The input is a cr2032 button battery, the output is 3.3V, the load works at about 10mA, and the standby is 10uA.
Boost chip TLV61225,
There is a normal output when there is no load, and when the load is added, the button battery is pulled down to 1V and there is no output.
There is no problem when using the voltage of the serial port board of the computer as the input power supply. If the power is supplied by the button battery, the load cannot be carried.
Ask how to solve it?
Phil 发表于 July 13, 2020
The button battery is dead. Generally, the minimum operating voltage of this type of IC is about 0.9V, and the battery is directly pulled down to this voltage.
Apple 发表于 July 13, 2020
The boost circuit of 10mA output, considering the boost efficiency, the current that the battery needs to provide is not less than 15mA. The internal resistance of the button battery should be relatively large. If the current of 15mA is provided, its own output voltage will be pulled to a very low level, so it is easy to understand that the booster cannot work normally.
If there is no size limit, it should be more reasonable to use iron-lithium batteries. It is also better to use 3.7v lithium batteries plus 3.3V ultra-low dropout linear regulator ICs.
Franz 发表于 July 13, 2020
There is no room for lithium batteries, otherwise it will not be so troublesome.
Henrique 发表于 July 13, 2020
It shows that the current of the button battery is greater than the output current of the chip, and the voltage drops below 1v when charging. Looking at the current when charging, you can find out the root cause.
