With all semiconductors, it is important to read the data sheet.
It would appear that the device you are using is a LM338T.
Your application wants to reduce a 15 volt supply down to 6 Volt at a current of 5 amp.
The device dissipation in this case is 45 Watt.
For this device, the TI data sheet says:
"These specifications are for power dissipation of 50W for the TO-3 package, and 25 Watt for the TO-220 package."
The thermal resistance (j-c) is specified at 4 degree C per watt.(note NO decimal place). So I will take that to be 4.4 degree C per watt for the TO-220 package.
The maximum junction temperature for the LM 338 is only 125 degree C. (for the TO-3 package, the thermal resistance (J to case) is 1.4 degree C per watt.)
Under the operating conditions you want, the temperature rise of the device will be 180 degrees C. There is NO WAY that the TO220 device you have selected will come anywhere near doing what you want it do unless you cool the case to -55 degree C.
You have several options:
1. include a series resistor between the power supply and the LM338. My feeling would be to use say 1.2 ohm to an in/out differential of 4 volt. The dissipation in the first case would be 25 Watt and in the second case would be 20 Watt. but remember the data sheet values are given for an input/output differential of 5 volt.
2. Use an external pass transistor, although in the data sheet, TI use multiple LM338's for current greater than 5 Amp.
3. reduce the in/out differential to say 2.5 volt but I'm not sure how the LM338 will behave at this differential.
4. Give away the plan to use a DC regulator and use a switching regulator instead.
The thermal design for the LM338, shows that in your case, the power dissipation is 25 Watt. Since the maximum junction temperature is 125 Deg C, then the temperature rise above ambient (say 40 deg C) is (125-40) = 85 degree C. The required thermal resistance (junction to ambient) is 85/25 =3.4 degree C per watt. Since the device has a thermal resistance (junction to case) of 4.4 degree C per watt, then the device is unsuitable for the application unless the case temperature is held at 15 degree C. Using the series ballast resistor, the power dissipation is 20 Watt, and the required thermal resistance is 4.3 degree C per watt. Even with the 1.2 ohm ballast resistor, the device will need an infinite heatsink (zero Degree C per watt). This is not possible, so you are going to have to use more than a single LM338.
Using the TO-3 package, the thermal resistance (J to case) is 1.4 degree C per watt). The required thermal resistance of 4.3 degree C per watt allows the heat sink thermal resistance to be 4.3 - 1.4 = say 3 degree C per watt.
Looking at your photo of the heat sink, my guess is that it is probably about 8 to 10 degree C per watt. To get the thermal resistance down to 3 Deg C per watt needs a LOT of area AND black to boot.
Check what the suppliers RS or Farnell (Element 14) have to say. These suppliers have decent trustworthy data sheets that can be relied on for DESIGN.
Hope this saves a lot of heart ache.
It would appear that the device you are using is a LM338T.
Your application wants to reduce a 15 volt supply down to 6 Volt at a current of 5 amp.
The device dissipation in this case is 45 Watt.
For this device, the TI data sheet says:
"These specifications are for power dissipation of 50W for the TO-3 package, and 25 Watt for the TO-220 package."
The thermal resistance (j-c) is specified at 4 degree C per watt.(note NO decimal place). So I will take that to be 4.4 degree C per watt for the TO-220 package.
The maximum junction temperature for the LM 338 is only 125 degree C. (for the TO-3 package, the thermal resistance (J to case) is 1.4 degree C per watt.)
Under the operating conditions you want, the temperature rise of the device will be 180 degrees C. There is NO WAY that the TO220 device you have selected will come anywhere near doing what you want it do unless you cool the case to -55 degree C.
You have several options:
1. include a series resistor between the power supply and the LM338. My feeling would be to use say 1.2 ohm to an in/out differential of 4 volt. The dissipation in the first case would be 25 Watt and in the second case would be 20 Watt. but remember the data sheet values are given for an input/output differential of 5 volt.
2. Use an external pass transistor, although in the data sheet, TI use multiple LM338's for current greater than 5 Amp.
3. reduce the in/out differential to say 2.5 volt but I'm not sure how the LM338 will behave at this differential.
4. Give away the plan to use a DC regulator and use a switching regulator instead.
The thermal design for the LM338, shows that in your case, the power dissipation is 25 Watt. Since the maximum junction temperature is 125 Deg C, then the temperature rise above ambient (say 40 deg C) is (125-40) = 85 degree C. The required thermal resistance (junction to ambient) is 85/25 =3.4 degree C per watt. Since the device has a thermal resistance (junction to case) of 4.4 degree C per watt, then the device is unsuitable for the application unless the case temperature is held at 15 degree C. Using the series ballast resistor, the power dissipation is 20 Watt, and the required thermal resistance is 4.3 degree C per watt. Even with the 1.2 ohm ballast resistor, the device will need an infinite heatsink (zero Degree C per watt). This is not possible, so you are going to have to use more than a single LM338.
Using the TO-3 package, the thermal resistance (J to case) is 1.4 degree C per watt). The required thermal resistance of 4.3 degree C per watt allows the heat sink thermal resistance to be 4.3 - 1.4 = say 3 degree C per watt.
Looking at your photo of the heat sink, my guess is that it is probably about 8 to 10 degree C per watt. To get the thermal resistance down to 3 Deg C per watt needs a LOT of area AND black to boot.
Check what the suppliers RS or Farnell (Element 14) have to say. These suppliers have decent trustworthy data sheets that can be relied on for DESIGN.
Hope this saves a lot of heart ache.