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Battery help needed

  1. Feb 23, 2009 #1

    I am attaching the schematic of a circuit. I need to power up this entire circuit using batteries. Basically i need two sets of voltages, "+-12 V for the opamps and +-6 for the 50Kohm resistor in the schematic.(This is not my design). If anyone has an idea how can I do this I would be very much greatful if you can shareit with me. I would really appreciate if you can explain the power circuitry design please.

    Attached Files:

  2. jcsd
  3. Feb 23, 2009 #2


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    How long a battery life do you need, how much space can you use?

    For a bench experiemnt I would be tempted to just use 2 x 12V lead acid batteries back-back, small alarm batteries are cheap and store lots of power.
    For a portable device you are probably going to have to use a chargepump DCD-DC converter to generate +-12V from 4x1.5V batteries.

    The the 50K I would get from the +-12V, just add a couple of fixed resistors each side of the variable one to limit the range to +-6, it's going into an op-amp so it's not taking any current.
  4. Feb 23, 2009 #3

    hm.. The space is not a problem for me. I would atleast want the batteries to last for 45min to an hour. I have thought of using 2*12V batteries.. but a bit confused as it is the first time i am trying to run something like this using batteries.

    Can you explain a bit in detail.

  5. Feb 23, 2009 #4


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    I don't know how much power the op-amps are going to use (it should be on the spec sheet)
    To generate +- power from batteries (without any sort of extra DC:DC converter) just connect them in series and take the 0v from the middle.

    So +ve terminal of battery 1 = +12V
    -ve terminal of battery 1 = 0V, connect to +ve terminal of battery 2
    -ve terminal of battery 2 = -12V

    (this obviously still applies if each 12V battery is really a box of 8x1.5V batteries connected in series)
  6. Mar 6, 2009 #5
    I would consider using a single 12V battery, referenced to center ground (i.e. +/- 6V), and a voltage doubler to supply the +/- 12V. Depending on total current draw, perhaps a small lead-acid battery (13.6V at full charge) and a low drop-out regulator. Circuits and chips readily searchable on the web.
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