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Is it ok doubler circuit for powering nRF52832 custom board

vishal

Hello Everyone,

I have developed my application using BLE for connect android mobile. I have design complete PCB using your reference design. But i have littele confused in power supply schematic because i have use MCP1640 doubler IC, and use only one AA 1.5V battery and converted to 3.3V. my question is i have this attached schematic to powering Nordic chip. Is it sufficient for continuously power to nRF52832 or i need to add any other components with this circuit. I just verify this circuit and hope it will not cause problem in future. I have tested this circuit in general purpose PCB its working fine and that's why i am using in my final design. Any extra energy consumption due this doubler circuit my device current consumption is 4mA-6mA. Please give your valuable suggestion for should i use thisimage description.

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  • +1

    It is basic electronics. The power supplied to your voltage converter is equal to the power flowing out of your converter (minus some loss inside the converter).

    But for the sake of argument, let us say that you have an ideal converter with no loss and that your nRF52 draws a constant current at a constant voltage. Power out of your converter is P=UI. The same goes for the power "pushed" into your converter. If you look at the equation you should see that if the battery voltage decreases then the current has to increase to deliver the same amount of power.

    In addition, if you read the datasheet of the MCP1640 you will see that different input voltage will give different efficiency. Lower Vin = lower efficiency. As a cause of this your battery will have to push even more current into the converter to meet the power demands of the nRF52.

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  • +1

    It is basic electronics. The power supplied to your voltage converter is equal to the power flowing out of your converter (minus some loss inside the converter).

    But for the sake of argument, let us say that you have an ideal converter with no loss and that your nRF52 draws a constant current at a constant voltage. Power out of your converter is P=UI. The same goes for the power "pushed" into your converter. If you look at the equation you should see that if the battery voltage decreases then the current has to increase to deliver the same amount of power.

    In addition, if you read the datasheet of the MCP1640 you will see that different input voltage will give different efficiency. Lower Vin = lower efficiency. As a cause of this your battery will have to push even more current into the converter to meet the power demands of the nRF52.

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