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I am designing a 3V battery-powered simple micro-controller to collect sensor data in RAM. How can I avoid a conflict with the 5V of a USB when it comes time to upload the data? Disconnecting the battery would lose the RAM. The solution needs to be very simple as the PCB is very small.

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  • $\begingroup$ Are you trying to manage both 5V and 3V supplies such that it will not affect your system. What type of micro controller are using? Please add additional detail. $\endgroup$ – Mahendra Gunawardena Apr 5 '17 at 12:15
  • $\begingroup$ I am powering a PIC with a 3V battery but then have to attach to 5V USB to get data. $\endgroup$ – Jim Lewis Apr 5 '17 at 14:22
  • $\begingroup$ What type of PIC? $\endgroup$ – user8055 Apr 6 '17 at 13:54
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There are two possible options based on the limited infomation you have provided

  1. Consider storing data in Non-Volatile Memory.
  2. Use a LDO such as TPS73633 that can support wide input voltage range TPS73633 support a range of 3.5V and 5.5V. The output is 3.3V. An provide both the supply voltage options.

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  • $\begingroup$ Non-volatile won't work for me. The LDO does not solve my problem. $\endgroup$ – Jim Lewis Apr 5 '17 at 14:24
  • $\begingroup$ @JimLewis Why will Non-volatile memory not work. Why does a LDO not solve the problem, $\endgroup$ – Mahendra Gunawardena Apr 5 '17 at 22:41
  • $\begingroup$ Because I need 16kb and don't want an extra chip. Where is the 3V battery and 5V USB in your LDO circuit? $\endgroup$ – Jim Lewis Apr 6 '17 at 8:36
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    $\begingroup$ With a couple of diodes to prevent backflow you can connect both the power sources to the +vbus $\endgroup$ – ratchet freak Apr 6 '17 at 10:23
  • $\begingroup$ Yes that might do the trick though the drop from 3V might be pushing limits. And I suppose I don't need a diode on the higher 5V supply. $\endgroup$ – Jim Lewis Apr 6 '17 at 10:39

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