I'm doing research into portable power generators for a university class and need help understanding how charging a device works. From my research, I've found certain methods of harvesting energy can yield an output of about .5mW of electricity. If you were to try to charge a device with this output, what would happen? For examples sake, lets say we were trying to charge my phone's battery. The battery is a 3200mAh, 3.8 V Li-Ion battery, and the default charger is a 5.3V, 2A charger.

On such a device, how does the amount of wattage being inputted affect charging? Is there a minimum amount of watts needed before the device will charge at all? What do the battery and charger voltage requirements mean?

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    \$\begingroup\$ In principle, you can charge a battery at any power level using appropriate voltage conversion. One thing I will point out to help you understand about the scale of things is this. Your battery is kind of like an energy storage tank. If you multiply the Ah * the Voltage, you will get an approximation of the energy storage capacity. 3.2 * 3.8 = 12 Watt hours. That means you need to charge it at 1 Watt for 12 hours (or 12 Watts for 1 hour) to fill it up. So, in order to fill it up when charging at 0.5mW, it will take around 3 years. Now in reality, this won't charge it at all, because... \$\endgroup\$ – mkeith Feb 5 '15 at 4:24
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    \$\begingroup\$ a battery is kind of like a leaky vessel. Depending on type and storage temperature, it has a certain amount of self-discharge. The leakage of your battery is likely somewhere around 0.5mW, so the battery might charge very slowly, discharge slowly (due to self-discharge) or just stay topped off. \$\endgroup\$ – mkeith Feb 5 '15 at 4:26
  • \$\begingroup\$ This is extremely helpful to know, thanks for the help! \$\endgroup\$ – master565 Feb 5 '15 at 5:02

Different chemistries require different charging techniques. But in general, as long as your source voltage is higher than your battery voltage charging will happen. If the source voltage is lower than battery voltage then charge will be drawn out of the battery. That's why your charger is 5.3V - your battery has a voltage of 4.2V when fully charged.

So all we need is to maintain voltage. What does current do?

You'll notice that your battery is rated as 3200mAh. Which is 3.2Ah. What this means is that if you fully discharge this battery over the period of one hour, the current consumed will be 3.2A. Similarly, if you want to fully charge this battery from empty (around 2.7V) and do it in one hour you'll need to supply 3.2A.

So, what happens if you can only generate 0.5mW? Well, we still need to maintain voltage above 4.2V to charge. So let's take your charger's 5.3V. At 5.3V, we'd be able to supply 0.094mA of current. Which is 0.000094A.

So, plugging the numbers in to Google:

3.2 amp hours / 0.049 milliamp = 3.88355924 years

So, using your charger, given 0.5mW power, assuming no other losses, it would take almost 4 years to charge that battery.

  • \$\begingroup\$ This is really helpful information, I just have 1 question. Why use a 5.3V charger when you can achieve a higher amperage with a lower voltage on the charger? If you used a 4.3V charger on a 4.2V battery, it will still be a higher voltage than the battery, but the amperage would be much higher. \$\endgroup\$ – master565 Feb 7 '15 at 18:56
  • \$\begingroup\$ You're assuming 100% efficiency between the wall socket and the chemicals in the battery. As things heat up (which they tend to do when batteries are being charged) the impedance/resistance of the system increases dropping the voltage. It's just a safety margin. It also depends on the chemistry. Some may stop charging earlier so you need to buffer against that as well. \$\endgroup\$ – slebetman Feb 8 '15 at 8:48

Recommend to read this article. http://batteryuniversity.com/learn/article/charging_lithium_ion_batteries The lithium battery charging is a 4-stages process based on the battery condition. See Fig 1. Now the industry uses the specific charging IC to control and monitor the battery charging job.


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