How does a 5v - 3.3v logic converter work?

Question

Arduino

I bought a converter: www.ebay.com/itm/5V-to-3-3V-IIC-I2C-Logic-Level-Co ...
, land to land.

And then the question arises: if you apply 3.3V to LV1, then 5V will come out to HV1, and if you apply 5V to HV1, then 3V will come out to LV1? Or is it only possible to apply for LV? Maybe the diagram is wrong? Briefly explain how it will be correct to make 3.3V from 5V (logic) Arduino for ESP8266.

Answer 1

Funny scheme.
In general terms: We
assume that the HV1 port works for reception and the "Floats" (HiZ) level, and is not pulled up by the port to 5V. So the only pullup is a resistor to 3.3V.
When there is a “1” on the low-voltage part, there will also be a “1” on the high-voltage part (so to speak). the transistor switch is closed.
In the case of "0" on the low-voltage part, the transistor opens and the current flows from the "upper" shoulder (on the right) from the high-voltage circuit to the low-voltage circuit (it is assumed that the "input" mode will be on the right, otherwise you can still burn the 3.3V part).
Now we assume that LV1 is reading data and the input is floating (pulled up with a resistor to 3.3V)
Similarly, the opposite is true - when the high-voltage part is "1", the transistor will be closed and does not make any changes. 0 on the high voltage part will cause the internal diode in the transistor to open and current will flow into the HV1 port (2 x 10kΩ resistors and possibly other loads on the "left").
Pretty simple circuit. Any P-MOSFET would do (yes, the same IRLML2402).
> explain how it will be correct to make 3.3V from 5V (logic) Arduino for ESP8266.
Arduino:
5V -> HV
TX -> HV1
RX -> HV2
ESP8266:
3.3V -> LV
RX -> LV1
TX -> LV2