I have a guitar amplifier (Blackstar ID Core 10 v2). Its power brick has gone bad last year, and I have been searching for a power brick ever since.
This device works with 6.5 V, 3 A DC. It is pretty uncommon as far as I can tell. The worst part is, I cannot get a replacement power brick here in Turkey. It just doesn't exist here and I cannot import one.
So what I am planning to do is: get a 12 V, 3 A or a 9 V, 4 A power brick, and an XL4015 DC/DC voltage (and current) regulator (I think it is a "switching" regulator, but I don't know the difference between switching and linear regulators).
I will desolder the power plug of the amp, solder the plug to the INPUT of the module, and the power cables of the amp to the OUTPUT of the module. So, the module's input will be 12 V, 3 A or 9 V, 4 A, and its output will be 6.5 V, 3 A (Don't worry, I will use a multimeter). I will put a small heatsink on the IC, but heat is still concerning me.
I read online that these modules tend to get hot to touch at 3 A even though they are rated for 5 A. This is when I realized I needed a heatsink.
Upon further research, I learned that if the voltage difference between input and output is too big, it might just burn the IC right away as it would get too hot. I have seen a temperature calculation for linear regulators, which is: voltage difference x ampere = temp difference. I don't know if this applies to my situation but I wanted to make sure because a small passive heatsink can cool 2 W of heat, but 16.5 W of heat? IDK and I don't think so.
Does the calculation above apply to my situation? If not, what kind of temperature would I get? Would a small heatsink be enough? Should I be doing this project?