yes, it's not an easy design. Even if you break it out into 28, I'm still not sure that's enough.
Assuming you want to hit 1 kW, then 1000 / 28 = 35 W per device. Let's do some more math:
First, let's talk about the heatsink.. The manufacturer says they characterized it at 0.3° C/W. You absolutely won't hit this number, and here's why:
The manufacturer will characterize this heatsink with a full on heat source across the entire planar surface of the milled sides.
This design, however, will have several point sources of heat on each side; the area of the heat source is much smaller compared to the area of the surface they are mounted to. The heat has to spread out, and this spreading encounters resistance, and takes time, and causes a localized heat zone at the point where the FET mounts onto the heatsink. This actually causes the FET to be hotter than a simple calculation shows.
Here's the heat spreading calculations
Let's assume just one FET dissipating 35W on the entire heatsink
The average temperature of the heatsink is simply the total power x the thermal resistance = 35W * 0.3° C/W = 10.5° above ambient.
The local point source heating will cause a localized rise in temperature above the average. The thermal resistance that governs that rise
is R
c, the constricting thermal resistance.
A
s is the contact area of the point source
A
p is the contact area of the baseplate
t is the thickness of the baseplate
k is the thermal conductivity (for aluminum, k = 200-250. lets pick 210)
R
? is the average thermal resistance of the heatsink ( the manufacturer's number)
} {1 + \lambda k A_p R_\theta \times\tanh(\lambda t)})
where
For this design:
A
s TO-220 = 7 x 13 mm (varies) = 91 mm
2 = 9.1*10
-5 m
2A
p = 80x254 mm (assumption) = 20320 mm
2 = 0.0203 m
2t = 2 mm (assumption) = .002 m
k = 210 W/m•K
R
? = 0.3° C/W
plugging it all into the equations:
(210)(0.0203)(0.3) = 184.04)
 = \tanh((143.91)(.002)) = 0.2801)
\pi\sqrt{(0.0203)(9.1*10^{-5})}} \times \frac{184.04+ 0.2801}{1+184.04\times0.2801} \\ &= 0.1483\times 3.51\\ &= 0.520 ^\circ C/W \end{align*})
thus the thermal resistance at the point heat source of a single TO-220 MOSFET all alone on the large heatsink is
To do it for more than 1 TO-220 device on the heatsink, just divide the area A
p by the number of devices. I did further calculations, with 4 devices, 8 devices, 10, 15 devices. The value for R
c gets closer and closer to about 0.200 °C/W. It gets even lower with more devices on the heatsink because more devices spreads the heat more evenly, and any one point heat source is not above the average temperature by very much any more. But at some point, you can't add more devices than the heatsink can actually dissipate, as seen in my previous posting.
So you'd be best to design with the assumption that the heatsink is actually going to yeild 0.3°C/W
(avg) + 0.2 °C/W
(local) = 0.5 °C/W
(total) per MOSFET, then calculate the maximum power you can dissipate from the heat sink to keep the heatsink average temperature below about 85C
Finally to wrap it up, going back to my previous post
T
sink = 150 - (Rth
jc + Rth
(mica+grease))*Watts
and
T
sink = (Rth
(avg) + Rth
(local))*Watts*Number_Of_FETS
therefore 150 - (Rth
jc + Rth
(mica+grease))*Watts = (Rth
(avg) + Rth
(local))*Watts*Number_of_FETS
150 - (1.11 + 1.3)*W = (0.3+0.2)*W*N
solving for Watts gives a formula to use to keep the Junction temperature below 150 °C : Watts
(per_fet) = 300/(4.82 + Number_of_FETS)
run it for 1-25 FETS and ask google to plot it
(remember this is all still on your original choice of heatsink, which is made up of 2 heatsinks from Farnel)
google.com/search?q=plot%20300%2F(x%2B4.82)%2C%20x*(300%2F(x%2B4.82))%2C%20from%200%20to%2025You get a nice graph. The blue line is the number of devices vs watts per device. If you multiply the number of devices * watts per device (the blue line) you get the red line, and the goal is to maximize the red line.
Conclusion : What you see is that this heatsink is only good up to about 300W. beyond that it gets too hot, and the junctions get too hot.
You can have any combination of number of devices (blue line, X-axis) * Power per device (blue line, Y axis). More devices, less power per device. Less devices, more power per device. You can find a spot around 25-30 devices and 10-8W per device that yeilds about 250-260W total on the heatsink. Then you need 4 channel heatsinks to get 1000W. you can put 25 devices (side 1 has 12, side 2 has 13, spaced 20mm apart or so). So you need 100 MOSFETS.
Here is where the more expensive heatsink from Farnell starts to make more sense. You only need 1. it can dissipate about 1000W. (0.06 °C/W)
finally, no one really needs to do all this math. Just always derate your heat sinks. about 20%, that might be a good rule of thumb.
--------------------
The blue line , X axis is number of devices. The Y axis is the Power per Device. The red line is the Devices*Power product.
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