I am busy designing a custom board based on Arduino's pro mini design.

In the official design, they have a MIC5205 voltage regulator supplying a constant 5V to the ATMEGA328 chip. However it is also possible to feed in 5V directly to the chip if you so wish, but when you do so, you are supplying 5V to the output of the MIC5205 chip.

In my design, I want to replace the MIC5205 chip with a LM7805 voltage regulator. But based on a bit of research it seems like you cannot supply 5V to the output of the LM780; based on a quick test, when you do so with 5V the chip doesn't seem to draw any current.

After going through each datasheet (MIC5205 and LM7805) I couldn't find any information about the chips' ability to withstand voltage applied to the output pins.

My question is finally:

  1. Can the MIC5205 actually handle voltage applied to its output?

  2. Is it bad to apply voltage to the output of the LM7805?

  3. How can I tell from looking at the datasheet if a chip can handle voltage applied to its output?

  • \$\begingroup\$ The MIC5205 is a PNP type LDO. The 7805 is an NPN type. Just plain different designs. I haven't looked at the details (lately), but I suspect that's a part-explanation for your observation. \$\endgroup\$
    – jonk
    Sep 21, 2017 at 7:36
  • \$\begingroup\$ Made a slight edit, if 5v is applied it all seems fine \$\endgroup\$
    – Makoto
    Sep 21, 2017 at 7:37
  • \$\begingroup\$ LDOs really don't like the voltage at their output to be higher than the voltage at their input. \$\endgroup\$ Sep 21, 2017 at 7:45
  • \$\begingroup\$ Most regulators can and should be able to handle a voltage at the output while there's no input voltage. This is a situation which will often occur when the input voltage is switched off but the output voltage remains (for some time) due to the decoupling capacitors at the output. \$\endgroup\$ Sep 21, 2017 at 7:45
  • \$\begingroup\$ 78xx regulators often show an external diode added output to input, to limit the differential voltage in the event that the input is unpowered with the output live. This is a normal state of affairs if the output has a large capacitor on it, and the input supply collapses. \$\endgroup\$
    – Neil_UK
    Sep 21, 2017 at 7:53

1 Answer 1


The problem of applying an external voltage to the output pin of an NPN-output type regulator is that this would apply a reverse bias to the emitter-base junction of the NPN. The emitter-base junction of a BJT typically has a very small breakdown voltage (the emitter is heavily doped).

In the datasheet you might find this: enter image description here

However, two considerations should be drawn:

  • You'll have a reverse current to the input, and this might cause some problems.
  • The datasheet states that the limit voltage is 7V. In a 7808 (or higher voltage, eg 7812), this would actually be a problem, as you would likely be applying 8V or more. However, in your case, you're likely to externally apply only 5V to the output (otherwise a larger voltage would possibly destroy the circuitry the 7805 is normally powering).

Therefore you might not need to insert the diode on a 7805.


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