Author Topic: lipo: boosting and charging  (Read 714 times)

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Offline MoriambarTopic starter

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lipo: boosting and charging
« on: June 25, 2022, 04:21:25 pm »
Hello.
I'm familiar (somewhat) with boost converter designs and I've been recently planning to create a lipo based psu.
What I'll do is basically boost the battery voltage to 5V (battery already has protection circuits etc.) but what I'd like to understand is how to create a charging circuit for… charging the lipo while it's still attached to the main circuit.
I really have no experience in this, can someone give me any pointers please?

Cheers
 

Offline Siwastaja

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Re: lipo: boosting and charging
« Reply #1 on: June 25, 2022, 04:34:09 pm »
Li-ion charger IC + application note example is a good start.
 
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Offline MoriambarTopic starter

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Re: lipo: boosting and charging
« Reply #2 on: June 25, 2022, 06:14:13 pm »
thanks, if there are any ics that you have in mind please tell me, otherwise I'll go rummaging through digikey's list ;)
cheers
 

Offline Buriedcode

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Re: lipo: boosting and charging
« Reply #3 on: June 25, 2022, 08:49:27 pm »
The charging method for lipos is constant current, causing the cell voltage to rise, then constant voltage - holding the voltage (usually 4.2V for most cells) whilst the current draw drops then terminating charge once the current draw has fallen below a preset value, such as 10% of the full charging current.

Powering a circuit from the cell whilst its being charged could mean that the current draw never falls below that 10% so it will never finish charging.  If the circuit draws more current than the charger provides, the cell will never charge. 

A way around this is to provide a path for the charging circuit input to power the circuit, as well as charge the cell.  I've seen the term "Power path" used for Texas instrument and microchip products.
At its most basic, the cell powers a circuit via a P-channel MOSFET, with its gate tied to ground so its on.  The gate is also connected to the charger power input, so when someone plugs in power to charge the cell, it turns off the MOSFET, and instead powers the circuit via a diode form the charger input.

This effectively isolates the cell whilst its charging, and the circuit will only draw power from the charge input, via the diode, rather than the cell.  See the attached image.
Note, the image uses a module with built in protection.  If you dont' need that, then the cell just connects to the OUT+ and OUT- points.



There are some caveats here.  The input is assumed to be 5V.  This means that, when charging, it can only provide ~4.4 - 4.8V to the circuit, because of a voltage drop across the diode.  If that circuit is a boost converter, it may (or may not) be too high for it.

If one uses a switching charger that takes, say, 12V input, to charge a 4.2V cell, then this system will provide ~11.4-11.8V to the circuit.

« Last Edit: June 25, 2022, 08:51:35 pm by Buriedcode »
 
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Offline Peabody

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Re: lipo: boosting and charging
« Reply #4 on: June 26, 2022, 02:36:24 pm »
That was my drawing for using a solar panel, and it doesn't actually work.  But it does work with a fixed 5V supply and without the second diode across the mosfet (see below).  Note the orientation of the mosfet, which some consider to be backwards, but is correct.  This is a "load sharing" circuit, aka "power path".

http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf

If possible you might want to consider using a 3.3V circuit rather than 5V.  Then it could make sense to use a simple LDO as the regulator.  Or, as in the case of something like the 3.3V 8MHz Arduino Pro Mini, you might not even need a regulator at all unless sensors, etc., require it.

Edit:  One of the best write ups on this is Zak Kemble's blog post.  He goes into the calculations needed to select the mosfet, the schottky, and the pulldown resistor:

http://blog.zakkemble.net/a-lithium-battery-charger-with-load-sharing/

And if you can stand it, you can watch my video on this subject, and skip over the modification process:


« Last Edit: June 26, 2022, 03:01:21 pm by Peabody »
 
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