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I am trying to design a breakout board for this STM32 Microcontroller https://www.digikey.com/en/products/detail/stmicroelectronics/STM32F038G6U6/5051266

I'd like to use the ADC's and I'm trying to figure out the power supply scheme. The description from the datasheet for Vdda is as follows

V DDA = from V DD to 3.6 V: external analog power supply for ADC, RCs and PLL (minimum voltage to be applied to V DDA is 2.4 V when the ADC is used). It is provided externally through VDDA pin. The V DDA voltage level must be always greater or equal to the V DD voltage level and must be established first.

I was looking at a couple of tables

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It seems like if I didn't care about the ADC's, I could use my Vdd for my Vdda but I'd like to reserve the ability to use the ADC's in case that should that become an application for this breakout board.

The question then becomes how do I set up this power supply? Do I need two different voltage sources?

My initial thought for Vdd was a 3V coin cell battery with a voltage regulator, assuming the .2mA would be sufficient for Vdd.

I guess I could do a separate 3V coin cell for Vdda as well?

Assuming these would be appropriate power sources for these two power domains, the next question is whether I need to connect the power domains in any way (other than obviously common ground)?

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  • \$\begingroup\$ Vdda and GNDa should be separated from the digital part. It is important because digital part is generating a lot of noise reducing accuracy of the analogue part. \$\endgroup\$ Aug 9, 2022 at 8:05

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Note that Vdda should always be > Vcc (or if lower, no more than absolute maximum 400mV) so you might need a Schottky diode to protect the chip if you have independent sources.

A simpler way would be to have one power source and two regulators, on for Vdda and one for Vcc. For example, you could use a 1.8V regulator and a 2.5V regulator from the same series such as MIC5259.

Idda is pretty low (hundreds of uA maximum). Idd (depending on your design) could be as high as the 20mA range which is way too much for a button cell, so you'd have to do some work to determine what is required for the power source.

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