pier Posted October 13, 2009 Report Share Posted October 13, 2009 Hello,I am facing a problem in my circuit where I made it for switching of a light when the battery voltage reaches 10.55 volt. The battery is 12 volt 3.4Ah SMF type. The problem is: When the battery reaches 10.55 volt (roughly) the battery is disconnected from the load. When the battery is disconnected from the load the battery voltage rises a bit and turns on the relay again. This repeats and repeats. I don't know if the circuit I made is right. Could somebody help me to fix this ? Quote Link to comment Share on other sites More sharing options...
KMoffett Posted October 13, 2009 Report Share Posted October 13, 2009 You need to add hysteresis feedback. This will require a higher voltage than 10.55 (you select) to switch the light back.http://www.intusoft.com/onsemipdfs/LM324-D.pdfLook at Figure 13 for the circuit and formulas.Ken Quote Link to comment Share on other sites More sharing options...
Hero999 Posted October 13, 2009 Report Share Posted October 13, 2009 Yes, there needs to be some hysteresis to prevent oscillation, just connect a 100k resistor between the op-amp's output and the + input. This will make the turn on voltage higher than the turn off voltage, see the data-sheet linked above for more information.I wouldn't use the LM324 anyway, the LM339, LM393 or LM311 would be much better because they're specifically designed for use as comparators. If you only need one op-amp/comparator then I'd recommend the LM311 which can drive a small relay without an additional transistor.Don't get me wrong, it'll be fine with the LM324, it just wouldn't be my first choice. Quote Link to comment Share on other sites More sharing options...
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