Designing With Beryllia

8/2/2019 Designing With Beryllia

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American Beryllia Inc.

Designing With Beryllia

Thermal Design of Beryllia Components

2009


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Thermal Design of Beryllium Oxide Components

Heat Flow

From the Second Law of Thermodynamics we know that heat will flow from a higher to a lowertemperature. If we attach a heat source to one surface of a beryllia component and keep theopposite surface at a lower ambient temperature, heat will flow in the direction indicated. Theheat will continue to flow as long as there is a temperature differential.

In designing a beryllia component, you will have to determine what rate of heat transfer isnecessary to keep your component within safe operational limits.

Fouriers Law of Heat Conduction states that the rate of heat flow equals the product of the areanormal to the heat flow path, the temperature gradient along the path, and the thermalconductivity of the medium.

This rate (q/t) is a unit of power and can be expressed in calories per second or the morefamiliar term: watts.


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Fouriers expression describes the maximum power that can be dissipated by conduction. Wecan increase the heat flow rate or power dissipation by increasing the area through which theheat flows, by increasing the temperature gradient, or by increasing thermal conductivity of thematerial.

Thermal Resistance

If we start with Fouriers law and assume unidirectional heat flow, we see that, on integrating, anexpression is realized which describes resistance to heat flow.

If we call x/KA=, then we have an expression for a term which relates power dissipation andtemperature gradient.

This term is called thermal resistance a property very similar in nature to electricalresistance. In fact, it is directly analogous to its electrical counterpart.


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Electrical Analogs

This brings up a very interesting point. All thermal equations have electrical analogs. Thus, adirect substitution of thermal terms into familiar electrical equations enables you to treat thermalnetwork analysis in an identical manner to electrical network analysis.

Electrical Expression Equivalent Thermal ExpressionVoltage Temperature Differential T (C)Current Power P (watts)Resistivity Thermal resistivity (inC/watt)Resistance Thermal Resistance (C/watt)Capacity Thermal Capacity c (cal/in3C)Capacitance Thermal Capacitance C (watt sec/C)Conductivity Thermal Conductivity K (watt/in - C)Conductance Thermal Conductance (watts/C)

Let us now look at the simplest of all electrical expressions: Ohms Law. Looking at its analogswe see the thermal expression that was just derived from Fouriers Law.

OHMS LAW FOURIERS LAW

IR = V P = T

The power being dissipated, or heat flow, is analogous to current flow. T, the temperaturedifference, is analogous to voltage. And both electrical and thermal resistances react the sameway. This formula shows what is happening thermally within a component. You can calculatethe power dissipated, the heat rise, or thermal resistance, provided two of the three factors areknown.

Design Example Power

For example, consider this beryllia substrate with a thermal resistance of 45C/watt when asemiconductor device is attached. If we do not wish the semiconductor to exceed a temperatureof 150 C and the opposite surface of the beryllia is kept at 25 C, we can calculate how muchpower can be applied to the semiconductor.


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Design Example Temperature

In our second example we will use the same system. But, now the device will be operated at 2watts and we will determine the maximum junction temperature. With thermal resistance andambient temperature the same, substitution shows that the junction temperature will be 115 Cfor a 2 watt heat load.

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Design Example Thermal Resistance

We can also find out the thermal resistance of the component if we know specific operatingconditions. For this example, the device is operated at 3 watts, the junction temperature ismeasured at 97 C, and the opposite face of the components will be held at 25 C ambient.

Substitution shows that the thermal resistance of this system is 24 C/watt.


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Thermal Resistances In Series

In the previous three examples we have used to represent the total thermal resistance of thesystem. A thermal network usually contains several elements through which the heat will flow.

In our semiconductor device example, there is a thermal resistance associated with the silicon,the bonding materials, and the beryllia substrate. These elements are in series and the thermalresistance of the network is the sum of each thermal resistance in the heat flow path.

Thermal Resistance In parallel

Our analogy with electrical resistance also holds for parallel heat flow paths. Multiple leads in anintegrated circuit package, for example, are parallel heat flow paths as are heat loss throughconvection and radiation.

The mathematics for handling thermal resistance in series and parallel is the same as forelectrical resistors.


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Thermal Resistance Network

Here is a basic example for calculating thermal resistance when both parallel and series heatflow paths are utilized.

Notice that a high thermal resistance in parallel with the heat flow path will have little effect; ahigh thermal resistance in series will seriously degrade performance.

Thermal Resistance Calculations

Now let us take a closer look at the thermal resistance of an element in a thermal network. Thefirst component, X, is length of heat flow path. Thermal resistance will decrease as the elementbecomes thinner. A problem one can run into is to design an element so thin that it is notmechanically strong enough to be handled or attached to another component.

The second element determining thermal resistance is thermal conductivity, K. This is aproperty of the material selected. Thermal conductivity values change with temperature.

The final element determining thermal resistance is the cross-sectional area, A. This area is noteasily measured because it depends on the size of the heat source and is affected by thermalspreading.


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Thermal Spreading

If a beryllia component is larger in diameter than the heat source, the heat will dispense asshown in this end view.

This thermal spreading increases the average area through which the heat flows. With Aincreased, the thermal resistance is less for a given thickness of material.

Consider a square heat source of side dimension, a, in intimate contact with a large slab ofberyllia of thickness X.

The equations that describe thermal spreading are very complicated. A simplification that worksis to assume that the heat spreads out at a 45 angle. We are then looking at a truncatedpyramid. Thus, a good approximation of cross-sectional area comes about through integrating.

Which for a square heat source gives:


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If the heat source is in the shape of a circle or a rectangle, the same principles apply to heatspreading. These formulas would be used to take into account the change in geometry.

In an optimal design, the beryllia should be large enough in diameter to allow full thermal

spreading. Increasing the diameter beyond this would have no further affect because the areanormal to the heat flow path would not increase. If the diameter is too small to allow full thermalspreading, thermal resistance will be higher because the effective cross sectional area will belessened.


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Minimizing Thermal Resistance

The first step in designing an efficient thermal system is to draw the thermal network. Include allcomponents in the system as series or parallel thermal resistances.

Now determine for each component in the system. Design a series path with minimized values. The four rules for minimizing thermal resistance are:

Maximize the heat source surface area.

Use a material with high thermal conductivity.

Make the component as thin as possible

Allow for full thermal spreading

Design Example Silicon Transistor.

Here is an example to demonstrate how to improve thermal design with thermal networkconcepts.


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Now we have the same semi-conductor bonded onto a beryllia substrate using a eutectic alloy.

The thermal resistance is now 2.8 C/watt or a factor of almost forty below that shown in theprevious example.

Design Evaluation

Poor Thermal Design Good Thermal DesignSilicon 1.2 C/watt Silicon 1.2C/wattEpoxy 98.0 C/watt Eutectic 0.2C/wattAlumina 9.8 C/watt Beryllia 1.4 C/watt

109.0 C/watt 2.8 C/watt

Even though we are promoting beryllia, we want you to see by just substituting beryllia, thethermal resistance would not be greatly reduced. For if we used beryllia in the example on theleft, with an epoxy bond, would only go down to 100.6 C/watt. If we kept the Al2O3 andsubstituted eutectic bonding, would reduce to only 11.2 C/watt. The total thermal networkmust be considered. Dont assume that by replacing only one component, you can archieveoptimum design.

If you are going to use beryllia, make sure that its effectiveness can be realized. A large thermalresistance in series will, for all practical purposes, negate the advantages of beryllia.


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Thermal Conductivity Values

This table lists the thermal conductivity for materials commonly used in thermal design.

Thermal Resistance Values

These charts provide thermal resistance values for a range of materials with differentthicknesses and cross-sectional areas. These values assume full thermal spreading. The valuesreported are calculated for 25 C and if higher temperatures are used, a simple inverse ratio canbe used.

Thermal Resistance of Materials

Beryllia

99.5% Alumina

265180

70 60 25 20050

100150200250300

Beryllium

Oxide (BeO)

Aluminum

Nitride (AlN)

Silicon Carbide

(SiC)

Boron Nitride

(BN) HP Grade

Aluminum

Oxide (Al2O3)

99%

Aluminum

Oxide (Al2O3)

96%

Thermal Conductivity (W/M K)

W/M K


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96 % Alumina

94% Alumina

Glass Borosilicate

Kovar


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Silicon Semiconductor Grade

Thermoplastic Thermosetting Alkyd Type

Thermal Resistance of Ceramic Metallizations

Moly Manganese

Thin Film


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Thermal Resistance of Bonding Materials

Epoxy Silicone Variety

Conductive Silver Filled Epoxy

Silicon Gold Eutectic at 25 C

Interface Thermal Resistance for a TO-3 Case

Semiconductor to Mounting Surface

1. No Insulator, No Thermal Grease .05 - .20 C/watt2. No Insulator, Thermal Grease .005 - .10 C/watt3. Beryllia Insulator .10 - .40 C/watt4. Anodized Aluminum Insulator .35 - .70 C/watt5. Plastic Film Insulator (2 mil mylar ) .55 - .80 C/watt


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Thermal Capacitance

All of the thermal networks we have discussed so far have been in terms of equilibriumconditions where there is a continuous source of heat.

As heat is firs applied, the temperature difference will increase as shown in this graph.

This shows that there is a finite time required to approach thermal equilibrium gradients.

The rate of temperature differential increase is identical to that of voltage build up in the morefamiliar resistance-capacitance network. Current flow, voltage, and capacitance are analogousto heat flow, temperature differential and thermal capacitance.


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Thermal capacitance is a term with which we can determine the rate at which equilibrium isachieved.

It is the product of the density, specific heat and volume of the heat conducting material.

Design Example Thermal Capacitance

In this example we will calculate the thermal capacitance of a 1 x1 x .025 beryllia substrate.The density of beryllia is 47.2 gm/in3 and its specific heat is .25 cal/gm-C.

Material Property Values For Thermal Capacitance Calculation

SpecificHeat(cal/gm-C)

Density(gm/cm3)

Gold 0.030 19.3Kovar 0.110 8.2Aluminum 0.220 2.7

Silicon 0.180 2.4Teflon 0.250 2.0Epoxy 0.200 2.2

Plastics 0.300 2.1Air 0.240 .0013

SpecificHeat

(cal/gm-C)

Density(gm/cm3)

BeO 99.5% 0.250 2.88Al2O3 99.5% 0.210 3.9Al2O3 96% 0.210 3.7Al2O3 94% 0.210 3.6Mica 0.200 3.2Glass 0.200 2.2

Silver 0.056 10.5

Copper 0.093 8.9


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Thermal Time Constant

In a thermal network the time constant, T, is equal to thermal resistance times thermal capacity.

The value of the time constant determines whether a system approaches equilibrium rapidly orslowly. If either or C increases, equilibrium is reached at a slower rate.


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Design Example Thermal Time Constant

The time constant for the 1 x1 x .025 beryllia substrate in the last example is determined bythe product of the thermal resistance and the thermal capacitance. If we assume a thermalresistance value, say = 1.12 C/watt, then T would be 1.38 sec.

Thermal Equilibrium

How long would it take for this beryllia substrate to reach equilibrium? From our equation, thetime would be infinite.

For a practical solution, consider the time required to reach 99.9% if the equilibrium temperaturedifferential.


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Substituting 1.38 sec. for T, and solving, gives t = 9.52 sec.

For comparison, an alumina substrate of the same dimensions would have a much longer timeconstant and would not reach equivalent value in 62 seconds.


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Intermittent Power

If power is intermittently applied, the following curve describes T.

For semiconductor devices, peak junction temperatures are a limiting factor in their application.If the power output desired results in a junction temperature higher than allowed, the devicemust be operated intermittently. The following examples are offered to calculate the time pulsedurations which allow safe operation.

These equations describe T during heating (power on) and cooling (power off)


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If we have a time constant, T, of 1.38 seconds and apply power for 0.7 seconds, we can find the T after 3 seconds of cooling.


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Thus at t = 3.7 seconds with power on for 0.7 seconds, the temperature differential within thesubstrate will have been raised to 4.5% of equilibrium T.

If we were to turn power on and off continuously, the temperature differential would approach

equilibrium value as shown by this series.

Graphically, the temperature differential would increase with each application of power asshown here.


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Parallel Thermal Paths Convection

The power radiated by convection equals the product of the convection coefficient (W/C-in2),convection surface area (A1) and the temperature difference from surface to ambient.

The convection factor, h, takes into account the properties of the convective medium. Withoutsubstantial air flow, this factor is quite small.

Typical Values For Convective Heat Transfer Coefficient Semiconductor Devices

Cold Plate Mounted h (watts/C in2) Free Convection h (watts/C in2)Still Air .01 to .05 Still Air 0.0001 to 0.001Forced Air 0.1 to 0.5 Forced Air 0.01 to 0.05

Liquid 2.5 to 3.0Evaporative 5.0

Calculation Of Convective Heat Transfer Coefficient

Forced AirV = Velocity of Air

h = B V0.75 / L 0.25 L = Length of Surface in Direction of FlowB = Constant for Air Properties and Surface Configuration

Free ConvectionD = Constant for Air Properties

h = DE (Ts T

A) 0.25 /L0.25 E = Constant for Surface Configuration

L = Length of Dissipator Force with Area FactorTs TA = Temperature Difference between Dissipator & Ambient Air

Parallel Thermal Paths Radiation

Heat losses through radiation are the product of the Stefan-Boltzmann constant for a blackbody, the emissivity of the radiating body, the radiating surface area, and the temperaturedifference.

As temperatures below 300 C, radiated heat loss is usually small compared to conduction.

Typical Emissivity Values

Beryllia 0.87 Gold 0.04Alumina 96% (white) 0.88 Nickel 0.05Copper 0.05 Kovar 0.05Aluminum 0.04 Silver 0.02


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Thermal Shock

There is a limit to the rate of temperature change that beryllia can withstand. If this rate isexceeded, the ceramic being brittle, will develop internal stresses that could exceed its tensilestrength and could result in cracking.

While the thermal strength of beryllia is high, it is a ceramic, and you must consider the effectsof thermal shock if your design calls for large temperature changes in short time durations.

Thermal Stress

The magnitude of the stress is given by this expression and is equal to Youngs modulus timesthe coefficient of thermal expansion and temperature differential, divided by one minusPoissons ration.

The thermal shock must be instantaneous for this formula to apply, and thus it is a worst casecondition.

We demonstrated the calculation for the time it takes to distribute the heat within a component.Thus, you can now calculate T for your particular component and determine stress levels fromthe given formula.

Your Source For Beryllia Technology

It is not possible in this written text to cover all the present and future capabilities ofberyllium oxide ceramics. Please contact us and share you design problems or productopportunities. We will be glad to work directly with you in applying beryllia technology toyour specific requirements.

Designing With Beryllia

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