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BQ24780S: PMON related question

Antony Lin
Genius 13480 points
Part Number: BQ24780S

Hi,

We have some PMON related questions as listed below.

  1. If we want to use RAC and RSR both as 20mohm, should I configure 0x3BH[13:12] as "00" (RAC  & RSR 1:1)?  And then I can choose 0x3BH[9] value as I want?
  2. Can you double confirm PMON output current is limited within 160uA?

Thanks!

Antony

  • 0
    TI__Expert 5330 points

    Hi Anthony,

    When you are using 20 mOhm for both RAC and RSR, you are correct that you should configure 0x3B[13:12] as 00b (1:1 ratio.)

    Because these are both 20 mOhm instead of 10 mOhm, the two selections of 0x3B[9] will provide twice the gain listed (which is listed for 10 mOhm sense resistors).  The gain will therefore be as follows:

    0b = 0.5 uA/Watt

    1b = 2.0 uA/Watt

    You are correct that PMON max output is 160 uA.  With a setting of 0b in register 0x3B[9], this will allow you to represent up to 320 Watt of system power with 20 mOhm resistors.  Keep in mind that 100 pF max capacitance is recommended for the PMON pin.  Using I = C dv/dt, this capacitance allows a full scale change from 0 to 3.3 V in 2 uS.

    Regards,

    Steve

  • 0 in reply to Steve Preissig
    TI__Genius 13480 points

    Hi Steve,

    When RAC and RSR are both configured as either 10mohm or 20mohm, 0x3B[13:12] value is configured with the same value as 00b.  So, how would the chip know it's the case of 10mohm or 20mohm to judge the gain should be the higher or lower set as defined in 0x3B[9] ?

    THanks!

    Antony

  • 0 in reply to Antony Lin
    TI__Expert 5330 points

    Hi Anthony,

    The chip doesn't actually need to know whether the sense resistors are both 10 mOhm or whether they are both 20 mOhm, it just needs to know their ratio.

    The calculation that the chip performs for PMON is (depending on the value of 0x3B[9]):

    Iout = (0.25 uA/Watt) * (V_ACN * (V_ACP - VACN) - V_SRN * (V_SRP - V_SRN)) / 10 mOhm 

    or

    Iout = (1 uA/Watt) * (V_ACN * (V_ACP - VACN) - V_SRN * (V_SRP - V_SRN)) / 10 mOhm 

    If you use 10 mOhm resistors, then this is equal to

    Iout = (0.25 uA/Watt) * (V_ADPT * I_ADPT - V_BATT * IBATT)  

    or

    Iout = (1 uA/Watt) * (V_ADPT * I_ADPT - V_BATT * IBATT) 

    But if you use 20 mOhm resistors, then this is equal to

    Iout = (0.5 uA/Watt) * (V_ADPT * I_ADPT - V_BATT * IBATT)  

    or

    Iout = (2 uA/Watt) * (V_ADPT * I_ADPT - V_BATT * IBATT) 

    So when you use 20 mOhm resistors, the gain factor of the PMON output will be doubled.  But the BQ24780s doesn't need to know the value of the sense resistors in order to correctly calculate PMON, it only needs to know their ratio.  Then the absolute value of the sense resistors will determine the gain, i.e. for default 0x3B[9]=1, 10 mOhm resistors will have gain of 1 uA/Watt, whereas 20 mOhm resistors will have gain of 2 uA/Watt.

    Regards,

    Steve