ESP8266 Community Forum

- Sat Jun 11, 2016 9:06 pm #49035 Hi , i would like to know how much time will my project can work on 3 * AA Battery 2500mA.
So i know the current consumption of each of my devices in the circuit for example

esp8266
distance sensor
LDO

My project going to weak up each 5 min run the distance sensor and send information to server if needed .

QUESTIONS :

1. How can i find out how much current will be decreased if i turn off the led on the esp ? is it possible ?

2. I know that the equation is :
Battery Life = Battery Capacity in Milli amps per hour / Load(Draw,consumption) Current in Mill amps * 0.70
What is the 0.7 stands for ?

3. Is the table below means mA per hour ? i mean if my program runs for 1 min each 5 min how should i calculate ?


4. is the LDO consuming current ? if yes which one can you advice that consumes minimum current ?
Or maybe i can use 2AA bataries which is 3V w/o LDO ?

5. What about current consumed by the esp when it goes from sleep to weak mode ,
Where can i find the current consuming due to this operation and how much time it takes?

6. Whats the meaning of 2500mA written on battery ? is it means that it can hold circuit where all devices in consumption
current in total is <=2.5A ? or it means that the battery supply's always 2500mA to the circuit ?
so for calculating total current consumption i should measure each device individually ?

Thanks !

- Sun Jun 12, 2016 2:46 am #49049 Hi @stavbodik,

1. How can i find out how much current will be decreased if i turn off the led on the esp ? is it possible ?

I would make a test with a multimeter by measuring current drawn by ESP when the LED is on and off.

2. I know that the equation is :
Battery Life = Battery Capacity in Milli amps per hour / Load(Draw,consumption) Current in Mill amps * 0.70
What is the 0.7 stands for ?

This formula looks to me like some estimate assuming certain way you discharge your battery. Check out this tutorial - https://www.youtube.com/watch?v=R8hTQXqURB4

3. Is the table below means mA per hour ? i mean if my program runs for 1 min each 5 min how should i calculate ?

This table does not mean mA per hour. It provides typical current in mA drawn by ESP in any given moment of time when running in certain mode. To calculate used battery capacity just multiply current by time ESP is running in this mode. For instance, if your ESP is running in Modem-Sleep mode for 5 min it will consume 5 min / 60 min/h x 15 mA = 1.25 mAh energy from battery.

4. is the LDO consuming current ? if yes which one can you advice that consumes minimum current ?
Or maybe i can use 2AA bataries which is 3V w/o LDO ?

5. What about current consumed by the esp when it goes from sleep to weak mode ,
Where can i find the current consuming due to this operation and how much time it takes?

There is a lot of information on the net how to minimise energy consumption / maximise battery life when running an ESP on it – to start with check out this “Classic” tutorial by Andreas Spiess - https://www.youtube.com/watch?v=6SdyImetbp8

6. Whats the meaning of 2500mA written on battery ?

Check out this tutorial - https://www.youtube.com/watch?v=R8hTQXqURB4

- Tue Jun 14, 2016 5:11 pm #49198 Hi and thanks for the great answer and tutorials , question about the battery tutorial .
According to battery discharge and power supply to use for project :



So in esp8266 case the voltage must be between 3.3v<E<3.6v , we cant use power supply which is higher then 3.6v right ?
Because he is saying if your project working voltage is 0.8v ( for the battery in example) the best is to use is 1 cell ,
But you cant do it when your device can burnout at 1v for example right ?
Even if i use 9v power supply and LDO it is not efficient because the esp will get his 3.3v which is bad , hmm did i misunderstood something ?


Hmm i saw now again the first tutorial where did the 14.8v came from ? and why he calculate like this and not battery capacity / device consumption ?

[RichardS]

- Tue Jun 14, 2016 6:20 pm #49200 I have had ESP8266's work down to 2V not sure if it will start again, however... it did function down to 2V

RichardS

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