# current limiting resistor for optocoupler

Discussion in 'The Projects Forum' started by photocs, Aug 27, 2013.

1. ### photocs Thread Starter New Member

Jun 1, 2013
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0
hi everyone!

well, i thought i had it figured out but things aren't working as they should; let me explain. i've bought an optocoupler (the LTV-826 http://www.mouser.com/ds/2/239/LTV-8x6-201828.pdf) which I'm using to trigger my camera by means of an Arduino Due (logic at 3.3V). I understand that I need a current limiting resistor on the input side of the optocoupler, so here is how I calculated it:

From the datasheet, we see that the forward voltage is (typically, @20mA) 1.2V; so subtracting 1.2V from 3.3V leaves us with 2.1V for us to deal with through the resistor. So, by Ohms law 2.1V / .02A = 105 Ohms. So, I put in a 100 Ohm resistor.

Here is how I've got things hooked up:
- straight off the digital pin (of the Arduino Due) I've got the resistor
- from the resistor I go to the anode of the optocoupler input (pin1 in the datasheet)
- i've got the cathode going to GND (pin2 on the datasheet)
(not that I write that, I wonder if it matter that the resistor is on the anode side and not the cathode??)

When I throw the pin connected to the input side of the optocoupler high (3.3V) and I check the resistance on the output side of the optocoupler, I'm not getting 0 resistance like I'm expecting.

I'm guessing I'm not using the proper value for the current limiting resistor; so I'm hoping someone could show me how to calculate for the proper one?

Thanks a million everyone!

2. ### #12 Expert

Nov 30, 2010
17,895
9,314
It doesn't matter whether the resistor goes from the driving voltage to pin 1 or from pin 2 to ground.

You are trying to measure a transistor, not a set of metal points. Transistors don't go to zero ohms. Try to run 10 milliamps of current through the transistor. The collector to emitter voltage should be collapsed to less than a volt.

I said 10 milliamps because the isolator promises a 50% current ratio. You are feeding it close to 20 ma on the diode side, so it should pass 10 ma on the transistor side with just a little voltage loss.

Jul 18, 2013
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The output is a similar to a transistor collector/emitter so reading resistance is going to be rather misleading using a standard ohmmeter that does not have the current capability on the ohms range to forward bias it.
Hook it up to a supply with a load resistor in the emitter or collector.
Max.

4. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,907
1,789
The resistor size depends on what you are loading the output side with, as this opto coupler has a "current transfer ratio" (CTR) of .5, so by driving with 20 mA you should be able to drive a 10mA load.

When you do this measure the voltage on the output of your Arduino and also where the resistor meets the opto. See if they agree with your expected values.

Now the output is a transistor, you may see a resistance change from off to on (probably will) *IF* your meter is connected with the proper polarity: + to collector, - to emitter.

If I was checking this I'd ground the emitter and put a 1K or 10K resistor from collector to 3.3V and use my voltmeter, but I have a rather large box of resistors to choose from.

The question I want to ask is how close have you looked at your camera trigger?
Is the trigger just a switch closing?
Can you use a momentary push button to trigger?
Assuming a simple switch works:
Is there a voltage when not triggered?
Is there a current when it is triggered?

5. ### photocs Thread Starter New Member

Jun 1, 2013
12
0
Cool, got it.

I'm losing you here. I thought the point of using an optocoupler was that you were isolating one circuit from the other; if current is passing through then what is the point? Do I need to take another approach to triggering my camera? Essentially, all I need is a circuit (in this case the output side of the optocoupler) to close when the pin on my Arduino goes high.

6. ### photocs Thread Starter New Member

Jun 1, 2013
12
0
Thanks for the explanation Ernie, this makes a bit more sense to me.

Yes, in order to trigger the camera, I just need the two leads going to the camera to short; so yes, you can use a momentary push button to trigger. No voltage is needed to trigger the camera, at least I shouldn't be supplying a voltage. The camera has a 10pin connecter on it, if I short the "focus pin" (on the 10 pin connector) to the "GND pin" (on the 10 pin connector) the camera will begin to focus; if I short the "shutter pin" (on the 10 pin connector) to the "GND pin" (on the 10 pin connector) the camera will take a shot.

Here is an example of the application and implementation by someone else. http://www.markfickett.com/stuff/artPage.php?id=375#schematic

Jul 18, 2013
13,160
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The only way you can usually test a semiconductor with a modern VOM is on the diode scale, the ohms scale cannot supply enough forward bias for the average junction, this is why if you measure a diode on the ohms scale it will usually show open or high resistance in both directions!.
The diode range actually passes enough current to forward bias and measures the junction volt drop.
Max.

8. ### #12 Expert

Nov 30, 2010
17,895
9,314
You are allowed to run the current from the camera through the transistor side of the optocoupler. Just be sure the polarity is right. Measure the pins for "shutter" and see which one is positive. Positive from the camera goes to pin 4 and the less positive voltage from the camera on pin 3.

The fact that current flows through both sides of the coupler does not mean it's useless. It means the two currents don't touch each other.

9. ### photocs Thread Starter New Member

Jun 1, 2013
12
0
So I checked the polarity of the pins on my camera by checking the voltage across the two leads I need shorted (or connected); I've been able to find the + and the - leads form this. I've taken the + side of the pins and connected it to the collector of the optocoupler; and I've taken the -V side and connected it to the emitter... is that correct?

I hate to ask; but what happens if I get them switched around?

10. ### #12 Expert

Nov 30, 2010
17,895
9,314
It just won't click. There is no external power being applied to the camera, so nothing but camera power is available on the transistor side of the optocoupler. It can't hurt itself by not getting a connection.

Look at that thing you pointed to in post #11
That means an optocoupler should work, even if it doesn't measure zero ohms.
There is still the possibility that YOUR camera is different, but we've got you aimed right. Try it.

Last edited: Aug 27, 2013
11. ### photocs Thread Starter New Member

Jun 1, 2013
12
0
Cool, perfect, everything seems to be working how it should. Thanks for the help everyone!

12. ### crutschow Expert

Mar 14, 2008
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That's true for a single PN junction, but not for the collector-emitter junction. A multimeter in the standard Ohms measurement range should register a low resistance when measuring between the collector and emitter when the transistor is turned on since Vce should be typically only a tenth of a volt or so under those conditions. You might have to try alternate polarities of the multimeter leads to get a low reading.

Oct 26, 2011
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