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Originally Posted by ricecrispi
http://img233.imageshack.us/img233/8...attpelt7tz.jpg
This is what' i've learned.
Qc is 100W (should be 90 W)
Find the point were Qc @100W crosses various current rating lines at bottom of the graph.
Draw a vertical line up to the same current rate line above. This gives voltage the pelt draws to get that DT
for 12.3 V= Dt is 34 C
Cpu temp =Th- DT
TH = 25 C degrees (is the hot side of the pelt at Q total of 346 W. TH varies depending on how good your setup is)
25-34= -9 C
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cpuT is not equal to Th - dT. Tc = Th - dT; Th is ~ equal to coolant temperature.
cpuT > Tc; under maximum thermal load, it may be greater by as much as 10°C to 20°C.
Th varies as a function of Ta, V, I, Q and RQ.
The hot side temperature (Th) will be equal to the ambient temperature (Ta) plus the rise in temperature across the heat sink from dissipating the heat load (Q) and the peltier input power (V x I) multiplied by the heat sink thermal resistance characteristics (RQ).
Th = Ta + (V x I + Q) RQ
RQ = thermal resistance of heat sink in C° temperature rise per Watt dissipated.
Using known values for Th, Ta, V, I, and Q, one can solve for RQ.
RQ = (Th - Ta)/(V x I +Q)