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Laser Burglar Alarm -- Need Help

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by Anonymous321, May 4, 2008.

  1. Anonymous321

    Anonymous321 New Member

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    Hi guys. My first post and thread.

    Ok, so I'm making a laser alarm. And I need help. The schematic of what I have is attached. This is what (I believe) happens:

    - Photocell is given voltage through laser beam AND 20 1.5 V batteries.
    - Voltage flows to the transistor. When there's enough voltage for the base of the transistor (it requires 30V), it opens and voltage flows through the emitter of the transistor. I'm a little sketchy here. So the current from the 9V battery cannot power the siren when the base of the transistor is getting 30V? But how? Like is it that b/c the switch is open, there is not a complete circuit or something for the 9V to power the siren?
    - And when the transistor is off (beam is broken), there is a complete circuit (I guess?) and the siren rings.

    The only part of that circuit that I haven't done yet is the 20 AA batteries connected to the photocell. They should give 1.5V each (so 30V in total) to the transistor so when I have the 9V clipped in, the alarm shouldn't ring, right? (I'm gonna try after this post to see if this happens or not)
    But then there's also the problem that if it does work correctly, I don't think that the beam would be doing anything useful b/c it's the 20 AA batteries providing all the 30V. So the only way to turn the transistor off and get the siren ringing is to remove one of the batteries or something. I couldn't tell how much voltage the laser pointer would be giving to the photocell with my multimeter on its lowest-10V-range (the needle just wouldn't move) so I don't know exactly how many batteries I would need. Am I even connecting it properly? I just attach the red and black leads of the multimeter to the 2 leads coming out of the photocell and shine the laser pointer onto the photocell.....The voltage the laser is giving is probably like 0.3V or something anyway. But even then, how would I get something to provide the rest of the 29.7V?? 19 AA batteries would provide 28.5V and the laser's voltage on that will still not be enough for the transistor to turn on.
    So I'm really confused on how to go on from here...I would really appreciate it if any of you helped out. Thank you in advance.
     

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  2. Anonymous321

    Anonymous321 New Member

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    Also, I may just be reading the transistor's ratings incorrectly.

    This is what is says:

    Silicon..........NPN
    Typical hFE...200

    Maximum ratings
    VCE.............30V
    IC................800mA
    Dissipation.....1.8W

    Does maximum rating mean it needs that much? Or you should only provide up to that much? I would think the latter but then even 1V should open the transistor's switch...
     
    Last edited: May 4, 2008
  3. Anonymous321

    Anonymous321 New Member

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    May someone please at least verify my schematic?
     
  4. dave

    Dave New Member

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  5. john1

    john1 Active Member

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    Maybe like this ....
    try to give a little more info.

    John :)
     

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  6. Anonymous321

    Anonymous321 New Member

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    Thank you for the reply, John. Ok so I see how you have 4 connections from the photocell: the resistor, the base of the transistor, the emitter of the transistor, and the ground. But how come you have the photocell connected to the emitter and base of the transistor? Shouldn't it just go to the base? Also, in my original circuit, I had only one wire coming from the photocell. It was to the base of the transistor. Does a photocell have a + and - system so that I'd have to have a wire from the other lead going to the ground? Also, how is the transistor powered? By the 9V as well? Also, for the resistor--do you know any specific resistor that I'd need to use? Or would any "dinky" one work?
     
  7. john1

    john1 Active Member

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    If your little siren doesn't take much current,
    then the circuit i've shown may work.

    The resistor has to be a low enough value to turn the
    transistor on, with no light on the sensor
    and a high enough value so as not to prevent the sensor
    turning the transistor off, when there is light on it.

    I would expect to see a darlington pair in this configuration,
    or two transistors in direct cascade instead,
    with the resistor around the 80K area.

    See what others say, before making the unit up.

    John :)
     
  8. john1

    john1 Active Member

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    Yes, almost all types of photo-cell are polarity concious.
     
  9. john1

    john1 Active Member

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    The resistor should be chosen by its Ohms.
    Usually colour-coded on the body of the resistor.
     
  10. john1

    john1 Active Member

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    Using a single transistor like that,
    I would guess at around 10 K,
    but there is really nothing to go on.

    See what others say.

    John :)
     
  11. Boncuk

    Boncuk New Member

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    Hi Anonymous321,

    John was right with his suggestion. I altered it to be an NPN-transistor.
    Using a photo resistor the circuit is really easy to make. A photo resistor has a high resistance in the dark (300KOhms and more) and a very low resistance at bright sunlight (represented by R4 in the schematic).

    This circuit activates alarm whenever the lightsource is interrupted. If you want it to react the opposite way just change the positions of R1 and R4.

    The switch is only for simlution of dark and bright. Omit it in your circuit.

    Use a self contained piezo buzzer (built in oscillator circuit). They have low power consumption and a noise level up to 105db.

    Regards

    Hans
     
    Last edited: Jul 7, 2008
  12. crutschow

    crutschow Well-Known Member Most Helpful Member

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    The maximum ratings (usually declared as "Absolute Maximum Ratings") are just that, the maximum at which you should operate the device. Anything beyond them may zap the device. Typically you want to operate well below (perhaps 50-75%) of the maximum for best reliability.
     
  13. Anonymous321

    Anonymous321 New Member

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    Thank you for your reply.
    So even 1 amp of current should open the transistor? On this website,
    http://www.kpsec.freeuk.com/components/tran.htm
    it tells me that Vce is the "Maximum voltage across the collector-emitter junction. You can ignore this rating in low voltage circuits." So I don't understand. How is the transistor supposed to work (what voltage does it require) so that the siren doesn't ring?
     
    Last edited: May 5, 2008
  14. john1

    john1 Active Member

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    I am trying to compose an explanation for you.
    But some things are puzzling me.
    Are you English ?
    I think you are.
    Is this a school project ?
    ... off line at a quarter to nine ? ...
    If so, then i think you should not stay up so late.

    I will try to compose an explanation for you, but for now
    i have to get on with removing an engine from an unhappy
    mini.
    Hopefully it will get happily fixed.

    Later, John :)
     
  15. crutschow

    crutschow Well-Known Member Most Helpful Member

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    1A of current through the transitor could fry it.

    The voltage across the collector-emitter is determined by the supply voltage and whether the transistor is on or off. When off, it will be equal to the supply voltage and when on, it typically is a few tenths of a volt. The collector current, when on, is determined by the value of the supply voltage and collector load resistance (in this case the resistance of the sounder/siren).

    The base current controls the transistor. A bipolar transistor is current controlled with the base-emitter drop being about 0.7V when the transistor is on. The required current to turn the transistor on depends upon the transistor gain (hfe). Good design practice is to use a gain of 10 to completely turn on the transistor when it's used as a switch. Thus, for example, if the collector current, when on, was 100mA, you would size the base resistor to provide 10mA of base current.

    When the photoresistor pulls the base-emitter voltage below about 0.6V the transistor will be in the off state and the siren is silent.
     
    Last edited: May 5, 2008
  16. Torben

    Torben Well-Known Member

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    What they mean is that in low-voltage circuits, you're not going to have voltages high enough to exceed the Absolute Maximum Vce, which is between 20V and 100V in the chart shown.

    The Absolute Maximum ratings are the highest levels at which the device is guaranteed to work. It's not guaranteed to work for long if one or more of the Maximums is met for very long, though. As crutschow noted, make sure you're operating the device with some headroom--you don't want to hit the Absolute Maximums in normal usage. Some people refer to these ratings as the Fry Points--go past them and you risk frying your device. :)

    I would second John1's suggestion of a Darlington pair--I find that I get a much sharper on/off. For a light-trigger alarm, what I would do is put the LDR from base to ground, a 10k* resistor from base to Vcc, hook the emitter to ground, and then hook up the buzzer/LED/whatever from the collector to Vcc.

    * - The value of the 10k resistor depends on the value of the LDR. I actually used a pot instead of the 10k resistor to find a good operating range and then measured the resistance of the pot, found it was near 10k, and then replaced the pot with the 10k resistor.

    Also, the alarm will only buzz as long as something is breaking the beam and will immediately go dark/quiet again once the beam is restored. You can hook up a 555 triggered via an inverter from the collector of the Darlington to control the length of time the alarm stays on for if the beam is broken (say something breaks the beam for 0.1 second; you can set the 555 to keep buzzing for, say, 5 seconds, so you have time to notice the alarm.

    Make any sense?


    Torben
     
  17. Anonymous321

    Anonymous321 New Member

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    Hi. Well if by English you mean that I am residing in England, then the answer is no. I am in Canada (eastern time zone). I am also in school. And I have no choice but to stay up late, John; this project is due on Wednesday (2 more days from now!) and I'm a nervous wreck here b/c I can't seem to get it to work (or understand exactly what I have to do to get it to work) and my teacher is going to destroy me during presentation time. My project is on different types of detection methods in burglar alarms (i.e. PIR, ultrasonic/microwave detectors, and photoelectric beams--laser alarms). And I just decided to make a laser alarm for my model (I should've just done a trebuchet for my project!). Thank you for the help all of you, though.

    crutschow -- I sort of understand what you are explaining but not really. So how am I supposed to turn the transistor on (ie. don't make the siren ring)? Supply 0.7V to it?

    Torben -- May you please tell me what you mean by the 'LDR' and the 'Vcc'? And by 'pot', I guess you mean potentiometer.

    But guys, I'm trying to read as much and fast as I can about these darlington transistors but I have a question: do they come premade in pairs? Or can one simply connect 2 transistors in series or something and that will work? Also, I won't be able to buy anything else most likely anymore b/c the The Source (Radio Shack in America) in my town doesn't sell electronic components anymore and I had to drive 1 hour away from here to another Source just to get the components that I have.

    At school, my friend made a suggestion that I make a very basic circuit utilizing just the buzzer, the photocell, and a battery. I'll post the schematic.

    I appreciate your guys' help. But like some of the things (like 555, inverters, etc.) that you guys are talking about I don't know about nor do I have access to (w/ the limited time I have and the resource being and hour drive away) and I really just want a simple design. I know the headache that I'm giving you guys and I apologize deeply. I thank you all once again for sticking with me (I hope).
     
    Last edited: May 5, 2008
  18. Anonymous321

    Anonymous321 New Member

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    Sorry, forgot to attach my friend's idea. It's in this post. Oh also, in the diagram, I wrote, "Then connect photocell to battery." I meant to write, "Then connect photocell to siren." Sorry...

    Also, if you guys don't mind, maybe I can explain this better talking over skype or something and then maybe I can understand better too.
     

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    Last edited: May 5, 2008
  19. blueroomelectronics

    blueroomelectronics Well-Known Member

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    That's worse.
     
  20. Anonymous321

    Anonymous321 New Member

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    So the use of a transistor and/or resistor is required?

    My siren's specifications:

    Voltage range: 3-20VDC/VCC
    Rated Voltage: 12VDC/VCC
    Current Consumption: 10mA max at 12 VDC/VCC
    Sound Pressure Level: 76 dM min. at 11 13/16" (30cm)/12VDC/VCC
    Resonant Frequency: 2,700(plus-minus symbol)500Hz

    I think the only relevant ones here are the voltage rand and the current consumption...
     
  21. john1

    john1 Active Member

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    Hi,
    back soon.
    that suggestion of using just the sensor is drawn incorrectly.
    that arrangement could damage it.

    Do you have a multimeter?
    if so please describe it. thank you.

    Do you have a bunch of resistors to choose one,
    please describe what you have. thank you.
     
    Last edited: May 5, 2008

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