I’m kinda new to the world of development boards and got a noob question. I couldn’t find it anywhere else on the forum, so here it goes.
I’m currently working on a project that uses an 18650 battery to power a NodeMCU (ESP8266). I’ve found a promising schematic somewhere on the web, but I’m a little unsure on how to read it.
As you can see on the provided picture, the circuit consists of an LDO regulator, an electrolytic capacitor and a ceramic capacitor. They are connected to each other, but I’m unsure how to connect the ceramic capacitor to the electrolytic capacitor.
Could you help me by explaining this? What would it look like if you’d put it on a breadboard? So I can exactly follow the wiring.
Put the MCP1700 into the breadboard so each of its 3 leads is in a different set of adjacent contact strips (the contact strips run vertically in your image.
Put the 2 leads from each capacitor into the same contact strips, but in different holes as the outer 2 connections to the MCP700.
Connect wires to join the contact strips the MCP1700 is using to the power rails at the top and to the power supply.
If you are in any doubt, wire it up, take a photo and post it here as per the instructions I gave you a link to. Don't connect the power until one of us has said it's OK.
The LDOs should have a ceramic capacitor and an electrolytic capacitor connected in parallel to GND and Vout to smooth the voltage peaks. Here we’re using a 100uF electrolytic capacitor, and a 100nF ceramic capacitor.
OK, so the ceramic capacitor should be closest to the regulator, in fact, one tenth of an inch away (next set of holes if on a breadboard, or perfboard or PCB design) while the electrolytic can be a little further away.
Have you checked in the datasheet whether you should have a capacitor close by on the input to the regulator?
As the NodeMCU already includes a voltage divider to the analog input, you do not need two additional resistors, only an extra 100k in series with the pin to read a 4.2 V battery.