0
\$\begingroup\$

i am using LM7805 whose output voltage 5V and Output Current 1A i have used this to feed 5V supply for my application Circuit and my application circuit draw 125mA current , even tho 125mA current is less than the 1000 mA my LM7805 is slightly heated and application circuit not behave like its original functionality so what should be the probable issue that my power supply circuit is having do you think and please let me know and sorry for my poor english !!! thank you !!!

enter image description here


Problem got solved and this issue is because of microcontroller oscillator frequency and after scaling down this frequency problem got solved !!!

\$\endgroup\$
  • 1
    \$\begingroup\$ if the 7805 is only 'slightly heated', it's unlikely to be malfunctioning due to heating. (12v-5v)*125mA is about 1 watt, so I'd expect 'slight heating'. You have capacitors before and after the 7805, which is good, forgetting the output capacitor is a frequent cause of problems. Can you measure the output voltage? Does your application work with an alternative 5v supply? \$\endgroup\$ – Neil_UK Dec 18 '17 at 6:32
  • \$\begingroup\$ I getting 5 volt at the output of LM7805 right now i don't have external 5 Volt power supply to test this \$\endgroup\$ – user374112 Dec 18 '17 at 6:44
  • \$\begingroup\$ sounds good. Does the app work with a different source of 5v? \$\endgroup\$ – Neil_UK Dec 18 '17 at 6:45
  • \$\begingroup\$ Does your application work with an alternative 5v supply? – No i dont have external supply !! \$\endgroup\$ – user374112 Dec 18 '17 at 6:46
  • \$\begingroup\$ Actually i dont have right capacitor value so i have used near by capacitor value in my power supply unit \$\endgroup\$ – user374112 Dec 18 '17 at 6:47
7
\$\begingroup\$

Making heat is how the 7805 works.

It is a dissipative voltage regulator. That means it sits in series with the load, and is more or less a variable resistor, making resistive heat so as to make the output voltage 5v.

If you input 12V, it causes a 7 volt voltage drop by making heat.

In your case that would be 7 volts x 125ma or 875mw, plus its own internal usage/heat Almost a watt of heat! More than your load is using. If you are on batteries it is wasting 7/12 of your battery capacity.

However it is simple and cheap. That's what makes 7805s special.

\$\endgroup\$
  • 3
    \$\begingroup\$ Hmm, I'm not sure I'd call it resistive, though if I squint I could see where you're coming from. Certainly 'dissipative'. While a 7805 has resistors in it, all the power stuff is done by a transistor, not a resistor. Even 'behaves as a series resistor' doesn't really quite cut it, in my book. 'Makes heat, the same way a resistor would', yes, I'd go for that. \$\endgroup\$ – Neil_UK Dec 18 '17 at 7:01
  • 5
    \$\begingroup\$ +1 "More than your load is using" Always does as long as the input voltage >2x the output voltage. \$\endgroup\$ – Trevor_G Dec 18 '17 at 7:19
  • 1
    \$\begingroup\$ +1 not much different, but add 60mW on top for the regulator itself (12V x 5mA typical) and you get 935mW, which will probably be too hot to hold your finger on for long (assuming no heatsink). \$\endgroup\$ – Spehro Pefhany Dec 18 '17 at 10:54
2
\$\begingroup\$

Harper already explained the theory behind it. Here is some additional info and stuff that you can do to improve the circuit:

As per datasheet:

Facts

1 A output current is possible only if you take away the generated heat. To achieve this, you can attach a heat sink to your 7805 IC. It looks something like this:

heat sink

This should improve your circuit. If it's still heating up, you can probably use a bigger heat sink:

bigger heat sink

However, you will still be wasting a decent amount of power. If you want to save that, you can go for a switching regulator rather than a linear one. The circuit is not as straightforward as 7805, though:

sw reg

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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