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keithg

New Member
I have made a three phase, permanent magnet generator for hand cranking to charge 4, 6, and 12 volt lead acid batteries. Using 22 ga wire, I assumed that, given high rpm, charging would not be a problem. All three batteries behave in the same manner; the generator quickly charges (5 min) beyond their relevant voltage but quickly dissipate under an LED load or any other for that matter, amounting to a 1:2 ratio of cranking/using the battery. Would this suggest that I am not cranking long enough? Not enough current? The generator easily rises to the proper voltage with .60 amps charging the six volt. I naively anticipated a much longer service time, just as if the battery was charged from mains. The six volt battery is rated 3 Ah. If I were to regulate the current with an LM317, how many miliamps are appropriate? Thank you for any advise.

MikeMl

Well-Known Member
Sounds to me that your batteries are so dead that they are not taking a charge. Measure the resting (open circuit) voltage after your batteries have been disconnected from any charger or loads for 12 hours. What do you read? (you need an accurate digital voltmeter for this reading).

smanches

New Member
It could also be that your generator is creating too high of a voltage with not enough current. What is the no load voltage on the generator?

Hero999

Banned
I'm with Mike here, it sounds like the batteries are dead.

larrybeaty

New Member
As MikeML said, measure the voltage some time after charge. If the measured voltage is less than 0.8 of the stated battery voltage, it is on the wrong side of use.

A good battery will hold the voltage down until the chemistry is complete.

To be safe, charge an unknown battery at 1/10 C. C is the Amp-hour rating of the battery. Liquid filled lead/acid batteries can take a much higher charge current. Other battery types do not do well with high charge rates.

Y'all be careful, they can explode if not careful.

Larry

keithg

New Member
Intermittent power up here so answering messages is not always possible. The generator is doing 80v dc without a load. I would like to reduce this to a workable 16 or so with more current. Is a step-down transformer the answer?

larrybeaty

New Member
Voltage measured without load us a useless measurement. Is your voltmeter measuring spikes and peaks? Is it measureing RMS? Load the generator and measure with an RMS meter. Now you can have some faith in the measurement. Yeah, it is a little more work, but good engineering takes work. Load the generator with some 120 light bulbs - incandescent variety. Or if the generator can take it, a heater.

Now, you said it was a DC generator. If you are to keep most of the power, you must step it down. This involves turning the DC to AC, put the AC to a transformer, step it down, then turning the AC to DC.

Is it in the realm of possibility of getting a 12 Volt generator/alternator?

Larry

keithg

New Member
Sorry, I should have said "AC" generator, not DC. It will light a bulb. I'll take a reading with this load and send the results. Thanks for the advise. Also, you mention "wrong side of use." I am inexperienced and don't understand the phrase. Would you please explain?

larrybeaty

New Member
Sorry, that assumed you had a background of American English. My apologies.

Battery use is from 0.8% of normal voltage up. Lower than that and use is poor. In this respece I set "use" at 0.8%. A better term would have been "critical voltage".

Larry

keithg

New Member
I am using a battery charge circuit file:///Users/keithgum1/Desktop/lead%20acid%20battery%20charger/Battery%20charger.webarchive to charge a 12v 3Ah battery. Current is only .05 to the battery using .50 ohm on R1. If all is correct, shouldn't the circuit be making 1 amp?

MikeMl

Well-Known Member
I am using a battery charge circuit file:///Users/keithgum1/Desktop/lead%20acid%20battery%20charger/Battery%20charger.webarchive to charge a 12v 3Ah battery. Current is only .05 to the battery using .50 ohm on R1. If all is correct, shouldn't the circuit be making 1 amp?

Assuming that it is a regulated, constant-voltage, intrinsically-current-limited type of charger, the charging current into a partially discharged battery should be initially be limited by the charger itself, and then as the battery accumulates charge and reaches a terminal voltage which is the same as the open-circuit voltage of the charger, the current will begin to taper, eventually reaching a steady-state value which might be between a few tens of mA to low hundreds of mA, depending on the charger voltage setting, the size and age of the battery, and the temperature.

larrybeaty

New Member
As Mike M. says. However, we are assuming the battery charge circuit has sufficient high voltage to charge the battery. If the charger has only 12 volts available that is not enough voltage to charge the battery. We would hope the internal voltage of the battery charger is more like 17 Volts.
Larry

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