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I'm adapting a breakout board for the LMS9DS1 IMU to suit my own needs. This breakout board has an input range of 3-5V and works on 3.3V. However, the circuit, as shown below, seems to have two ways of dropping the voltage. The MIC5225 (bottom left) is a voltage regulator, but don't MOSFETs (BS138, top left) also serve this purpose? What is the purpose of the extra MOSFETs? And can I drop both the regulator and the MOSFETs if I already have 3.3V?

Sorry if this is a silly question.

LMS9DS1 breakout board schematic

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U2 is a voltage regulator providing a +3.3V supply. Q1 and Q2 are being used as logic level shifters to translate +5V signals to +3.3V levels. They do not behave the same.

If you have another +3.3V supply you can eliminate U2, assuming your +3.3V source provides the same or higher current, same or better switching noise (if any), etc.

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  • \$\begingroup\$ Thank you for the clarification. Does this mean that if I have a chip that can work with +3.3V logic levels, I can eliminate the +5V connection and drop Q1 and Q2? \$\endgroup\$ – user207636 Dec 19 '18 at 15:14
  • \$\begingroup\$ @O.H.Bother I'm not sure what's on either side of SDA_3V and SDA_5V, but if you replace them with chips whose logic levels are compatible then you don't need to include Q1 and Q2. If you select some chips and you're not sure if their levels are compatible, you can ask a new question. \$\endgroup\$ – Kevin Kruse Dec 19 '18 at 16:31
  • \$\begingroup\$ SDA is part of the I2C lines, together with SCL. Already confirmed that the chip will work with 3.3V logic, so my question is answered. Thank you very, very much! \$\endgroup\$ – user207636 Dec 20 '18 at 10:53

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