That is 12 watts, spread across 2 silicon diodes.
So may I use two large silicon diodes in series?I thought of LDO or linear regulators since the voltage which I want to sink is almost the same voltage drop in LDO or linear regulators so it wont be so power loss and heat
So yes, if you use two diodes rated at 10 Amps, it will drop approximately 1.5 V, from 14.5 to (less than) 13. At low current draw, it will rise a little.
A bridge rectifier has two diodes in series, and a second pair in parallel, so is a good solution.10-25Amp ones (low voltage) are common, and have a hole in the middle to screw them to a heatsink.Only use the + and - terminals, not the AC terminals.Leo..
An alternative suggestion is to use a buck-boost converter.
A possible problem with what you are suggesting is that if your supply is a battery then while the voltage might be 14V5 when the battery is fully charged it will be somewhat less as the battery discharges. A buck-boost converter will take care of this.
Please, it is "silicon" not "silicone"
Nice suggestion, but, it's not easy to find 8 to 10 Amps buck-boost converter which reliable and safe!
Do you see another possible problem which I can face in future?
Any battery capable of supplying 10A is also capable of supplying enough current into a short circuit to start a fire. Include a HRC (high rupture capacity) fuse in your circuit as close to the battery as physically possible, between 2 cells if possible.
Yup, automotive fuses prevent wiring fires just as well as other types!
Good idea, but, the bridge rectifier isn't PCB-mounting friendly! I'm thinking of using two STTH2002C in series with one big heat sink (which I didn't calculate yet).
I changed it... thank you Any more ideas?