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USB to TP4056 charger - why is the diode there?

This is the relevant part of the schematic of an 18650 battery charger and boost converter (to 5V) module:

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And here's the module:

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Coming from the USB power pin, there's a series Schottky diode and a very low value resistor.

The resistor is found in the TP4056 datasheet typical application schematic, and I think I understand why it's there. The TP4056 is a linear part, and when the battery is mostly discharged, the charger will be supplying 600mA (in this case), and the voltage drop will be at its greatest. While the TP4056 has a built-in current reduction if the die gets too hot, the resistor can take part the voltage drop, and part of the heat, and allow the charger to continue delivering 600mA.

But I don't understand why the diode is there. Of course it would also drop the voltage, but if that's its only purpose, the value of the resistor could have been adjusted to accomplish the same thing - without adding another part. If it's there as a reverse polarity protection, I would just say that as a practical matter you aren't going to get a polarity reversal from a USB port. Of course the 18650 being charged could be inserted backwards, I don't see how that would affect the USB port.

Can anyone suggest why the diode is there, or if they've seen this in other charger circuits? I ask because I'm planning to modify the module to add a load sharing circuit, and the diode would come in handy for that if it's not really needed in its original position. Thanks very much.

Reply to
Peabody
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Microchip's In Circuit Serial Programming (ICSP) implementation:

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uses a Schottky-type diode. The diode isolates the MCLR/Vpp pin from the application circuit when the chip is programmed.

Thank you,

--
Don Kuenz KB7RPU 
There was a young lady named Bright Whose speed was far faster than light; 
She set out one day In a relative way And returned on the previous night.
Reply to
Don Kuenz

The diode is there to prevent the charger from powering the USB host and blowing something up.

CH

Reply to
Clifford Heath

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the datasheet says no diode is required, I think the diode and resistor is simply there to reduce dissipation in the IC

Reply to
Lasse Langwadt Christensen

Would it have been possible to eliminate the diode and instead increase the value and wattage of the resistor to achieve the same voltage drop? That would have eliminated a relatively expensive part from the device.

The biggest voltage drop at full charging current is when the battery is at 3V (below that there will be low-current trickle charging). And the datasheet says the minimum Vcc is 4V. So using only a resistor you would have to drop 1V at 600mA, which would be 1.67R and 600mW. Is a chunky resistor like that difficult or expensive to source as SMD? Even so, it seems using two snaller resistors would be less costly than using the Schottky diode.

Well, I guess I need to figure the voltage drop and dissipation curves for the two alternatives and see what the difference is. But I suspect there won't be much difference.

Reply to
Peabody

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