MP3Car.com - View Single Post - Mastero MK-4 ATX PSU with features.

rando · 05-05-2004, 08:44 PM

Quote: Originally Posted by MrPerfectionest

Also, I am wondering what kind of heat sinking is going to be on that regulator so that it can really meet a 5A spec. If someone decides to hang something off that line that only draws an amp of current then you are looking at (13.8V-5V)*1 = ~9W of heat that has to be dissipated, 3A and you are in the 30W range which would pretty much take up half 1/4 of the PSU case with a heat sink and a necessary fan to keep it cool.

I have no idea of this is correct.

Pd = (Vin - Vout)Iload + VinIg

Eff = Pout/Pin => Pin = Pout/Eff
Pd = Pout/Eff - Pout = Pout(1/Eff - 1) = VoutIout(1/Eff - 1)

Assume Vout = 5V; Iout = 5A; Eff = 80% @ operating Vin and Temp
=> Pd = 5*5(1/0.8 - 1) = 25(1.25 - 1) = 6.25W

05-05-2004, 08:44 PM	#70
rando Raw Wave Join Date: Mar 2004 Location: Redondo Beach, CA Posts: 1,969	Quote: Originally Posted by MrPerfectionest Also, I am wondering what kind of heat sinking is going to be on that regulator so that it can really meet a 5A spec. If someone decides to hang something off that line that only draws an amp of current then you are looking at (13.8V-5V)1 = ~9W of heat that has to be dissipated, 3A and you are in the 30W range which would pretty much take up half 1/4 of the PSU case with a heat sink and a necessary fan to keep it cool. I have no idea of this is correct. Pd = (Vin - Vout)Iload + VinIg* Eff = Pout/Pin => Pin = Pout/Eff Pd = Pout/Eff - Pout = Pout(1/Eff - 1) = VoutIout(1/Eff - 1) Assume Vout = 5V; Iout = 5A; Eff = 80% @ operating Vin and Temp => Pd = 55(1/0.8 - 1) = 25(1.25 - 1) = 6.25W __________________ 2004 4runner*