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Basic Electronics » Opto Interrupter once again

September 16, 2013
by Ralphxyz
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I tried to get an opto isolator working last February but never succeeded.

Now here is the circuit I am trying (unsuccessfully).

I do not understand this circuit. It seems like the pull-up resistor RL would always put a positive voltage on Output.

I have tried a 10k resistor for Rl.

I also tried a 2.7k resistor.

I am seeing a steady voltage at Output.

Here is my breadboard:

I was hoping to light the LED as Output, I'll go to the mcu once I see it working hopefully but there might not be enough voltage to light the LED.

But I should be able to measure a higher value when the IR led is not impeded right?

Of course I got the cheapest opto interrupter I could find off ebay so it did not come with any information :-(

There is 301-07 on the body and LITON has a 301-07 photo interrupter.

There is something about the opto interrupters I am just not getting.

Of course I do not really know which of the two pins is the collector and which is the emitter. There is a HY over the pins. Currently I have 5v at the H pin with the series resistor.

Ralph

September 16, 2013
by Ralphxyz
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Well tried reversing my connections right after I posted and low and behold it works!!

So Vcc goes to Y not H!! Y is the Collector!

I had "thought" (which I know is always dangerous) that Output would pulse high when the IR led was uninterrupted but Output is high "when" the IR led is interrupted.

Maybe I'll try my previous attempt again.

Ralph

September 16, 2013
by BobaMosfet
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Ralph,

Here's how the schematic works--

Essentially, as long as the base diode applies power to the transistor, the emitter of the transistor is grounded-- which also grounds the output pin. When you cut off the base, the transistor shuts off, and the potential across the pull-up diode comes into play-- the output signal goes high.

BM

September 16, 2013
by scootergarrett
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On a similar note I tried this Infrared heart beat dector because it looked so simple. I used this set of Infrared Emitters and Detectors with this circuit diagram with no luck. So the idea is that the blood from your finger reflects a different amount of IR depending on flow or something pulse related, but when I try it there is no measurable variation in the signal. set up picture you can tell the IR emitter is on using a digital camera. Does anyone have any experience with this stuff? I haven’t gotten much time into it but this post looked similar, and I don’t know where to go to start figuring out this problem. Thanks

September 17, 2013
by Ralphxyz
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scooter, try aiming the led directly at the detector, don't use your finger.

Does that work?

This is great I have to make up one, thanks.

Ralph

September 17, 2013
by BobaMosfet
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scootergarrett-

Yes, this is for pulse oximetry. This working principle of this is a wavelength change in the frequency of infrared being generated and received.

BM

September 17, 2013
by BobaMosfet
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scootergarrett-

I meant to say, it can also be used with pulse-oximetry. Principle is generally the same. Wavelength based.

BM

September 17, 2013
by Ralphxyz
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Thanks BobaMosfet, for the circuit explanation.

Now this does exactly the opposite:

I am seeing 4.88v when the IR led is unimpeded and .23v when blocked.

Damm I love it when things actually work.

Ralph

September 18, 2013
by BobaMosfet
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Ralph-

Yes, but do you understand why/how it does the opposite? :P

BM

September 18, 2013
by Ralphxyz
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Hi BM, yeah sorta but the .23v when blocked I do not fully understand.

Ralph

September 18, 2013
by BobaMosfet
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Ralph,

Do you have a datasheet for the transistor-- what you're seeing is likely a drop across the transistor and/or due to leakage current.

Don't you find the values 4.88 and .23 interesting?

BM

September 18, 2013
by pcbolt
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Ralph -

When the photo transistor is in your circuit, it will behave like a resistor with two different values, one for when the light is on and one for when it is off (actually there is a small zone of light intensity that will make the resistance somewhere between the two but we can neglect that for now). If you re-draw your last circuit and substitute a resistor for the photo transistor, you can figure out what the ohms are for "light ON" and "light OFF", by using all the information your measured. Of course you could just measure the resistance directly with your meter w/ and w/o light, but where's the fun in that. (Turns out when the light is on the ohms drop to 1/1000 of the "light off" state).

September 19, 2013
by Ralphxyz
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BM, I think this is the opto interrupter.

This has been really informative I wish I had the time to really understand it, but it works!!

Hey scootergarrett, did you get your circuit to work?

Ralph

September 19, 2013
by scootergarrett
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I haven't had time to put into that little side project, but I will return.

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