How can I run 3,6V MCU on a fully loaded 4,25V Li-ion battery? Normally the battery would be 3,6V, but on fully charged it is 4,2V, it would kill the microcontroller.
Load is 10mA, minimum voltage is 3,0V.
Is there a solution without any fancy LDO/IC? Zener calculator said 3V is too low for Min input: http://www.calculatoredge.com/electronics/zener.htm
Answer
If you build your own regulator, there may be problems with load response, current draw, regulation, and/or oscillation. In a MCU application, this can lead to hard-to-find bugs. I would strongly recommend a small LDO regulator like the Microchip MCP1700-3302. It costs less than 50 cents.
If you are still keen on building your own regulator, I would start with a design something like this:
Q1 supplies current to the MCU. Should the MCU voltage exceed 3.2-3.5V (5 diode drops at low current), Q3 will conduct and reduce the current flowing through Q2 which draws current operating Q1. This circuit is untested, so do more research and testing before building it. Designing a good LDO Regulator is not easy.
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