Thermal Management: MCPCBs for LED Applications

MCPCBs for LED Applications
Thermal Management Material Specifications

Purpose of the Webinar
Defining your needs
To break down your needs for thermal management materials, specifically metal core printed circuit
board (MCPCB), in LED Applications by technical requirements in order to make more effective
callouts.

Controlling the process
The end result should be shorter lead times, lower cost, and more reliable product.

Non-browning soldermasks
Introduce new non-browning white LED soldermasks.

Metal Core Boards
Definition
The MCPCB commonly consists of a metal core layer (typically aluminum or copper), a continuous dielectric
layer and a copper circuit layer.

Part One
Specifying Materials for LED Applications

Survey Question
How do you specify materials in MCPCB LED applications?
a) Company Name and/or Part Number (e.g. Bergquist, Laird, DuPont, etc.)
b) Thickness of dielectric
c) Tg or Td
d) “Metal Core”
e) Insulation Resistance & Thermal Conductivity Values

Survey Answer
How do you specify materials in metal core LED applications?

a) Company Name and/or Part Number (e.g. Bergquist, Laird, DuPont, etc.)

Comment
Specifying by brand name locks you into a particular product produced by a particular manufacturer.

Risks Include:
• Locked into pricing
• Subject to lead times
• Preventing new materials from being used on your product

Survey Answer

b) Thickness of dielectric,
c) Tg,
d) Td

Specifying only these items may not fully address your needs.
• Does not address thermal conductivity
• Does not address electrical insulation resistance
• Does not address type of dielectric

Answer

e) “Metal Core”

Comment:
Specifying “metal core” does address anything whatsoever.

Answer

f) Insulation Resistance & Thermal Conductivity Values

Comment:
You are a rock star! This is the whole point of our webinar.

Consider your needs when bringing the
bare board into the equation

Survey Question
What Are Your Needs When Selecting Materials? (Choose all that apply)
a) Transfer heat (Thermal Conductivity)
b) Prevent short circuiting to base metal core (Electrical Insulation Resistance)
c) Thickness of dielectric
d) Brand name of the material?

Cost Drivers For MCPCBs
Dielectric Substrate
The # 1 cost component of the MCPCB is the dielectric substrate between the copper traces and the
metal heatsink / core.

Definition
An insulating medium which occupies the region between two conductors. In this case, the copper
circuits and the metal core heat sink.

Cost Drivers For MCPCBs (cont.)
Introducing competition
The most effective way to reduce cost of dielectric is to introduce competition:
 Laird
 Bergquist
 Dupont
 Uniplus
 Iteq
 Insulectro
 Plus future innovators

Definition and Applications
Thermal Transfer
Product Application
Electrical Insulation
Product Reliability

Purpose #1
Transfer Heat
Common Callouts include Thermal Impedance / Resistance (°C in2/W) and Thermal Conductivity / (°W/m-K).

Why Kapton® / Polyimide?
Heat Resistance
Stability
Flexibility
Dielectric Properties
Density/Weight

CooLamTM MCPCB (metal core PCB) Offers:
• Very Low Thermal Impedance
• Excellent Reliability Performance
• Excellent Durability and Stability at High Temperature
• Uniform Thermal, Mechanical & Electrical Properties Under Environmental Stress
• Lead Free Solder and Wirebond Process Compatibility
• Halogen Free
• Meets UL 94 V-0
• Construction Variations to Meet Thermal Management Needs
• 3D Shapes

Thermal Measurement
Characterize the Thermal Performance of Materials
Objective:
Measuring thermal performance per ASTM D5470.

Equipment:
1) Steady State “Thermal Interface Material Tester” (TIM)
(Analysis Tech. Inc.)

Factors to Consider:
1) Reducing contact resistance between sample and test unit
2) Repeatability of measured values
3) Identify any equipment and/or material limitations

Power (Watts)
Basic Principles of ASTM D5470
Heat Source T1 What is Thermal Resistance ?
Thermal Resistance (Rth) is defined as the difference in
temperature between two closed isothermal surfaces
divided by the total heat flow between them.
T2
Rth = (T(A) – T(B))  Power
(surface A)
(surface B)

T3

Heat Sink ∆C
T4 Watt (V*I)
* Present system does a good job of accounting for all heat and monitoring temperature but nothing is perfect.

Power (Watts)
Output from TIM unit:
Thermal Resistance (Rth): ∆C
Heat Source T1 Watt
= (∆ Temp.  power)

T2

Thermal Impedance (RthI): C - in2
Watt
Test Sample = (Thermal Resistance) x area (of test sample)

and
T3
Thermal Conductivity: Watt
Heat Sink m -C
= thickness (material)  thermal impedance
T4

Basic Principles of ASTM D5470 (cont.)

Power (Watts)
• Total Thermal Resistance (Rth total) is the sum of
components and its thermal resistance value.

“thermal grease” Grease

Copper

Test Sample Dielectric

Aluminum

“thermal grease” Grease

A B C

ASTM D5470 Thermal Measurement Technique

ASTM D5470 Thermal Measurement Technique
Measured Thermal Impedance (C-in2/Watt):

RthI (grease_top)
“thermal
Grease”

+
RthI (sample)
Test
Sample Total RthI
+
RthI (grease_bottom)
“thermal
Grease”

Total RthI B A

RthI (sample) = Total RthI - (RthI (grease_top) + RthI (grease-bottom) )

Thermal Impedance
Electronics industry defines thermal impedance as the following:

thickness
impedancethermal 
k
This does not include an area through which the heat flows!

Thermal Impedance Thermal Resistance

tDiel tDiel
I R
kDiel kDielA
Units: Units:
m m2 K m K
I  R 2

mK m W
W
W
mK W
22

Thermal Impedance
Calculating the thermal conductivity of a material
Plot Ra vs. thickness from TIM tester
1/k =thermal conductivity k = is the slope of the line

Fitted Line Plot
K-in^2/W = 0.01366 + 39.35 Thickness (in)
0.225 S 0.0010461
R-Sq 100.0%
R-Sq(adj) 100.0%
Thermal Impedance K-in^2/W

0.200

0.175

0.150 Thermal Conductivity LG:
0.25 W/M-K
0.125

0.100 = 1.0 in / 39.35 K-in2/W
0.075
= .006394 W/in-K x 39.4
in/M (convert to metric)
0.050
0.001 0.002 0.003 0.004 0.005
= 1.0 W/M-K
Thickness (in)LC
23
Data Plotted from multiple readings

Example problems
What is the thermal resistance through the thickness of a 0.3 meter by 0.3 meter piece of
stainless steel that is 1.5 cm thick? Assume the thermal conductivity of stainless steel is 15
W/mK.

k=15 W/mK 1.5cm

0.3m
t
R 0.3m
kA
t 0.015m
R 
kA (15W mK )(0.3m)(0.3m)

R  0.011 K W

Example Problems
This same piece of stainless steel now has a
heater attached to one side that generates
1.5cm
100 Watts of heat into the plate. Ttop

If the bottom of the plate measures 45°C k=15 W/mK 0.3m
(318K), what is the temperature of the top Tbottom 0.3m
side of the plate?

(Ttop  Tbottom) (Ttop  Tbottom)  q  Rthermal
q
Rthermal

Ttop  q  Rthermal  Tbottom
Ttop  100W  0.011 K W  318K
Ttop  319.1K  46.1C 25

Basic Question Room Temp = 30°C
Same LED and power input

MCPCB 1
LED
MCPCB 2 LED
Cu Cu
Dielectric Dielectric
Al/Cu Base Al/Cu Base
Heat Sink Heat Sink

Tjunction = 65°C Tjunction = 70°C

WHY???
Heat Transfer 26

Thermal Management for LED Applications
What is the role of the PCB?

Presented by Clemens Lasance
Clemens is a former Principal Scientist Emeritus with Philips Research, the Netherlands, with a 30 year +
focus on thermal management of electronic systems.
He is now a consultant for Somelikeit Cool, contact info: lasance@onsnet.nu

Motivation
Providing the right information the first time
The goal is to provide the LED application engineer with the right information to select the most
optimal MCPCB from a thermal point of view. Thus, ensuring the best decision is made for a
specific application in regards to thermal performance and cost.

Main Goal of Thermal Management
The Calculation of application-specific junction temperatures

Reasons / Key Issues
 Lifetime
 Color point
 Efficiency

All these key issues are significantly dependent on the junction temperature.

Basics of Heat Transfer
Fundamentals of Critical Temperature Calculation
Determining critical temperatures is contingent upon a critical understanding of:

• The electrothermal analogue
• Thermal conductivity k
• Heat transfer coefficient h
• Thermal resistance Rth

Electro-Thermal Analogue

∆T = q * Rth  ∆V = I * R

Temperature drop (°C)  Voltage drop (V)

Power dissipation (W)  Current (A)

Heat flux (W/m2)  Current density (A/m2)

Thermal resistance (K/W)  Resistance (Ohm)

Electro-Thermal Analogue
A heat transfer path can be described by an electrical network

T source

Rth T ambient
q

ΔT = Tsource-Tambient

Rth (K/W) = thermal resistance

Series Circuit
Example for two resistances in series

T intermediate
T source R1 R2

q T ambient

DT = Tsource - Tambient = q * Rtotal

Rtotal = R1 + R2

Consequence of Series Circuit
Often only the largest R is relevant!

T intermediate
R1
T source R2

q T ambient

Designer’s job: find the largest resistance!

Thermal Conductivity: Fourier’s Experiment (1822)
Heat insulation
Cross Sectional Area A

hot Heat flow q cold

Heated block L Water cooled block

Result of Fourier’s experiment:
A
q ~ ∆T = Thot – Tcold
q ~ A = cross-sectional area q  k   DT
q ~ L-1 L
The proportionality constant k is called the ‘thermal conductivity’

Typical Thermal Conductivity Values

The Heat Transfer Coefficient h
The heat transfer from a solid wall into a fluid (air, water) is called convection, and to first order this
heat transfer is proportional to the area and the temperature difference between the wall and the
fluid:

q  h  A  DT

The proportionality constant h is called the
‘heat transfer coefficient’

Typical values:
Natural convection: 10 W/m2K
Forced convection: 50 W/m2K

Practice
Suppose a Designer needs 5W to reach a required light output
What procedure is most optimal?
DT
q
Information required for first order guess: Rthtotal
• Maximum junction temperature: e.g. 120 °C
• Maximum ambient temperature: e.g. 40 °C
• An estimation of all thermal resistances in the total heat transfer path

Practice (2)
Next steps

 Check if required power is manageable by suitable heat sinks & convection

 Check which thermal resistances are dominant
 Then focus on them to reduce the total thermal resistance

 Final step should always be a detailed analysis
 Recall that often we don’t talk one-dimensional heat transfer but heat spreading,
which is a rather complicated issue
 For more information please see Heat Spreading, Not a Trivial Problem in the May 2008
issue of ElectronicsCooling

Practical Example
A L
q  k   DT Rth 
L kA
1
q  h  A  DT Rth 
h A
Relevant thermal resistances, assume 1 cm2 PCB

K/W
Rth-LED 16 Luxeon Rebel
Rth-MCPCB 0.5 t = 100 µm, k = 2 W/mK
Rth-TIM 1 t = 100 µm, k = 1 W/mK

Rth-heatsink 50 A = 20 cm2, h = 10 W/m2K

Practical Example
We need 5W for our light output
Can we find a suitable heat sink?

We have:
Tjunction = 120 °C
Tambient = 40 °C
Rth-LED + Rth-MCPCB + Rth-TIM = 17.5 K/W; Rth-heatsink = ?

Can we dissipate 5W? No way: DT
q = 80 / (17.5 + Rth-heatsink) q
Rthtotal
Conclusion:
Even with an ideal heat sink (Rth-heatsink = 0) we cannot dissipate the required 5W. In this case we
need an LED with less thermal resistance.

Practical Example
We need 1W for our light output

We have again:
Tjunction = 120 °C
Tambient = 40 °C
Rth-LED + Rth-MCPCB + Rth-TIM = 17.5 K/W; Rth-heatsink = ?

Which heat sink is OK? DT
q = 80 / (17.5 + Rth-heatsink) ; Rth-heatsink = 62.5 K/W q
Rthtotal
We found earlier:
A=20 cm2, h=10 W/m2K  Rth-heatsink = 50 K/W

Conclusion:
We can use a simple heat sink with natural convection.

Practical Example Conclusion
Thermal resistance of the MCPCB does not play any role

 In our first case, the LED thermal resistance was dominant; while in the last case, the
convective resistance was dominant.

 Hence, from a thermal point only, you can choose MCPCBs with a much lower thermal
performance and, hence, lower cost.

 The thermal performance of the PCB is relevant only for very high heat flux cases (e.g. liquid
cooling) and for top-of-the-bill LEDs.

Nomenclature
The confusing situation regarding ‘thermal impedance’
Fact:
‘Electrical impedance’ is historically reserved to describe time-dependent electrical resistance. In the
limit of steady state, thermal impedance equals thermal resistance  hence, units should be the
same!

Hence,
‘Thermal impedance’ , as used by U.S. vendors, violates the electro-thermal analogy, because:
– Unit does not correspond (K/W vs. m2K/W)
– Definition does not correspond (time-dependent vs. steady state)

Is this a problem?
Yes, because time-dependent (dynamic) test methods will be increasingly used, one output of which is
the ‘correct’ thermal impedance.

Proposal:
Use thermal resistance per unit area, or unit Rth.

Electrical Properties
Presented by Michael Gay (Isola) and Norm Berry (Insulectro)

45

Common Tests
• Insulation Resistance (IR)
• This test is used to provide a quantifiable resistance value for all of a product's insulation

• Dielectric Breakdown Strength Test
• The test voltage is increased until the dielectric fails, or breaks down, allowing too much
current to flow.

• Dielectric Withstand Voltage Test (DWS) or High Potential Test (HiPot)
• A standard test voltage is applied (below the established Breakdown Voltage) and the
resulting leakage current is monitored

• Conductive Anodic Filament Test (CAF)
• This test is used to determine the capability of the material for various design feature spacing
under high humidity conditions with applied voltage

46

Typical Proof Test (VDC)
Proof Test or High Potential Test (HiPot)
A standard non-destructive test voltage is applied (below the established Breakdown
voltage) and the resulting leakage current is monitored. The test is done to evaluate the
integrity of the dielectric material.

Typical Proof Test (VDC)
High Potential Dielectric Testing (HiPot)
This non-destructive testing methodology is used to characterize the laminate material insulation
capability under a specified potential. Leakage current is measured and compared to predetermined
limits

Typical Test Conditions:
500-2500 Volts DC (VDC) Applied +
700-1200 Volts AC @50-60 Hz (VAC)
Potential
Up to 30 seconds
10 MΩ – 500 MΩ
10 μA – 100 μA

-

Breakdown (kV AC)
Breakdown Voltage Test
Breakdown voltage is the stress at failure when AC at power frequency is applied using a rate of rise is
500 Volts/second. Breakdown voltage testing does not relate to proof stress “Proof Testing”.
Breakdown voltage is the potential difference at which dielectric failure occurs under prescribed
conditions in an electrical insulating material located between two electrodes.

Breakdown (kV AC)
Dielectric Breakdown Voltage
This testing methodology is used to characterize the
laminate material insulation capability to failure.
Applied +
Typical Test Conditions: Potential
Ramp Volts AC @50-60 Hz (VAC) until failure

Example:
Determine the electric strength in volts per mil for each
specimen by dividing the breakdown voltage expressed in
kilovolts by the thickness express in inches. -

ES = (6.8 KV/.005 inch) X (1000 V/KV) X (1 inch/1000 mils) = 1360 v/mil

Permittivity (Dk)
Typical thermal management materials are modified to have higher thermal conductivity. The
modifications of the dielectric material raise the Dk of the material to the range of 6 or higher.

Product Reliability
Presented by Richard A. Wessel of DuPont

Simplified LED Tjunction
q

Tjunction
LED Tsolder point RJ-SP
Cu Tsolder point
Dielectric
Tboard RSP-B
Al/Cu Base
Tboard
Heat Sink
RB-A

Tambient Tambient

Big concern: what is Tjunction?

Lower T junction Better
Increased relative luminous flux for lower
junction or thermal pad temperature

Cree® XLamp® MC-E LED

Luxeon®
Rebel

54

Increased Lifetimes
Lower Junction Temperature
Increased lifetimes for lower junction temperature

Luxeon®
Rebel

55

Suggested Fabrication Notes
Taking Control of Your Design!

56

Suggested Fabrication Notes
Purpose of Fab Notes
Take your requirements for each characteristic that you deem most important and specify (values for
thermal conductivity and electrical insulation resistance).

We also suggest including a list of pre-approved materials.

Suggested Fabrication Notes (cont.)
Example of a Fab Note
Copper Weight: 2 oz.
Metal Core: 0.060” Al 5051
Dielectric:
Thermal Conductivity: Must meet or exceed 1.3 (°W/m-K)
Thermal Resistance: Maximum 0.9 (°C in2/W)
Electrical Insulation: Must meet or exceed 2,000 VDC
Any other characteristic you feel is important: XYZ
Currently Approved Materials:
 Bergquist HT-07006
 Bergquist LTI-06005
 Laird T-lam SS 1KA06
 DuPont CooLam

Suggested Fabrication Notes (cont.)
Benefits of Fab Notes
Allows supplier to quote multiple vendors.

Subsequent Results
• Lower Cost
• Quicker Lead Times
• Adaptability to take advantage of new technologies down the road
• Introduce competition from the raw materials and finished PCB suppliers

Part Two
Non-browning White Soldermask

White Soldermask
Current Resin Systems
 Focus on initial thermal shock, not constant heat

 Tend to Brown over time

 Not developed to be a Reflecting Surface for LEDs

White Soldermask (cont.)
Desired Resin Systems
 Remain white through Assembly and Lifespan of LED final product

 Have high reflectivity to enhance LED performance
 Some believe high reflection soldermask results in lower power consumption for same light
output

What are the Solutions?
New White High Temperature Liquid Photo-Imageable Soldermasks
 Past two years a number of “LED” white soldermasks have been introduced
 Unfortunately we have not seen improvements in all of them

 Through numerous studies we have so far found two offerings that show significant improvement:
 Peters (SD2491SM TSW)
 Sun Chemicals (CAWN 2589/2591)

What are the Solutions? (cont.)
New White High Temperature Thermal Soldermasks
If Bright white soldermask is the objective, these solutions may not suffice

 Suggest base coat of LPI White then topcoat of High-Temp thermal white soldermask

 After numerous thermal whites, one does not brown after multiple cycles (Peters SD2496TSW)

 This is the formula behind “proprietary” white soldermasks offered by other PCB vendors

Cost Expanders
New White High Temperature Liquid Photo-Imageable Soldermasks
 To date, we’ve seen cost increases over standard soldermask price of 30% - 100%
 Approximately .20 to .30 cents per panel (18” x 24”)

New White High Temperature Thermal Soldermasks
To date, we’ve seen cost increases over standard soldermask price of > 100%
 Approximately $1 per panel (18” x 24”)
 Includes labor and set up for secondary application process

Thermal Webinar Conclusion

your needs for thermal performance of the dielectric

your needs for the electrical performance of the dielectric

your needs in context of entire product, not just PCB (when applicable)

that you have white soldermask choices
but keep your fab notes open to take advantage of future developments

Thermal Management: MCPCBs for LED Applications

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