Darkness Detector-LM393N

Discussion in 'The Projects Forum' started by JDR04, Oct 29, 2012.

  1. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Hi Guys, could somebody help me out with this one please. (I've attached the schematics)

    The circuit works great and the cameras trigger etc etc so that end of it I think is sorted out.(Provided battery is connected up correctly-of course)

    What I've noticed is that the LM393N quickly burns out (and I mean really hot!!!) if the 9V battery is connected up the wrong way round for a brief moment. After that the LM393N POWER PIN (8) and Ground PIN (4) are dead shorted.

    Is there a simple way of protecting against a reverse polarity connection without adding too many components as I am limited for space.

    Could I also take this opportunity to wish all you guys in the USA luck with Hurricane Sandy. I hope all works out good for you.
    Thanks - JDR04
     
  2. tracecom

    AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    Put a 1N400x diode in series with the +9V lead. It will drop the voltage about .7V, but that shouldn't be a problem. You will likely have to readjust VR1.

    BTW, R5 isn't really 200k, is it?
     
    Last edited: Oct 29, 2012
  3. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Hi tracecom, thanks for the quick reply.

    Thanks for spotting my error with R5, it should be 200R.

    Would something along the lines of a 1N4001 diode suffice?

    Thanks again

    Cheers-JDR04
     
  4. tracecom

    AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    That would be perfect; install it forward biased in the +9V lead, i.e., cathode toward the circuit and anode toward the power source.
     
    JDR04 likes this.
  5. ScottWang

    Moderator

    Aug 23, 2012
    4,850
    767
    The circuit was changed as posted, it was according to the datasheet, you can adjust the value of resistor when you testing.

    The changed things as below:
    The indicator LED that I used 80% duty cycle, as 10mA x 80% = 8mA.
    Changed VR1 from 10K to 100K, because that is easy to get the half of 9V, and compare to P.3(+) of LM393.
    I were seprate SFH618A from the series circuit, the LED inside of SFH618A has Vf 1.1~1.5V, I were choose 1.5V to calculate, and get R5=820.

    I calculate the R6 that I got 1.2K, the math is (9V-1.5V-1.5V-0.2V)/5mA=1.16K.

    [​IMG]
     
  6. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks ScottWang, I appreciate your time on this.-JDR04
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Be aware that the guaranteed current sinking capability of LM393 is only 6mA. Typical is 16-18mA, depending on mfr, so you can probably get away with more on a one-off.
    Personally, I design to worst case specs.
     
    JDR04 likes this.
  8. campeck

    Active Member

    Sep 5, 2009
    194
    3
  9. tracecom

    AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
  10. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,157
    You could use a diiode bridge ... then it doesn't matter how the battery is installed.
     
    JDR04 likes this.
  11. JDR04

    Thread Starter Active Member

    May 5, 2011
    339
    4
    Thanks Campeck, have booked your link. Some interesting stuff. JDR04
     
Loading...