I have got a new problem here: the power dissipation in L7812 and L7805 is a little bit higher than I expected making the heat sink to reach about 47 C.
I know that currently it will probably be OK, but I am wondering what would happen if the temperature in my workshop will reach 30 C (currently the room temperature is 21 C).
If I will replace the diode D1 (1N4004) by a zener diode in reverse, will the circuit work the same ?
I mean that there will be advantages and disadvantages, for example, the voltage on the input of 7812 will be reduced, while there will be power dissipation on the diode... which will make the Zener to dissipate heat.
I measured the current through 7812 using a multimeter and it was about 0.11A. If I will use a 3V3 / 1.3W Zener diode (just one, or I can use 2 in series), then the power dissipation should be around 0.4W / diode.
I am asking this because I cannot introduce into the case a larger heatsink for L7812 and L7805, and I need somehow to reduce the power dissipation on these 2 components, and I cannot redesign the PCB.
Would it be sufficient to use 3V3 / 1.3W diode ?
Are there any better solutions for this problem ?
Edit: Also, I am thinking about using a resistor instead of D1, for example a 47 R / 2 W or 5 W will reduce the voltage by 5.17 V and will dissipate 0.57 W.