Battery Capacity

Discussion in 'The Projects Forum' started by blah2222, Oct 31, 2013.

  1. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
    554
    33
    Hi all,

    In short, I am working on a device that is to take a pulse train from a microcontroller output (Duty Cycle < 0.5% @ 5V, 30 Hz), give it a current boost using a Darlington pair, and increase the signal of the voltage by a factor of 12 with a step-up transformer.

    In order to get an output on-time current pulse of around 40 mA it would mean that my Darlington pair current through the transformer primary windows would need to have an on-time value of around 500 mA.

    Ideally I would like to have this unit battery powered, and am considering using series AAs or a 9V, but not sure if they can handle the load of the Darlington pairs for a significant duration of use. The device is only operating for around 3 seconds.

    From wikipedia it mentions that the RMS value of a pulse train is:

    <br />
I_{DPrms} = I_{pk}\sqrt{Duty Cycle} = 500mA\sqrt{0.005} ~= 35 mA<br />

    That seems like a relatively small amount for the battery to provide, but I feel as though the 500 mA draw might be too much for the battery to take.

    Hope that someone might be able to lend a hand, for deciding on a power supply method.

    Thanks,
    JP
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    when dealing with bursts or sudden demand on power, general strategy is to use capacitor (just put it in parallel with your circuit). so when there is a sudden demand, most of the current (at least in the instant) will come from capacitor and smaller part from battery. when demand (short pulse) is gone, capacitor will slowly charge up. so capacitor is doing the stressful task (ESR is a measure of capacitors ability to suddenly change charge) while battery sees more or less constant load (or one with comparatively little change). this means that average battery current is considerably smaller than peak current demanded by circuit.

    9V battery is really weak, particularly for loads exceeding some 10mA, but it may be reasonable choice for this application since it operates only briefly.

    I would not use Darlington, they have large Vce when on (1.1-1.5V or more). if you use two transistors but not connect collectors together, you can get better switch and Vce of only 0.1V or so.

    the larger Vce, the less voltage across load, and lower efficiency.
     
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  3. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
    554
    33
    Thank you for the reply.

    What is the configuration for connecting twos BJTs to my transformer's primary, so that I can obtain 500 mA peak for the pulse?

    Cheers,
    JP
     
  4. wayneh

    Expert

    Sep 9, 2010
    12,153
    3,058
    You might look into using a MOSFET instead of a darlington. It doesn't require any base current to operate. (OK, a tiny bit at low Hz.)

    The only concern would be gate voltage - you have to be sure to fully turn the gate on. A normal MOSFET needs ~10V but logic level MOSFETs are available that turn fully on well below 5V.

    Put the N-channel MOSFET as the last step in the path to ground for the primary current. Gate to signal, source pin to ground, drain pin to the transformer. This is a standard "low side switch".
     
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  5. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
    554
    33
    Are these MOSFETs rated for ~500 mA and do you think a battery can hold up to this kind of pulse switching?

    Thanks!
     
  6. wayneh

    Expert

    Sep 9, 2010
    12,153
    3,058
    Yes and yes. The MOSFET I use for generic chores (IRF540N) is rated to 33A. You obviously don't need all that but you'll have no problem finding what you need.

    As mentioned, use a capacitor to handle the pulsing. I'd calculate it out so that the cap holds at least about 90% (wild guess) of the burst energy, so that the battery is not expected to contribute much of it.

    I built a TENS device that does something very similar to what you're doing. Let me see if I can find the schematic for you.

    [edit] Here it is. Look at the lower right hand side of the schematic. It shows how a clock pulse is used to spike the gate of a MOSFET to allow current from a 9V battery to pulse a transformer where it is stepped up to over 100V.
     
    Last edited: Nov 1, 2013
    blah2222 likes this.
  7. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
    554
    33
    This is exactly what I was looking for!

    Thank you.
    JP
     
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