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MSP430FR6877: Current consumption when MCLK = HFXT

Cruijff
Expert 2760 points
Part Number: MSP430FR6877

Hi community member,

In MSP430 datasheet, all current consumption is specified as fMCLK = fDCO.
Do you think that the current consumption when fMCLK = fHFXT is larger than fMCLK = fDCO?
I think as follows.

Current consumption of HFXT when fHFXT = 4MHz: IDVCC, HFXT = 75uA (Typ.)
Current consumption when DCO is 1MHz: IAM, FRAM_uni = 210uA (Typ.)
When fMCLK = fHFXT: IDVCC, HFXT + IAM, FRAM_uni

Is this idea wrong?
Please tell me the correct way of idea.

Best regard.
Cruijff

  • 0
    TI__Mastermind 27560 points

    Hey Cruijff,

    Everything looks right to me. The HFXT will consume more current than the DCO, as the circuitry to drive and manage the HF crystal has to be enabled. 

    Your interruption of the tables also appears correct.  The IAM table is based on the DCO frequencies.   Then the IDVCC.HFXT would need to be added to this when using an HF crystal.  

    If you were feeding in a square wave and had HXFTBYPASS set, then the HXFT circuitry would be shut down and this power would not be consumed.  

    Thanks,

    JD

  • 0 in reply to JD Crutchfield
    Expert 2760 points

    Hi JD,

    Thank you for your reply!

    I understood that my idea is correctlly.

    Thank you

    Cruijff

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