You can see here that the power rating is based on mounting to a sheet metal box with dimensions 4" x 6" x 2" x 0.040"(thickness). That is the rating for either the 5W or 10W. This should give you an idea that they are assuming quite a bit of surface area to dissipate the heat. If you put the heat into the lower triple it would be even better.
So unless you mount solidly (using heat sink paste) to something substantial (i.e. don't just float it in the air) these devices will get hot. The derating is from 25 deg C where the device will operate at full rated power down to 0% at 275 degC. So properly mounted at any ambient temp you are going to be no more than 10% derated (i.e. maximum of 4.5 watts for 5W or 9 watts for 10W device)
5 ohms 10W is fine if you don't find the size objectionable.
Dan,
The 6 ohm resistor will actually will run hotter than the 5 ohm. I posted a set of power calculations in a matrix of voltage v.s. resistance . Even though this is counter intuitive it is true. You have to do the calculations to convince your self.
I = V / R_total
R_total = 12.5 + R_ballast + R_switch (use 1 ohm is you like for R_switch)
P_ballast = I^2*R_ballast
Also the switch seems to be dropping quite a bit of voltage (I calculate 1 ohm using Age's values) the 5 ohms is keeping the total current in check along with the switch.
Jim
So unless you mount solidly (using heat sink paste) to something substantial (i.e. don't just float it in the air) these devices will get hot. The derating is from 25 deg C where the device will operate at full rated power down to 0% at 275 degC. So properly mounted at any ambient temp you are going to be no more than 10% derated (i.e. maximum of 4.5 watts for 5W or 9 watts for 10W device)
5 ohms 10W is fine if you don't find the size objectionable.
Dan,
The 6 ohm resistor will actually will run hotter than the 5 ohm. I posted a set of power calculations in a matrix of voltage v.s. resistance . Even though this is counter intuitive it is true. You have to do the calculations to convince your self.
I = V / R_total
R_total = 12.5 + R_ballast + R_switch (use 1 ohm is you like for R_switch)
P_ballast = I^2*R_ballast
Also the switch seems to be dropping quite a bit of voltage (I calculate 1 ohm using Age's values) the 5 ohms is keeping the total current in check along with the switch.
Jim
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