I'm trying to charge 18650 cells from LG which have a capacity of 3200 mAh. I read the datasheet of the cells and it said 4.2 V and 0.5C of nominal charging. I'm not an expert so I did some research and found the cells should be charged with CCCV.

So I picked up a LM2596 CCCV module. It has 3 settings for: current and voltage limiting and threshold current for charge-complete LED. I also picked a 1s BMS since, from what I understand, it is necessary for safety especially above 1s configs.

So this is what I understand of CCCV:

1.) We set the voltage limit on LM2596 to 4.2 V for single cell charging.

2.) We short the output terminals and set the current limit to 1.6 A as 0.5C.

3.) With these settings, cell would charge to 4.2 V at 1.6 A CC; when it reaches 4.2 V, CV mode would engage and keep the cell at 4.2 V by decreasing the current until it reaches the cut-off current which is 50 mA according to the datasheet. After that, which I also set with the potentiometer as 50 mA, the charge-complete LED will illuminate.

With this knowledge I hooked up my 19 V, 3.6 A laptop charger to the LM2596 as source and connected the output to the BMS, then connected the BMS to a single cell in a 18650 case.

Initially I started with a cell that was at 3.52 V and it charged in CC mode in the beginning.

Now here is where the problem begins: Unlike the graphs of CCCV explanations and demonstrations on the internet where current stays the same until 4.2 V and starts decreasing in CV mode, in my setup current starts to decrease long before 4.2 V, generally around 3.9 to 4 V.

I hook up my ampmeters and measure the current dropping constantly as the voltage increases slightly. At the end when it reaches the cut-off current of 50 mA, because it starts to decrease in current too soon, the charging completes with the cell voltage at around 4.05 to 4.1 V. At the end I can't charge any of the cells to 4.2 V, the current does not stay the same and decreases before reaching 4.2 V.

I tried to solve the problem by removing the BMS since I was already using 1s. However, it did not solve the issue, the pattern of charging and end voltage is the same: around 4.08-4.1 V.

What is the problem in here that prevents the CC to run until 4.2 V?

  • 1
    \$\begingroup\$ Are you sure the LM2596 module you use operates as a proper CCCV charger, and is not simply a DC/DC converter that lets you set an output voltage and a maximum current? Do you have specs/documentation for that module? \$\endgroup\$
    – ocrdu
    Aug 9, 2022 at 19:10
  • 3
    \$\begingroup\$ The LM2596 is not a lithium charger chip so whatever module you have it needs other components on it to work as a CC-CV module or as a lithium charger, and it may not be intended or even suitable to be a lithium charger. Please use the module according to manual. We don't know what module you have and what it is suitable for. \$\endgroup\$
    – Justme
    Aug 9, 2022 at 19:22
  • \$\begingroup\$ What BMS did you choose? Can you share a link/product info? \$\endgroup\$ Aug 9, 2022 at 19:29
  • \$\begingroup\$ What is the small IC on the module - this may give some clue as to how CC is achieved. It is reasonably likely that the circuit does not achieve true CC when the voltage difference between Vtarget and Vout is small. \$\endgroup\$
    – Russell McMahon
    Aug 16, 2022 at 13:47

5 Answers 5


enter image description here

The charging current depends on the output voltage of the DC-DC you use as a source, the battery voltage, and the resistances in series with the circuit. That's the output impedance of your DC-DC converter, and the internal resistance of the battery. It's just ohm's law: I=(V1-V2)/(sum of all resistances).

Now if it looks like this:

enter image description here

This will be the usual, with a counterfeit LM2596. Check the frequency, LM2596 switches at 150kHz, if yours switches around 50k it's a fake. The capacitors will most likely be the cheapest available, high ESR, unable to properly filter the converter's ripple current. If the inductor was sized for a real LM2596, at the fake chip's frequency it will saturate.

All this means the output voltage is not stable and drops quite a lot under load. In other words, this has quite high output impedance... and that answers the question.

  • \$\begingroup\$ Thanks for the comment. Yes my board looks exactly like that. Many of you asked for the documentations or manual, I couldn't find any but this board is sold under the name "Li-ion charger" everywhere in my country. So I would assume It comes down to board simply being cheap or fake? \$\endgroup\$
    – Expose
    Aug 10, 2022 at 8:42
  • \$\begingroup\$ If you have an oscilloscope you could measure the switching frequency to check if it is the same as a real LM2596. Another test is to run it at its supposed rated current and check how much the output voltage drops, and output voltage ripple. On the fake boards, the capacitors get very hot. \$\endgroup\$
    – bobflux
    Aug 10, 2022 at 9:24
  • \$\begingroup\$ I dont have an oscilloscope but conducted the other test you mentioned. Since I dont have the knowledge to interpret the results, I'm writing the results: Had a battery at 3.52V and module's output at 4.2V. Inserted the battery to case and immidiately the voltage rose to 3.62V when measured directly on the cell's terminals. However when measured at the module's output terminals it measured 3.82V. Are these numbers normal? By thre way resistors did get hot, was hard to touch but not to an extreme degree. \$\endgroup\$
    – Expose
    Aug 10, 2022 at 10:49
  • \$\begingroup\$ Resistors? You mean capacitors? \$\endgroup\$
    – bobflux
    Aug 10, 2022 at 11:00
  • \$\begingroup\$ Yes sorry my bad, I did check capacitors they werent too hot. \$\endgroup\$
    – Expose
    Aug 10, 2022 at 11:08

The LM2596 you chose is not a charge controller IC, it is a DC-DC step-down converter. It may be suitable as a primary converter for a downstream charge controller, such as a BQ25181 (since you like TI).

There are hundreds of different single-cell Li-ion battery charge controllers with a variety of features, it is up to you to determine which ones you want (eg. integrated battery temperature monitor, fault monitoring, voltage input range, etc.)

  • \$\begingroup\$ It seems there are 2 models of LM2596: one is just CV source and the other one which I picked up goes by "CCCV LED driver and battery charger" module and there are people on Youtube charging 18650's with that. So that means It still isnt enough for charging 18650's? \$\endgroup\$
    – Expose
    Aug 9, 2022 at 19:51
  • 4
    \$\begingroup\$ @Expose We still don't know what module you have and what are the specs for it. And people do all kinds of stuff on Youtube that might best be left undone. Even if people on Youtube connect some random things together to charge their batteries, it does not mean they or anyone else should, unless they know what they are doing. Or at least, they should know that they don't know what they are doing, and thus be prepared for consequences, such as lithium batteries exploding or bursting into flames by incorrect charging. \$\endgroup\$
    – Justme
    Aug 9, 2022 at 20:06
  • \$\begingroup\$ @Expose I have no doubt that a LM2596 could charge a battery, but I do not think that it is a good idea. As Justme said, people do things on YT all the time that they shouldn't. \$\endgroup\$ Aug 9, 2022 at 20:15

I had this as a comment but I think it may actually be an answer.

Exactly how are you measuring the current and voltage? If you are using a DMM with a 2A range, it may drop 160 mV at 1.6 A, so the charger will see 4.2 V while the cell has terminal voltage of 4.04 V. Also, your DMM leads may have an additional significant voltage drop. You may need to use a Hall effect current sensor, or maybe a 10 mΩ shunt that will read 16 mV at 1.6 A. And use short, heavy wires from the charger to the cell. You could also compensate by setting the float voltage to 4.3 V, but that could be risky.

  • 1
    \$\begingroup\$ Or keep it at 4.2 V and just don't charge the cell fully which will prolong its lifespan. \$\endgroup\$
    – Oskar Skog
    Aug 10, 2022 at 6:35
  • \$\begingroup\$ Thanks for the response. I'm using a cheap multimeter to measure but I dont need to use it to realize CC is disengaged and current starts to decrease. The module has a built in CC/CV LED that turns off when current drops below the set value so no any additional resistance or measurement error from my DMM anyway. @OskarSkog for your comment, thanks. Yes it is possible to do that but we are trying to build RC plane with 9 of that cells in 3s3p config so every mAh counts for a decent flight time. \$\endgroup\$
    – Expose
    Aug 10, 2022 at 8:40
  • \$\begingroup\$ A cheap multimeter may not be accurately calibrated, so you may not know if the actual voltage is 4.20 V. A 1% error could mean the actual voltage is 4.16 V or 4.24 V. \$\endgroup\$
    – PStechPaul
    Aug 10, 2022 at 20:13

Give or take doubts about whether the regulator is authentic or not, I would assume its output characteristic is not sharply CC-CV.

In particular, that line of SimpleSwitcher regulators use a transconductance error amp, with very limited gain -- about 80 if I recall correctly. This most obviously affects regulation: for the control output (error) to vary over say a 5V range (or whatever it might be internally), the feedback pin (or internal equivalent) needs to change by 5V/80 = 62mV. Which, out of a 2.5V reference and 1:1 feedback divider for a 5V output say, is 125mV, or 0.125V/5V = 2.5% load regulation. It's not a high performance part; it'll get you there, it won't win any prizes.

A similar principle applies as CC mode engages, which in that series should be due to current sensing and slope compensation. Partly due to limited gain of the sense amp, partly due to amplitude-dependent propagation delay to the PWM latch.

The overall effect is adding slope to the otherwise-ideal straight vertical/horizontal CV/CC regions, and a rounding-off of the otherwise ideal square corner between them.

If the devices are other, or not authentic, a similar analysis may apply (this is generally true of any peak current mode style control, with the amount of sloping/rounding depending on loop gain, propagation delay and the like), or they may be completely different types (e.g. a separate DC current loop regulates the CC region very precisely, at least at DC; but may incur a slow response during the transition).


Actually after sometime, the charging of Lithium-ion Cells / Battery switches to CV(Constant Voltage Mode) from initial Constant Current mode (CC)

I came across an article from Digikey website citing

"The battery cell will have most of its charge when the battery voltage reaches 4.1 V or 4.2 V. At this point, the current going into the battery gradually decreases."

You may read the article here :


Cheers ;) Vrajj VTEC

  • 3
    \$\begingroup\$ This is not an answer to the question. The question was about too early current drop. \$\endgroup\$
    – Jens
    Aug 9, 2022 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.