A part like that may be mounted to a Printed-Circuit-Board (PCB), or in a pinch, all of the supporting components (stiffening caps on the output, decoupling caps on the input, and a return diode to protect the regulator from the charge stored in the stiffening caps) and the package itself can be hot glued to the back of an instrument cluster.
As that part says in its description - a heatsink may be needed in some cases. This mostly depends on the draw of the load on the voltage regulator. With a parts like the classic 7812 (for +12VDC), the 7809 (for +9VDC), or the 7805 (for +5VDC), in a TO-220 package the power limit is kept at about 15Watts. The voltage regulator is considered "linear" since it is a transistor which is limiting the voltage that is present at the output based on an internal reference. The transistor conducts more or less based on the circumstances of the output pin in relation to that internal reference. Since the transistor is directly limiting the output power it will be exposed to the current which is being demanded of the device.
[Edit] The "dropout" of a 7812's transistor with a 1Amp load (its rating) is 2Volts. This is the voltage difference from input to output which must exist before the part will correctly begin regulating its output voltage. All of the rest of the difference is lost to the regulator as heat. To figure out the power dissapated by the voltage regulator use the following equation:
) * AmpsLoad
) + (VoltsInput
The power to ground in Linear regulators is negligable (less than 1/20Watt) so we can just omit that for simplicity. Using the 12volt lightbulb example - we will assume a 5Watt light bulb at 12Volts. If those of you who have read the post on diodes across a battery bank solenoid recall an equation:
Volts x Amps = Watts
We can work backwards too: 12volts / 5Watts = ~417mA (for the lightbulb)
So looking at the Linear power equation, we can calculate (I'm assuming a solid 28.8Volt alternator charge voltage):
) * .417AmpsLoad
Of course this all assumes a consistent ambient temperature of 25degC - if your temperature around the part is higer there is typically a "de-rating" (so it won't be able to support a full 1Amp ouput if the temperatures are high). The bigger the voltage difference and higher the load curent, the harder the voltage regulator has to work to keep the voltage right. A plastic part like this will typically have a maximum temperature of 150degC for operation at the silicon. One must find if the cooling is enough for the temerature rise of the part. There is a specification usually for this measurement "n
There is some thermal resistance to the ambient air and some to the metal tab, with a TO-220, you can expect between 4 to 7degC/Watt to the tab, and about 50 to 100degC/Watt to the ambient air (through the plastic package).
Using those values, we can use another simple equation:
or 7Watts * 75C/W = 525degCSilicon
temperature rise (or "poof" in the silicon world... since the temperature limit for that part is about 150degC). You'd need to run this part in a -375degC ambient to avoid burning out the part!
Before looking for a heatsink it's a good idea to find out how much temperature rise your part can allow in its environment. If you have a part that's rated for 150degC and an ambient temperature of 70degC - the allowable rise of the heatsink is:
or in this case: 150 - 70 = 80degC
All of the thermal resistances must be accounted for - tab resistance, glue or thermal paste resistance, and heatsink resistance. The way thermal resistances are summed is really just as easy as series resistors:
If you have an thermal paste which has a thermal resistance of .0045DegC/Watt (like "Arctic Silver") and a heatsink which has a thermal resistance to ambient of 3degC/Watt with no moving air you'd get roughly:
5.5C/W + .0045C/W + 3C/W = 8.545C/W
Running the heat-rise numbers again, we see:
7Watts * 8.545C/W = 59.815degC of heat rise to the heatsink
Since we know that the maximum allowable heat rise is 80C, 60C is a good 20deg margin of safety, and the part should run fine for long periods of time.
We can also look at the efficiency of this solution with respect to power consumption. The bulb I used for this example is 5Watts. The V-Reg in this example will dissapte 7Watts as heat as a "cost" of the regulation. The total power consumed to light up this bulb is 12Watts - with only 42% of the power going to the bulb. If you are boondocking - notice of his type of detail can help squeeze the most run-time out of your batteries. Replacing an incandescent bulb drawing 5Watts with an LED which only draws 0.58Watts (28.8volt charging voltage - 3.5Volt @ 20mA blue LED with a current limiting resistor) is a savings of 11.42Watts. Even with an LED, running one 3.5volt part from a 28.8volt supply with a current limiting resistor means that about 88% of that power is still being wasted as heat by the resistor. Keep in mind the total loss of 510mW to heat is very small, but the useful work is relatively low as well - the LED is simply more efficient at outputting light at the specified spectrum than an incandescent bulb is - where upwards of 90% of that 5Watts is being emitted as IR "heat" whereas an LED may only emit about 10% of its 70mW as IR "heat". While the v-reg to light-bulb solution has moer of the power going to the load than the "regulator" the overall power draw is higher than that of the LED solution (and you can get more usefull work out of the LED option by simply putting more LEDs in series - for instance 6 LEDs of the above specs gives you 73% useful work and only about 27% waste as heat in the resistor).
Thanks to "pvcces" (Tom Caffrey) for pointing out my absurd miss-posting of info earlier today (this is the second time I've done that...
For applications which require more than 1Amp of output - linear regulators are not likely to be the most efficient way to go (due to the drop-out there is a lot of waste once the power goes up). For things like running 110Watts of 12Volt headlights off of 24Volts, there are switching regulators (like those found in common PC power supplies) which are more efficient (>60%).