Depending on the nRF52832 supply voltage, can the quiescent current vary?~

hello.

We are considering an external LDO that supplies VDD to the nRF52832.

I'm reviewing an LDO called TPS7A02, but there's a phenomenon I don't understand.

When using TPS7A0228 (LDO output 2.8V), the standby current of the board with nRF52832 is measured to be very low, at 1uA ~ 20uA.

On the other hand, when using TPS7A0230 (LDO output 3.0V), 20uA is added and 21uA ~ 40uA is measured.

TPS7A0228 : https://www.mouser.kr/ProductDetail/Texas-Instruments/TPS7A0228PDBVR?qs=sGAEpiMZZMug9GoBKXZ751mAF56Mq1BpFfw3x6s4ZMIdLDEJ2sm0Ow%3D%3D

TPS7A0230 : https://www.mouser.kr/ProductDetail/Texas-Instruments/TPS7A0230PDBVR?qs=sGAEpiMZZMug9GoBKXZ751mAF56Mq1BpAesZnDPtYLWRF%2FNPd%2FJnPQ%3D%3D

There is no current difference in the TPS7A02 LDO Datasheet.

Does the standby current consumed by the nRF52832 vary depending on the VDD voltage (LDO output)?

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  • I just measured the standby current on a bare nRF52832 (no port i/o pins connected to anything) at both 2.800 volts and 3.000 volts, no change in current measured with a 5-digit meter with 10nA resolution:

     1.12uA  at 2.8 and 3.0 volts - some errata applied in SystemInit()

     1.01uA  at 2.8 and 3.0 volts - no errata applied

    I should add that at this low level of current the hysteretic mode of the LDO is enabled, and there is no information that I know of regarding the current at which LDO hysteretic mode is disabled and normal continuous LDO mode takes over (this is not related to DC/DC mode, which is not enabled in this test).

  • I forced the LDO into normal continuous LDO mode by driving output ports low against the internal pull-up using high drive H0H1 and find there is no increase when the consumption is a steady 1mA approx (5 output ports driven low against internal 13.33k pull-up). Since current into such a constant resistance depends on voltage, lowering the voltage lowers the current, but it is as expected with no change in LDO Iq.

    #define NUMBER_PINS_TO_DRIVE 5 //32
    static void TestPower(void)
    {
       // Set up to 32 output pins
       for (uint32_t i=0; i<NUMBER_PINS_TO_DRIVE; i++)
       {
            // Configuration      Direction    Input            Pullup       Drive Level      Sense Level
            // ===============    ==========   ==============   ==========   ==============   =============
            NRF_P0->PIN_CNF[i] = (PIN_OUTPUT | PIN_DISCONNECT | PIN_PULLUP | PIN_DRIVE_H0H1 | PIN_SENSE_OFF);
            NRF_P0->OUTCLR = (1UL << i);
       }
       // Sleep
       while (1)
          {
             __WFE();
             __NOP(); __NOP(); __NOP(); __NOP();
          }
    }

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  • I forced the LDO into normal continuous LDO mode by driving output ports low against the internal pull-up using high drive H0H1 and find there is no increase when the consumption is a steady 1mA approx (5 output ports driven low against internal 13.33k pull-up). Since current into such a constant resistance depends on voltage, lowering the voltage lowers the current, but it is as expected with no change in LDO Iq.

    #define NUMBER_PINS_TO_DRIVE 5 //32
    static void TestPower(void)
    {
       // Set up to 32 output pins
       for (uint32_t i=0; i<NUMBER_PINS_TO_DRIVE; i++)
       {
            // Configuration      Direction    Input            Pullup       Drive Level      Sense Level
            // ===============    ==========   ==============   ==========   ==============   =============
            NRF_P0->PIN_CNF[i] = (PIN_OUTPUT | PIN_DISCONNECT | PIN_PULLUP | PIN_DRIVE_H0H1 | PIN_SENSE_OFF);
            NRF_P0->OUTCLR = (1UL << i);
       }
       // Sleep
       while (1)
          {
             __WFE();
             __NOP(); __NOP(); __NOP(); __NOP();
          }
    }

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