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how to convert the signal received by Tsop1738 ir receiver to steady output?

Discussion in 'General Electronics Chat' started by Mashare, Jun 10, 2018.

  1. Mashare

    Mashare New Member

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    Hi All ,,

    I want to convert the signal received by IR Receiver Tsop1738 to steady output level for example 1 if there is ir signal 0 if no Ir signal received(an object cross Ir beam).


    the signal received by the receiver as the following:


    hhh.jpg

    >>as you see in figure above the received signal is square wave 32hz(2ms off/29ms on).
    >> if there is no signal received due to object cross the ir beam the output of the tsop1738 goes high and still high while the object cross the beam.
    >>I see i can do what i need using 555 timer and other ic's like 74123 but i don't want to use them
    >>minimum time for crossing the beam 20ms.



    >>The transmitter circuit

    jjjjjjjjjjj.jpg
    as you see the circuit send 38khz for 2ms and off for 29ms.

    Regards
     
    Last edited: Jun 10, 2018
  2. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    You could use another 555 timer, google for '555 missing pulse detector'.
     
  3. gophert

    gophert Active Member

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    Do you need a 0-5V output from the additional circuit or would a 0-3.3V output work?
     
  4. dave miyares

    Dave New Member

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  5. Mashare

    Mashare New Member

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    0-5 v
     
  6. Mashare

    Mashare New Member

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    I don't need to use 555 timer :)
     
  7. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Will this do?
    MissingPulse.PNG
     
    • Agree Agree x 1
  8. dave miyares

    Dave New Member

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  9. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    You don't 'need' to, but it's an effective and simple solution.

    Personally I'd have used a PIC processor in the first place, greatly simplifying the project :D
     
  10. Mashare

    Mashare New Member

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    I connect the output to 4518 counter..
    No increment when I cross the beam :(
     
  11. Mashare

    Mashare New Member

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    No need to MCU ..

    OK..IN missing pulse circuit using 555 how I choose R,C ?
     
  12. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    Google it like I said - the operation of it is very easy to understand.
     
  13. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Here are a couple missing pulse detector circuits as alternatives to a 555 circuit.
     
  14. gophert

    gophert Active Member

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    Here is 0-5V out. No 555 but a NPN and PNP to keep from inverting and avoiding a diode voltage drop at the output.

    The envelope filter will react to a loss of signal in 0.5 mSec (5 time constants).



    7C6BD455-4597-439D-97DF-48BC6AFD8469.png
     
  15. Mashare

    Mashare New Member

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    I see 38khz ..
    The received signal 32 Hz..94% DC
    Does your circuit work if 32hz used ?
     
  16. Mashare

    Mashare New Member

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    Thank you I got it :)
     
  17. Mashare

    Mashare New Member

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    Screenshot_20180611-002757.png

    *Can I use lm358 instead of lm393 ?
    **For my received signal should I change the value of capacitors and resistors ?
    ***what is the purpose of c2 and D1?
    ****which the two circuit better using lm393 or using 4093 ?

    Regards:)
     
  18. gophert

    gophert Active Member

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    Yes, works fine.
     
  19. crutschow

    crutschow Well-Known Member Most Helpful Member

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    No.
    The circuit requires the open-collector output of the comparator to operate properly.
    As stated in my writeup, you select the R1C1 value to be somewhat longer than the pulse period.
    C2 is to differentiate the pulse edge, giving a short pulse trigger the circuit, and also block the DC bias from the input.
    D1 clamps the differentiated pulse from going significantly below ground.
    The LM393 circuit's trigger levels, determined by resistors, are more accurate than the 4093's. which depend upon the internal hysteresis value of the 4093 Schmitt trigger input.
    But either circuit should be more than adequate to do the job.
     
  20. Mashare

    Mashare New Member

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    I test the your circuit the output if the object cross the ir signal high and if there is no object cross the ir signal the output square wave with short off time ^^
     
  21. Mashare

    Mashare New Member

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    Thank you :)

    The received signal period about 31 ms .
    Is this mean Any object cross the ir signal in time below 31ms can't detect it ?
     
  22. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Not 100%.
    If one pulse is interrupted, then the output will go high for one pulse.
    If the object passes so fast that it doesn't block at least one pulse, then obviously it can't be detected.

    A block of one pulse is shown in the simulation below for an R1C1 of 31ms.
    The output goes high about 33ms after the loss of one input pulse and stays high until the next pulse is received.

    upload_2018-6-11_20-2-41.png
     

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