3
\$\begingroup\$

I'm working on a fairly complicated project and in the process I encountered a fairly simple problem. My project has a microcontroller which will, of course require programming. However, the PCB will draw its power from a custom designed motherboard, and for the purpose of testing and writing the firmware, it would be grossly inconvenient to require that it be connected to the motherboard simply to program it.

My solution was to simply stick a header onto the 5 V rail and power it externally. However, the 5 V rail is supplied by a linear regulator (LDL1117), which bucks 6 V from the mobo down to 5 V. This means that if I power the rail externally, the regulator will have 5 V at the output, while the input would be floating. Is this inherently a problem? Could this damage the regulator?

\$\endgroup\$
3
  • \$\begingroup\$ It could damage it. What does the datasheet say? Which exact part number it is and can you link to a datasheet? What other components there are connected to the input side, like is there a huge bulk capacitor? Any protection diode? Can you post schematics? \$\endgroup\$
    – Justme
    Commented Oct 11, 2020 at 22:10
  • \$\begingroup\$ Datasheet says that the voltage at the OUT pin should not exceed the input voltage by +0.3V. In your application, the OUT pin will see +5V while the input voltage is zero. Is this enough to answer your question? \$\endgroup\$
    – Rohat Kılıç
    Commented Oct 11, 2020 at 22:18
  • \$\begingroup\$ I did see in the datasheet that Vout should not be greater than Vin + 0.3, but that's usually when the chip is operating. In this case it would be idle. I could apply +5 V to the 6V rail and the regulator would buck it down to 4.65 V. The microcontroller should still be able to program itself at 4.65 V. \$\endgroup\$
    – AMacDonald
    Commented Oct 14, 2020 at 3:37

1 Answer 1

Reset to default
2
\$\begingroup\$

I'm not familiar with the regulator but a common trick on PSU design is to feed back from the output to the input using a diode. This way the input can only be one diode-drop below the output. Using a Schottky diode will minimise the voltage drop. You'll have to decide if this is enough.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The scheme - if the +6V is available on the header - would be simple.

Judging by @Rohat's comment this is risky.

The obvious solution then would be to apply +5 V to the +6 V input too. This eliminates the problem if that pin is easily accessible.

\$\endgroup\$
1
  • \$\begingroup\$ If having Vin<Vout isn't inherently a problem, then I don't need anything. Because even with a Schottky, Vin will be slightly less than Vout. Again, I just need to power the microcontroller to program it. I'm not going to use this for anything else. \$\endgroup\$
    – AMacDonald
    Commented Oct 12, 2020 at 6:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.