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I have a 12 volt power source which powers an LM7805 regulator, in order to power a circuit board. If I were to accidentally connect the power source backwards (reverse polarity,) which would be true (the circuit board is not protected against reverse polarity):

  1. The 7805 is protected against reverse polarity and I can just change it to the proper polarity and everything will be fine. No damage to the 7805 no circuit board will result.
  2. The 7805 will be destroyed, but it won't affect the circuit board.
  3. Both the 7805 and the circuit board will be destroyed.

I am asking this because as an afterthought, I wish I would have put a blocking diode at the input of the 7805. However, to add one now is not an easy task. So I am doing a risk assessment.

The battery is a gel battery which will need to be removed from time to time so there is a risk of improperly connecting it.

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  • \$\begingroup\$ How much current would the 7805 carry under normal usage? Can't be much anyways, but it does sound like adding an external diode in front has nice effects, heat-wise, because you're distributing the wasting of 58% of your power from only the 7805 to series diodes + the 7805. \$\endgroup\$ Mar 10, 2022 at 16:32
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    \$\begingroup\$ Does anyone know if the switching 78xx drop-in replacements have intrinsic reverse-polarity protection? I couldn't find anything for the Murata ones, but I may have skimmed over the data sheet too quickly. At least they would have the bonus of not wasting so much battery capacity. \$\endgroup\$ Mar 10, 2022 at 16:58
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    \$\begingroup\$ @AndrewMorton not that I can see. None of the usual suspects (Murata, Traco, Recom) list it as a protection feature, and it would be something I think they'd list fairly prominently on their datasheeet. CUI specifically says they don't have it on theirs. \$\endgroup\$
    – vir
    Mar 10, 2022 at 17:56
  • \$\begingroup\$ @AndrewMorton If doesn’t explicitly state so listed in the features on the first page, it’s safe to assume every power supply, IC or similar won’t like reverse polarity at all. \$\endgroup\$
    – winny
    Mar 11, 2022 at 16:28

5 Answers 5

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Most probably It will be destroyed (see edit at the end of the answer).

However a simple solution in your case exists: add a series diode (e.g 1N4007) before the input of the IC. Since you are already dropping 12V-5V=7V across the 7805, this won't have any drawback.

On the contrary, it will help offload a bit of power dissipation from the IC, since now you'll have 7V-0.6V=6.3V across the 7805.

You should have a setup like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The resistor is there to offload another bit of power from the IC, to keep it cooler, so that it doesn't heat up so much. Depending on your load current, the IC could dissipate up to 1A times the Vin-Vout voltage, so if you really need the resistor depends on the max current you deliver to the load.

If your max current is near 1A, probably you could use a 3A diode like the 1N5408, instead of the 1A 1N4007, just to have some margin and avoid overheating the diode.

You could even use more than one diode in series: it will increase the voltage dropped on the diodes, offloading more power from the IC. Don't go overboard, though, because the 7805 needs an input voltage that is at least ~3V higher than the output (5V), i.e. Vin>~8V.

EDIT (Alternative cheaper, more efficient solution)

You don't say much about your scenario and what are your constraints. However nowadays there are lots of cheap buck (step-down) DC-DC converter modules that are extremely efficient, like this one (first hit I found on Amazon.com):

https://www.amazon.com/Zixtec-LM2596-Converter-Module-1-25V-30V/dp/B07VVXF7YX

enter image description here

Just add a diode for reverse polarity protection as I showed before (and the fuse for good measure) and you won't need the power resistor. It could even cost you less than the 7805 solution and won't get warm at all (assuming your load draws max 1A - those are 3A max modules).

This will also help you extend the battery life, since the 7805 solution will waste lots of power: at 1A current you'll waste 7W in heat, whereas with the buck module you'll waste ~0.6W in the diode and at most 1.2W in the module (assuming 80% or better efficiency).

EDIT (to address your question explicitly)

Yes, it will burn. To see why, here's an excerpt of the LM7805 datasheet by TI (modifications by me in red):

enter image description here

There is a low-impedance path when the GND and Vin terminals are reversed (in red): it is comprised of D1 and the BC junction of Q12. It's two forward-biased PN junction in series without anything limiting the current except the internal resistance of your battery. The chip WILL die horribly and could possibly start a fire.

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    \$\begingroup\$ This is the right answer. With a gel battery, you really want a fuse to prevent fireworks if there's a serious fault. A diode isn't a big efficiency hit, and prevents damage from a polarization error. With the 7805, use the extra headroom to reduce its dissipation with a resistor, but the 21st century solution is to use DC-DC conversion. \$\endgroup\$
    – John Doty
    Mar 12, 2022 at 16:42
  • \$\begingroup\$ I had considered this product, but it is a much larger footprint than a TO-220 package, and I'm trying to keep the footprint down. And especially since I need both 3.3V and 5V, this would now require 2 of them. If they made one with dual outputs it would be more appealing. But yes, given that one has the space, I think this would be a better solution. \$\endgroup\$
    – KevinHJ
    Mar 13, 2022 at 13:20
  • \$\begingroup\$ @KevinHJ There are smaller modules. This is a 3A rated one. BTW, you didn't mention the max current you will need to deliver to the load. I've never seen dual output ones, though. OTOH there are smaller modules with max 0.5A (or so) whose dimensions rival those of a TO220 package. \$\endgroup\$ Mar 13, 2022 at 23:04
  • \$\begingroup\$ @KevinHJ Just a quick search on Amazon.co,. Look at these. Just 22x17 mm and 1.8A continuous! \$\endgroup\$ Mar 13, 2022 at 23:08
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The LM7805 will definitely be destroyed and a lot of current will flow.

With only about -225mA flowing into the input the output will exceed -800mV typically (about -1.5V in). That's before whatever fireworks happens when the chip melts, and will likely exceed the rating of any bipolar or CMOS chip connected to it.

The downstream circuitry may survive or not, but it should not be counted on and even if it does appear to survive, it may be subtly damaged.


There are pin-compatible linear regulators with reverse-polarity protection. Eg. LM2937. However, the ESR requirements for the output capacitor in particular are not as liberal as for the LM7805. In particular, they are prone to oscillation if used with a ceramic output capacitor (minimum ESR is 10m\$\Omega\$ for the LM2937) and they may also oscillate if used with some aluminum electrolytic capacitors (maximum ESR is a few ohms). The current capability and other characteristics vary, and if you consider swapping the parts out you should carefully consider all possible issues.

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I am asking this because as an afterthought, I wish I would have put a blocking diode at the input of the 7805. However, to add one now is not an easy task. So I am doing a risk assessment.

First of all, you should have a series fuse protecting the PCB from the short circuit current of the gel battery, which can be in tens of amperes. In most cases, short circuits on unprotected PCBs fed by such batteries are catastrophic: some copper gets vaporized, and the PCB material gets scorched.

When you have the fuse, the diode can be anti-parallel instead of series. The diode will blow the fuse if the battery will be connected backwards. The diode can be a Schottky diode to limit the reverse voltage.

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  • \$\begingroup\$ Very good suggestion, inserting the diode that way will make options for me much easier; I could even place it at the battery connections. I am limited to putting it in series downstream in a now not so convenient location as placing it upstream results in a drop thus power loss to the 12V circuit. \$\endgroup\$
    – KevinHJ
    Mar 14, 2022 at 22:26
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The 7805 data sheet provides no information on reverse polarity, so we have to assume it will be destroyed, particularly if fed from a low-impedance source such as a battery.

Then it is a lottery as to whether the rest of your circuitry is destroyed as well, but I wouldn't gamble on it surviving.

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The 7805 was introduced more than 50 years ago. It's widely sourced - many manufacturers - and the only thing they promise is what's in their individual data sheets. Some might be destroyed by reverse polarity and others not. You should key the connectors to prevent misconnection if it's possible to do so. If you need consistent behavior, you need to restrict your purchases to manufacturers that meet your requirements. There's no such thing as a 7805 that's the same from everyone, any more than there's a 741.

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  • \$\begingroup\$ Yes, I later thought of keying the connectors so they can't be reversed, which I did. \$\endgroup\$
    – KevinHJ
    Mar 11, 2022 at 17:18

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