How do I calculate the battery life of ESP8266-based products?

Does anyone share the considerations for calculating the battery life of ESP8266-based products? Does LDO consume current? if yes which LDO and battery is advisable to use? @Hardware Integration
21 Replies
RR
RR8mo ago
Yes, LDO consumes current
RR
RR8mo ago
You can use RT9013 3.3 V LDO
kmulier
kmulier8mo ago
An LDO doesn't really "consume" current. The current going in at the entrance is the same current that leaves the exit of the LDO. It just brings the voltage down (eg. from 5V to 3V3) and turns the surplus energy into heat. A switched voltage converter would be more efficient. It can bring the voltage down, while preserving the ingoing energy. For example: a 1A going in at 5V could then theoretically result in a 1.5A at 3V3 at the exit (if efficiency would be 100%, which is of course impossible). So the efficiency of a switched voltage converter is considerably higher compared to an LDO. However, the downside is that it injects noise in your circuit.
Umesh Lokhande
Umesh Lokhande8mo ago
Thank you @RR Let me check the datasheet
Aditya thakekar
Aditya thakekar8mo ago
Depends on code execution time, sleep time
Navadeep
Navadeep8mo ago
LDOs drop power,infact it depends on who much current is drawn. If V1 is input voltage and V2 is output, the power loss is (V1-V2) x I + (V1 x Iquiscent) so LDOs suit apps drawing low current and lower voltage drop band. This calculated, you can arrive at the battery life adding MCU power consumption(very dynamic depending on power modes, peripherals used etc). Common one is LM1117 and they come in both fixed and adjustable versions.
Petr Dvořák
Petr Dvořák8mo ago
@Navadeep , yes. Each device has its quiescent current, otherwise it would not work. Efficiency of switching mode converters is sometimes hard to calculate because the efficiency does not depend only on input and output voltages and currents but the big portion of it are switching losses and conduction losses in inductors. The situation must be considered individually and based on measurement all the time.
Navadeep
Navadeep8mo ago
True, Petr. But LDO's loss is straight right? It's drop is linearly proportional to the in/out voltage level and current.
Petr Dvořák
Petr Dvořák8mo ago
Yes, it is.
Petr Dvořák
Petr Dvořák8mo ago
I aggree with that you wrote.
Petr Dvořák
Petr Dvořák8mo ago
I just reacted on the previous comments.
Petr Dvořák
Petr Dvořák8mo ago
This is what describes the efficiency of a typical LDO (V1-V2) x I + (V1 x Iquiscent). As you correctly wrote.
Navadeep
Navadeep8mo ago
Navadeep
Navadeep8mo ago
@Umesh Lokhande you might want to try this out for you IoT designs.
Navadeep
Navadeep8mo ago
My cursor goes to 'buy free sample'. Do anyone have experience getting samples form ST?
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