can we use 7805 regulator in a circuit that work 24 hours a day?

Joël Huser

Joined Jun 30, 2019
Now in 2019, we use more and more rarely 7805 anymore..... in other words "LDO" with high current output (but we use yet LDO for small load consumption)..... because it's often getting hot, and so takes more place on the board because of the heatsink.... AND you should add some capacitors at the input and at the output ! But it's full analog.... and some people as me, decide sometimes to keep doing things according to the "analog" way of life....

Nowadays, we are using more and more often DC-DC converters. And Murata had a BIG idea ! I let you discover it with the attached datasheet : the OKI78SR33, OKI78SR05 or OKI78SR12.... DC-DC converters which have following advantages :
  • It's DC-DC so it becomes not hot, so no need of a heatsink, while saving place.
  • It has exactly the same footprint than the 7805, is PIN-compatible and can replace a 7805.
  • It has exactly the same output current of 1.5 [A].
  • Capacitors are already included. But I advise you to put 1 more at input and 1 more at output.
  • 36 [VDC] input is very large.
I am using oki components since 2012.


Last edited:


Joined Jul 1, 2009
can we use 7805 regulator in a circuit that work 24 hours a day?
Yes, if you also do the required calculations to ensure the thermal operating range, and output current based on that range is adequate for your needs.

In order to properly calculate whether or not it can be used for your project at all, let alone long-term, you need to know what voltage your power-supply is, what voltage you want to output (regulated output voltage), what is the desired maximum current your load might need, because the LM7805 will have to be able to provide up to that much on demand. You also need to know what your desired operating temperature of the regulator is (is it safe to touch, or does it need a heatsink? Does it need shielded?)

In a real-world example, let's say that we are using an LM7805 regulator to provide power to a 4-digit display system. That display system requires 400mA for it's triode heaters (100mA each), and about 15mA additional for each of the anode/cathode aspects of each display.

Input voltage is 9VDC
Output is 5VDC.
Desired Max current output from LM7805 is 500mA.
Desired top temperature is 38.8C - it's touchable, and we will be using a Heatsink, so we also have to calculate that dissipation in from the datasheet.

The regulator has to dissipate 4V (9-5 = 4). And at the current we need, it's going to dissipate 2W. Our junction temperature operating range is between -40 and 125C. R(thjA) is 50C, R(thjC) is 5C, and Max Allowed Power Dissipation before thermal shutdown of the regulator is 2.424W. The heatsink offsets by 10C.

Knowing these things, we know that we can safely operate the LM7805 satisfying out current and voltage requirements safely with a little head-room because we can draw as much as 606mA if needed.


Joined Jan 23, 2018
With the 7805, if you use the recommended bypass capacitors and an adequate heat sink, and use it within typical, not max, ratings it will indeed last many years. I have a bunch of laser measurement systems, each with a 7805, and no regulator failures in 15 years. Of course the current is 500Ma and the input voltage is 12 volts, so they are well within ratings.