THERMAL TRANSFER CHARACTERISTICS ON HEATER

Jorge Hernandez PhD. Stundent DIMES -UNICAL

Heat Transfer

• Temperature conduction is performed from a hot body to a cold body.

• Thermal Equilibrium happens in two bodies have the same temperature.

Thermal Transfer Variables

Difference Temperature

• Contact surface of two bodies experiences a difference in temperature.

• While thermal equilibrium come to happen, the difference go to zero.

Material thermal characteristic

• In previous example, heat was transferred from water through the metal to water.

• The rate of heat transfer depends on material through heat is transferred.

• Heat transfer coefficient (k) express thermal conduction characteristic of a material.

• Large value of K means thermal conductor and a lower value means thermal isulators.

Material thermal characteristic

• k is determined experimentally and its unit is 𝑊

𝑚. °𝐶

Source: http://www.roymech.co.uk/Related/Thermos/Thermos_HeatTransfer.html

Material k Material k

Aluminum 237 Porcelain 1.05

Brass 110 Wood 0.13

Copper 398 Water 0.58

Gold 315 Air 0.024

Material body characteristics

• Area (A) – Wider areas conduct more heat and this means

greater heat dissipation. Not good for our project.

• Thickness (d) – Thin walls conduct heat in faster way. This affect

the rate thermal transfer.

Mathematical Equation

Mathematical Equation • Variables:

– Temperature difference: (T1 – T2) – Thermal Conductivity Coefficient of the Material: k – Area: A – Thickness: d

(1) 𝑃𝑟𝑟𝑟𝑟 = 𝑄𝑟

= 𝜎𝑟𝑡𝑡 (𝑇𝑇 − 𝑇𝑇) [𝐽𝐽𝐽𝐽𝑟𝐽𝐽

= 𝑊]

(2) 𝜎𝑟𝑡𝑡 = 𝑘 .𝐴𝑑

[𝑊°𝐶

] Average Thermal Conductance (3) 𝑅𝑟𝑡𝑡 = 1

𝜎𝑡𝑡𝑡 [°𝐶

𝑊] Average Thermal Resistance

Information extracted from reference [1], assuming that areas are flat and regular.

Application Problem Case

Heater V1

Practical Considerations – Contact area of SafeFET

(heat source size) is smaller than the opposite side area and even the DUT contact area.

– Location of heat source is important to note because of thermal distribution on sink.

– Also, thermal resistance of electrical isulator must be considered.

Mathematical Resolution

• Calculation of Total Thermal Resistance based on reference [1] and [2]. 1. Calculate 𝑅𝑟𝑡𝑡. 2. Calculate 𝑅𝑅𝑅𝑐 3. Calculate 𝑅𝑅𝑅𝑅.

Calculation 𝑅𝑟𝑡𝑡 according to [1] 𝐴 = 0.0𝑇8 𝑥 0.0𝑇𝑇 [𝑚2]

𝐴 = 3.96𝑥𝑇0−4 [𝑚2]

𝑑 = 0.003 [𝑚]

𝑘 = 𝑇𝑇0 [𝑊

𝑚. °𝐶]

From (𝑇) and (3):

𝑅𝑟𝑡𝑡 = 𝑑

𝑘. 𝐴=

0.003(𝑇𝑇0)(3.96𝑥𝑇0−4)

°𝐶𝑊

𝑹𝒕𝒕𝒕 = 𝟔𝟔. 𝟔𝟖𝟖𝟏𝒕−𝟑[°𝑪𝑾

]

Equations for 𝑅𝑟𝑡𝑡 • In [2], Spreading Thermal Resistance (𝑅𝑟𝑡𝑡) is stablished

in function of 𝑅𝑟𝑡𝑡 and the difference between size of the heating areas in each side of the sink.

• 𝑅𝑟𝑡𝑡 is an additional quantity that is needed for determining the maximum heat sink temperature.

• Also, [2] considers the location of heating source on the sink surface. 𝑪𝒇 [ 𝟏

𝒎]

• Ideally, this should be in the center of the same. 𝑪𝒇 = 𝟏

Equations for 𝑅𝑟𝑡𝑡 and 𝑅𝑟𝑡𝑟

(4) 𝑅𝑟𝑡𝑡 = 𝐶𝑓 𝑥 𝐴𝑝. 𝐴𝐽

𝑘 . 𝜋. 𝐴𝑝. 𝐴𝐽𝑥

λ. 𝑘. 𝐴𝑝. 𝑅𝑟𝑡𝑡 + tanh (λ. 𝑑)𝑇 + λ. 𝑘. 𝐴𝑝. 𝑅𝑟𝑡𝑡. tanh (λ. 𝑑)

[°𝐶𝑊

]

Where: λ=

𝜋32�

𝐴𝑝+

𝑇𝐴𝐽

• 𝐴𝑝: Footprint area of the heat sink base-plate • 𝐴𝐽: Contact area of the heat source

(5) 𝑅𝑟𝑡𝑟= 𝑅𝑟𝑡𝑡 + 𝑅𝑟𝑡𝑡

Calculation of 𝑅𝑟𝑡𝑡 according to [2]

𝐴𝑝 = 0.0𝑇8 𝑥 0.0𝑇𝑇 = 3.96𝑥𝑇0−4[𝑚2] 𝐴𝐽 = 0.0𝑇04 𝑥 0.0𝑇0𝑇 = 𝑇05.04𝑥𝑇0−6[𝑚2]

λ = 𝜋3

2�

𝐴𝑝+

𝑇𝐴𝐽

= 377.39 [𝑇𝑚

]

Calculation of 𝑅𝑟𝑡𝑡 according to [2] • From (4) 𝑎𝑛𝑑 assuming that heat source is placed in the center of

the base-plate:

𝑅𝑟𝑡𝑡 = 𝐶𝑓 𝑥 𝐴𝑝. 𝐴𝐽

𝑘 . 𝜋. 𝐴𝑝. 𝐴𝐽𝑥

λ. 𝑘. 𝐴𝑝. 𝑅𝑟𝑡𝑡 + tanh (λ. 𝑑)𝑇 + λ. 𝑘. 𝐴𝑝. 𝑅𝑟𝑡𝑡. tanh (λ. 𝑑)

[°𝐶𝑊

]

tanh 377.39 𝑥 0.003 = 0.8𝑇𝑇76 λ. 𝑘. 𝐴𝑝. 𝑅𝑟𝑡𝑡 = 377.39 𝑥 𝑇𝑇0 𝑥 3.96𝑥𝑇0−4 𝑥 68.87𝑥𝑇0−3 = 𝑇.𝑇3𝑇𝑇

𝑅𝑟𝑡𝑡 = 𝑇 𝑥 3.96𝑥𝑇0−4. 𝑇05.04𝑥𝑇0−6

𝑇𝑇0 . 𝜋(3.96𝑥𝑇0−4)(𝑇05.04𝑥𝑇0−6)𝑥

𝑇.944𝑇.9𝑇9

[°𝐶𝑊

]

𝑹𝒕𝒕𝒕 = 𝟓. 𝟐𝟖𝟏𝒕−𝟑 [°𝑪𝑾

]

Calculation of 𝑅𝑟𝑡𝑟 according to [2] • From (5):

𝑅𝑟𝑡𝑟 = 𝑅𝑟𝑡𝑡 + 𝑅𝑟𝑡𝑡

𝑅𝑟𝑡𝑟 = 68.87𝑥𝑇0−3 + 5.𝑇𝑥𝑇0−3 [°𝐶𝑊

]

𝑹𝒕𝒕𝒕 = 𝟖𝟕. 𝒕𝟖𝟖𝟏𝒕−𝟑 [°𝑪𝑾

]

Note.- Line graph shows the Cf variations in function of the distance from the center of the heat sink to the heat source placed along the center line at y=0 and -37.5 < x < 37.5 [mm]. Cf is case dependent.

THERMAL TRANSFER SAFEFET - DUT

How does the heat flow?

• Considerations: – Contact temperature

between case SafeFET and right side of sink is 𝑇𝐶𝐶.

– Contact temperature between case DUT and left side of sink is 𝑇𝐶𝐶.

– Thermal resistence of the electrical isulator must be considered.

Case study extracted from Project • Both SafeFET and DUT are heat

sources, but in practice SafeFET deliveres more heat than DUT.

• All considerations are referenced to regime work. – SafeFET (DUT 2)

• 𝑇𝑗𝐽 = 𝑇56.8𝑇 °𝐶 • 𝑃𝐶𝐶 = 6.04 𝑊

– DUT 2 • 𝑃𝐶𝐶 = 0.𝑇4 𝑚𝑊 • 𝑇𝐶𝐶 = 𝑇50 °𝐶 (measured with LM35

and It will be demonstrated mathematically)

Analog focus for resolution

Case Temperature on SafeFET Adapting (𝑇) for this situation:

𝑇𝑗𝐽 − 𝑇𝐶𝐶 = 𝑅𝑟𝑡𝑗𝑡𝐽 ∗ 𝑃𝐶

Where, from datasheet SafeFET : 𝑅𝑟𝑡𝑗𝑡𝐽 = 0.6[°𝐶

𝑊] : Thermal resistance junction-case of SafeFET

𝑇𝑗𝐽 = 𝑇56.8𝑇 °𝐶 It is important to note that because both devices are heat sources due to their power consumption:

𝑃𝐶 = 𝑃𝐶𝐶 + 𝑃𝐶𝐶 𝑃𝐶 = 6.040𝑇4 [𝑊]

Therefore: 𝑇56.8𝑇 °𝐶 − 𝑇𝐶𝐶 = 0.6

°𝐶𝑊

∗ 6.040𝑇4 [𝑊]

𝑻𝑪𝑪 = 𝟏𝟓𝟑. 𝟐 °𝑪

Transfer Temperature between heat sink sides

Adapting (𝑇) for modified value of thermal resistance of the heat sink: 𝑃𝐶 = 𝜎𝑟𝑡𝑟 (𝑇𝐶𝐶 − 𝑇𝐶𝑆) [𝑊]

Where: 𝜎𝑟𝑡𝑟 = 1𝑅𝑡𝑡𝑡

= 𝑇3.5 𝑊°𝐶

𝑇𝐶𝐶 = 𝑇53.𝑇 °𝐶 𝑇𝐶𝑆 ≡ Temperature in side contact between heat sink and insulator 𝑃𝐶 = 𝑃𝐶𝐶 + 𝑃𝐶𝐶 = 6.04𝑊 + 0.𝑇4𝑚𝑊 = 6.040𝑇4𝑊

𝑇𝐶𝑆 = 𝑇𝐶𝐶 − 𝑃𝑟𝑟𝑟𝑟

𝜎𝑟𝑡𝑟

𝑻𝑪𝑺 = 𝟏𝟓𝟐. 𝟖𝟓 °𝑪

Case Temperature on DUT transferred through insulator

Adapting (𝑇) for this situation: 𝑃𝐽 = 𝜎𝑟𝑡𝑖 (𝑇𝐶𝑆 − 𝑇𝐶𝐶) [𝑊]

Where: 𝑇𝐶𝑆 = 𝑇5𝑇.75 °𝐶 𝑇𝐶𝐶 ≡ Case temperature of DUT 𝑷𝑪 = 𝑷𝑫𝑪 + 𝑷𝑫𝑫 = 𝟔. 𝒕𝟕𝒕𝟏𝟕𝑾 Whit:

𝜎𝑟𝑡𝑖 = 𝑘𝑖 𝑥 𝐴𝑖

𝑅𝑖= 𝑇.6𝑇

𝑊°𝐶

From the vendor site: 𝑘𝑖 = 3.5 [ 𝑊

𝑚.°𝐶]

𝐴𝑖 = 0.0𝑇5 𝑥 0.0𝑇9 = 𝑇85𝑥𝑇0−6 [𝑚2] 𝑅𝑖 = 38𝑇𝑥𝑇0−6 [𝑚] Therefore:

𝑻𝑪𝑫 = 𝑻𝑪𝑺 −𝑷𝑪

𝜎𝑟𝑡𝑖= 𝟏𝟓𝒕. 𝟕𝟕 °𝑪

Junction Temperature on DUT Adapting (𝑇) for this situation, and considering that heat source is the SafeFET:

𝑇𝐶𝐶 − 𝑇𝑗𝐶 = 𝑅𝑟𝑡𝑗𝑡𝐽 ∗ 𝑃𝐶 Where: 𝑇𝐶𝐶 = 𝑇50.44 °𝐶 𝑇𝑗𝐶 ≡ Junction temperature of DUT 𝑷𝑪 = 𝑷𝑫𝑪 + 𝑷𝑫𝑫 = 𝟔. 𝒕𝟕𝒕𝟏𝟕𝑾 𝑅𝑟𝑡𝑗𝑡𝐶 = 0.5 °𝐶

°𝑊

Therefore:

𝑻𝒋𝑫 = 𝑻𝑪𝑫 − 𝑷𝒔. 𝑹𝒕𝒕𝒋𝒕𝑫 = 𝟏𝟕𝟖. 𝟕𝟐 °𝑪

Conclusions • Difference of junction temperatures for this case study analyzed.

∆𝑇𝑗𝐶𝐶 = 𝑇𝑗𝐶 − 𝑇𝑗𝐶 = 𝑇56.8𝑇 °𝐶 − 𝑇47.4𝑇 °𝐶 ∆𝑻𝒋𝑪𝑫 = 𝟗. 𝟕 °𝑪

• Error between temperature setpoint and 𝑇𝑗𝐶

%𝐸𝐸𝐸𝐸𝐸 =𝑇50°𝐶 − 𝑇47.4𝑇°𝐶

𝑇50°𝐶∗ 𝑇00 = 𝟏. 𝟖𝟐 %

• Thermal Resistance of Heat Sink

𝑹𝒕𝒕𝒕 = 𝟔𝟔. 𝟔𝟖𝟖𝟏𝒕−𝟑[°𝑪𝑾

]

• Total Thermal Resistance of Heat Sink according location of heat source criteria.

𝑹𝒕𝒕𝒕 = 𝟖𝟕. 𝒕𝟖𝟖𝟏𝒕−𝟑 [°𝑪𝑾

]

• Temperature aquisition of the system was adjusted with an LM35, which

performs a sensing of its case temperature, in this way this mathematic process shows this right adjustment with the case temperature value on DUT:

𝑻𝑪𝑫𝑴 = 𝟏𝟓𝒕. 𝟕𝟕 °𝑪 (Mathematical value) 𝑻𝑪𝑫𝑬 = 𝟏𝟓𝒕. 𝒕 °𝑪 (Experimental value)

References

• The Physics Classroom http://www.physicsclassroom.com/class/thermalP/u18l1f.cfm

• Calculating spreading resistance in heat sinks http://www.electronics-cooling.com/1998/01/calculating-spreading-resistance-in-heat-sinks/

¿QUESTIONS?