[Help] Battery Monitor Circuit

Discussion in 'The Projects Forum' started by Slasherbaven, Jul 23, 2008.

To improve..

  1. Thread Starter needs to read more.

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  2. Thread Starter needs to experiment more.

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  1. Slasherbaven

    Thread Starter Member

    Jun 23, 2008
    13
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    I've created a schematic of a Battery Monitor, where it monitors three voltage ranges:

    0~5V
    5V~9V
    9V~12V


    However.. I am unsure of my circuit.
    So yeah, well before I build the circuit, can any of you guys double check this for me?
    Really sorry for the bother

    Thanks in advance. =]


    (Attached to this post is the Battery Monitor schematic)
     
  2. blocco a spirale

    AAC Fanatic!

    Jun 18, 2008
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    This won't work. I think it would be best to start again rather than try and fix it.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, it needs work.
    1) Use comparators instead of operational amplifiers. Op amps should not be used as comparators; they weren't really designed to be run in open-loop mode indefinitely. Power consumption will be far higher than comparators, as the outputs of the opamp will be in constant saturation. Consider using an LM339 quad comparator, or better yet use an IC that's really designed for this type of thing, an LM3914N. Comparators and 10 LED drivers are built right into this IC.

    2) Your LEDs do not have current limiting resistors.

    3) The op amps have only negative supply and ground; no positive supply.

    4) Although you've provided a 2.5v reference to the resistor ladder for the inverting inputs, the noninverting inputs see full battery voltage. As a result, the outputs will not change states until the battery voltage falls below the minimum input voltage of the TLE2425, at which point the outputs will be unpredictable.

    5) Battery voltage remains remarkably constant from when they are fully charged until when they are nearly discharged. An automotive 12.6v battery is considered fully discharged at 11.4v. Dry cells have somewhat different levels, but it's a similar concept.
     
  4. Slasherbaven

    Thread Starter Member

    Jun 23, 2008
    13
    0
    I'm a bit confused...

    Sergant Wookie, I am required to use an LM324 quad (a low-power quad Op-Amp) for the circuit.


    And here, I re-did the whole thing.
    Hopefully it's better than the one before.
    [​IMG]

    The Vcc's purpose is to power up the op-amps.
     
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    There is an open circuit. There is no current path through the resistors. There will be no voltage across them.
     
  6. Slasherbaven

    Thread Starter Member

    Jun 23, 2008
    13
    0
    >_< Alright, here's another one.
    It seems quite right.

    [​IMG]

    I'll start fixing things from here.


    =EDIT=
    The earlier post was deleted by accident.
    Sorry about that.
     
    Last edited: Jul 31, 2008
  7. KMoffett

    AAC Fanatic!

    Dec 19, 2007
    2,576
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    U1 (78L12) will not work for this because the battery on the input is only 10V. You were thinking of a 78L05 maybe? And the + terminals on the op amps are hooked to the (-) end of the 10V battery.

    Ken
     
  8. Slasherbaven

    Thread Starter Member

    Jun 23, 2008
    13
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    Ah, my bad on the 10v battery and it's connection.
    Here, I made the minor adjustment:
    [​IMG]

    And as for the 78L12, why would using a 78L05 be better?
     
  9. KMoffett

    AAC Fanatic!

    Dec 19, 2007
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    1. For the 78L12 to regulate it needs an input of +14V or more. If your battery is 12v then the regulator has have an output of 10v or less to work. 78L05 is a common IC, but could be a 78L09 or 78L06.

    2. The negative end of your battery should be hooked to ground, not the regulator's input. The regulators input should be tied to the battery's (+) terminal.

    For the reference voltage, you need to research: three-terminal voltage regulators...how they work.

    For level sensing you need to research: voltage dividers...for both of the comparator's inputs.

    Ken
     
  10. Slasherbaven

    Thread Starter Member

    Jun 23, 2008
    13
    0
    Thank you for being patient, Ken.

    I just read up on it, and it's pretty interesting.
    Seems that the excess voltage dissipates as heat, and the hole on a regulator has to be attached to a heat sink.

    Since I'm putting in a 12V, there'll be an excess of up to 5V, so I shall put a heat sink to be safe.

    Thank you.


    I'm sorry, but level sensing is..?


    There we go:
    [​IMG]

    I'm still a bit confused as to why, but will reading up more details on regulators help?
     
  11. bertus

    Administrator

    Apr 5, 2008
    15,649
    2,348
    Hello,

    Here I have a schematic that I used for a thermometer.
    You have to recalculate the resistors in the voltage divider.
    There is a constant currentsource of 10 mA for the resistors.

    Greetings,
    Bertus
     
  12. Tahmid

    Active Member

    Jul 2, 2008
    344
    25
    Hi,
    Why don't you just use TL431?
     
  13. KMoffett

    AAC Fanatic!

    Dec 19, 2007
    2,576
    231
    You're getting there. :) We can deal with the heat sinking when you get a working design.


    You also need to hook the op amps (+) power terminals to the Battery (+) terminal, so the op amps are powered.


    With +5 V into the top of your string of 1K resistors, you will get +5V into the inverting input of U2, +3.33V to the input of U3, and +1.66v to U4. Now what you need to do is scale down the battery voltage with a voltage divider (2 resistors) so the non-inverting inputs of IC's can compare the declining, scaled voltage of the battery, as it discharges, to these three fixed voltages. When the battery is fully charged, you want the voltage at U2's non-inverting input to be a little greater than +5V.

    Once you have the circuit laid out correctly, we can work on the values for the 1k Resistor string. As it is now the lowest trip value would put the battery at < 4V, so the 5V regulator would no longer work.

    Ken
     
  14. Slasherbaven

    Thread Starter Member

    Jun 23, 2008
    13
    0
    Thanks, I will definitely look it through and compare once I'm done with mine =]
    I'm still, well, a noob at this.
    I'll stick with 7805 for the time being. Thanks though.
    Fix'd :)
    [​IMG]
    Hmm, if it has to work like that, to scale down the voltage of the battery, if I were to set the resistor values differently, ie 3 ohms, 2 ohms and 1 ohms, voltage at U2's non-inverting input will be a +6V.

    Will that do?

    =EDIT=
    U3 gets a +2V only though..
    But if I used three resistors of the same value, U2 will get a +8, and U3 will get a +4...
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    Well, you don't want to use really low-value resistors if you can avoid it. They'll drain too much current from the battery.

    Try keeping the total current through the resistive divider network down to about 1mA. Since it's a 12v battery, that means 12.6k Ohms total.

    Also, a 12.6v lead/acid battery is considered fully discharged at 11.4v. That leaves you a range of just 1.2v for switching the three outputs.

    You would get much less current consumption in your circuit if you used comparators instead of operational amplifiers.
    Have a look at LM339's. They're in Circuitmaker under:
    Devices -> Browse -> Comparators -> Comparator5
    The LM339's output does not source current, but it can sink up to 10mA.

    If you're using red LED's with a 1.7v rating, use 330 Ohm resistors from the output of the 7805 to the anode of each LED, and connect the cathodes to the outputs of the LM339.
     
  16. Slasherbaven

    Thread Starter Member

    Jun 23, 2008
    13
    0
    So that would mean it would be better if I used 4.3k Ohms each for R4, R5 and R6.

    I don't really get this part.
    Can you elaborate a little bit further?

    I understand, however for this I am required to use an LM324 for my design, which would mean using op-amps is a must.

    Well, here's the modified version of it so far:
    [​IMG]
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    No, it means that the total for all of the resistors should be about in that range.

    OK.
    When a lead-acid battery is fully charged, it measures 12.6V.
    When a lead-acid battery is fully discharged, it measures 11.4V.
    I'm going to have to get back to this tomorrow, I'm really exhausted at the moment.
    In the meantime, please read through this chapter in our E-books:
    http://www.allaboutcircuits.com/vol_1/chpt_6/1.html

    I see, I didn't realise this was a homework assignment?
     
  18. Audioguru

    New Member

    Dec 20, 2007
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    I think you need to learn about electronic circuits so that you can design them without serious errors.
     
  19. SgtWookie

    Expert

    Jul 17, 2007
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    OK, on the one side you have a fixed 5v reference. As long as the battery stays above 8v, you'll have a constant 5v.

    On the other side, you have a resistor "ladder", or voltage divider network.
    You need to work out the voltage divider so that when the battery measures 12.6v, that "tap" point in the resistor network measures slightly more than your reference voltage, or 5v.

    A 2nd point that is worthy of consideration is 12.4v. When a lead-acid battery's voltage drops below 12.4v, sulphation begins. This causes a layer of gook to accumulate on the plates of the battery, eventually causing it to not be able to accept a charge, or release stored energy. So, I suggest that 12.4v is also a good voltage to indicate.

    As I mentioned yesterday, 11.4v is the voltage at which a 12.6v lead-acid battery is considered fully discharged. So, you also need a tap from your resistor divider network that will output 5v when the battery voltage is just slightly higher than 11.4v.

    Your network will thus consist of 5 resistors; one on the top of the network, one on the bottom, and three in the middle. If you read through the E-book section that I pointed out above, it will be very helpful in understanding how to figure this out.
     
  20. allenchuah

    New Member

    Sep 2, 2008
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