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CC2340R5: BLE MCU MicroController Capacitance Question

Chris Willis
Intellectual 290 points
Part Number: CC2340R5


In Table 8.3 of the datasheet, we will be powering the MCU with coin cell batteries, it recommends a 10-µF cap to the ground from VDDS.

 It recommends 0.1µF on each of the VDDS which is connected to our battery positive, so we have 4 x 0.1 = 0.4µF plus a 4.7µF and 1µF

 Should we add another cap to make the total 10µF? 

 Can you clarify what is meant by slew rate? Is this in reference to the fall time of a digital output including SPI communication? 

Due to the battery performance deteriorating at its end of life, the 10µF acts as a power buffer storage for when an output changes states creating an increase in current draw? (without the capacitor the battery could struggle to meet the demand causing the output rise time to increase?)

Thank you

  • +1
    TI__Expert 6520 points

    Hi Chris,

    See our reference design for an example. There are the 4 x 0.1 = 0.4µF for local decoupling and one 10µF on VDDS. That's the total we recommend.

    For slew rate, it's how fast the battery voltage drops in response to high current draw during TX or active events. It's not related to the digital output, it's the voltage seen at the VDDS terminals. We have an older app note about lithium coin cell battery effectiveness in pulsed current applications like a Bluetooth Low Energy radio. See https://www.ti.com/lit/wp/swra349/swra349.pdf The information in general is still valid for newer part. You are correct that it's due to the battery performance deteriorating at end of life. The capacitor is there to mitigate the lower voltage and higher internal resistance. Without this additional bulk capacitance you could lower VDDS to the extent to cause a brown out (VDDS < 1.71 V) during a high current draw event.

    Best Regards,

    Jake