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I've found this useful explanation for a load sharing circuit based on the established TP4056 charger IC:

LINK: https://www.youtube.com/watch?v=T70mBHeIOZA&ab_channel=WellDoneTips

SCREENSHOT: enter image description here

Now I would like to know why there is a SS14 Schottky diode and a 0.4 ohm resistor between the USB connector and the TP4056 charger IC.

And would it work without it too?

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It is presumably to avoid power from the cell discharging into the USB port.

Particularly of the host/charger is powered off.

The 0.4 ohm resistor will help reduce the power dissipated within the IC at high charging currents.

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  • \$\begingroup\$ I already heard that this module could get very hot, the more it is interesting how to solve this/such problem(s). Thank you \$\endgroup\$ – playmobox Feb 19 at 21:32
  • \$\begingroup\$ @playmobox I believe it regulates the output current to avoid the device getting above the max temperature of ~125C. \$\endgroup\$ – Kevin White Feb 19 at 21:34
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And would it work without it too?

Probably, depending on what is inside the 'TP4056 charger' block.

The resistor won't be needed if the USB cable has sufficient resistance, and even if it doesn't all that will happen is the TP4056 charger IC gets hotter and has to reduce the charging current so the battery takes longer to charge.

The diode stops current back-feeding from the charger into the 10 k resistor attached to the FET Gate, ensuring that it turns on when there is no USB power. The TP4056 itself has less than 2 μA of back current, which produces less than 20 mV across a 10 k &ohm resistor.

However the charger module may have a large capacitor on the USB input which could take a while to discharge, causing a relatively long delay before the FET turns on. Before the FET turns it has a voltage drop of ~0.7 V across its internal body diode. If the battery has a low charge state then the device being powered might not get enough voltage during the changeover, causing it to glitch.

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  • \$\begingroup\$ Does this mean that the diode solves the current back-feeding UNLESS there is an internal capacitance in the charger block? \$\endgroup\$ – playmobox Feb 19 at 21:27
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    \$\begingroup\$ No, it means the diode solves the current back-feeding IF it would otherwise occur (whether due to internal capacitance or some other reason). So if that could be a problem you should leave the diode in. \$\endgroup\$ – Bruce Abbott Feb 20 at 2:50
  • \$\begingroup\$ Okay, thank you for clearifying this. Now I know that it is better to add those components to such a circuit in order to prevent several negative side effects. \$\endgroup\$ – playmobox Feb 20 at 5:12

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