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Comparator - Battery Cutoff

Discussion in 'Power Electronics' started by Xenobius, Nov 11, 2017.

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  1. Xenobius


    May 15, 2012
    Hi all,

    I have a simple requirement and I had a go at it using a simulator and it worked. I thought I understood what was going on so I built it and .... nothing worked. I would like to share it with you so that you may help me understand better what I did wrong.

    I am using an LM258, power directly from battery which may be anywhere from 3.7v down to 3.1v.
    What I need is to output low when the voltage input reaches anywhere between 3.1v to 3.3v.

    The diode in the circuit is a 3.3v zener diode.


    Attached Files:

  2. kellys_eye


    Jun 25, 2010
    R6 is too high a value. The zener requires a minimum current to reach the zener voltage. 100k mean only 33μA flowing..... get it to the mA range.
  3. Alec_t


    Jul 7, 2015
    If you read the datasheet for the LM258 you will see that the common mode input range is from 0V to Vcc-1.5V. If your supply voltage (Vcc) is 3.7V or less, then the voltages at the inverting and non-inverting inputs must be 3.7V-1.5V = 2.2V or less for the opamp to work correctly. You need a voltage reference less than 2.2V when Vcc is 3.7V, but if Vcc drops to 3V then the reference would have to be less than 1.5V.
  4. dorke


    Jun 20, 2015
    I agree with the above notes.

    More than that,
    I would say there is no need to reinvent the wheel...;)

    Here is 3 pin under-voltage detector that will do what you need in every way better:
    smaller,cheaper,more accurate,less power and way simpler.

    There are many others,
  5. Xenobius


    May 15, 2012
    I took it down to 10k but still same thing
  6. Xenobius


    May 15, 2012
    Oh dear ... if only I knew before... this is what I need. Only the boards are ready now :(
    I was trying to order some of these, specifically the 3.1v but I cannot find it much. Is this brand new from ST?
  7. dorke


    Jun 20, 2015
    It isn't new,the datasheet is from 2006.
    The design you have is faulty.

    You can't expect to have 3.1V on the zener when you feed it 3.1V through a 10k resistor.
    lets' say you feed it with 3.15 V,the current would be (3.15-3.1)/10k=5μA,that is way too low.

    You should use a lower voltage zener, say 2.5V, and bias it properly(a few mA),
    and scale down the "sampled voltage" on the other leg of the op-amp.

    Another thing ,
    you should use a rail-to-rail(at least the upper rail) op.amp that can work down to 3V (or a bit lower if possible) .
  8. Xenobius


    May 15, 2012
    Yes I agree that it is faulty :) I just want to learn and when you recommend rail-to-rail I am still a bit lost. I will do some more research.
  9. Audioguru


    Sep 24, 2016
    The 1N4728A zener diode is 3.3V only when it has a whopping 76mA(!) of current in it that will quickly kill the battery. There are many 3.3V zener diodes available that draw only 5mA. This is a horrible faulty circuit.

    On another thread or another website there is a 3 pins IC that draws a very low current and gives an output when a battery voltage is low.
  10. Xenobius


    May 15, 2012
    Yeah horrible indeed I used 100k pull ups just to lower the current consumption as much as possible only to put a zener like that. I am at the moment removing this part of the circuit and finding an alternative ready made solution.

    This was going to drive a low power voltage regulator which has a power on and off feature. When LOW it shuts down power and this is why I need a LOW from this circuit.

    Thanks everyone for your valuable input.
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