I'm working on a project where I want to program a dsPIC33E. My board integrates an UA78M33CDCY linear regulator and I want to program the controller by powering the target from PICKit3.

Is it safe for the regulator to apply 3.25V from the PICKit to its output?

I mention that the regulator is connected as recommended in the datasheet (with two decoupling capacitors), without any additional components and the board is powered only by the PICKit. The functional block diagram of the regulator is given at page 10 of the datasheet.


Most likely (but not guaranteed) yes, as long as there is nothing connected to board power input.

The regulator datasheet implies some voltage difference is OK:

In the event of an input short circuit or another case where the output voltage can be higher than the input, a diode shunt can be connected across the device with the anode at the output and cathode at the input

This means that the regulator can tolerate input being 0.5v (diode drop) lower than the output.

Can you run it without extra diode? In my experience, yes. Giving 3.3 v to regulator output raises input to 2.8 v or so. If there is nothing else on that regulator input, everything works fine. While I never had problems with this method, you can always add a reverse diode as datasheet says if you are worried.

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  • \$\begingroup\$ Input is 2.5V. I thought it would be OK, but I wanted a second and experienced opinion. Thank you! \$\endgroup\$ – Cristian M Jan 26 '17 at 18:47
  • \$\begingroup\$ For the reference, that person with other deleted answer posted very cool link to infineon appnote infineon.com/cms/en/product/promopages/aim-mc/… which shows that most regulators already act like they have a built-in reverse diode. \$\endgroup\$ – theamk Jan 26 '17 at 19:56

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