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Is this homemade battery charger circuit dangerous?

Discussion in 'Electronic Basics' started by Kanon Kubose, Jul 17, 2013.

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  1. Kanon Kubose

    Kanon Kubose Guest

    The other day I found plans for a $3 battery charger that was just a diode and light bulb in series with the battery, using mains electricity. I posted a question about modifying it with a dimmer switch here:!topic/

    Everyone there said I was a candidate for a darwin award and that I should come to sci.electronics.basics before I kill myself.

    Taking their advice to heart, I have not built this darwin battery charger,but I do still have batteries to charge, namely a 7.2v nicd battery pack for a jigsaw. I found an 8.7vdc 360mA wall wart, and by putting a 12v car heater/fan in series with the battery, I got the amps low enough to trickle charge it. However, I suspect the 8.7v isn't enough, and also, as the chargegoes on, the current falls. After a few hours, I measured C/24, and overnight it was C/60. I've read that nicads need at least C/10 to fully charge (though another source says C/16 will work), so this wall wart seems to be insufficient. (Strangely, when the current reached C/24 and C/60, removing the heater didn't affect the amp rate at all. I wonder why?)

    My new idea is to take a 24vac wall wart that I have and add a diode to remove reverse current, then put it in series with the battery and some current-limiting load. This is essentially the same $3 dangerous battery charger except using 24vac instead of 110 mains voltage. Is it still dangerous? People in the other group said even 25 mA is lethal if it crosses the heart, like would happen if you stupidly grabbed both leads of unshielded alligatorclips, and it doesn't matter that the current is limited. It's not limitedenough, especially when the lightbulb begins cold.

    Other questions, assuming this circuit isn't dangerous:

    1. Could I use 12v appliances or light bulbs as the current limiter or willthe 24v burn them out? Since it's attached to a 7.2v battery, the voltage will be lower, right?

    2. Would a dimmer switch work to make it finely tuneable? That way I could monitor it and increase it as it falls, keeping the C/10 rate.


  2. We purchased a number of cheap cordless drills at work. The chargers
    that come with them are nothing more than a wall wart, diode and
    resistor. They work for a while, but eventually the batteries all died,
    some of them simply would not take a charge, others got hot and melted
    while in the charger.

    Save yourself a lot of grief and just purchase a proper charger.
  3. There is nothing wrong with the theory, but I agree with those that
    say it is too dangerous.

    When the battery is not connected, there will be full mains voltage on
    the connector, since the light bulb has much lower resistance than
    your body.
    C/10 will charge it in about 14 hours. C/20 will charge it in about 28
    hours, and so on.
    A fully charged NiCd cell can easily reach 1.4V. In this case, there
    simply is so little voltage difference between the battery and the
    power supply, that, even if there is almost no resistance, only a very
    small current flows.
    24V is very safe. Most people won't even feel anything if they touch
    the poles directly to the skin. On your tongue, it will sting a bit,
    since your tongue is wet and has very thin skin.
    Ohm's law says current is determined by two factors, voltage and
    resistance. More voltage gives more current, while less resistance
    gives more current. Your body (the skin, mostly) has quite a high
    resistance, so the 24V will not be able to drive a current that is
    high enough to be dangerous.
    Appliances are not good. They will have a very unpredictable
    resistance. A brushed motor, for example, may have a different
    resistance depending on where in its commutation cycle it is parked.

    Light bulbs, on the other hand, are excellent for this application.
    They have a strong current regulating effect. If you increase the
    voltage, the filament heats up and the resistance increases, so the
    current will not increase as much as the voltage increase would have
    caused in a linear resistor. Here's an illustration:

    In fact, I used to have a commercial battery charger that did use
    light bulbs as the regulating element. 20-30 years ago, they were
    quite common.

    Since you are using a half wave rectifier, the 12V bulbs will probably
    be OK. If you sprung for another three diodes and made a full bridge
    rectifier, the 12V bulbs may be overstressed. When your battery is
    fully discharged, it will probably be about 6V. That would leave 18V
    for light bulb.
    Yes, a dimmer should work. However, if you can find the right bulb,
    you will not need to adjust the current during a charge cycle. It will
    remain plenty constant enough.

    Note, however, that this type if charger is fully manual. It will keep
    charging the battery until you disconnect it. In order to know how
    long to charge, you must know the battery's initial charge state,
    which in practice means that you must fully discharge the battery
    before you can charge it.

    Personally, I'd consider buying an automatic charger. It is
    tremendously convenient:

    - You can charge the battery without knowing its charge state in

    - You do not need to remember to stop the charging at the correct

    Such a charger does not have to be expensive. Look for the term "delta
    peak". That indicates the automatic charge state detection. Here's an

    You must, of course, pick one that is suitable for your specific
  4. ehsjr

    ehsjr Guest

    On 7/17/2013 2:40 PM, Kanon Kubose wrote:

    Because the closer to fully charged, the lower the current
    the battery will draw, as you observed. The lower the current
    drawn through your heater, the lower the affect the heater
    will have on the circuit. The effect the heater has can be
    computed by V = I * R where V is the voltage dropped in the
    heater, I is the current drawn through it, and R is the
    resistance of the heater.

    Your circuit looks like this:

    8.7 Vin----[Heater]---+
    Gnd --------------+

    As the charge continues, the NiCd voltage rises, so the voltage
    drop across the heater decreases. If the pack voltage reaches 8.7 volts,
    there is no current drawn, and no voltage drop across the heater. It
    would make no difference if the heater was in the circuit
    or not.

    Also, measuring small currents presents a challenge, even to
    expensive multimeters. When used in series with the circuit
    to measure current, they cause a voltage drop that affects
    circuit performance. That can sometimes cause a significant
    error in the measurement.

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