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Discovery current consumption
I have a question regarding the current consumption of the discovery mode.
The system that I am designing will use have two BLE devices, one stationary and one mobile. When I will want to connect the devices I will press a button on the mobile device, whereas the stationary device will have no buttons.
The first obvious solution is to use the stationary device as an advertising device and when the button is pressed on the mobile device, it would discover the stationary device and connect to it.
But it is crucial that the battery on the stationary device lasts as long as possible, so I am also considering the alternative, where the stationary device is constantly discovering devices, and the mobile device starts advertising when the button is pressed.
Now the question is, which solution will use less current and thus prolong the battery life of the stationary device?
If I understand correctly the discovery of devices consists only of listening on the advertisement channel, and when an advertisement is received, then the scan request is sent. So if this is true, I presume only listening on the advertisement consumes less current than advertising, or does the advertising actually consume less current than listening?
from the CC2540 datasheet:
Please click the Verify Answer button on a post if it answers your question! :)
Thank you for your answer, I have of course checked the datasheet and the numbers, so I will rephrase the question.
Let's suppose we have a stationary advertising device that advertises once every 5 seconds and a mobile scanning device which begins scanning when the button is pressed, and the duration of scanning is 20 seconds. If it finds the advertising device, it connects to it.
So the advertising power consumption is understandable to me, every 5 seconds a short transmission of data will occur, which will use ~25mA for the duration of the transmission.
What I do not quite understand is the power consumption of the scanning device. Will the scanning device
a) consume ~19.6 mA for the entire duration of 20 seconds, or
b) will it only consume ~19.6 mA when it is receiving the advertisement (a short duration).
I presume the answer is b), please correct me if I am wrong. If the answer is b) then what I would like to know is, what is the current consumption of the device when it is scanning (listening on the advertisement channel) but not receiving an advertisement.
If the answer is a), then I have to make the stationary device the advertiser, since short periodic transmissions will use less power than continuous scanning (listening). If the answer is b), I need more data, before I can decide.
I will send you a private message, explaining what this will be used for.
Much clearer now :)
First of all, have you read this? It is quite complete, it will take no more than an hour to read it.
Yeah, YOU Matej can send me a private message. If you are this kind, please give me a few indications about the project nature as well as what this project is part of (Master thesis, PhD, commercial development...). If you can also indicate me which institution are you in, it would be very nice :)
Thank you, I have not yet seen this.
As long as You turn on the radio Your device will draw ~19,6 mA. It has nothing to do with the actual transmitting or receiving actions but rather with internal circuitry current consumption. So if I understood correctly I believe you are faced with option a)
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