Question (newbie): voltage divider values in circuit PDF
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Hello.
Sorry if the question may appear stupid: in the sample circuit.pdf there is what seems a voltage divider to protect the ESP32 input pin. Shouldn't the values be 1K/2K (instead of 1K/3.3K) to get from 5V to 3.3V)?
Thanks
The voltage divider equation is v_out = v_in * ( r2/ (r1+r2) ). If you plug in 5*(3300/4300) you get 3.8v which is a little too high but close. 3.3k is pretty common, I have plenty in my parts box, and that is why I used it. You are correct that 1k/2k would yield exactly 3.3v. Even 2.2k would be closer (yielding 3.4v).
There's conflicting information out about the voltage tolerance of ESP32 pins. The voltage divider is for insurance. I have people swear that they connect ESP32 input pins to 5v signals without a problem. But, I'm more cautious, hence the added resistors.
leaving this open so people can see it...