Phototransistor question

Discussion in 'General Electronics Chat' started by dsp_redux, Jul 18, 2010.

  1. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    Hello everyone,

    Got a simple question about a phototransistor I bought in a local shop. I plugged an infrared emitting LED on one pin of a uC and it does flashes fine on a 1 second interval. The phototransistor is pointing completely in the other direction about 1 inch of the LED. The problem is: I don't have ANY info on the phototransistor else than
    a) It's a phototransistor
    b) It's the golden one type with two pins
    c) It's written 5mm on the description, which should be the diameter of the opening for receiving light.

    I've connected what I think is the collector (flat pin on the frame) to a 2.2M (after many tries) and the emitter to ground to get what I think are the best value high/low values for now. I get something like 2.35V and 1.15V on a 5V supply. That is not enough to make any difference on a hex schmitt inverter (always low). Is there a way to improve the on/off voltage so it's more sensitive are it is normal behavior of a phototransistor and I must get a comparator to get that mighty 5V/0V back in my uC?
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    Try connecting a 4.7K resistor in series with photo instead of a 2.2M.
    It will work.
     
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  3. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    Gives me a constant 4.94V Vce with 4.7k.
     
  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    When the photo switches on it should be almost zero, if not then the photo is not working.
     
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  5. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    I just changed the transistor for a new one but got the same result with 4.7k. It seems like the higher the resistance, the higher the sensitivity is. Or maybe I'm not push enough current in the emitting LED (I can see it flashing ok with a digital camera).
     
  6. Ghar

    Active Member

    Mar 8, 2010
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    Even in an optoisolator (LED and phototransistor in one package) the transfer ratio between current can be less than 50%. That is, transistor current = 50% x LED current

    If the LED and transistor are far apart you should expect a much, much lower percentage. You need to pick a resistor such that this transistor current through that resistor is at least 5V.

    If you are only swinging between 1.15V and 2.35V do you mean that you get 2.35 V output with the LED off and 1.15V with your LED on?

    If so then the resistor is too large because the ambient light is dropping your output by over 2V. The resistor needs to be such that ambient light doesn't create significant current and then you need LED current * transfer * resistance to be 5V or more.
    If your situation is what I think it is it means the transfer ratio is too small, which means either the LED and transistor aren't matched, you're not running enough current through the LED, or they're simply too far apart.

    What's your LED current?
     
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  7. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    30mA in emitting LED. I don't think they are matched. Yes, 2.30V when off, 1.15V when on. I tried with a remote controller and same thing. If I try with smaller resistance, I get 4.94V on/4.95V off. I asked the guy at the local shop for a datasheet, he as none.
     
  8. Ghar

    Active Member

    Mar 8, 2010
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    I think 30 mA is going to be approaching the limit you can have so not much you can do there I guess.
    Just as a confirmation have you tried moving the LED and transistor closer together to see if it works, or adding a paper tube around the light to channel more to the transistor?
    I realize it's not a solution but it would be an insightful check.
     
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  9. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    Yes, to the point where they are touching each others. What are standard values for 5V supply on a LED emitter IR and a phototransistor? The thing is, at 2.2M and higher, I can make the transistor switch, but not on the whole 5V (only a difference of 1V between on/off state). I'm thinking of using a comparator, but I have the feeling I'm not using the ptransistor the right way.
     
  10. bc108

    New Member

    Jul 18, 2010
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    Is it possible that the device is round the wrong way, and what is being measured is reverse biased emitter base photocurrent leakage, rather than normal phototransistor action, hence needing a big resistor rather than the standard 4k7?
     
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  11. Ghar

    Active Member

    Mar 8, 2010
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    I would think that putting them so close together should have made it work if it worked at all. The transfer ratio should hugely go up with distance.

    Aside from what bc108 said maybe the devices are just bad.
     
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  12. Potato Pudding

    Well-Known Member

    Jun 11, 2010
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    The problem is that it is an unidentified device.

    Pictures please and a pinout.
     
  13. tom66

    Senior Member

    May 9, 2009
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    Is a higher voltage unavailable?
     
  14. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    I tried the transistor on the other side (only two pins, so two possibilities), and Vce = 1V all the time. A higher voltage is unavailable for the project unfortunately. Still banging my head here. Guess I'll have to order a KNOWN phototransistor from digikey, but then I should also buy other parts since 7$ for shipping for a phototransistor is quite expensive. My local shop doesn't have anything else than this unknown part.
     
  15. Potato Pudding

    Well-Known Member

    Jun 11, 2010
    684
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    Can you test it with a higher voltage because a simple chopper power supply to get 50μA at 25 Volts is a very minor challenge. Look at it this way. Higher voltage on the Photodevice can allow you more slack with the sensitivity since you will be limiting the output back to 5V.

    I am also thinking that you might have been sold a goat when you wanted to buy a pony.

    This link is not really a possibility because you say your device has 2 pins but the idea is that there are other kinds of phototransistors.

    http://www.linearsystems.com/datasheets/LS627.pdf

    A photofet, photomosfet, or photoresistor would make as much sense as this device being a dead phototransistor.
     
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  16. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    The thing is, the circuit I'm building must fit in a really small area, and I can't afford the space on the board for an other supply. I guess I'll just buy another phototransistor with a datasheet. Thanks for the help.
     
  17. Audioguru

    New Member

    Dec 20, 2007
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    Maybe the unknown "photo-transistor" is actually a photo-diode. Thay have a very low sensitivity. Connected in the forward direction it will have a continuous voltage of about 0.6V.

    I agree that you must buy a real photo-transistor that has a datasheet and a pinout.
     
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  18. dsp_redux

    Thread Starter Active Member

    Apr 11, 2009
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    I guess I'll go with the PNZ109L from Panasonic (much more expensive than the 0.59$ unknown ptrans.) and at the same time I'll replace my 74LS14 Hex Schmitt Inverter (wich is a quad inverter) with a single 74AHC1G14... or maybe two MOSFETs (NMOS with PMOS inverting CMOS style)?
     
    Last edited: Jul 21, 2010
  19. Audioguru

    New Member

    Dec 20, 2007
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    The 74LS14 is an old fashioned TTL IC with a fairly high input current. It will kill the sensitivity of a photo-transistor.

    A CD40106 is a Cmos hex Schmitt trigger inverter (hex means 6 in a package, not quad) but has no input current so the photo-transistor can be as sensitive as its collector resistor allows.

    A 74AHCxxxx is also Cmos but is very high speed.
     
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