nrf54l15-DK Power Consumption Test

RE: nrf54l15-DK Power Consumption Test

Håkon Alseth — Thu, 13 Feb 2025 09:37:22 GMT

Hi,

You can adjust the intervals for both advertising and your connection. See the bluetooth dev academy course for how to do this:

https://academy.nordicsemi.com/courses/bluetooth-low-energy-fundamentals/

Kind regards,

Håkon

RE: nrf54l15-DK Power Consumption Test

pintai — Thu, 13 Feb 2025 09:17:44 GMT

Hi,

Yes, the VDD measurement is also 1.8V.Sorry, I probably didn't express my needs clearly before, I was trying to test nrf54l15 lowest power consumption.I need the nrf54l15 to stay connected to the phone all the time and the power consumption needs to be 5uA.But I can't find a project that I can use as a reference right now.

RE: nrf54l15-DK Power Consumption Test

Håkon Alseth — Wed, 12 Feb 2025 09:45:25 GMT

Hi,

You are assuming no payload and no scan responses from nearby scanners. If you set the payload to > 20 bytes, and assume that there will be a scanner or two, meaning that the peripheral will send a scan_response back, it will extend the current consumption closer to 100 uA, depending on how much activity there is within range.

Remember to check what your voltage is, by measuring the VDD here:

On mine, it is 1.8V, so adjust that accordingly in the calculator, and you get approx. 100 uA with a 20 byte payload.

Kind regards,

Håkon

RE: nrf54l15-DK Power Consumption Test

pintai — Wed, 12 Feb 2025 05:08:22 GMT

Hi，

Thank you for your reply.With Uart disble,advertising power consumption is 110uA and the connected power consumption is 210uA, which is much higher than the results of the online power calculation tool.I've set the connection interval to 100ms.What do I have to do to get the lowest power consumption.

RE: nrf54l15-DK Power Consumption Test

Håkon Alseth — Tue, 11 Feb 2025 09:54:35 GMT

Hi,

This is likely due to the UART being enabled. Can you try to set "CONFIG_SERIAL=n" in your prj.conf file and re-do the measurement?

Due to the bluetooth advertisement, you should then see approx. 100 uA on average.

Kind regards,

Håkon