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I am using a 36 V battey pack (10s 2.5 A each cell 3.6V.) Its maximum voltage is 42 V. I am using a STM32 microcontroller whose input is 3.3 V.

I am designing a PCB in which the power supply is drawn from the battery pack (42 V.) I would like to power the STM32 microcontroller from the battery pack power supply.

I would like to regulate 42 to 3.3 V so that I can power my STM32 microcontroller.

Which method is best to convert it either using regulator or buck converter based on the pcb designing?

Battery Specification: 10s1p

  • Battery pack nominal voltage: 36 V
  • Battery pack maximum voltage: 42 V
  • Battery pack current: 2500 mAh

Microcontroller specfication: STM32F401RCT6

  • Supply voltage: 3.3 to 3.6 V
  • Supply current: 100 to 160 mA

enter image description here

Which method or component is best to regulate 42 V so that I can power my microcontroller with 3.3 V?

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    \$\begingroup\$ Have you looked into buck converters? If yes, what did you find? Have you tried TI webench? \$\endgroup\$ – winny Apr 9 at 9:49
  • \$\begingroup\$ If you have an opportunity to tap the voltage between bayteries, that would be the best choice. \$\endgroup\$ – Sridhar C Apr 11 at 3:12
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I quite like the LT8631 (1 amp at 3.3 volts and input voltage range up to 70 volts): -

enter image description here

Or maybe the LT8630: -

enter image description here

Or possibly the LTC7138: -

enter image description here

Or choose your own buck converter using the Analog Device's selection tool.

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  • \$\begingroup\$ Thank you so much i have selected LTC3638. May i know what are the main things we need to know by selecting the buck converter. \$\endgroup\$ – Muthu Apr 9 at 19:10
  • \$\begingroup\$ The main thing is that it has a very decent power efficiency compared to a linear regulator. \$\endgroup\$ – Andy aka Apr 9 at 21:14
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    \$\begingroup\$ @Muthu I've always found that Analog Devices have the most expensive stuff, and they certainly dont disappoint here. For less money, you can get the MAXM15062 which doesn't require any external components and is still a very compact package \$\endgroup\$ – BeB00 Apr 9 at 22:04
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    \$\begingroup\$ You pay for what you get in my experience and yes, they are more expensive (and so what Linear tech before ADI acquired them) but, I'm a big believer that when it comes to power supply chips it's worth forking out for higher performance than what is expected to get reliability. \$\endgroup\$ – Andy aka Apr 10 at 7:36
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A buck converter is the best approach

A linear regulator circuit in the worst case is going to have to drop 42V down to 3.6V = 38.4V. At 160mA the power dissipated by the regulator will be 6.14W.

On the other hand a buck convertor will drop voltage to the required level with possibly 90% efficiency (less energy wasted as heat) and likely reduce the current consumption to 3.6/42*160mA = 13.7mA allowing your battery to power the circuit 11.67 times as long.

https://www.monolithicpower.com/en/products/dc-dc-power-conversion/switching-regulators/step-down-buck/converters/vin-max-48v/mp2492.html or similar could do this.

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  • \$\begingroup\$ At that dissipation, the heat rejection is going to be an even bigger problem than battery life. \$\endgroup\$ – chrylis -cautiouslyoptimistic- Apr 10 at 6:35
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The best method is to use a buck-topology switching regulator.

For example, Analog Devices LT3437 has an reference circuit/example for your application

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A buck converter is the right solution, and many of the chips suggested in other answers are easy to use as the datasheets already have designs that you can just use. But it does still require a PCB layout and other design work.

SHR05 diagram from datasheet

The easiest solution is to use a premade buck converter module. For example SRH05S3V3 can provide 3.3V at up to 500 mA from an input voltage anywhere between 9V and 72V. It has three pins only: Vin, Vout, GND. Even the input and output capacitors are optional, as the module has small capacitors integrated.

SHR05 photo

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  • \$\begingroup\$ is there any disadvantages in it like efficiency wise or power wise \$\endgroup\$ – Muthu Apr 11 at 7:05
  • \$\begingroup\$ Be wary with such devices. They tend to be very noisy - in this example ~75mVpp ripple on the output. Probably be fine for an MCU, but wouldn't be good for any analogue part (include PLLs, ADCs, DACs, etc.). They also tend to exhibit poor efficiency \$\endgroup\$ – Tom Carpenter Apr 11 at 22:31
  • \$\begingroup\$ @TomCarpenter: How about using such a device to produce 5.0V and then using a linear regulator regulator on that? Efficiency wouldn't be great, but would still probably be 5x that of a linear regulator from 36V. \$\endgroup\$ – supercat Apr 12 at 1:40
  • \$\begingroup\$ @supercat You can, but mind the dropout and PSRR on the linear regulator. Switching harmonics are high enough that the linear regulator's control loop cannot respond in time and pass straight through. So if low noise is what you are after caps and ferrites are needed. \$\endgroup\$ – DKNguyen Apr 12 at 3:10
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If you're making a bunch of them, in particular, and are concerned about cost, you might consider the XL Semi XL7015, which costs only about 25 cents in 100's, about 1/20 the cost of the LTC boutique parts.

Typical efficiency is only about 70% with 36V in and 160mA out vs 85% for the LTC part, so there is a cost in battery consumption (about 0.08W more loss). There's also less voltage margin and it's a physically larger TO-252-5 part. On the plus side, it's capable of a lot more output current.

enter image description here

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I think you can find off the shelf BECs for 10S - or even higher. If you plan on building your own thingy, you must account for the voltage drop which appears with the battery's depletion. BECs are specially built for this purpose, hence the name (battery eliminating circuit).

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As your input battery pack is 36V then linear voltage regulator is out of question (although it is possible due low consumption amperage).

With linear voltage regulator power dissipation would be (Vin-Vout)*Iout. For example if consumption amperage would be 1A then (36-3.3)*1=32.7W of wasted power.

The best option would be a buck converter (switching power supply) which monitors voltage on output capacitor and recharges it when it will be required. Efficiency of such power converters is quite high.

If you are looking at very inexpensive solution then check LM2596 -- datasheet.

Power consumption of MCU is very small and there plenty of other solutions on the market.

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i think by making between the two components resistance with value equal to 15 - 16 K (ohm)

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    \$\begingroup\$ Welcome to the site :-) However this is a very short answer for Stack Exchange. You didn't explain how using a such a resistor would produce a stable "output" voltage, with varying load (hint: it wouldn't), nor did you explain any other problems with that approach e.g. efficiency and power loss in the resistor etc. I recommend that you read the higher-voted answers, to see the quality and level of detail which is well-received here. Do some calculations to see how your answer would work in practice, and you will see some big problems with it :-( Please read the tour and help center too. Thanks \$\endgroup\$ – SamGibson Apr 10 at 16:36
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The easiest answer according to my knowledge is to use a zener diode with 3.3Volts. You also need a resistor to limit the current that flows through the diode and to the microcontroller!

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    \$\begingroup\$ That's a simple way, but certainly not a good way. Did you notice discussions about efficiency in the other answers? Linear regulators are very inefficient, which is why everyone has recommended switching regulators. Your suggested zener diode regulator is less efficient than a linear regulator. \$\endgroup\$ – JRE Apr 10 at 7:12
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    \$\begingroup\$ And with a voltage drop of up to 41 volts or so, a linear regulator will become hot. \$\endgroup\$ – NomadMaker Apr 10 at 10:14

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