Nick said:
Hi,
What is the minimum current for a regulator from there onwards a heat
sink should be used?
I am using a 1Ampere LM7815 regulator with 45mAmps current consumption.
Graham gave you the specific answer, but in the hope that you might want to
understand the principle involved:
The regulator dissipates power P = (voltage drop) * (current). The voltage
drop is the input voltage minus the output voltage, which is 21V - 15V in
your case. So it dissipates 6V * 45mA = 270mW.
That power is dissipated through the "thermal resistance" of the regulator
case. If you look at the datasheet, there is a rating for theta-ja, the
thermal resistance from junction to air (that's without any additional heat
sink).
Thermals can be modeled just like electric circuits: thermal resistance is
like resistance, power is like current, and temperature drop is like voltage
drop. So using Ohm's Law (V = I * R), and substituting in the the thermal
equivalents (T = P * theta), you can calculate the amount of temperature
drop between the regulator chip and the air, for a given power.
Looking on the National Semiconductor datasheet for an LM7815T, I see
theta-ja for the TO220 case is typically 50 C/W (that's degrees Celsius per
watt). So, we can calculate T = P * theta = 270mW * 50 C/W = 13.5C. In
other words, the chip will be running at 13.5 degrees above ambient, or
about 40C. I also see that the max operating temp is 150C, so you are quite
safe.
....Or, you could just look at the graphs in the datasheet, where it shows a
chart of how much power the chip can safely dissipate, with various kinds of
heat sinks and at various ambient temps.
