I must preface by saying I’m new to this whole world, but love to tinker with new things. I’m building a IOT monitoring device that will be remote and needs to be able to recharge itself. I have a Particle 3G board with 2000 mAh battery, that I need recharged by a DC motor. The DC Motor is connected to a propeller with wind scoops on it (similar to what you would see on a mobile weather station). Unfortunately solar is not a good option in the location I’m in due to shadows.
Can I use something like this (https://www.adafruit.com/product/390) to attach to the DC motor to charge the LiPo, or would I need another application? What are the specifications I would need on the DC motor?
The device is deep sleep/sleep majority of the time and wakes once a day to gather data from the sensor (no cellular needed) and then send me a text file with some data (cellular on). The text file is a few kb’s. I’m trying to design this that even without any wind, the device can function for approx 14 days, given it’s remote location (basically over engineering it because I can’t check on it regularly)
You could use a solar controller with any dc motor (solar controllers aren’t just for solar panels) see if you can get one that will take in the voltage of your motor, and give 3.3v or 5vdc out. Then input that voltage to the 3.3v or 5v input.
Make sure your motor doesn't exceed 6.5V when you're turning it on this charger. Beyond that will probably damage the circuit. Motor voltage is proportional to speed, so if it doesn't go past that at your max speed, you should be good. Might want some capacitors to help clean up the motor output?
Make sure your motor doesn’t exceed 6.5V when you’re turning it on this charger. Beyond that will probably damage the circuit. Motor voltage is proportional to speed, so if it doesn’t go past that at your max speed, you should be good. Might want some capacitors to help clean up the motor output?
So if I get the board I linked and I get a 5V DC motor, would the circuit board protect me in case of super high winds? Or would the 5V DC motor never output more than 5V no matter how fast the shaft was spun?
That’s just the rated voltage. It’s somewhat more complicated, but put 5V on the motor and however fast it spins, don’t force it to spin faster than that. That should keep you out of trouble.
Any more than that (like circuit protection) would be a good bit more to explain and figure out.
I’m using the DC motor to charge, so I wanted to run it in reverse (to recharge the battery). I don’t know what the maximum speed of the shaft will be since it will be wind and there is no way of telling what a gust of wind can be.
I suggest that you get someone to drive a car for you while you hold the wind turbine out the window, and measure the voltage with a volt meter. That should get you the answer to how much voltage it will produce at high wind speeds.
A 7805 would need about 7V in to comfortably output 5V. In the real world, wind turbines park in high winds & also have a gearbox & control systems, overkill perhaps for a small system. If your turbine is prone too high winds a high voltage is not the only issue, high revs could damage the motor bearings (if there are any) & coils too, of course a motor might be so cheap its not worth protecting.
There are dedicated over voltage protection devices you could use as well as circuits, or any number of mechanical systems you could invent or copy…
Careful with a 7805 as they are “dumb” devices. They are inefficient, they effectively ‘burn’ the surplus voltage to reach the 5V.
Looking at this stackexchange, it looks like adding a capacitor to smooth out the ripple and feed that to a DC-DC converter (in the correct polarity!) is close to what you want to achieve. Keep in mind that the converter takes 6.5-32VDC as input so your motor will have to operate in this range.
Perhaps a safer way to test the output voltage (apply a load to the motor for accurate measurement) is to spin it with another motor
Would using this board do the same https://www.adafruit.com/product/390? Maybe overkill, but this would limit output current to 500ma. I want to make sure I don’t damage the battery as that is more of a fire risk.
