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[on:]

I made my own Arduino-like board. I have a 7805 on it to provide +5V, but sometimes I need to power the board by the +5V/GND pins, bypassing the regulator (using an external one). This works perfectly, but I was wondering: what happens to the onboard regulator when I feed +5V on its output pin? Can it get ruined? Should I somehow protect it? Maybe with a diode?

As a side question: my board will also need +3.3V. I have another regulator for that: is it better to power it in parallel with the 7805 or should I power it *through* the 7805? Its dropout voltage is low so it can also turn +5V into +3.3V without issues.

Thanks in advance .

[Reply #1 on:]

While not ideal a 7805 can look after itself in this situation. I have not had any trouble doing this.

Yes I would regulate the 3V3 from the 5V, there is enough voltage drop.

[Reply #2 on:]

Thanks for your reply, Mike!

I think I have noticed that when +5V is provided on the 7805 output pin, ~7.8V seems to present at its input pin. I'm not sure I measured this correctly (I was looking for some incorrect wiring), so that's why I was in fear of ruining the 7805.

Regulating the 3V3 from the 5V has the advantage that I also get 3V3 when powering through the Vcc/GND pins, so I'll go that way if it causes no issues. But if "protecting" the 7805 is only the matter of adding a diode on its output pin, I might just throw that in.

[Reply #3 on:]

I think I have noticed that when +5V is provided on the 7805 output pin, ~7.8V seems to present at its input pin

No that is not only wrong it is impossible. You can't get more voltage than you put in with this kind of circuit.

But if "protecting" the 7805 is only the matter of adding a diode on its output pin

No it is not. Adding a diode will drop the voltage it produces by 0.7V so don't add one.

[Reply #4 on:]

I think I have noticed that when +5V is provided on the 7805 output pin, ~7.8V seems to present at its input pin

No that is not only wrong it is impossible. You can't get more voltage than you put in with this kind of circuit.

Right, that should've been obvious . Now I recall that that happened when I provided +9V on the output pin.

Thanks again!

[Reply #5 on:]

Put a Schottky diode in the output pin, like it's been said, so your voltage drop will be minimal.

[Reply #6 on:]

Can it get ruined?

It depends on the regulator. 1117's for example don't like that much. What you should do is to put a diode from the regulator's output pin to its input pin. Any diode will do.

[Reply #7 on:]

1117's for example don't like that much.

True but the OP was talking about an 7805.

[Reply #8 on:]

Interesting conversation, I regularly power my self made boards through the 5V pin bypassing the regulator. I mostly use LM2940's, I have not had any problems so far. Can I ruin the LM2940 that way?

[Reply #9 on:]

Very unlikely. In general, the ldos tend to have issues with that, and 1117 is probably an extreme example: it gets hot very quickly if powered reversely.

[Reply #10 on:]

I don't know about the LM78xx...

Where I work, we sell a board that can be powered by 5V, or with a higher voltage using an on-board LM2741 regulator (similar to an Arduino). 

There is a "regulator bypass" jumper that simply shorts-out the 2741 input & output pins for a direct (externally regulated) 5V connection.     Sometimes during test, we bypass and "back-power" the regulator (without the shorting-jumper connecting back to the regulator input).  With this chip, there is no problem back-powering in either case.

Some of our other products use different (usually switching) regulators (and without the bypass jumper on the board), and when there's a "power supply problem", it's often helpful to temporarily bypass the on-board regulator with a bench supply (to see if there's a 5V to ground short, etc.) and I've never had a problem back-powering the regulator in those cases either.