Photoelectric Optical Smoke Sensor

Discussion in 'General Electronics Chat' started by Sam Matthews, May 21, 2016.

  1. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    I'm attempting to create a 'chamber' that holds a custom made reflective optical sensor to sense the presence of smoke inside. My current attempts include a IR LED and a photo-transistor that has a peak sensitivity is that of my LED's wavelength. However, i can't seem to get the set up to work properly. I know that they work together because when i have them facing each other i can get a analog reading from the transistor when connected as the following circuit.

    [​IMG]
    However, once i attempt to create some sort of DIY chamber to test this using cardboard, to look like the following it just doesn't seem to pick up the light that reflects from the smoke.

    [​IMG]

    The following are the datasheets of the IR LED and the photo-transistor, please let me know if you feel i have the incorrect hardware for the job. I have got a photo-diode on order to test this out in case the transistor is the issue.

    IR LED - http://www.farnell.com/datasheets/1683593.pdf
    Photo-transistor - http://www.farnell.com/datasheets/1683591.pdf
     
  2. Sensacell

    Well-Known Member

    Jun 19, 2012
    1,124
    266
    The dispersed light is going to be many orders of magnitude weaker than the direct beam, slim chance of detecting it with the simple circuit shown.

    Modulate the LED, use an AC coupled amplifier with lots of gain, then you should be able to see the smoke.
     
  3. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    All of the circuits i have been looking at are the simple design that i have been attempting to implicate. I do not see a slight jump in readings at all. I'm guessing that it would change it even by the tiniest of readings surely?

    So are you proposing i put the transistor into a OP-AMP circuit to amplify the readings?
     
  4. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,766
    1,101
    Not surprising. The IR will bounce all round the chamber walls. You are looking to detect a tiny amount of scattered light against a background of reflected light from a wall surface area thousands of times bigger than the smoke particle surface area. Needles (very small ones) and haystacks (humongous ones) come to mind.
     
  5. BobTPH

    Active Member

    Jun 5, 2013
    782
    114
    Have you at least painted the inside of the chamber a dull black? And made sure that no direct light from the LED gets to the sensor?

    You might also want to try a higher collector resistor, which will amplify the voltage of the detector.

    What sensacell was saying is to use a scheme similar to that used by IR remotes. The LED is modulated at a specific frequency, then you build an amplifier and filter to detect that frequency. This eliminates the background light reaching the sensor and concentrates on the light from the LED.

    Bob
     
  6. Sensacell

    Well-Known Member

    Jun 19, 2012
    1,124
    266
    The design of the optical path is critical for sure, you need to go to great lengths to prevent non-dispersed light from reaching the sensor.

    Lots of baffles and light traps! if you modulate the light, you don't need to fret too much about ambient light sources, otherwise how does the smoke ever get in there!
     
  7. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    So from reading all of these comments i'm beginning to realise that what i thought and didn't want to hear is true. I will have to go with making the chamber properly, 3d print one out of black filament maybe, then test the circuit again.

    It seems i was wrong in assuming that i would get a difference between holding my fingers over the transistor to stop any light touching it at all then removing my fingers to let ambient light hit the transistor.

    Do you think i will still need an amplification OP AMP circuit to detect the reflected light?
     
  8. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,766
    1,101
    I don't know if modulation would be any advantage in this application. If the sensor is in a sealed chamber there shouldn't be any background (in the sense of unintentionally generated) light. I think a bigger problem would be stabilisation of the IR LED emission, since changes in that could far outweigh the minute changes in received light due to smoke scattering.
     
  9. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    The more i read, the more i seem to think that i shouldn't need a OP AMP with this sort of setup. Please correct me if i'm wrong but i thought that a Photo-Transistor was a photodiode connected to the base of a transistor? Then therefor it is acting like a amplified photodiode. Are you guys basically saying that even though it is essentially amplified through the transistor inside, it still isn't amplified enough?

    The following is extracted from: http://www.electronicshub.org/light-sensors/.

    "In a photo transistor, the collector – base junction acts as a photo diode. The collector – base junction is reverse biased exposing it to a light source. The current at this junction is amplified by normal transistor action and hence the collector current is large."
     
  10. crutschow

    Expert

    Mar 14, 2008
    12,977
    3,220
    If you want dispersion from smoke particles, then a shorter wavelength light source may be better.
    The shorter the wavelength, the more a given particle size will reflect light.
    IR has a longer wavelength than visible light and thus is reflected less by the small smoke particles.
    That's why fog lights on vehicles use a filter to remove the shorter wavelengths of light (leaving a yellowish light) which allows the light to penetrate fog better with less light reflected back to your eyes from the fog.
    So you might try an LED towards the blue end of the spectrum to see if that helps.

    A Si sensor does have a lower sensitivity towards the blue end of the spectrum but the increased scattering should make up for that.
     
    cmartinez likes this.
  11. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    Would you still use an OP AMP with the phototransistor?
     
  12. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    So i'm experimenting with a block pill tub and have the phototransistor and the LED sitting in the led facing down into the container (Which is black inside and i'm keeping all external light away). They are both facing the same way, towards the bottom of the container, and the op amp circuit is giving me a reading of 3.9v, and i tested the circuit by letting light hit the phototransistor and it drops to 0v. However there is no in between, is this whats expected? Please correct my thoughts here, but i'm sure a transistor is not either open or closed, i thought that there is a correlation between the current at the gate and the power let flow from collector to emitter? Since the reading on my voltmeter is 3.9v it means there is no light inside of the pill tub and therefor i added smoke and... nothing.

    So firstly, is my op amp circuit working as intended?

    I have tried many different circuits but they all seem to be performing in the same characteristics.
     
  13. crutschow

    Expert

    Mar 14, 2008
    12,977
    3,220
    Well, firstly, we need a schematic of your circuit to determine that. ;)
     
  14. Sensacell

    Well-Known Member

    Jun 19, 2012
    1,124
    266
    The problem is keeping the circuit from "saturating" - too much or too little signal and the amplifier will "go to the rail" and stay there, you will not see any change under these conditions.

    This problem is much more difficult than one would imagine- you need lots of gain to detect the tiny signal from the smoke, but you also need to prevent the output from saturating, this is a contradictory state.

    You need a way to bias the signal in the linear region, otherwise you cannot see the minute smoke signal.

    One approach would be to use modulated light and have some kind of feedback at DC that servos the output to Vcc/2.
    Utilize an AC coupled amp to process the resulting signal, this way you can have a widely varying gain, but always keep your signal in the linear region, separating the the problem of gain Vs. operating point.
     
  15. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    So this is basically what a smoke detector IC chip does, one such as http://www.electroschematics.com/wp-content/uploads/2013/01/lm1801-datasheet.pdf does. I sort of understand the method that you're trying to explain to me but because i'm no where near to your level of knowledge i'm finding it overwhelming if i'm perfectly honest with you.

    I configured my op amp and transistor like that on this image, but a multimeter in place of the ADC to monitor the output pin: http://web.eng.gla.ac.uk/rpi/2013/4/189879Circuit.png
     
  16. bertus

    Administrator

    Apr 5, 2008
    15,638
    2,344
    Hello,

    I see that there is a TL071 used in the schematic.
    That will not work on 3.3 Volts.
    Here is a selection table for single supply opamps:

    Single_supply_opamps.jpg

    Bertus
     
  17. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    I'm sorry, i didn't mean the exact parts. The op amp i'm using at the moment is a TLC272 (datasheet). Below is the current circuit:

    Phototransistor.jpg
     
    Last edited: May 22, 2016
  18. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    I forgot to mention that i added the schematic to the above post. Sorry on late notification.
     
  19. crutschow

    Expert

    Mar 14, 2008
    12,977
    3,220
    As noted, you need to bias the circuit properly to detect the small current from the sensor.
    Try connecting the phototransistor collector directly to the 3.3V .
    Connect a resistor from the emitter to ground.
    Connect the op amp (+) input to the resistor-emitter junction.
    To easily determine the proper resistor value, use a 10 megohm pot.
    Adjust (or select) the resistor value to give you a few tenths of a volt output from the op amp with no smoke.
    You may also have to increase the gain of the op amp by reducing the value of R23.
     
  20. Sam Matthews

    Thread Starter Member

    Jan 16, 2016
    178
    3
    Okay, thank you for that. So setting it to be 0.2/0.3v with no smoke and no light will be the bias needed. Then just using a pot instead of the fixed resistor in the position specified will allow me to change the bias to get whats needed. Then changing R23 lower it would increase the gain.

    I don't have a 1Mohm pot, i will get some ordered tonight. I'm guessing that a 20k would be miles off? Also, are you saying put the phototransistor to 3.3v because then the voltage that flows through is smaller and therefor wont hit the rails so easily? I have it connected to 5v.
     
Loading...