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I'm going through this application note - because it is said that the MCP73833 device does not allow to charge the battery and drive the load simultaneously.

Hence, they've given the below circuit in the above application note. I'm having trouble finding understanding the working of the protection circuit marked, which allows the input voltage to disconnect the load from the battery while simultaneously driving the load itself.

Can someone explain how the protection circuit works?

enter image description here

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1 Answer 1

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Pg.10 of the application note mentions that when power is supplied, it will charge the battery and the load will be supported by the power source as well.

power path management

When a power source is present, the gate of the P-Channel MOSFET Q1 is pulled high which turns Q1 off disconnecting the battery from the load while the source also supports the load through D1. MCP73833 charges the battery through VBAT at the same time.

When a power source is not present, the gate is pulled down through the resistor Rpull which turns Q1 on and the battery supplies power to the load through Q1.

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  • \$\begingroup\$ Thank you for your answer. Could you please let me know the Vgs of the MOSFET and the voltages at the source and the gate as an example along with the purpose of D1? \$\endgroup\$
    – user220456
    Commented Feb 3, 2023 at 10:21
  • \$\begingroup\$ Also, even if the input is present that charges the battery and supplies current to the load, there is a leakage path through the body diode of the MOSFET, right? Wouldn't that cause a problem? \$\endgroup\$
    – user220456
    Commented Feb 3, 2023 at 10:22
  • \$\begingroup\$ The Vg would be equal to the Vin while the Vs would be Vin - forward voltage of D1. D1 prevents the flow of current to gate when supply is not present. Body diode won't cause a problem as Vs will be higher than the battery voltage. \$\endgroup\$ Commented Feb 3, 2023 at 10:47
  • \$\begingroup\$ Thank you for the answer. Could you please tell me a value of Rpull? \$\endgroup\$
    – user220456
    Commented Feb 5, 2023 at 17:18

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