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How to configure cell battery and USB power

Serafino Convertini

Hi to all,

this is my first public ticket ;)

I would power a nRF52840 on my custom board both with a cell battery (specifically CR2032) and the VBUS provided by the USB connection.
I was thinking to use two schottky diode in the way showed in the attached image with using the internal DCDC of the nRF52840 as in the development board.

But I'm not sure to use schottky diode since the voltage drop on the diode. In the proposed configuration for the nRF52840, I cannot find any solution that could fit with this question.

Has anyone done this task before?

Thanks in advance

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  • 0

    My grandfather used schottky diodes, nowadays there are better low-drop solutions. Have a look at these:

    max40203 Ideal Diode

  • 0 in reply to hmolesworth

    Let's assume that using MAX40203 or TCK106 (as used in nRF52840-DK). Does make sense to connect the VBUS (when present) to VDDH and VBATT (qhen present) on VDD, or should I connect the output voltage (VBUS or VBATT) to VDDH?

    In both case I think I configure the nRF power stage using internal DCDC.

  • 0 in reply to Serafino Convertini

    VBUS from USB complicates things as it is +5V and therefore must connect to VDDH. VDD when VBUS is present will output the programmed voltage on REG0 and so REG0 must be set to the maximum expected CR2032 voltage to ensure no power is taken via the MAX40203 by the CR2032. This implies i/o pins and external circuitry will be at (say) 3.4 volts with VBUS present; internal MCU and RADIO will be 1.3V. With VBUS absent the CR2032 will provide power via MAX40203 to VDD; this implies i/o pins and external circuitry will be between (say) 3.4 volts and 2.6 volts with VBUS not present; internal MCU and RADIO will always be 1.3V. Pulse currents will cause even lower voltages on the CR2032-driven VDD, and so a lot of bulk capacitance is required (ceramic, 3 x 3.4V or say 10V rating). In this scenario, typically the MAX40203 is only required on the CR2032 battery assuming that when VBUS is absent nothing connects to the USB. Adding a MAX40203 to the VBUS input provides better protection against potential back-drive of stuff on the USB port.

    The alternative is to use an external 3.4V DC-DC on VBUS and only attach that to VDD with a MAX40203 together with the CR2032 battery; probably a less power-efficient (and more expensive) solution

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  • 0 in reply to Serafino Convertini

    VBUS from USB complicates things as it is +5V and therefore must connect to VDDH. VDD when VBUS is present will output the programmed voltage on REG0 and so REG0 must be set to the maximum expected CR2032 voltage to ensure no power is taken via the MAX40203 by the CR2032. This implies i/o pins and external circuitry will be at (say) 3.4 volts with VBUS present; internal MCU and RADIO will be 1.3V. With VBUS absent the CR2032 will provide power via MAX40203 to VDD; this implies i/o pins and external circuitry will be between (say) 3.4 volts and 2.6 volts with VBUS not present; internal MCU and RADIO will always be 1.3V. Pulse currents will cause even lower voltages on the CR2032-driven VDD, and so a lot of bulk capacitance is required (ceramic, 3 x 3.4V or say 10V rating). In this scenario, typically the MAX40203 is only required on the CR2032 battery assuming that when VBUS is absent nothing connects to the USB. Adding a MAX40203 to the VBUS input provides better protection against potential back-drive of stuff on the USB port.

    The alternative is to use an external 3.4V DC-DC on VBUS and only attach that to VDD with a MAX40203 together with the CR2032 battery; probably a less power-efficient (and more expensive) solution

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