Darkness Detector-LM393N

Thread Starter

JDR04

Joined May 5, 2011
356
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
 

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tracecom

Joined Apr 16, 2010
3,944
Is there a simple way of protecting against a reverse polarity connection without adding too many components as I am limited for space.
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?
 
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Thread Starter

JDR04

Joined May 5, 2011
356
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
 

tracecom

Joined Apr 16, 2010
3,944
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
That would be perfect; install it forward biased in the +9V lead, i.e., cathode toward the circuit and anode toward the power source.
 

ScottWang

Joined Aug 23, 2012
7,080
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.

 

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Ron H

Joined Apr 14, 2005
7,014
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.
 
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