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12v battery protector circuit

Discussion in 'Electronic Basics' started by andy, Aug 26, 2004.

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  1. andy

    andy Guest

    I'm looking for a circuit that will protect a lead acid battery against
    excessive discharge - i.e. cut off the supply to the main circuit when the
    voltage from the battery goes below about 11V. Ideally, once the circuit
    has switched off, the protection circuit shouldn't use any current itself,
    either, which is the tricky bit. The best I've come up with so far is:


    Vin o--------------------o------------------------------o
    |
    |<
    +------|
    | |\
    R1 .-. | Vg
    100k | | o---------+
    | | | |
    '-' .-. |
    | | | R2 |
    | | |100k |
    - '-' |
    zener, 9.5v ^ | |
    +--------o |
    | |
    .-. |
    | | R3 |
    | | 3.3k |
    '-' ===
    | |^|
    Gns o-------------------o---------+|+-----------------o

    I built it without the mosfet, and it does work - the circuit switches
    right off at Vin=~10V, and Vg goes high at Vin=~10.5V, which would switch
    the mosfet on. But there's still the 0.5V gap between the two, and it's
    not very adjustable - is there a better way of doing this?
     
  2. andy

    andy Guest

    P.S. R1 is just to stop the base current going too high when the
    transistor saturates - it doesn't have much effect on the switching (with
    Vg=~10V, the voltage across R1 was about 0.4 V)
     
  3. Gene

    Gene Guest

    What about a low-power comparator (always connected to the battery) with a
    suitable hysteresis whose output drives a NPN transistor which de/energizes
    a relay?
    Gene
     
  4. Rich Grise

    Rich Grise Guest

    What's wrong with 0.5V hysteresis, especially in a low voltage cutout? Or
    do you mean that you need more?

    Thanks,
    Rich
     
  5. andy

    andy Guest

    It's not hysteresis - the mosfet having switched on or off doesn't make
    any different to the switching point of the transistor circuit. A bit of
    hysteresis would be good, but that's not what's happening - it's that the
    mosfet and hence the main circuit will switch off at Vin=~10.5V, but the
    transistor circuit won't switch right off until Vin drops to ~10V. It
    would probably do the job OK (in practice, switching off the main circuit
    would probably be enough, because the battery would then get charged back
    up from the solar panel on the next day's sunshine), but I'm wondering if
    there's a better way, where the whole thing switches off at a single
    cutoff voltage. (Ideally with a bit of hysteresis to stop it dithering or
    browning out).
     
  6. andy

    andy Guest

    Could do, but I'm still looking for a way where there's no current draw at
    all apart from leakage once the circuit has turned off.
     
  7. Gene

    Gene Guest

    I dare to say it is extremely difficult, because once the circuit is turned
    off, one needs some form of energy to turn it on. This "energy" has to come
    from somewhere, either external to the system (external battery, manual
    reset, etc.) or internal. In any case, micro-power comparators or OpAmps
    draw ~50 microA, this would be the only device, together with the voltage
    sensing and reference circuit, that uses power from the battery in the
    off-status. Probably you could achieve confortably less than 100microA
    "leakage". (BTW, is cost an issue?)
    Gene
     
  8. John Fields

    John Fields Guest

    ---
    Yes, check "Low voltage cutoff circuit" on
    alt.binaries.schematics.electronic. Jim Thompson and I have both
    posted circuits which should work.

    On mine you'll have to change R4 to 12.1k because of your 11V cutoff
    requirement and on Jim's, you'll need to change R3 to 2.21k.
     
  9. Andy, your current circuit draws current because the BJT applies some gain to
    the small leakage current through the zener (perhaps 100X or so) and this is
    then allowed to flow through the collector. The zener leakage is probably
    something like a microamp or so, so this suggests something like 100uA of
    consumption for your circuit.

    Frankly, I'm not even sure what your circuit is supposed to be doing. Those
    100k resistors seem very big. For one thing, zeners are spec'd to operate at
    much higher currents than you'll get with your 100k resistor. For another, your
    zener will leak current to the base of the BJT and the BJT will apply gain to it
    to yield collector currents that are some 100X higher.

    In the case where the zener hasn't yet reached its zener voltage, it will just
    leak at say, 1uA. Thus, your Vg will be:

    Izener ~= 1uA (rough guess)
    V(R1) = Izener * R1 ~= 0.1V (negligible)
    so, Ib(BJT) ~= Izener, Ic(BJT) = beta * Ib(BJT) ~= 100uA (estimate, for now)

    With an Ic of 100uA and an (R2+R3) of slightly over 100k, you'd expect to see
    about 10V across them. Since we already know as a rough guess that the voltage
    supply is below the zener voltage and thus below 10V, we can be pretty sure that
    V(CE) is small and that the transistor is probably saturated (reducing the beta
    so that the actual Ic matches up with (R2+R3) to yield about the supply voltage.

    So, in effect Vg would basically just track your supply voltage before the zener
    voltage kicks in, placing the supply voltage directly on your MOSFET gate.
    Mostly because of that huge R2 value, though.

    But what happens after the zener has kicked in? Well, Ib will be even higher of
    course. And V(CE) will just be that much less (not much less, because it was
    already saturated to begin with.) So Vg would be about the same then, too.

    So you might as well just hook up the supply to Vg and be done with it.

    This all changes some if you make R2 a smaller value. I'm not sure if you are
    telling us correctly about the 100k value on R2, or not. But if R2 is a lot
    smaller than 100k, then the case before breakdown and the case after breakdown
    look different, which is probably what you are after. Then Vg is held down to a
    low value by the much smaller R2 and R3 against the relatively smaller looking
    Ic=beta*(Ib=leakage) until the zener breakdown when the Ib rises (although
    again, I suspect your R1 should be something less than 100k as even these tiny
    Ib currents present a real voltage drop across something that big and your zener
    really needs more substantial currents after breakdown, anyway) and then more
    substantial Ic values can begin to work against R2+R3 and cause Vg to rise up
    towards the supply voltage.

    But any reasonable way you work it, your Ic is going to be at least beta times
    the zener leakage and that's going to set the total leakage of your circuit,
    when off.

    I also think Gene's point about having to have some kind of comparison circuit
    operating, even when the MOSFET is disabled, is true enough. You'll need
    something running. But I believe you can get this down to about a microamp or
    two with predictable performance, with some careful design. (I'm not a good or
    professional designer, but I do have an idea how this might be achieved.)

    Jon
     
  10. chuck h

    chuck h Guest

    www.homepower.com/files/lvdhp60.pdf
     
  11. Yes, at:



    and,



    I just looked.
    Both of these appear to require a switch to turn them back on, though, since
    power to the rest of the circuit is blocked by both of these once the source to
    gate resistor takes over and holds the MOSFET off. I'm guessing, but Andy is
    looking for something that does this automatically when the voltage rises *AND*
    where it has very low Iq when the power is disconnected from the load.

    Jon
     
  12. andy

    andy Guest

    I hadn't thought of that, I admit.
    I was trying to keep the current down as much as possible - the main
    circuit only uses 0.8 mA, so I'd like this part to use maybe up to 0.2 or
    0.3mA when it's on, and under 10 uA when it's off.

    The idea of it is -

    - when Vin is below Vzener plus 0.7 volts, there's no current except for
    leakage, because both devices are off (but I forgot about this being
    amplified, as you said).
    - when Vin goes above Vz+0.7, then current will start to flow in the
    zener. This will create a collector current, which will start to increase
    the voltage across R2 and R3. R3 is meant to give a negative feedback - as
    the collector current increases, this will push the base voltage on the
    transistor closer to Vin again, tending to switch it off.
    - so Vg will rise until the transistor saturates. The maths I worked out
    has this happening when Vin=(Vz+0.7)*(R2+R3)/R2
    - then the voltage across R1 starts to rise, which stops the base current
    going too high after the transistor has saturated.
    It didn't do that, honest. I'll try building again to be sure.
    The idea of that circuit was that you don't, because it's relying on the
    transistor and zener shutting off when Vin goes below their combined
    switch off voltage. But I hadn't thought about the transistor amplifying
    the leakage, as you've said.
    I did build it as in the diagram, and it worked as I said with those
    resistor values - Vg stuck at 0V for Vin = 0 to ~10V, and swung up to near
    Vin as Vin went from ~10 to ~10.5V. I probably am using the devices
    outside the specs though, so maybe I was just lucky.
     
  13. andy

    andy Guest

    That is what I'm looking for. I'll have a proper look at the circuits and
    the other comments when my head's a bit clearer, and see if I can come up
    with something that does what I want. The TL431 looks like a
    possibly-useful component apart from the 1mA minimum current - The whole
    circuit only uses 0.8 mA, so this is a bit high for what I want.
     
  14. andy

    andy Guest

    Thanks - I finally found a free news server that carries that group
    (news.tehnicom.net). What I want is a circuit that doesn't need resetting,
    so I'll have a proper look at both those designs when my head's a bit
    clearer, and see if I can come up with something.

    R3 is meant to be a pot, like you said in your reply on abse - I just
    tested with a static resistor because it was easier.
     
  15. How about using something a bit more straightforward:

    P-MOSFET Q3
    -----------o---------o------------o-------+^+-----------
    VIN | | | ||| VOUT
    .-. | .-. ===
    | |47k | | |220k |
    | |R1 | | |R4 |
    '-' | '-' |
    | ___ |< | |
    o-|___|-| PNP o-------'
    | 100k |\ Q1 |
    | R2 | ___ |/
    | o---|___|--| NPN
    10V - | 47k |> Q2
    Zener ^ .-. R5 |
    D1 | | | |
    | | |220k |
    | '-'R3 |
    | | |
    | | |
    GND | | |
    -----------o---------o------------o----------------------

    D1 = 1N4740A (10V Zener)
    Q1 = 2N3906 (PNP)
    Q2 = 2N3904 (NPN)
    Q3 = IRF9Z34S for example

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    When VCC dips below Vzener + 0.7, Q1 turns off, making Q2 turn off and
    Q3 turn off!

    This circuit also has a fairly sharp knee, meaning it won't drag VOUT
    through an extended period of low voltage, it'll drop to zero fairly
    quickly. This is affected adversely by high impedance of the voltage
    source.

    It also uses very little current; at VIN = 15V, it uses around 600uA.
    At VIN = 10, it uses almost none.

    The zener should get more current to be accurate, but it doesn't
    really matter; it'll probably be within 5%. If you are worried about
    accuracy, you should instead use a voltage reference and a comparator.

    Regards,
    Bob Monsen
     
  16. John Fields

    John Fields Guest

    ---
    Try this:

    PCH
    BAT+>----+-------+-----+-----------------S D------+
    | | | G |
    | [845K] | | |
    [360K] | | | |
    | +----|-\ | |
    | | | >--------------+ | |
    +-------|----|+/ | | |
    | | U1A 4001 A--+ | |
    | | LMC6762 +--Y U2A | |
    | [22.6K] | B-----+ [LOAD]
    | | | | |
    | | U1B +--B | |
    +-------|----|-\ U2B Y-----+ |
    | | | >-----+--A | |
    | +----|+/ | | |
    [LM385-2.5] | | +--A | |
    | [226K] | | U2C Y-----+ |
    | | | +--B | |
    | | | | | |
    | | | +--A | |
    | | | | U2D Y-----+ |
    | | | +--B |
    | | | |
    BAT------+-------+-----+----------------------------+

    All of the resistors are +/- 1% except for the 360k, which is +/- 5%.

    The resistor string values were selected for a VBAT cutoff of <=11V
    and a turn-on of >=12V. You'll have to recalculate them if you want
    something else.

    Operating currents at 11V are: for the LM385, 24µA
    for the resistor string, 10µA
    for the comparator, 25µA
    for the 4001, 1µA
    for the MOSFET, 0
    -----
    For a total of ....................................... 60µA

    I haven't added any hysteresis because the latch makes it unnecessary,
    but if the circuit chatters and you find that objectionable, connect
    0.1µF ceramic caps across the inputs of each comparator.
     
  17. CFoley1064

    CFoley1064 Guest

    Subject: Re: 12v battery protector circuit
    Hi, Andy.
    You got very lucky. There have been an amazing number of really well
    thought-out answers to your original post, especially the two circuits posted
    by Mr. Thompson and Mr. Fields in binaries. If you start out with an
    inadequate problem description and keep adding to it as you go along, you're
    kind of wasting the time of the people who respond, as well as an opportunity
    to learn something. But...

    If you're going for fleapower both when on and when off, and you can't deal
    with a reset switch, look at the available art. There are a number of battery
    power management ICs, including some by Maxim. One IC that comes to mind is
    the High-Voltage, Low-Current Voltage Monitor in SOT Package made by Maxim,
    specifically the MAX6460. Of course, this is a surface mount part, but it
    should fill the bill here. (View in fixed font or M$ Notepad):

    Fleapower Undervoltage Lockout
    Vbat+ Vout
    o-----o--------------------o-----------o--+^+--------o
    | | |||
    | | | ===
    .-. | .-. |
    2M | | | 47K| | |
    | | | | | |
    '-' | '-' |
    | .----o------. | |
    .-. | Vcc | | |
    | |<-------------oIN- | | |
    25K | | | | | |
    '-' | OUTo----o----'
    | .-------oREF | |
    .-. | | | |
    | | .-. | | |
    220K | | | | | MAX6460 | |
    '-' | |1M | | .-.
    | '-' | | | |
    | | | | | |10M
    | .---o-------oIN+ | '-'
    | | | | | |
    | | .-. | GND | |
    | | | | '-----o-----' |
    Vbat- | | | |1M | |
    o-----o | '-' | |
    | | | | |
    === | === | |
    GND | GND | |
    | | |
    '-----------------o----------'

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    The IC has a micropower voltage reference and a comparator with open drain
    output which is optimized for this type of thing. One gotcha is that, since
    it's primarily made as an overvoltage switch, the output defaults to low at
    less than 4V, but by that time, your battery is cashed anyway, and 4V shouldn't
    be enough to turn on the MOSFET anyway, if you choose wisely. This circuit has
    current use of a shade more than 10uA when off, and when on will be dependent
    primarily on the pullup resistor. You can tweak that resistor if you can live
    with a slower turn-off.

    Note in the circuit the extra 10M feedback resistor. That will give you about
    a half volt of hysteresis, so your turn off should be about 10.5V, and once
    off, your turn-on will be at about 11V. That's important because, if you turn
    off the power to the load, the battery voltage will go up, which might lead to
    power cycling (and might anyway -- if it's a problem, tweak that resistor
    value. Speaking of which, the 25K pot will allow you to tweak shutdown point
    to just where you want it.

    (By the way, there aren't any small lead-acid batteries. You're being way too
    conservative on the off-state current requirement of 10uA. A standard circuit
    with an LM10 voltage comparator/micropower reference will draw only 270uA
    typical, and should be able to run for many hundreds or even thousands of hours
    on an 11V nearly discharged lead-acid battery before it goes to deep discharge.
    How many hours depends on the size of the battery. If you tweak shutoff
    voltage up a little, you can make that time even longer.

    By the way, make sure to add a good sized capacitor to Vbat at the IC to reduce
    load transients, which may cause false tripping.

    If you're going with the Maxim 6460 circuit, take a good look at the data sheet
    before you start. Then, if you don't like something, you can tweak it
    yourself...

    http://www.maxim-ic.com/quick_view2.cfm?qv_pk=3492

    Maxim has these in stock. You can order samples with a credit card and get
    them in next-day, if you want.

    Would you by chance have any more project revisions, sir?

    Chris
     
  18. John Fields

    John Fields Guest

    ---
    Oops... Miswired the NORs

    Should be :

    PCH
    BAT+>----+-------+-----+-----------------S D------+
    | | | G |
    | [845K] | | |
    [360K] | | | |
    | +----|-\ | |
    | | | >--------------+ | |
    +-------|----|+/ | | |
    | | U1A 4001 A--+ | |
    | | LMC6762 +--Y U2A | |
    | [22.6K] | B-----+ [LOAD]
    | | | | |
    | | U1B +--A | |
    +-------|----|-\ | U2B Y-----+ |
    | | | >--+--|--B | |
    | +----|+/ | | | |
    [LM385-2.5] | | | +--A | |
    | [226K] | | | U2C Y-----+ |
    | | | +--|--B | |
    | | | | | | |
    | | | | +--A | |
    | | | | U2D Y-----+ |
    | | | +-----B |
    | | | |
    BAT------+-------+-----+----------------------------+
     
  19. CFoley1064

    CFoley1064 Guest

    Subject: Re: 12v battery protector circuit
    OK. That circuit should do the job. I guess you've got a good idea of where
    you're going. A few comments:

    * There are few solar power problems that more panels can't help at least a
    little.

    * If this beast is going to be outside, you need to be careful about things.
    In the summer, overheating the battery can cause it to go flat. In winter and
    in conditions of fog or high humidity (you never get that on your side of the
    pond, right?), the possibility of condensation is going to wreak havoc with
    your best-laid plans of microamp current consumption with leakage currents
    across the surface of the board and components, unless you either hermetically
    seal the circuit from the atmosphere or apply some kind of sealant on the
    board. You might have the best luck with a gasketed enclosure which is
    shielded from direct sunlight, and is somewhere that convection cooling and any
    breeze will help. Along these lines, also make sure any wires enter/leave the
    enclosure through gasketed/sealed glands. Resist the temptation to use
    all-temp outdoor silicone sealant here - it corrodes wires.

    Good luck
    Chris
     
  20. John Fields

    John Fields Guest

    ---
    "That's looking promising. Thank you." would be so much nicer, don't
    you think?
    ---
    Maybe, maybe not. the LMC6762 has push-pull outputs, so if you need an
    open-collector output, you're out of luck. But, you could use an
    LMC6772. ISTR that I posted a circuit using it once but you blew it
    off for some reason. Oh, well...
    ---
    ---
    Jesus, Andy, do some of your own leg work for a change. It's easy
    enough to do, just point your browser around and find out who's got
    what in stock. Since National makes the part that would be the
    logical place to start, and here's what they have on it:

    http://www.national.com/search/psearch.cgi?keywords=LMC6762
     
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