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Maximum average power consumption / current when transmitting close to bandwidth (nRF52)

The power estimation tool on https://devzone.nordicsemi.com/power/ maxes out at a minimum interval of 7.5ms and a maximum payload of 27 bytes, with 824mA average current. As discussed on the blog post that was published with the tool, this is nowhere near the bandwidth of BLE 4.2 or BLE 5, but the tool has not been updated yet to reflect higher throughput.

Can someone who has tested this tell me what current or power consumption you see closer to the maximum bandwidth of BLE 4.2 or BLE 5?

Thank you!

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  • Hi, I'm working on updating the online power profiler now, and it should be out soon with BLE 5 support. You can also use the tool as it is now to calculate for higher bandwidths if you do some calculations on your own. For example if you want to enable data length extension in your application you can extend the radio lengths with 1 us for each bit (1Mbps). Then you have additional_length*radio_average_current = additional_event_charge. And additional_event_charge / ble_interval = additional_average_current.

  • Thank you! I should dig deeper into that...

    Looking at the max. packet size (27 bytes in BLE 4.0) the max. in BLE 5 is 257 bytes (251 payload). Playing with the power profiler, at the minimum 7.5 ms theoretical connection interval, the total average current seems to be a pretty linear function that's around (627+[num_bytes]*9) µA. So at 257 bytes, that'll be 2.94 mA. That would be 17kB/s with ACK; but I don't know how to do the math at the higher bitrate without ACK, which is probably the more relevant and lower current.

    Am I going in the right direction here, or am I way off?

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  • Thank you! I should dig deeper into that...

    Looking at the max. packet size (27 bytes in BLE 4.0) the max. in BLE 5 is 257 bytes (251 payload). Playing with the power profiler, at the minimum 7.5 ms theoretical connection interval, the total average current seems to be a pretty linear function that's around (627+[num_bytes]*9) µA. So at 257 bytes, that'll be 2.94 mA. That would be 17kB/s with ACK; but I don't know how to do the math at the higher bitrate without ACK, which is probably the more relevant and lower current.

    Am I going in the right direction here, or am I way off?

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