Series/Parallel Battery Question

Discussion in 'The Projects Forum' started by N0ctrnl, Jan 15, 2010.

  1. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    Hello all!

    I have a question I haven't yet been able to find the answer to. I currently have 4 - 1.2v NiMH batteries wired in a series/parallel configuration (configured for 3.6v). I essentially have 3 of them wired in series with a fourth wired parallel with one of them. It looks like this:

    [​IMG]

    Now, clearly this gives an output of 3.6v. When I put a 3.6v load on this, what rate will each of the batteries drain at? Will the 2 non-parallel batteries end up dead while each of the ones in parallel still have 50% charge? Will the singles drain slower because of the larger output capacity of the parallel wired batteries?

    Thank you very much for your help. I've looked for the answer to this question, but I simply haven't been able to find a setup exactly like what I've got going.
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Basically, you are thinking about it correctly. It is unlikely that the parallel batteries will have exactly 50% charge, but it will be in that ball park. The parallel batteries will drain at about half the total load current. However, the higher current in the series batteries will reduce their effective capacacity just a little. I'd expect the end result to be that the actual remaining charge in the parallel batteries will be just a little bit more than 50 % because the series batteries will be draining a little more than twice as fast (relative to the available capacity at that current) as the parallel batteries.

    Generally it is not recommened to do this kind of thing. It is safer to use the exact same battery type in a symmetrical arrangement of parallel and series cells.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You wrote "...a 3.6v load..." - that does not make sense.
    It may be a load that is rated for a 3.6v supply.

    What you need to tell us is what is the wattage of the load, or the current requirement of the load when supplied with 3.6v.

    The batteries in parallel will discharge at roughly half the rate of the other two single batteries, so they will still have roughly half their charge remaining when the other two are completely discharged.

    [eta]
    Sorry Steveb, we cross-posted. ;)
     
  4. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    Thanks for the replies. I'm glad to see what I was thinking is actually mostly correct.

    My situation is this; The load is a string of parallel LEDs. They are rated at various voltages ranging from 2.0v to 3.6v. Naturally I have resistors on all of them to compensate for the higher voltage. For the sake of this discussion, I'll just say there are 10 of them rated at 3.2v (22-ohm resistor) and 20mA.

    If I were to wire all 4 batteries in series (bringing the source to 4.8v) and increase my resistor value for each LED to compensate, will I still end up with an increased battery life or will the additional resistor load essentially negate that?

    I know some of these questions are probably pretty simple, but for some reason I just can't seem to wrap my mind around this like I used to.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    If you have 10 LEDs in parallel, and want to run 20mA through them, your total load current will be 200mA.

    Your batteries are rated for 2300mAh, but that is at a 20 hour rate of discharge, or 115mA. Your 200mA load will cause the batteries to discharge much more quickly than 11.5 hours, as more power will be dissipated across the internal resistance of the batteries.

    The resistors will dissipate more power, but the light output from the LEDs will be more even over the life of the batteries.

    If you want the batteries to last a lot longer, you might consider reducing the number of LEDs in parallel, and/or decreasing the number of LEDs in your circuit.

    If you wish for more efficiency, you can use a boost-type LED driver. There are devices like National's LM3421/LM3423 that in conjunction with a number of external components, can take the output from your four batteries and boost the voltage up high enough so that you can run them in series. It's more complex than simply using limiting resistors though.
     
  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    No matter how many batteries you stack in series, the total current through all batteries will be the same. Hence, the batteries will drain in the same amount of time.

    The only way around this is to use a DC/DC converter which would convert scale both voltage and current to maintain equal power. However, there is an efficiency loss and this is much more complex as well.

    However, keep in mind that using 4 batteries with a larger current limiting resistor will result in a more stable control of current. LED properties change with temperature, and higher voltage will stabilize current changes.
     
  7. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    Ok, so, any suggestions for a circuit to use? I could have up to 40 LEDs in the parallel string. Space is an issue, but I can probably work that out depending on the complexity of the circuit.

    I'm guessing I'll make them symmetric and go with a 4 battery series/parallel combination at 2.4 volts. The space constraints just won't allow me to go series.

    Thanks!
     
  8. Redbelly98

    New Member

    Jan 16, 2010
    5
    0
    What about the 3.6V LED's in your circuit? They'll never light if your source is only 2.4, unless you use a boost converter like SgtWookie suggested.

    I don't understand how space constraints would prevent series wiring. If you can fit 4 batteries in the space, it's all in how you run the wires, which take up very little space.
     
  9. N0ctrnl

    Thread Starter Member

    Aug 14, 2009
    14
    0
    I guess it's not so much the space as it is the logistics of wiring it that way. And yes, going with 2.4v assumed the use of a voltage stepping circuit, as I said.

    Every circuit I've found that looks workable just won't support the load I need. I could conceivably have up to 1.2A load.
     
Loading...