+/-12V from 9V battery?

Discussion in 'The Projects Forum' started by GRNDPNDR, Apr 3, 2012.

  1. GRNDPNDR

    Thread Starter Member

    Mar 1, 2012
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    to keep it short, what I'm hoping to achieve here is to use a single 9V battery and get positive and negative voltages around 12VDC.

    Is this possible? cheap? without using a MAX1044?
     
  2. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    What current load?
     
  3. GRNDPNDR

    Thread Starter Member

    Mar 1, 2012
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    I don't honestly know, I haven't done the math on it yet TBH. it's for a 2xLM324 circuit using all 8 op-amps as non-inverting amplifiers for a piezo transducer.

    It will also provide power to;
    an LM386/LM555 that converts a piezo transducer signal to an on/off signal.
    a D3009 reed relay
    probably 10 LEDs (they flash when the piezo produces a signal so not always on)
    a few transistors (maybe 10)
    1 diode and about 35 resistors, mostly 100K pots.
     
  4. rfredel

    New Member

    Jan 23, 2011
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  5. GRNDPNDR

    Thread Starter Member

    Mar 1, 2012
    435
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    I don't think that's quite what I'm looking for, I don't even know where to begin using that thing.

    Besides I don't think this circuit is really loaded perse, it's a signal amplifier for the most part.
     
  6. Audioguru

    New Member

    Dec 20, 2007
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    A little 9V battery has a very low capacity. If you use a boost-the-voltage circuit then its current will be high and its capacity will be 3 times less than if the circuit is designed for 9V instead of plus and minus 12V.
     
  7. GRNDPNDR

    Thread Starter Member

    Mar 1, 2012
    435
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    Ya I'm kind of stuck between a rock and a hard place, I want to give the op-amps a higher voltage BUT would like to avoid using a wall-wart and keep costs down.

    I was considering some type of battery pack and small recharging circuit so it could just be charged or run from the wall, but again it comes down to cost, battery packs are on the expensive side.....and large.

    On the other hand the circuit isn't doing a lot so maybe the cost of a battery pack wouldn't be too bad.

    but then how do I build a charging circuit easily that will run from the wall or battery.

    MOST IMPORTANTLY, how can I get dual polaritys from a single polarity supply? that's what I'm really after here. I can work out the rest later.
     
    Last edited: Apr 3, 2012
  8. Audioguru

    New Member

    Dec 20, 2007
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    ANY circuit that uses a dual-polarity supply can be simply modified to use a single supply.
    Why do you need 24V?
    Why don't you post the schematic because now we do not know anything about it.
     
  9. BillB3857

    Senior Member

    Feb 28, 2009
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    Since +/- 9V is well within the operating range of the LM324, why not use 2X9V batteries connected in series with the center tie representing ground?
     
  10. GRNDPNDR

    Thread Starter Member

    Mar 1, 2012
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    That's basicallt what I'm doing now. but for a final product I would like to reduce it to a single 9V battery. I was hoping for 12V due to the number of items being powered by the supply as a whole.
     
  11. BillB3857

    Senior Member

    Feb 28, 2009
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    Do the items "being powered by the supply as a whole" require 12V? If not, the number of devices being powered will relate to the current requirement (which as already stated isn't much from a 9V battery). As for keeping the cost down by using batteries instead of a wa-wart, have you thought about the on-going cost of constant replacement of the batteries? If this device is to be used for more than the life of 10 batteries, a wal-wart will start to look like a bargain compared to the cost of replacement batteries.
     
  12. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    I would do a simple boost circuit from the 9V battery up to 24V then stack a pair of 12V zeners across the 24V. Use the center node for circuit ground and you have +/-12V supplies.

    An alkaline 9V has maybe 500mA-hr capacity. Boosted to 24V (with losses) it drops to maybe 150mA-hr. If your circuit draws 30 mA, you get maybe 5 hours life. Not great.
     
  13. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    A wall wart costs about $2, the price of about one 9V battery. I would use the wall wart unless the device must be portable.
     
    Last edited: Apr 4, 2012
  14. GRNDPNDR

    Thread Starter Member

    Mar 1, 2012
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    I have a bunch of warts, but its more a concern of practicality. there is nowhere close or easily accessible to plug this in each time I move my drumset over to play.

    The devices do not need 12V because it's mostly op-amps (there will be a reed relay) BUT they seem to function better with 12V.

    The plan WAS to use 2x9V batteries and LM7809/7909 to regulate the voltage to keep the rails "balanced" . The LM's would work better with 12V as well.

    Reducing it to a single battery is again a matter of convenience.

    Dying batteries? this is where the rechargable batteries come in. Another part of "my grand master plan" was to use (hopefully) a single 9V rechargable, and have the ability to power this whole unit from the single battery but to build a charging circuit that will charge the battery AND give the ability to run from wall power.

    As I am in the design/research stage this is why I'm trying to explore all of my options within my abilities and comprehension.

    I'm trying to find a comprimise that;

    A) I can build, because I lack the ability and understanding to design from scratch

    B) will minimize the amount of components in the circuit/minimize complexity unless it's absolutly needed

    C) Will keep costs as low as possible, there is a chance I could retail this device I'm building, so I would like to keep my costs low because if someone did want to buy one of these, I want to be able to offer the lowest price possible (not trying to sound like a sales guy here cause I'm really not)

    This design is for myself, so I want it to work correctly, and I will be offering up the final WORKING design for free to anyone, but in the chance that someone wants to pay me to build one of these for them I would like to be the lowest price (I dont charge for my labour on a hobby level, I just enjoy doing it)
     
  15. nemuikuma

    New Member

    Apr 4, 2012
    7
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    There are multiple ways to do this, but the single battery approach is the least convenient.

    You can easily do a voltage multiplier with a 555 IC, you can get regulated +/-12, but it will waste a LOT of energy and need relatively lot of parts also, not very efficient so a battery won't last long and it won't be able to supply a lot of current either.

    The best, most efficient method is to use 2 9 volt batteries and you already have +/- 9 volt without any components (or 4 batteries to get +/- 18 volts)
    An extra 9 volt battery will need about the same space as the extra circuit needed for voltage conversion, or even less (unless you are working with SMD)
    [​IMG]

    I advise against voltage regulators on battery operated circuits. because the are "burning up" excessive voltage, wasting energy on heating.
    The most important role of the voltage regulator is not to regulate the volt but to filter out AC ripple, and since you are operating from battery you don't have any AC ripple.

    So for opamps,you don't need to regulate voltages from batteries.
     
  16. GRNDPNDR

    Thread Starter Member

    Mar 1, 2012
    435
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    I'm pickin' up what you're layin' down :)

    Maybe I'll just stick with 2x9v batteries, but I wanted to find a solution to the problem of the battieries getting low and causing the circuit to go nuts.

    The biggest reason for the regulators ( forgot this lastnight) was to simply allow the circuit to be on or off. When the batteries got too low the regulators would cut power, period.

    Also the reason for the bypass so that when it did die, one could complete the song they were playing without interupption or the circuit going nuts because the voltage is too low.
     
  17. Audioguru

    New Member

    Dec 20, 2007
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    It sounds like you have way too much circuitry to be powered by little 9V rechargeable batteries. Each charge might last for only an hour with the voltage dropping all the time.
    Here is a graph of the voltage dropping on an Energizer 9V Ni-MH rechargeable battery at various currents:
     
  18. Audioguru

    New Member

    Dec 20, 2007
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    No.
    A voltage regulator does not cut off the power, it simply stops regulating and allows the reduced voltage to bounce all over the place.
    Your 7809 and 7909 regulators WILL NOT WORK from a 9V battery anyway because their minimum input is 11V. With 9V as the input then the output is about 7V. When the 9V battery drops to 6V then the output of the regulator will be about 4V.

    With 9V batteries you need low-dropout 5V regulators that still produce a regulated 5V when the battery drops to 6V.
     
  19. nemuikuma

    New Member

    Apr 4, 2012
    7
    0
    one solution maybe is to make an indicator!
    The simplest is to put 4 red LED in series with a 270ohm resistor it will consume very little current about 2ma (even less with a 500 Ohm resistor) and it will go dark is the voltage drops below about 8.5 volts.
    So it will be still a few minutes before the battery dies completely
    The color is important because different colors have different forward voltages, for red is about 1.8-2 volt.
    I never tried it, but in theory it should work!
    But there are many kinds of circuits that gives you feedback about the voltage.

    A regulator won't shut off, it will be always a few volts lower than the input!
     
  20. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    No, they don't unless you add some level detecting comparators and also have IC regulators that have ON/OFF pin capability.
     
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