Electronics > Beginners
4S Li-ion pack charger and protection complete circuit
VEGETA:
Hello,
I have been working on a circuit that allows charging and protecting a 4S Li-ion battery pack assuming the use of non-protected cells. I plan to use this circuit in another future project thus I like to take your opinion on it.
Input is 5V USB and outputs around 16.8V with current limiting circuit by pulling the EN pin of the DC-DC boost converter to ground to shut-off the boost converter and so on. This is the only solution that I got so far so what do you think? Assuming the cheapest parts available.
Looking forward to your comments!
mvs:
Your diff amp looks a bit strange. 5v supply is not sufficient for high side operation. It may work with "over the top" opamp like LT1637, but not with general use opamps.
VEGETA:
--- Quote from: mvs on February 13, 2020, 04:51:15 pm ---Your diff amp looks a bit strange. 5v supply is not sufficient for high side operation. It may work with "over the top" opamp like LT1637, but not with general use opamps.
--- End quote ---
What do you suggest then? I am still gonna use this cheap op-amp. I thought it would take it ok but maybe you mean the inputs will be higher than its rail voltages??
I am more interested in the rest of the circuit to be ok, what do you think?
As for current limiting, I am open at making a hard constant current limit like 1A using any cheap method. No need for adjustable current limit, any hard one is ok with me and that is why I choose this one.
fcb:
You aren't using the CTL pin to control charge/discharge? Just relying on the cell voltage state machine?
Where are you connecting the negative side of the battery to? I can guess, but it's not on drawing?
Why not use the sense resistor 'Rsense' (which is 100||100 - to high?) to do your current sensing for charging.
I'm lost why you have a constant voltage boost (HX3608) setup for charging. Especially with that mad 7 resistor potential divider. It'll just sit there and hiccup. Have you read how to charge Lithium cells?
Going to ask where do V+ and V- (for the opamp) come from before I declare that won't work.
What does "charger(-16.8v)" mean? Why negative?
mvs:
--- Quote from: VEGETA on February 13, 2020, 07:19:35 pm ---I thought it would take it ok but maybe you mean the inputs will be higher than its rail voltages??
--- End quote ---
Yes, max. common mode voltage on opamp inputs is 16.8V*R22/(R21+R22)=16.8V*10K/11K=15.3V, which is a lot higher then Vcc of 5V. You may reduce gain to get it right.
--- Quote ---As for current limiting, I am open at making a hard constant current limit like 1A using any cheap method. No need for adjustable current limit, any hard one is ok with me and that is why I choose this one.
--- End quote ---
Cheap? If you do not need any precision, 4A switch current limit of HX3608 may do the job. Lets estimate...
Duty cycle D=(Vo+Vin)/(Vo)=(16.8-5)/16.8=70%, Iout(max)=Isw*(1-D)-dI/2=4A*30% - dI/2= 1.2A - dI/2
Inductor ripple current dI is typicaly around 30% of Iout, so you get just around 1A.
In discharged state, say at 3V per cell, D=(12-5)/12=58%, Iout(max)=4A*42% - dI/2= 1.7A - dI/2.
In this case you will get around 1.5A of your boost converter.
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