# Simplest possible circuit for charging a li-ion cell

I want to create an ad-hoc single cell li-ion charger. I have a buck step-down that can supply 4.2 volts. If I connect a 1 ohm resistor in series with the lithium cell, the current should go down to 0 when the battery is also at 4.2 volts. A 1 ohm resistor should supply a maximum of 500 mA when the battery is at 3.7 volts.

The battery has protection. Will this be safe? Do I really need that tiny resistor?

Forgot to add: 500 mA is well within the charging capability of the battery 1C (2200 mAh)

Also, I am aware that this will not fully charge the battery, since I am sort of skipping the constant voltage phase.

• The "simplest possible circuit" is not to be recommended for dealing with Li type batteries. There is a reason that specialized chips and modules have been developed for charging Li types of batteries. I strongly suggest that you use one of them and stop trying to hack out a solution like you have suggested. May 12 '14 at 15:17
• Well, I am working in a safe environment with this and don't really care about that for this project. Also, ordering one will take a month where I am living. May 12 '14 at 15:21
• Then I would say that you are entirely on your own. Li type batteries can be very dangerous if not handled correctly. At least you've heard it from me. If you are in a remote region all the more reason to exercise caution with life and limb. May 12 '14 at 15:28
• So, you're basically saying that I can experience a fast powerful explosion (and risk life and limbs) if I hook one battery (with its protection circuits) up to a 4.2 volt source? May 12 '14 at 15:36
• @frodeborli batteryuniversity.com - please read through that web site. May 12 '14 at 15:44

If you really want to charge battery from time to time - charging with resistor and constant voltage 4.2V or less will work and battery will not blow up if you choose proper resistor. You have to use resistor to limit current.

If you build circuit like this - keep in mind that charging current will be higher when battery is closer to empty state. When battery is closer to empty state - it heats up more, because internal resistance is higher. So - less charge in battery - more current, more power, more heat. When battery is empty - is not ready for fast charging.

When your battery is ready to be charged with high current - your circuit will feed it with with very low current. Everything upside down. Charging thru resistor from 4.1-4.2V constant voltage is most inefficient way to utilise Li-Ion battery.

Thats why there are specialised IC's and CC/CV charging methods. To charge it in efficient way. Many people think, that all Li-Ion need special special care, complicated chargers, advanced alghoritms etc. It's just about being sure that voltage never goes outside 3.2-4.2V range and heat produced on battery internal resistance.

Battery is rated for 1C (2200mA)? Sure, but if you read battery manual carefully you may find something like "you can't charge empty battery with 1C because it will overheat and you should monitor temperature".

Current source based on LM317 and one resistor would be much better.

If you need serious charger with very low cost and/or if you have no time - build something like this:

Image source: MCP-73831 datasheet.

MCP-73831 is one of many cheap integrated Li-Ion chargers available on the market. SOT-23 package, cost: < 1USD.

• thank you for a very "to the point" answer. Just a extra question: If I set input voltage to say 4.5 volts, but stop charging before the battery reaches 4.2 volts, would that be bad? I have a feeling a proper CC charger actually regulates the voltage down to just above the battery voltage to maintain a constant current. But with a resistor, constant current can only be achieved by having a constant potential between source and battery. May 13 '14 at 8:10
• It's just about keeping voltage below 4.2V. Doesn't matter if you do it manually or circuit will do that. However - if you accidently put more than 4.2V on battery - irreversible chemical reaction will happen inside battery will loose capacity permanently. May 13 '14 at 8:16
• You also have to be sure, if voltage is ALWAYS below 4.2V and you have no ripple above that voltage. You can have 4.0V reading on multimeter, but in fact - there may be ripple from 3.5V to 4.5V and battery will be damaged. May 13 '14 at 8:22
• You're right. Without a calibrated voltage, I'll have a big problem doing this. May 13 '14 at 11:27
• I meant that, if voltage is not constant and there are spikes from converter - you be not able to see this on multimeter. May 13 '14 at 12:41

I've been using this for quite a while and it works. I charge pairs of two Li-Ion cells in parallel with it.

Source: Electronics DIY

A major disadvantage is the large heatsink required by LM317. Mine is 10 by 10 cm and still gets quite hot. Instead of R1 - R2 I used single 2.2 ohm / 5 W resistor (that's what I had).

• Thats waaay more complicated than using Li-Ion charger IC. May 12 '14 at 18:02
• MCP73831 is pretty popular and cost is similar to LM317. May 12 '14 at 18:54
• @Kamil The OP has indicated that it is difficult to order specialized parts where he/she lives. I don't see anything wrong with suggesting a solution that uses more readily available components. In your own answer you suggested something like this, so why not let this answer stand on its own merits? May 12 '14 at 19:45
• I know what you mean, but Microchip distribution network is pretty good. If I can buy that part in Poland and price here is equivalent to 0.5EUR - it should be available anywhere. It may be harder only in south part of Africa :) May 12 '14 at 19:54