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  • Case ID: 328431
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nRF52805-Power consumption decreases with the addition of a 10uF capacitor

sunmnengnan

HI  all,

  I've found that adding a 10uF capacitor to the 52805 beacon reduces the power consumption, especially after speeding up the broadcast interval (100ms broadcast interval).

 The conclusion is that the reason for the high power consumption without capacitors is the high total power consumption during broadcast (average current of one transmit wave peak * peak duration)

 btw, beacon with chip and peripherals only,the firmware is Nordic official routine, it can broadcast only, 100ms, 4dBm.

 We also tested the 52810 beacon and found the same phenomenon, but we don't have a more accurate power consumption test instrument, I'm not quite sure if it's the addition of this capacitor that reduces the power consumption, or if the power consumption board (PPK2) can't measure it.

Nordic 52805 beacon带不带10uF电容功耗对比测试.xlsx

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  • 0

    Hi,

     

    Adding a capacitor will smooth out the peak currents, which might be the reason why the PPK2 has slight variations in this case. Adding such a capacitor shall not improve the overall average current consumption.

    I would recommend that you use another multimeter to verify the setup here.

     

    In addition; could you try to connect the PPK2 either directly to the USB port in the PC, or via a hub?

    Do you then see any differences between the two scenarios? USB noise can propagate to the actual ppk2 measurements.

     

    Kind regards,

    Håkon

  • 0 in reply to Håkon Alseth

    HI

      I measured the power consumption of the PPK2 with a direct connection to the computer and it was about the same as plugging it into the hub.

      We don't have any other current test tool here, can you help us to test it (using ppk2 and power score to test the broadcast average and peak current of the Nordic52 chip and the surrounding circuits respectively), it will help us a lot.!

  • 0 in reply to sunmnengnan

    I would recommend that you try to get ahold of a multimeter or similar and check the current consumption to see if this also shows a ~5-10 % difference when using a capacitor or not.

    It could be related to floating inputs on the design, or the state of any external ICs that you might have on your custom board.

     

    Kind regards,

    Håkon

  • 0 in reply to Håkon Alseth

    Sorry, we don't have any other instruments to test the current.

    The beacon only has the ic and the capacitance and inductance around it.

  • 0 in reply to sunmnengnan

    The expected current consumption would be approx. 116 uA for the scenario of 3.0V, DCDC ON, +4 dBm, and 27 byte payload in TX only scenario:

     

    I used this calculator:

     Online Power Profiler for Bluetooth LE 

    I do now know how many bytes you transmit, but please plot in your payload length, and compare against that.

     

    I would still recommend that you obtain a multimeter and measure your average current using this, just to see if there's anything problematic with your PPK2.

    PS: Do the PPK2 measurements with the highest sampling rate. You currently have it at 10kHz, try 100 kHz sampling rate to see if this affects the measurement.  

     

    Kind regards,

    Håkon

  • 0 in reply to Håkon Alseth

    HI! There is no difference in average power consumption when adjusting PPK2 sample rate.Also I've found that when I cover the pcba's antenna with an object (whether it's metal or not), the power consumption increases.I would like to know if the transmitter side of the chip regulates the transmit power when the antenna reflection loss is high, and if this is the reason for this increase in power consumption.

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  • 0 in reply to Håkon Alseth

    HI! There is no difference in average power consumption when adjusting PPK2 sample rate.Also I've found that when I cover the pcba's antenna with an object (whether it's metal or not), the power consumption increases.I would like to know if the transmitter side of the chip regulates the transmit power when the antenna reflection loss is high, and if this is the reason for this increase in power consumption.

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