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This is my first time posting in this forum. It was recommended to me by a friend. I have a background in mechanical engineering and am looking for some help regarding battery charging.

I have developed an electro-mechanical system that generates an AC signal that is 11.3V peak to peak with no load.

After AC to DC rectification, I am able to produce 5.29Vdc with no load into a small bank of capacitors.

I purchased a 3.7V 6000mAh Rechargeable Lithium Polymer Battery Pack on Amazon. When I first received it, it had a charge of 3.84Vdc.

I tried to charge it just a hair more with my system and could see that the peaks of the sinusoidal AC signal were chopped when connected to the battery with the measured peak to peak voltage reduced from 11.3V to 9.45V.

After roughly three hours with no increase in the measure battery voltage, I stopped.

I then discharged the battery down to 3.71Vdc. I also adjusted my system to generate an AC signal that is 10.2V peak to peak with no load.

After AC to DC rectification, I am producing 4.77Vdc with no load in the capacitor bank.

I then tried to charge the battery with my system. Again, I could see that the peaks of the sinusoidal AC signal were chopped when connected to the battery with the measured peak to peak voltage now 8.85V.

I was able to raise the battery voltage to 3.73Vdc, but no more.

It is probably worth noting that I have seven hand wound coils connected in series that have a total measured resistance of 312 ohms.

It is probably also worth noting that the AC signal has a frequency of ~5.5 Hz.

I’ve considered discharging the battery further and then trying to charge it back up to 3.73V, but fear that I might be missing something and ought to seek some help/guidance first.

Please let me know how to improve the charging system to raise the battery voltage closer to the value when received (3.84Vdc).

Thanks in advance.

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  • \$\begingroup\$ you cannot apply AC to a battery and expect it to charge \$\endgroup\$ Commented Dec 28, 2024 at 2:45
  • \$\begingroup\$ Understood. I have a full bridge rectifier for AC to DC and then am charging a bank of capacitors that are then connected to the battery. \$\endgroup\$ Commented Dec 28, 2024 at 2:52
  • \$\begingroup\$ Missing is the spec of your charger, but 3.7 is near 100% SoC. Measure the Vbat after a light load for 1 minute. \$\endgroup\$ Commented Dec 28, 2024 at 3:31

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One can simply use a dedicated lithium-ion charger module—after rectifying the low-frequency AC to DC—to provide a constant current (CC) until the battery reaches 4.2 V, then switch to a constant voltage (CV) stage that holds Vbat = 4.2 V. First, rectify the ~5.5 Hz AC signal (peak-to-peak voltage near 10–11.3 V unloaded) through a diode bridge and place a suitably large capacitor to smooth it. Let this create a DC rail, Vrail. Then feed Vrail into a standard LiPo charger IC that demands an input higher than the maximum battery voltage but typically less than 12 V. The charger imposes the necessary two-stage protocol: it limits the current ICC (e.g., ICC = m/hour if the battery capacity is m mAh, usually m/2 or less for longer battery life), and once the battery reaches 4.2 V, it provides the CV mode so that the charging current tapers to near zero. Mathematically, during CC, one has I = ICC and the battery voltage rises according to dVbat/dt = ICC/Cbat. In CV, the charger holds Vbat = 4.2 V and solves dI/dt ≈ -1/rint dVbat/dt → 0. A compact module such as a TP4056 (if the rectified rail is below 8 V) with a suitable pre-regulator) or an adjustable buck converter with integrated charger ensures correct charging without risking over-voltage or under-current, and that simple approach generally outperforms ad-hoc attempts to drive the battery directly.

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  • \$\begingroup\$ Thank you for the detailed response. If I understand you correctly, I would add the TP4056 module into my circuit downstream of the capacitors and prior to the battery? Do you have any good schematics that you would recommend on how to make use of the TP4056? \$\endgroup\$ Commented Dec 28, 2024 at 3:55
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With 312 ohm coil resistance you are limited to few mA charging current.

Considering 6000mAh battery it will take very long time to rise the battery voltage even a little.

Btw, a ~3.7V battery voltage is the point where battery “keeps” most of energy, so too much energy is needed for even 100mV rise.

Edit: enter image description here

Source of image: https://siliconlightworks.com/li-ion-voltage

At 3.5V you should see the voltage rise more obviously, at 3.2V more better.

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  • \$\begingroup\$ Thank you for the updated info. That helps! Do you have any suggestions on how to safely discharge the battery down to 3.5V or 3.2V? \$\endgroup\$ Commented Dec 28, 2024 at 4:35
  • \$\begingroup\$ Drain the battery with 10 ohm/5W resistor. \$\endgroup\$ Commented Dec 28, 2024 at 5:16
  • \$\begingroup\$ Thanks again Michal. \$\endgroup\$ Commented Dec 28, 2024 at 13:25
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The challenge you're facing is that so little of your alternator energy is making it to the battery - only the peaks. You're also losing some energy due to the rectifier's forward voltage drop.

I looked into this problem a while ago, and came up with a rectifier solution that works better for situations like this. See here: Bike power supply system

The circuits shown take an AC source and double the voltage, using low-loss 'ideal' diodes built from FETs and BJT's. With a very low frequency like you mention, you may need to upsize the capacitors, but it should deliver more power to your battery charger than a simple bridge.

If you are willing to consider an IC, the LTC3330 is a nanopower energy harvester / battery charger / controller that has an integrated diode bridge. Its output current is limited to 50mA, but may be adequate. It could be used with one of the proposed doublers I linked above.

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