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Posttime: 19 Apr, 2012 Views: 183 Author: Jacky

Calculation Model for LED Lighting Heating Transfer|LED Thermal Resistance Algorithm

Take a calculation for example, this shows the equivalent circuit of the LED thermal resistance algorithm.

Following is the basic structure that lamps requirements, LED was enclosed in a sealed plastic enclosure, the right side of the insulation blanket is thicker, and the heat conduction basic can be neglected comparing the other part of the thermal conductivity. Heat is mainly through supporting frame, plastic shell, rubber coat, and then through the external air dispersion of the air.

thermal resistance algorithm

1. Simplified model:

(1) Aluminum plate was regarded as an isothermal heat source;

(2) There is an extra heat conduction layer between the support plate and and aluminum plate;

(3) Because thermal conductivity of the plastic is much higher than air thermal conductivity, so the air thermal conductivity can be neglected;

(4) There is an extra heat conduction layer between the support plate and and plastic shell;

(5) Plastic shell and rubber skin connect closely;

(6) The heat convection between aluminum plate and shell can be neglected

So total thermal resistance, among them:

R=R1 +R2 +R3 +R4 +R5 +R6

R1 : thermal resistance of extra layer between support plate and aluminum plate;

R2 : thermal resistance of support plate;

R3 : thermal resistance of extra layer between heat sink and plastic shell;

R4 : thermal resistance of plastic shell;

R5 : thermal resistance of rubber cortical;

R6 : the thermal resistance when rubber skin is in air convection heat transfer.

2. Calculation

Below are calculated respectively of each part thermal resistance:

Thermal resistance calculation

Above all,

ki(i=1, 2, 3, 4, 5): the coefficient of thermal conductivity of each media

Ai (i=1, 2, 3, 4, 5): heat conduction equivalent cross-sectional area of each media

di (i=1, 2, 3, 4, 5): each medium length for heat conduction

R6

Above, h6 is the average heat transfer coefficient.

L: qualitative length, in large cylindrical convective heat transfer case, usually take cylindrical diameter;

Grl and Pr: respectively for the dimensionless Grashof number and Prandtl number, different conditions in the numerical can check table obtain;

C: adaptation coefficient, in the cases usually laminar take 0.53 ~ 0.54;

A6: convection heat transfer of the effective area;

k6:  is the air coefficient of thermal conductivity.

So tota heat resistance is

R=R1 +R2 +R3 +R4 +R5 +R6=86.37    (W/K)

As for LED when there is about 1 W power transfer into heat, the rise of temperature of the aluminum plate is

T

T2 is temperature of aluminum plate, T1 is environmental temperature.

If the environment temperature is 40 ℃, the aluminum plate temperature will reach 126 ℃, and at the time the LED junction temperature is up to 166 ℃, following is the relationship between the failure and the temperature of Luxeon LED:

Luxeon LED

The failure rate of Luxeon in such a high temperature is 92854 times, close to 100000 times than junction temperature 120 ℃. The operation reliability is very poor in such a high temperature, so this thermal conductivity structure is not feasible. Comparing various thermal resistance components, the main thermal resistance is the conductive thermal resistance of support plate, improve the structure of it is needed.

In another situation, we use another structure of heat transfer, cancel plastic support stands, with 0.3 mm thick electrolytic copper cooling plate.

The other parts are constant, the thermal resistance of electrolytic copper cooling plate is:

R2

The edagefold of electrolytic copper cooling plate is6 mm, the equivalent thermal resistance of this part is:

R7

So, total thermal resistance is:

R

If the environment temperature is 40 ℃, the aluminum plate temperature will reach 64.6 ℃, and the junction temperature of LED is up to 104.6℃ at the time, the heat transfer structure is feasible in theory.

The following chart is results comparison between two kinds of structural temperature test and theoretical calculation.

Plastic Frame Structure

Electrolytic Copper Cooling Plate

Environ T (℃)

Al Plate T (℃)

Environ T (℃)

Al Plate T (℃)

In Theory

40

126

40

64.6

In Experiment

39.5

127.3

39.8

72.3

It can be seen from the calculation above, use the resistance that equivalent to the thermal resistance of circuit calculating method, select the appropriate simplified model. For different heat transfer structure, the rise of temperature of the focus on computation calculating, can judge advantages and disadvantages of the structure before making mold in the heat transfer, and can according to each part of the thermal resistance of the calculation results of the main structure improving direction of judgment, this to guide and improving the structure design of the actual significance.

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