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I am currently trying to choose a charger IC for Lithium-ion cells batteries and thereby came across the datasheet for the MAX1737:

https://datasheets.maximintegrated.com/en/ds/MAX1737.pdf

The datasheet does not seem to mention the maximum charging current this IC is able to supply.

Hence my question: Did I miss something, or could it be that the most important characteristic of any charger IC indeed is not mentioned in that datasheet?

EDIT 1

I have understood that the charging current is driven by an external MOSFET. But when you go to

https://www.maximintegrated.com/en/products/power/battery-management/MAX1737.html/tb_tab1#tab1

and then switch to "key specs", you can see that the maximum charging current is 4 A.

So I am rewording my question:

Why is the charging current limited at all (given that charging is driven by an external MOSFET), and why is that limitation not mentioned in the datasheet?

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The MAX1737 uses external MOSFETs for charging output. It is essentially a controller chip. Therefore the result depends on transistors (and magnetics).

EDIT 1: The limitation in marketing materials is likely because the controller can't drive more powerful MOSFETs efficiently, and 4-A was a practical limit. Power MOSFETs can have 10 nF gate capacitance, and with 7-Ohm driver inside the MAX1737 the turn-on transients can be 100-200 ns, which, at 300 kHz operations (pulses can be sub-1-us in some regimes of charging) would lead to likely unacceptable losses and transistor overheating. See Section "Application Information", MOSFET selection. The fast charge current, however, is not limited by any particular restrictions on the value of set resistor, it can be anything.

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  • \$\begingroup\$ Thank you very much. But if you go to maximintegrated.com/en/products/power/battery-management/… and then switch to "key specs", the page shows that the maximum charging current is 4 A. Given that the charging current is driven by an external MOSFET, why is it limited at all? \$\endgroup\$ – Binarus Apr 26 '18 at 6:56
  • \$\begingroup\$ @Binarus, I would guess that the drivers inside the IC are not powerful enough (7 Ohms impedance) to drive more powerful MOSFETs with proper efficiency, so only MOSFETs of certain size can be driven, see Application Infromation, MOSFET selection, page 16. \$\endgroup\$ – Ale..chenski Apr 26 '18 at 7:07
  • \$\begingroup\$ Thanks for the helpful explanation. I am giving the answer to you. \$\endgroup\$ – Binarus Apr 26 '18 at 7:40
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The DC-DC converter uses an external dual N-channel
MOSFET as a switch and a synchronous rectifier to
convert the input voltage to the charging current or voltage.
The typical application circuit is shown in Figure 1.
Figure 2 shows a typical charging sequence and
Figure 3 shows the block diagram. Charging current is
set by the voltage at ISETOUT and the voltage across
R18.

enter image description here

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  • \$\begingroup\$ At first, thanks for taking the time. What you write is true, but is not an answer :-). As you correctly state, R18 determines the charging current. But my question was about the maximum charging current. In other words: What is the lowest value which is allowed for R18? The data sheet does not seem to answer that question. The diagram you have shown is for R18=0.1 Ohm only. But how could I know if I am allowed to use even lower values for R18? \$\endgroup\$ – Binarus Apr 26 '18 at 6:47
  • \$\begingroup\$ Actually, my worries arise from the following page: maximintegrated.com/en/products/power/battery-management/…. On this page, please switch to "key specs"; the page then shows that the maximum charging current is 4 A. So if that IC just drives an external MOSFET, why is the charging current limited at all? \$\endgroup\$ – Binarus Apr 26 '18 at 6:54
  • \$\begingroup\$ @Binarus, so, your question was loaded. Why didn't you say so upfront? \$\endgroup\$ – Ale..chenski Apr 26 '18 at 7:12
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    \$\begingroup\$ @Binarus The charging current is probably mostly limited because of safety reasons with LI-ION batteries. In typical use, most people won't need a max charging current of over 4A. It was probably just more cost effective during the design of the chip to limit the design to typical use then to extend the gate drivers in the IC for even more powerful mosfets electronics.stackexchange.com/questions/22851/… \$\endgroup\$ – Remco Vink Apr 26 '18 at 7:19

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