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So long story short I'm attempting to build a charger for batteries using a DC motor. The batteries I'm planning on charging are old smartphone batteries at 3.7V and 1000mAh. The DC motor that I am currently looking at can be found here. From what I've been able to puzzle out I'll probably need to rectify the output of the motor and soften the wave with a capacitor. Beyond that, I've heard that I probably will want to use a voltage regulator and have a CC,CV circuit. Working off of this post I found a trickle charger that would lead me to this:

schematic

simulate this circuit – Schematic created using CircuitLab

I would like to use something other than a trickle charger, as a trickle charger will probably take too long. If anyone has a better circuit I'd love to see it. Additionally, if anyone could walk me through this so I can evaluate the values of the resistors, caps, and diodes. My first instinct on the diodes is to use schottky's so I don't lose as much voltage and to rate them at about 3A. As for the rest, any help would be appreciated. Additionally, any pointers on modelling the behavior of this circuit would be highly appreciated. Magically I didn't find a DC motor in spice, so I don't know.

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MJAFort
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    The worst thing you can do is overcharge a Lithium-ion battery. There are many IC's dedicated to properly charging Li-ion batteries and monitoring for overcharge. They are mostly self contained (except those for smart phones) so the parts count is low and is much more efficient than U1. –  Jul 07 '16 at 22:19
  • The complexity of what your doing and the variables involved make this question too broad in scope for a simple answer, and too time consuming for a more complete answer. Break this project down into smaller steps and end your questions with a '?'. –  Jul 07 '16 at 22:30
  • 1) That charger circuit was for a NiMh battery. Do not use it for a lithium battery (listen to Sparky). 2) If the motor is only turning in one direction, you don't need the bridge rectifier. 3) You do need a high-frequency capacitor for the commutation noise, but shouldn't need a filter capacitor (although having one wouldn't hurt). 4) You would need to spin that motor at least at 3000 RPM (maybe 2000, depending on your charger chip) to get a usable voltage from it. And even then, it wouldn't be much current. – Mark Jul 07 '16 at 22:46
  • Thanks, I'll look into a fast charger for li-ion and add an update to the schematic. However, the motor may or may not switch directions. I'm making the circuit for anyone to make their own mechanical components, so it needs to be pretty robust and deal with things like quick switches of direction. – MJAFort Jul 07 '16 at 22:48
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    To generate 12 volts with this motor, you need to turn the shaft at about 6000 RPM. The output voltage at a lower speed would be approximately 12 V X actual speed / 6000 RPM. Since the seller gives very little information about the motor, it is impossible to tell how much current can be safely generated. The order of magnitude would probably be 1 amp. –  Jul 07 '16 at 23:25
  • @Charles Cowie - The seller states (on the questions page) that the no load current draw is 30ma (and I'm assuming that is at 12v). So that's only about a third of a watt. So, at 6000 RPM, would it take about 33 hours to charge a 1000mAh battery? If the step-down from 12 volts to 3.7 volts is lossless (which it's not), would it take about 10 hours? – Mark Jul 07 '16 at 23:49
  • The 30ma at no load means that the mechanical losses are about 1/3 W at 6000 RPM. 1000mA-hours at 3.7 V that is equal to 3.7 watt-hours of energy. There will be losses at every stage of the process: friction in the mechanical system that is driving the generator, electrical and mechanical losses in the generator, electrical losses in the charging control/voltage conversion circuit, losses in the batteries. –  Jul 08 '16 at 00:57
  • @Mark - No load current is what the motor draws when powered externally. It is required to overcome internal losses, and won't be seen in generator mode. The important factor is armature resistance, which wastes voltage and therefore limits the current that can be drawn. – Bruce Abbott Jul 08 '16 at 01:05

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Battery negative should be connected to the top of R1/2/3, not the bottom. The idea is that charging current flows through the resistors and creates a voltage drop. When it reaches 0.6V, Q1 turns on and pulls the ADJ terminal down to reduce the voltage and limit current.

The value of R1/2/3 must be chosen to drop 0.6V at the charging current you want. If your cellphone batteries are old then they probably should be charged relatively slowly, eg. at 0.5C (2 hour rate) which is 0.5A for a 1000mAh battery. 0.6V/0.5A = 1.2Ω. Maximum power loss in the resistor will be 0.6V * 0.5A = 0.3W, so you should either use a 0.6W (or higher) rated resistor, or several smaller wattage resistors which make 1.2Ω when combined.

You don't need a bridge rectifier because the DC motor already rectifies the voltage via its commutator. However a diode is required prevent the battery from discharging back into the motor when it is not generating sufficient voltage. Use a rectifier diode rated for at least 1A (preferably a Schottky type for lowest voltage drop).

The output of a 3 pole DC generator is 3 overlapping rectified sine waves so there is some ripple in the 'DC' voltage, as well as some higher frequency commutator noise. If you can spin the motor fast enough to always stay above ~8V then theoretically you only need a small filter capacitor to reduce commutation noise and maintain regulator stability. However a larger capacitor (up to several thousand microfarads) will help to reduce ripple and hold the voltage up during any momentary rpm reductions that might occur.

If you do use a large filter capacitor then you should wire a rectifier diode (eg. 1N4001) across the regulator to prevent it from being damaged by reverse current flowing into the capacitor if a battery is plugged in without the generator going. Anode goes to OUT and Cathode goes to IN, so the diode is normally reverse biased but conducts when the input voltage is lower than the output voltage.

Q1 can be any general purpose NPN transistor. I would use one that can take at least 0.5A peak, eg. BC337 or 2N2222A.

R5 needs to provide sufficient feedback current for the LM317. The recommended value is 240Ω, but up to 470Ω should be OK. Normally it has 1.25V across it (from the regulator's internal voltage reference) so R4 needs to drop 4.2V - 1.25V = 2.95V. If R5 is 240Ω then it passes 1.25V / 240Ω = 5.2mA, and since R4 carries the same current its value should be 2.95V / 5.2mA = 566Ω.

The reference voltage may not be precise so R4 should be made adjustable so the output voltage can be set to exactly 4.2V without a battery connected. Only a fine adjustment is necessary, so I would use something like 330Ω in series with a 470Ω trimpot (which is ~235Ω in the center position) to make the required total of ~566Ω without the adjustment being too sensitive.

If insufficient voltage is available to charge the battery then it will slowly discharge through these resistors. Therefore the battery should be disconnected if the generator is expected to be idle for a long period of time.

Bruce Abbott
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  • Thanks, you answered most of my questions for how to evaluate this. I'll probably end up having to switch circuits since this circuit is for NiMH batteries rather than lithium ions. – MJAFort Jul 08 '16 at 01:40
  • With the correct resistor values to get 4.2V it will be fine. Here's what I used when I started experimenting with li-ion cells:- http://shdesigns.org/lionchg.shtml – Bruce Abbott Jul 08 '16 at 02:41
  • Thanks Bruce, that's exactly what I was looking for. You're a life saver. – MJAFort Jul 08 '16 at 15:32