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LPW3 power: different between 1.8 and 3.6V supply?

Other Parts Discussed in Thread: MSP430F2274

Hi,

Power consumption usually is proportional to the square of voltage, and to frequency. However, device F2274 has a wide operating range of 1.8V to 3.6V, between the two end points the square differs by 4 times.

In operating mode, does MSP430F2274’s power consumption depend on power supply voltage?

In LPW3 mode, does MSP430 power consumption depend on its power supply voltage?

And do the answers above also apply to G series devices?

 

Matt

  • Matt Shira said:

    In operating mode, does MSP430F2274’s power consumption depend on power supply voltage?

    Yes.

     

    Matt Shira said:

    In LPW3 mode, does MSP430 power consumption depend on its power supply voltage?

    Yes.  Total power consumption generally has 2 components, static and dynamic.  When in Low Power Mode 3, the static power consumption due to quiescent current draw would be the dominant factor.

    Ptotal = Pstatic + Pdyn
    Pstatic = Istatic * V
    Pdyn = Idyn * V

     

    Matt Shira said:

    And do the answers above also apply to G series devices?

    Yes.

  • Brandon,

    Got it, thanks a lot.

  • BrandonAzbell said:
    Ptotal = Pstatic + Pdyn
    Pstatic = Istatic * V
    Pdyn = Idyn * V

    Well, I might add that Istatic isn't static. While the change is less significant as on a pure resistive system, the leakage current still somewhat depends on supply voltage.

    On a resistive system Pstatic = Istatic * V = V * V / R.
    On semiconductors, the term I=V/R isn't true anymore. Because R is something non-linear depending on meterial, temperature, even the current itself, sorrounding ME fields etc.

    However, the smaller V is, the smaller the leakage currents are - just that it is a highly non-linear relation.

    So a smaller voltage does not only decrease power because power is I*V, but also decreases I itself, because I a little bit depends on V too.

    Thios is true for all semiconductors. But due to the large numer of influences, this is difficult to calculate.

  • Jens,

    We don’t really have this knowledge. Thanks for the comment!

     

    Matt

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