I'm planning an ESP8266 project, where I connect a soil moisture sensor to an ESP8266 chip (only the chip, not a board like the NodeMCU) (Analog Input). Since the whole setup will be battery-powered, I'm trying to minimize the quiescent current as much as possible. Now my question is: is there a quiescent current through the analog input pin of the ESP8266? and if yes, how large is this current approximately?
Thanks in advance
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If your moisture sensor uses dc, consider watching this video: youtube.com/watch?v=udmJyncDvw0Seamus– Seamus05/25/2020 00:28:21Commented May 25, 2020 at 0:28
1 Answer 1
That depends on if you're using a NodeMCU or a "naked" ESP8266.
The NodeMCU has a fixed voltage divider with 220K/100K resistors to ground, contributing about 10 µA to quiescent current (plus the much smaller current caused by the ADC input impedance).
If you're using a NodeMCU you also have to worry about the USB-to-Serial chip and the on-board voltage stabilizer.
The input impedance of the ADC input of a "naked" ESP8266 is (probably) a lot higher, 1 or more MΩ, although it is almost impossible to find definitive numbers.
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1Thanks for your answer. I was also looking for definitive numbers, but couldn't find any. I'm currently switching from the NodeMCU to the naked ESP8266, so your answer fits perfectly :) However if anyone finds out the definitive numbers, please post them here :)MarayJay– MarayJay05/24/2020 17:41:58Commented May 24, 2020 at 17:41