Greetings,
I am new to electronics but interested in learning, I searched and learnt on various blogs on Google that 7805 can reduce a12V input supply to 5V. Looks like 7805 is heating up a lot.
I have added a 1000uf capacitor in parallel with input and a 100uf capacitor in parallel with output. I learnt this also in a YouTube video.
Are there any other alternatives to 7805 that heats less? PS, since I am a novice, it should be easy for me to solder/setup.
That is why heat sinks are made and used! Do you know how to compute watts? Volts time amps. Your 7805 MUST dissipate 12 volts minus 5 volts = 7 volts times the current passing through the 7805. All that heat must go somewhere else the 7805 will continue to get hotter!
The heat generated in a linear regulator depends on the voltage dropped across it and the current through it.
The wattage (which is converted to heat) is calculated as Voltage X Current. For example, 12V dropped to 5v and 1 Amp, the regulator is dissipating 7 Watts, with the 5V load using 5W.
The LM7805 is rated up to 1.5 Amps, but that requires a BIG heatsink and it may depend on how much voltage is dropped across it so at a higher input voltage you might never get 1.5A, even with a big heatsink.
If it overheats it will shut-down or oscillate. (It's thermally protected and it should shut-down before it burns-up and dies.)
A switch-mode (or "switching") voltage regulator can be almost 100% efficient, which means it doesn't get nearly as hot, and you can actually get more current-out (at lower voltage) than you feed-in.
Switch-mode regulators can be tricky to build so it's usually best if you buy a pre-made module/circuit board.
Yes, use a switching step down regulator. They are much more efficient, except when the voltage drop is very low (like producing regulated 3.3V from a 3.7V LiPO battery).