Why do my lm317T and lm78XX regulators die so easily?

While using my last 7808 voltage regulator as nothing more than a voltage reference for a breadboarded linear power supply design, I noticed that after a few small modifications, that the 7808 was no longer outputting 8V, and was instead floating 14.8V (kinda close to the +15V supply rail). At no point did the regulator overheat, and it was simply connected to the a 200K potentiometer to ground and a noninverting input of a LT1112.

In the past, I have killed at least 3 LM317T regulators in my velleman 3-in-1 lab thingy, (it is a relatively simple circuit and was the first board I have ever fully reverse engineered.) The heatsink never did get too hot, but I used to always connect heavy loads to it that caused the overload lamp to come on. That LED was basically in series with a large resistor to the input of the LM317 so when the LED has enough bias voltage, it comes on when around 1A is being pulled.

Anyways, I have killed countless other small 78xx regulators, one way or another, and even made some of them literally explode right away. (generally by getting the pinout wrong.) I blow up these regulators so commonly I even have video evidence lol! click to around 3:30 for the FAIL!


So is it just me and my luck, or are these regulators which supposedly have all sorts of thermal protection, current limiting, and short circuit protection just not as rugged as they seem they should be?

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One thing they don't like is reverse biasing, so never arrange that the input is lower than the output, which can happen when the supply shuts down. Put a diode across input and output. I've generally found that negative regulators, like the 79XX series are more sensitive to abuse than positive ones.

-max- (author)  steveastrouk1 year ago

OK, so that makes sense, So would connecting and disconnecting power to the regulators be bad for them if there are large filter capacitors (>1000uF) on the output?


-max- (author)  steveastrouk1 year ago

Is there a reason why there is no protection for this built into the regulator? It seems like a plausible scenario if say the main power supply fails for whatever reason and a 5V regulator would experience negative dropout potential while the filter capacitors are charged.

The designs are ancient, and by adding excess reservoir capacitance AFTER the regulator, you are creating a situation the makers didn't anticipate, except in the app notes where the diodes are shown.

-max- (author)  steveastrouk1 year ago

Ahh, so I will definitely stop adding huge capacitors to the outputs of my regulators! Maybe someday I will stop being so clumsy in my breadboarding techniques too ;P

Also, why is it that so many makers use (what to me seems inferior) components, all my electronics books use parts in projects like the 2N3904 and LM741, and other parts that just seem like they would be obsolete if there was no hobbyist market!

Because to learn about the alternatives, and why they're better, and when to use them is the realm of the professional engineer and not the amateur.

-max- (author)  steveastrouk1 year ago

So in that case, what would make a good 'rugged' alternative to the old 7805? Or is there no such thing?

Just put a diode on it. I've never blown a 7805 in 35 years. 79XX, sure quite a few, but the 78 is pretty damned solid, if you don't stick stupid reservoir caps on it....

-max- (author)  steveastrouk1 year ago

I'll certainly be doing that from now on! BTW what is the purpose of the reverse-biased diodes I commonly find directly across the output of a linear regulator? My only guess is that it is there to shunt negative voltage spikes.

Yup. For 78 series, they're belt-and-braces, but for 79 series...

-max- (author)  steveastrouk1 year ago

What is it about the negative voltage rail, the PNP transistors seem to be
"weaker" in the same package style, the negative voltage regulators are less "rugged" overall, Is it some inherent thing with the manufacturing process? Or is it that positive voltages are far more common so development is focused there? Is the same true for vacuum tube stuff?

It will be to do with the way the PNP junction structure is fabricated into the silicon.

Why does the diode do if you put it where you said?

Imagine putting a big capacitor on the output of the regulator, so big, it can hold the regulator output 'Up' at its rated voltage. Switch off the INPUT and now the regulator sees '-' rated output across its terminals, and can be damaged or die.

Put a diode across the regulator, with its Anode to the output, and its cathode to the input. Under normal operation, the cathode is higher than the anode, and the diode won't conduct. When the power is cut, the diode conducts, because its anode is now higher in voltage than the cathode, and the voltage across the diode, and therefore the regulator is now only 0.6-0.75 or so volts, and it lives

-max- (author)  steveastrouk1 year ago

Interesting. I might bodge on a few idiot diodes to the pins if it helps lol! Also one of the things I notice a lot with many linear power supplies is that there is always a large rectifier reverse bias on the output. My guess is that it there to prevent current flowing the wrong way, back into the power supply if it was perhaps in series with another, more powerful power supply. Is that correct?

The video did not show your power supply.

What is your ripple voltage?

Pretty much all regulators must be 2 volts under the ripple voltage.

As an example transformer 35 volts out to full bridge rectifier to 2200 uF capacitor at 1 amp draw might give you a ripple 35 to 31 volts so not a problem to a 7812 or even a 7824 since you are 7 volts and more under the ripple.

But transformer 10 volts out to full bridge rectifier to 1000 uF capacitor at 1 amp draw might give you a ripple 10 volts to 8 volts, 7805 is Ok 7808 gets fried because it is at ripple not 2 volts or more below ripple.

-max- (author)  Josehf Murchison1 year ago
The power supply was 3 transformers all in series to give me 50V open circuit (peak voltage). I realised moments later that that was beyond the absolute maximum ratings for it.

I am not sure if I understand you correctly, why would a voltage regulator fail if the instantaneous voltage on the input is too close to the dropout voltage? The pass element will simply bias and the output voltage will follow the input voltage, minus that dropout.

I have never blown a voltage regulator.

You asked why you have bad luck with voltage regulators. Answer ripple voltage.


If you look at the data sheets most require 2 to 5 volts below ripple voltage.

The large electrolytic capacitor should go between the voltage regulators input and the rectifier keeping the voltage above a certain voltage on the input but it still has ripple.

Regulators should remove ripple if used right, if used wrong the regulators get trashed.

-max- (author)  Josehf Murchison1 year ago

I understand that operating below the dropout voltage regulator will cause the output to also not be regulated, and I'm familiar with ripple voltage and current, heck thats is why I hoard the largest electrolytic capacitors i can find in stuff I take apart, my current 12V power supply uses this 25,000uF capacitor!!!

How can too low an instantaneous voltage cause a regulator to be damaged? My only guess is like what steve mentioned, that if the power supply was to abruptly fail in such a way the voltage rail becomes a low impedance path to ground, then energy stored in large, low ESR capacitors on the output of a linear regulator would flow backwards through the regulator, causing reverse polarity on the error amplifiers internally, destroying the regulator.

I have experimented a LOT with homemade voltage regulator designs, my latest design is below. I was using the 7808 as a voltage reference since LEDs and zener diodes did not give me the line regulation I hoped for. I also need a good current source but that's another topic in another active question of mine.

Screenshot from 2015-11-20 15:35:58.png12179591_1007593432625461_674120266_n.jpg

Ok look at it this way.

Electricity is like water it flows down hill.

When the ripple is at its peak on the input side of a 12 volt regulator, lets say 14 volts, it charges the capacitors on the output side of the regulator to 12 volts.

Now the ripple voltage drops to its minimum lets say 10 volts.

So now you have 10 volts on the input and 12 volts stored in your capacitors on the output of your regulator, reverse biasing your regulator as the electricity tries to flow from the 12 volts on the output to the 10 volts on the input. This will blow your regulator.

Now you can use a diode on the output to prevent the reverse flow or to by pass the regulator but it isn't perfect and the regulator gets hit 60 to 120 times a second until it cant take it any more blowing the regulator. A diode is not perfect and there is reverse flow called leakage so your regulator does get hit.

If your input is always above the output of your regulator, it never gets hit with reverse bias.

-max- (author)  -max-1 year ago

Note: That schematic is a bit out of date now, I more recently realised that the op amps are being WAY overvolted, with 44V differential peak between Vcc and Vss!!! :O I discovered this when I started to breadboard this circuit up and realised that my LM324 is only rated for up to 36V across the power pins. I revised it by connecting the negative power input of the op amp to ground for the top, positive half, and the positive voltage rail for the negative rail to ground.

iceng1 year ago

The only thing that can kill a 317 is voltage. oH yea bad wiring too...

Over current, over heat will simply cause the unit to turn off.

Your supply could be generating a start high voltage spike.

-max- (author)  iceng1 year ago
Would a unregulated transformer, full bridge recifier, and large capacitor which has 22V peak voltages cause issues? It might be possible when I was rearranging the circuit that the output might have contacted -22V rail.