you said 1.5A is not enough

List the Reply # where I said that .

Not only did I NOT say that , I said 1.3A was enough (1.2 for the solenoid,+0.1 for the buck converter.

so the current is 1.63A, @ 7V

I said 1.65W, NOT 1.65A.

1.65W/12V = 0.1375A@12V, NOT 1.65A.

Do you even know what Watts is ?

Watts is "P" in "P=I*V". (Again, Ohm's Law)

The solenoid draws 1.2A @12V.

Everything you said you planned to use draws

LESS than 333mA @5V which is ONLY 237mA @7V, which is only 0.138A@12V.

(I may have said 0.1A but it I meant to say 0.138A@12V , which, while different , is still a far cry from

1.2A@12V (which you claim I said) for the buck converter)

from the buck converter. Your math is all wrong.

The buck converter efficiency is probably 95% or better..

The power dissipated by the buck converter is the output voltage (7V) times the output current 237mA, (1.65W) , because it based on what you

are powering (the devices you listed) snd NOT

the arduino maximum (800mA@5V), so 1.2A@12V=14W + 237mA@7V (1.65W) = 15.65W

15.65W/12V= 1.3A (as I stated before).

The buck converter current is only 100mA @12V.

You don't need 4A@12V because YOUR

application (what you SAID you plan to use

only draws 100mA (0.10A) @12V.

0.1A+1.2A=1.3A@12V, and as Georg has tod us,

P=I*V

power =1.3A *12V =15.65W

current I = P/V

=15.65/12V

=1.3A (@12V)

NOT 4A

After everything you STILL Have NOT

Googled "Ohm's Law".

Instead you choose to consult the vendor.