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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.

  • 0

    Hi Sir, Finally we again decide to use momentary switch in my device so now as per my previous question i added circuit diagram here Please sir see again as per your suggestion for use high value resistor i have few question on same following:

    1. If i use 10K for protecting internal diode in nRF52832 SoC is it Ok, then by adding this resistor will be consume more current or this will not affect on current consumption.
    2. Also my confusion is if instead of resistor i use diode for protection reverse voltage at the time of user press switch is it work for long time, or it will consume more current or not due to use diode.

    Please sir let me know the above question asap, now we are in final stage design. Thanks.....

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    1. If anything, a resistor should help you save current.
    2. That might be an even better solution.
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    Thanks sir for your reply, I use 10K resistor it is better for me, One of the other problem is for doubler IC MCP1640 i test the current consumption of my single AA battery. The result is different consumption like if i put full charge battery i measure current at MCP1640 input side is 15mA. 2. If i put 1.2V battery current consume is 22mA. 3. If i put low means 1 V battery charge battery the current consumption is increase near about 40mA. Why is happen i think it depend on MCP1640 efficiency, is it possible to increase doubler efficiency upto 90%. As per my measurements as battery voltage discharge the current consumption is increase. Please sir how i can avoid this problem. Because we need to use single AA 1.5v Battery for power supply due easily accessible in India and also battery life is important. Thanks........

  • +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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