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Current consumption increases with decreasing the input voltage

Hi Team,

We are using nRF52840 DK with sensors. We developed a custom code that is working fine. But the issue is current consumption. Showing below the readings.

The supply is given through the external input supply(P21), moved the SW10 to ON, Moved the SW6 to nRF only

  • With 3.3 input voltage observed peak 4.53 mA
  • With 1.8 input voltage observed peak 6.85 mA

Then why the current consumption increases with decrease in input supply. Not only peak the average current consumption also increases.

Then I read the answer regarding this issue at https://devzone.nordicsemi.com/f/nordic-q-a/30320/nrf52-current-consumption-in-relation-to-supply-voltage.

But now I would like to know that we will measure the battery life based on the numbers which we get at 3 v, if as the battery discharges the board consumes more power where the battery is critical and the life span we expected will decreases drastically. so how we can conclude this?

As Nordic mentioned that operating voltage is 1.8 to 3.6 so when we give the supply with in the range the current consumption should not alter right. 

Thanks,

venkatesh

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  • Hi

    This is the nature of how a DCDC regulator is working. A DCDC regulator will be more efficient the higher the input voltage is, as it'll need to consume more current in order to reach the same power levels using a lower voltage according to the Power [P]= Current [I] * Voltage [U] equation. This curve won't be as steep using the LDO regulator I believe, as it has a constant drop-out voltage, making it more stable. But the DCDC regulator will be more power-efficient in most scenarios if I'm not mistaken. 

    Best regards,

    Simon

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  • Hi

    This is the nature of how a DCDC regulator is working. A DCDC regulator will be more efficient the higher the input voltage is, as it'll need to consume more current in order to reach the same power levels using a lower voltage according to the Power [P]= Current [I] * Voltage [U] equation. This curve won't be as steep using the LDO regulator I believe, as it has a constant drop-out voltage, making it more stable. But the DCDC regulator will be more power-efficient in most scenarios if I'm not mistaken. 

    Best regards,

    Simon

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