Electronic – How to raise voltage output of charge controller IC due to cable loss

charginglithium ionvoltage measurement

I am using a CN3703 IC to charge a 3 cell li-ion pack (layout as below.)

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The charge current is programmed using a resistor between its CSP and BAT pins which develops a voltage drop of 200mV. The output voltage on the BAT pin is preset to 12.6V (+/-1%.)

Due to cable and connector voltage loss, I want to raise the output voltage so that 12.6V (instead of 12.4V currently in my previous layout) reaches the battery. I am not sure exactly how to do this as there is no mention in the datasheet, but I came up with the following circuit (circled in red.)

The idea is is that 99% of the voltage is fed back to the IC so that it raises its output by 1% to around 12.72V.

  • Would this modification likely work?
  • How large can I make these resistor values without them introducing significant error?

EXTRA INFO: I should mention that I have already modified (U6) INA219 breakout by swapping the 0.1ohm resistor with a 0.01ohm resistor.

Full Schematic from working prototype and picture of PCB
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schematic

UPDATE: Some very helpful comments. I have redone the layout to put the shunt before the battery sense so that the chip sees the voltage of the battery directly. I have also switched to a XT30 connector for the battery so that i am not using the same connector for the input power and the output charging cable. I have also moved the battery connector closer to the IC and used a short wide plane to connect them. Finally I have avoided using vias for the high current path by placing the low current signals on the other side.

Revised Layout

Best Answer

I believe your problem is in those absolutely unnecessary voltage dividers R7/R8, R11/R12. You basically adding 500 Ohm load to the charger's output and also shifting feedback voltages, messing up all the charging profiles the chip is designed to follow.

Remove R8, R12, C10 and put jumpers instead of R7, R11.

Also, what on Earth is R2 doing there? If you want a discharge path (again, absolutely unnecessary) for that tiny 10uF capacitor, put 1M there, not 22k

Finally, don't use VIAs in high current path. You have through-hole pad on VIN+. Why would you take a trace on one side only to bring it to the other with VIA if you already have same pad on the other side?

UPDATE:

One more thing. The whole premise of this post is an attempt to tweak output of the charger to accommodate the voltage loss in some (supposedly) too long or too thin charging cable. This is like attaching RV trailer to Fiat 500 and then tweaking the carburetor hoping it will go.

What if you change the cable tomorrow? What if somebody uses different gauge cable? IMHO, you are fighting self-created problem here.

UPDATE 2:

is there a way to compensate for the lower than expected output voltage?

There might be, but IMHO it is not worth doing. First, the standard way to detect end of charge of lithium battery is to monitor current while in constant voltage stage. If you use correct shunt resistor and use short appropriately sized feedback traces then battery will probably be charged OK.

Second, you should focus on minimizing voltage losses in your wiring. Heavy and as short as possible PCB traces, no VIAs, good connector, heavy and short cable. See if you can find alternative to INA219 with lower than 20uA input bias. Then you might be able to share shunt, further reducing voltage drop.

Also, I'd definitely try another chip or two to see if this lower than expected voltage is a fluke or a norm for this IC.

Regarding discharge resistor - similar ICs from TI and AD with built-in power path technology often use ideal diodes internally, to minimize both forward drop and leakage current. Maybe it worth checking those out. In any case, I would definitely use larger value. Keep in mind that it essentially drains the battery when it left connected with no power applied.

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