Nrf51 Radio Peak TX (DC DC enabled at 0dbm) which is 10.5mA

[MARTIN FR]

Hello Everyone,

I am doing the rough calculation of the power consumption of NRF51 where in the radio part of the datasheet it is indicating 10.5mA whether it is denoted as 10.5mAh or can i cnsider in this way 0.0029mAh if am wrong please correct me about my assumptions does it means it will consume 10.5mA when it is transmitting ? please clarify this point

[Stian Røed Hafskjold]

Well, it simply means that the run current for the radio in TX mode (no DCDC, 0dBm) is 10.5mA. Nothing more than that. If you want to get the total current consumption (mAh) you need to include the time parameter in your calculations. How long the radio is in TX mode.

So lets say you are transmitting for 0.1ms, you will get a total charge of:

10.5 mA * 0.1 ms = 1.05 µC (or µAs if you want) (or even: 1.05µAs -> 0.00000029 mAh if that's more practical)

Your number, 0.0029mAh, is only valid when transmitting for a whole second.

EDIT

I think you are mixing up average current (mA) and total current consumption (mAh) here.

1. First you need to find the total current consumption for ONE advertising event. This is complicated to calculate. You have TX, RX, CPU, HFXO, LFXO, RTC... And it is dependent on voltage, TX power and payload (bytes you want to send). Please see this link to get an idea on all the peripherals included in an advertising event. But I can tell you that for the nRF51 this would be about 20 µC (µAs) (Full 20 byte payload, 0dBm @3V)

2. Then you will have to find the average current for your advertising interval. Let us say this is 500ms. 20µC / 0,5 s = 40µA. So now the chip will draw 40µA in average when you have an advertising interval of 500ms. If you choose an advertising interval of 1 second, you will get 20 µC / 1 s = 20µA in average.

3. Now you can calculate the total consumption over 3 years. 3 years equal 26280 hours. 40µA * 26280 h = 1051.2 mAh. If you choose 1 second advertising interval you will get 20µA * 26280 h = 525,6 mAh.

You can play with the power calculator for nRF52 to get an idea of how the different parameters affect the average current: devzone.nordicsemi.com/.../

[MARTIN FR]

yes thank for our kind reply where in the datasheet i found this information on datasheet (TX +I_CPU)(mA) =10.5mA + 4.1mA so totally it will consume 14.6 mA when simply advertising so could you tell me am transmitting continously so i calculated ((14.6mA *1sec)/3600)= 0.0040mAH my practical value is correct? radio power consumption during advertsing for 1sec

[Stian Røed Hafskjold]

Where did you get the TX + I_CPU from? Using Softdevice there should be no CPU running while the radio is transmitting. But if you are adding those two together you will get 14.6mA as you say, and 0,0040 mAh for a whole second.
BUT, when advertising, the radio is never in TX mode for a whole second, just a couple of hundred micro seconds every advertising interval.
infocenter.nordicsemi.com/.../advertising_event.html

[Stian Røed Hafskjold]

Is your question how much the average current consumption will be when advertising with a 1 second interval?

[MARTIN FR]

thank ou for your detailed reply so when radio is transmitting CPU will be in sleep mode so there wont be necessary to include cpu power when caluclation average power consumption during radio advertsing mode so can i consider 500ms for advertsing and then (10.5m*500ms/3600)so i will get round about 0.00145mAh does it seems meaningful calculations for considering the average power consumption because am trying to power my device for consecutive three years so am doing this rough calculations please suggest me what are the other things do i need to consider when calculating average power consumptions