output will be low when the LED is on, so it's actually an inverter. Since you're asking the question, I guess that's not what you want.

(end of edit)

I was also thinking about Polynomial's solution: swap the optocoupler's output transistor and the load.

enter image description here

When the transistor is driven to saturation the voltage between collector and emitter will be a few hundred mV. That means that the output will be near V+ when the optocoupler is driven. When the LED and therefore the transistor are off the output resistor will pull the output low. So you get your positive logic: LED on = output high.

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So this would be figure 2 in noah1989's link: optekinc.com/pdf/App%20Bulletin%20213-Opto%20Assemblies.pdf right? – Feynman Apr 11 '12 at 16:15
@Feynman - Yes. I'll add the image to my answer. – stevenvh Apr 11 '12 at 16:18
I thought that OP asked for inverting logic. LED on = output low (no current flow through load between output and GND). did I miss something here? – noah1989 Apr 12 '12 at 7:10
I just realize that you could just put the load between VCC and Vout in your solution to get the same result. – noah1989 Apr 12 '12 at 7:13
@noah1989 - I interpreted it the other way around. The question isn't very clear here. From your accepted answer it appears he wants the inverted transfer function. In that case the question is a bit silly, because that's how 99% of optocouplers is configured. – stevenvh Apr 12 '12 at 7:15

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