It mean, it always use 3V3 for communication with eeprom. As far as I understand, that mean, that chip have internal 3.3V regulator and IO pins are NOT 5V tolerant. The operating voltage of other circuits connected to the chip cannot exceed 3.3V. When working with 3.3V voltage, the V3 pin of the CH341 chip should be connected to the VCC pin, and an external 3.3V power supply is input, and the CH341 core is connected. When using a 5V supply voltage, the VCC pin of the CH341 chip inputs an external 5V power supply, and the V3 pin should be externally connected with a power supply decoupling capacitor of 0.01uF to 0.1uF. It can be translated: The CH341 chip supports a 5V supply voltage or a 3.3V supply voltage. At end of chapter 5.2 硬件说明 (Hardware description) is stated. The trick is, that you miss 2 nd LDO, that is inside CH341A chip. The operating voltage of other circuits connected to the chip cannot exceed 3.3V.Īs far as I understand, that mean, that chip have internal 3.3V regulator and IO pins are NOT 5V tolerant. My question is, is it safe to just do the short-circuit without removing the 1117 linear regulator? Hi, The trick is, that you miss 2 nd LDO, that is inside CH341A chip. I can use the hot air station to remove the 1117 and then short-circuit the input with the output. One simple solution to fix this issue is to simply bypass the 1117-3.3V regulator. There are no limiting series resistors on those signals. So all I2C and SPI signals are 5V while the external memory is powered at 3.3V. Source: All Vcc connections in the ZIF socket are 3.3V but the CH341A is powered at 5V. Someone already did the schematic of this board (atached). I bought one cheap chinese CH341A serial memory programmer black PCB (see attached picture).
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