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I'm designing a battery powered circuit where I would like to charge the battery via USB, and monitor the remaining battery capacity during operation.

Schematic

I'm using the following IC's:

  • MCP73831 (charger)
  • BQ27441 (gauge)
  • AP7217 (LDO)

The LDO can be driven from the battery or USB. If the both battery and USB are connected, the MOSFET Q3 should disable the battery from the LDO and let the USB drive the LDO while also charging the battery. Everything works fine except one thing that I don't understand.

If I only connect a USB cable to the device, the yellow diode D11 is active, meaning that MCP73831 is charging the (disconnected) battery. Looking in the datasheet of MCP73831, description of battery detection:

"A 6 μA (typical) current is sourced by the VBAT pin to determine if a battery is present or not. If the voltage at VBAT rises to VREG + 100 mV (typical), the device assumes that a battery is not present. If the voltage stays below VREG + 100 mV (typical), the device assumes that a battery is detected. In order to correctly detect a battery insertion, the impedance seen by the VBAT pin before the battery is connected must be greater than 2 MΩ."

I have two questions. The first one is about the theory of the quote above. Why would V_BATT rise above VREG(4.2V) + 100mV when the circuit is sinking this small current and no battery is connected?

Secondly, why would V_BATT in my circuit NOT rise above this voltage and thus make the charger believe that a battery is (falsely) detected?

Thanks :)

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  • \$\begingroup\$ If you look at the operating current spec for U9 (Fuel Guage) you will see why the VBAT net is being loaded down. \$\endgroup\$ – sstobbe Oct 12 '18 at 14:49
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I once had the same problem with other chip. In my case it was a bulky cap I used on bat pin that made the chip think that there was a bat even when it's not. try remove C33 in your circuit and it may solve your problem

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