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

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

Stian Røed Hafskjold — Mon, 30 May 2016 13:57:52 GMT

Yes, you can add the idle current like you are doing. But you must look at the System ON IDLE current (I_ON,16k or I_ON,32k) 2.6 or 3.8µA (depending on how much RAM retention you need) to get the sleep current. System OFF (0.6 µA) will turn off the chip and it can only be turned on again by an external device.

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

MARTIN FR — Mon, 30 May 2016 13:43:12 GMT

Thank you for your detailed response so i saw in the datasheet 0.6uA in the off mode so sleep current will remains same or it will change accordingly so i can aadd average current consumption 20uA plus 0.6uA sleep curent totally 20.6uA curretn will be consumed for one advertsing event

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

Stian Røed Hafskjold — Mon, 30 May 2016 13:41:54 GMT

Yes, you should add the sleep current as well to the average. About 4uA. Not sure why the link is not working. It works here. You can find the page here: "nRF51 Series > SoftDevices > S130 SoftDevice > S130 SoftDevice Specification v2.0 > BLE power profiles"

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

MARTIN FR — Mon, 30 May 2016 13:31:21 GMT

Thank you for your detailed answer and we dont need to calculate the sleep current when considering the average pwoer consumption? could you please send the detailed link because the above link which you have mentioned to see the parameters for advertsing event it goes to general page am not able to find the exact topic could you show me the link in detail please

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

Stian Røed Hafskjold — Mon, 30 May 2016 13:04:12 GMT

Please see my edited answer

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

MARTIN FR — Mon, 30 May 2016 12:29:00 GMT

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

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

Stian Røed Hafskjold — Mon, 30 May 2016 12:26:20 GMT

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

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

Stian Røed Hafskjold — Mon, 30 May 2016 12:21:55 GMT

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

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

MARTIN FR — Mon, 30 May 2016 12:08:43 GMT

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

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

Stian Røed Hafskjold — Mon, 30 May 2016 11:56:16 GMT

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.

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)
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.
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/.../