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I have an "ultra-low-power" MCU STM32L011F4. According to the specs its powering voltage range is from 1.65 V to 3.6 V. I want to power it with a LiPo battery with nominal voltage of 3.7 V. Which is slightly larger.
But according to the table of absolute maximum ratings from the datasheet its maximal value of VDD–VSS is 4.0 V.

How bad would it be to power the MCU directly from the battery? Should I put a diode to drop the voltage a bit? Or would it be beneficial to use a proper voltage regulator?

Teivaz
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    Will you be charging the battery while it is connected to the MCU? The voltage could be significantly higher in that case. – Elliot Alderson Feb 16 '20 at 00:03
  • @ElliotAlderson It's a great point! I was not planning to charge the battery while it's connected to the circuit. But should really consider it in further versions. – Teivaz Feb 16 '20 at 00:05
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    STM32L absolute maximum voltage 4.0V. Fully charged Lipo voltage 4.2V. Result - boom! – Bruce Abbott Feb 16 '20 at 02:17

2 Answers2

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Absolute maximum ratings are stress ratings after which you get permanent damage.

Operating of the device is not guaranteed above the normal operating range, and the long-term reliability may suffer. The datasheet conditions apply for being within the normal operating range.

It means, the device does not have to work properly above normal operating range.

The 3.7V is only nominal voltage for the LiPo cell. It can have more than 4.0V when fully charged. Even if you don't charge it in-circuit. Do not connect to battery directly. Diode is not the best either - depending on diode, with extremely light load, it won't drop much voltage.

Justme
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  • So what do you suggest then? Using linear voltage regulator? – Teivaz Feb 16 '20 at 00:28
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    It depends on a lot of factors what to suggest and there is little known. It depends on what supply voltage range and current you need for the MCU and other chips. And ripple specs for them. And battery capacity and time of operation with one charge. Maybe a linear regulator is enough. Maybe one with low quiescent current draw. Maybe you need a buck-boost switch mode power supply. Maybe a charge pump. – Justme Feb 16 '20 at 00:41
  • A "low dropout" (LDO) linear regulator will supply (for example) 3.3V when Vbatt > 3.4V and something like (Vbatt - 0.1V) otherwise. But the LDO draws additional current of its own. Exact details depend on current draw : read datasheets. –  Feb 16 '20 at 01:46
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    MCP1700 has quiescent current 1.6uA, max output 250mA, 0.178V drop at 250mA. – Bruce Abbott Feb 16 '20 at 02:21
  • @BruceAbbott thank you for the hint – Teivaz Feb 16 '20 at 12:57
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Going for a low drop out regulator with a low quiescent current is the best option with a option to monitor the battery voltage too.

  • This way all the peripherals will work with the common stable voltage. Helps in interfacing to daughter boards.
  • The battery monitoring will ensure that the user indication about low battery will be still possible/MCU operations can be planned based on battery voltage.
  • Since the drop across the linear regulator isn't that much, and the current too is less.. the energy wasted is considerably less across the regulator.
  • If diode solution is for the low cost approach, choose a diode with ultra low leakage current. These diodes will have higher forward voltage drop compared to normal diodes even at very light loads.
User323693
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