Electronic Swear said:
Thank you for your suggestion.
Basically, I have tried to use IRFP450 before. Of course there is
a heatsink behind the body of the transistor. However, it is also
failure in that operation. At 40A current 1-2 seconds, the transistor
cannot switch to cut-off. Then the transistor finally burnt out with
short the junction.
If I use for the IRFP460, what size of heatsink should I need for?
Topically, the IRFP460 is not much affordable a higher current then
IRFP450. So, is it also afford that high current at 40A?
A single IRFP450 device will not be adequate, regardless of the size of
heatsink. Although better, a single IRFP460 will also be inadequate. You
will likely need to use at least three IRFP460 devices in parallel all
mounted to a single or three separate heat sinks. I'll provide the link to
the IRFP460P datasheet again for reference:
http://www.irf.com/product-info/datasheets/data/irfp460p.pdf
We notice from figure 11 in the datasheet (transient thermal impedance
curves) that for a two second event we will receive no benefit from the
thermal inertia of the IRFP460 die. Therefore we will need to parallel
enough IRFP460 devices such that they could handle 40A of current
indefinitely (provided they had enough heatsinking).
So lets assume we use three IRFP460P devices in parallel for your solution.
How big should the heat sink be?
First find the steady state dissipation at 5A. I^2 * R at 25 deg. C with
three devices in parallel will be around 2.25 Watts. Each of the three
devices would therefore be dissipating around 750mW of heat. Giving this is
a fairly small number, and we will need a heatsink anyway (for it's thermal
inertia), we will make the approximate assumption that the MOSFET dies will
be operating at say 50 deg. C prior to an overcurrent event. This is a
fairly arbitrary assumption, but it gives us a starting point to work with.
We will therefore also assume the heatsink temperature is around 50 deg. C
prior to an overcurrent event. Since the ultimate objective of these
calculations it to insure a design that will keep the MOSFET die
temperatures at or below 150 deg. C (their rated maximum), we will need to
know the thermal resistance of the MOSFET dies to heat sink. From the
IRFP460P datasheet the thermal resistance should be (without electrically
isolating thermal pad, but with a flat heatsink that has been greased and
appropriately torqued) around 0.45 + 0.24 deg. C/W = ~0.7 deg. C/W.
So now how much heat must the MOSFETs dissipate at the end of the 40A
overcurrent event? Let us assume we allow them to reach the maximum die
temperature allowed (150 deg. C) after two seconds. This means their on
resistance will be around 0.27*2.5 (from figure 4) = 0.675 Ohms. For three
in parallel the total MOSFET set resistance is 0.225 Ohms. So I^2 * R at
40A and 150 deg. C is 360W. Each device must dissipate one third of this or
120W.
Now we can approximate the thermal rise due to the thermal resistance
between die and heatsink. 120W * (0.7 deg. C/W) = 84 deg. C. So 150 deg.
C - 84 deg. C = 66 deg. C. Since we assumed the heatsink would start at a
temperature of 50 deg. C (mainly due to ambient heat), this means we have
66 - 50 = 16 deg. C heatsink temperature rise to work with. So we need to
select a heatsink with enough thermal mass so as not to rise more than 16
deg. C given our anticipated total energy lost during the overcurrent event.
During the overcurrent event we assumed the MOSFET set is dissipating 360W
(given a die temperature of 150 deg. C, which is probably fairly accurate
since the MOSFET dies will heat up to near 150 deg. C very quickly compared
to the two second event since most of the thermal rise will occur between
die and heatsink). Since the overcurrent event can last up to two seconds,
the total energy dissipated should be no more than about 360 W * 2 s = 720
Joules. So we need a heat sink capable of taking 720 Joules with only a 16
deg. C temperature rise.
Assume we select aluminum as our heatsink material. Aluminum has a specific
heat of 900 J/(kg*K). Solving for the heat sink mass (720 J) * (kg*K/900 J)
* (1/16 K) = 0.05 kg. Since aluminum has a density of 2700 kg/m^3, we need
a heatsink that has a volume of at least 18.5 cm^3 and mass of 0.05kg. You
could use either one heatsink of 18.5 cm^3 or three heatsinks of 6.2 cm^3.
The optimal heatsink shape in this case would not have large fins, but be
one large rectangular block, since a solid block would have less thermal
resistance between the TO-247 contact point and the rest of its thermal
mass.
The heat sink/spreader does not need to dissipate much steady state heat, so
you could pot the whole heatsink or cover it with something electrically
insulating if desired to increase technician safety (and/or comply with
possible safety organization specifications). Unfortunately when using only
three IRFP460P devices you will not be able to use electrically isolating
thermal pads (unless they are extremely good) since they will increase the
thermal resistance from package to heatsink too much. SOT-227 packaged
devices are usually extremely expensive, but it does have the very novel
feature that you get an electrically isolated heat sink tab.
So does this help answer your questions?