PART-E

Analysis of thermal failure of electronic

components

25. Analysis of thermal stresses and strain in electronic components

• Substantial efforts made in the fabrication methods, mounting methods, and cooling techniques of the electronic devices to reduce the hot spot temperatures below 100 °C. This has produced a significant improvement in the reliability and effective operating life of the electronic equipment.

• Experience has shown that most of these failures are produced by a mismatch in the thermal coefficients of expansion (TCE) of the different types of materials typically used in electronic assemblies. The mismatch often generates high forces and stresses, which produce fractures and cracks in the electronic components and assemblies.

Thermal expansion effects in electronic equipments

• Temperature changes will produce dimensional changes in almost all materials normally used in the assembly of electronic chassis and PCBs.

• Temperature changes within the electronic assembly can occur due to power cycling, where the power is turned on and off.

• These dimensional changes, which can occur along the X, Y, or Z axes of the electronic assemblies, can produce a wide variety of failures in the structural elements of these assemblies.

Thermal expansion effects in electronic equipments

• Due to thermal cycles; thermal expansion differences between the component and the PCB along the X and Y axes can produce failures in this subassembly.

• The failures will not be in the lead wires. Instead, the failures will occur in the solder pads, which will be lifted off the surface of the PCB by overturning moments in the lead wires.

Example • Determine the deflections and stresses expected in

the lead wires and solder joints of the surface mounted transformer shown in figure, when it is mounted on an aluminum composite PCB which experiences in plane (X and Y) thermal ex pansion during rapid temperature cycling tests over a temperature range from -55 to +95 °C, with no electrical operation.

Solution• 1) Determine the expansion differences between the transformer and the

PCB in X-Y planes is X= (aT - aP) b ΔtWhere:

- aT = average TCE of transformer, considering a mixture of epoxy potting copper, and iron core in and PCB in X-Y plane (Z axis expansion are ignored here)

- aT = 35 x 10-6 in/in/°C or 35 parts per million/°C (35 ppm/°C)

- aP = average TCE of composite PCB with epoxy fiberglass and aluminum heat sink core in X-Y planes = 20 x 10-6 in/in/°C or 20 parts per million/°C (20 ppm/°C)

- b = 1.2/2 = 0.6 in (effective length of transformer, including wire length

with the transformer)- Δt = 95-(-55) = 150 °C ( peak to peak temperature range )- Δt = 150/2 = 75 °C (neutral point to high and low temperature )• Then the expansion difference is

X= (35-20) x10-6(0.6) (75) = 0.000675 in

Solution

• 2) Determine the horizontal force induced in the wire as it is forced to bend through this deflection. The wire geometry is shown in the following Figure.

Solution• The horizontal displacement of a square frame with clamped

ends, with bending of both wire legs due to the action of the lateral force (P), can be determined from the following equation

• Where:• X = 0.000675 in (wire displacement in X-Y plane)• LW = 0.1 in (vertical and horizontal wire length)• • IW = = (wire inertia)

• EW = psi (modulus of elasticity, copper wire)• Substituting in Equation 25.2 yields to

WW

W

IE

PLX

5.7

3

64

4d 46-4

inx10051.064

)032.0(

Ib31.4)1.0(

)10x051.0)(10x16)(000675.0(5.73

66

P

6106

Solution

• 3) Determine the bending stress in the lead wire and the shear stress in the solder joint.

• The bending moment (P) in the wire at the solder joint can be determined from Figure 25.2, by summing up the bending moments for the wire frame.

M = 1.2PLW (wire bending moment) = 1.2(4.13x0.10) = 0.495 lb in

• Then the bending stress (Sb) in the wire can be obtained as in the following Equation 25.4.

WI

KMCbS

Solution• Where:

K = stress concentration factor = 1 hereC = Wire radius to neutral axis = 0.032/2 = 0.016 in

• Substituting in Equation 25.4 yields to

• This far exceeds the ultimate tensile stress of 45,000 psi for the copper lead wire, which means that the wire will be in the plastic bending range. However, testing experience with this condition shows that the probability of a wire failure is low (if there are no sharp cuts in the wire) due to the low number of stress cycles normally expected for this type of environment.

26-

Ib/in155294x10051.0

016)(0.459)(0.bS

Solution

• The direct shear stress (Ss) in the solder joint can be obtained from the solder pad area estimated to be about 0.09 in x 0.032 in. This direct shear stress does not include solder joint stresses produced by the overturning moment. Both stresses may be combined to obtain the maximum or the von Mises stress. Only the shear stresses were used here to determine the approximate fatigue life of the solder joint. Stress concentrations are not considered here because the solder is so plastic.

• Where: P= 4.13 Ib A= 0.09 in x 0.032 in = 0.00288 in2

• Then the direct shear stress is

A

PsS

2s Ib/in1434

00288.0

13.4S

Reducing the thermal expansion forces and stresses

a) Decreasing the moment of inertia of the wire

b) Decreasing the deflection of the wire X c) Increasing the length of the wire L.

Methods for increasing the wire length to decrease the forces and stresses in the solder joints.

26. Effect of PCB bending stiffness on lead wire stresses

• When axial leaded devices on a PCB are exposed to thermal cycling environments, overturning moments can occur which may force the PCB to bend as shown in the following figure.

Effect of PCB bending stiffness on lead wire stresses

• The horizontal displacement expected at the top of the wire will be the sum of the wire bending and the PCB rotation, as shown in the following equation 26.1, when the horizontal and vertical legs of the wire are the same length.

R5.7

3

WW

W

IE

PLX

Fatigue life and vibration environments

Introduction to fatigue generation• Materials can fracture when they are subjected to repeated

stresses that are considerably less than their ultimate static strength. The failure appears to be due to submicroscopic cracks that grow into visible cracks, which then leads to a complete rupture under repeated loadings.

• The turn-on and turn-off process introduces alternating stresses in the structural elements as the assembly heats up and then cools down. Every stress cycle experienced by the electronic system will use up a small part of its total life. When enough stress cycles have been experienced, the fatigue life will be used up and cracks will develop in structural elements such as solder joints, plated throughholes, and electrical lead wires, resulting in failures.

• Fatigue can also be generated in electronic systems by shock and vibration.

Slow cycle fatigue and rapid cycle fatigue

• Test data on solders shows that the frequency of the applied alternating load has a significant affect on the fatigue life, as shown in figure below. Where the effects of slow cycle fatigue and rapid cycle fatigue appears.

Solder subjected to slow cycle fatigue is weaker than solder in rapid cycle fatigue

Slow cycle fatigue and rapid cycle fatigue

• The fatigue life can be estimated from the sloped portion of the curved based on the relation.

• Where:• N= Number of stress cycles to produce a

fatigue failure• S = Stress level at which these failures will

occur • b = Fatigue exponent related to the type of

fatigue

b22

b11 SNSN

Slow cycle fatigue and rapid cycle fatigue

Vibration and thermal cycle fatigue, (vibration at room temperature)

Example

• Determine the approximate fatigue life expected for the solder joints on the surface mounted transformer shown in the foregoing sample solved example. For two different conditions as follows:

• A) Original rapid temperature cycling from -55 to +95 °C, which resulted in a solder joint shear stress of 1434 psi.

• B) Revised rapid temperature cycling from -25 to +75 °C.

Solution:

• PART (A)• An examination of the solder joint fatigue curve has

shown in Figure 26.4 for thermal cycling conditions with 1434 psi solder shear stress.

Approximate solder fatigue life = 650 cycles

• At 650 the cracks may be expected in some of the solder joints when the solder shear stress level is about 1434 psi.

• This does not mean that electrical failures will occur instantly. It means that visible cracks may have developed and that these cracks can continue to grow in this environment, so a catastrophic failure is not far away.

Solution:• PART (B)• When the temperature cycling range is changed, the fatigue

life of the solder joint can be approximated by assuming a linear system, so the stress is directly propor tional to the temperature change. The high and low temperatures to the neutral points are:

• Condition (A)

• Condition (B) • Using a linear ratio of the temperature change, the solder joint

shear stress for the 50°C temperature change will be:

• By Figure26.4 the approximate fatigue life is.Life = 1600 cycles to fail

C752

(-55)-95

2S Ib/in 956(1434)

75

50S

C502

(-25)-75

Solution:• Another method can be used to obtain the approximate fatigue

life of the solder joint using Equation 26.4, along with the exponent b of 2.5, which represents the slope of the thermal fatigue curve for solder. A reference point must be obtained from Figure 26.4 to start the process. Any convenient starting point can be selected, such as 200 psi, where the fatigue life is 80,000 cycles to fail. This will be selected as point 2 on the fatigue curve.

• Changing Equation 26.4 slightly to solve for N1 cycles to fail and using the slow cycle fatigue exponent b with a value of 2.5:

• Where:• N2 = 80,000 (cycles to fail at reference point 2)• S2 = 200 lb/in2 (stress to fail at reference point 2)• S1 = 956 lb/in2 (stress resulting from 50°C temperature change

from condition B)• Substitute into Equation 26.4, to get the fatigue life for condition

B is:

5.21221 )S/(SNN

fail tocycles1601)956/80000(200N 5.21

27. Vibration fatigue in lead wires and solder joints

Introduction• There are two basic types of vibration: sinusoidal (or

sine) and random excitation: Sine vibration, or simple harmonic motion, repeats itself, but random motion does not.

• Vibration-induced failures are often caused by the relative motion that develops between the electrical lead wires and the PCB, when the PCB is excited at its resonant frequency, as shown in the following Figure 27.1. The resonant frequency of the PCB must be determined in order to obtain the approximate fatigue life relations.

Introduction (cont.)

Relative motion in the lead wires of a large component due tothe flexing of the PCB at its resonant frequency

PCB resonant frequency

• Where:• E = modulus of elasticity, Ib/in2

• h = thickness of PCB, in• μ = Poisson's ratio, dimensionless• W = Weight of assembly, Ib• g = 386 in/sec2, acceleration of gravity• a = PCB length, in • b = PCB width, in

frequencyresonant PCB Expected11

2 22

ba

Dfn

are)unit per (mass

in

stiffness flexural

)1(12 2

3

gab

W

EhD

Example

• Determine the resonant frequency of a rectangular plug-in epoxy fiberglass PCB simply supported (or hinged) on all four sides, 0.080 in thick, with a total weight of 1.2 pounds, as shown in Figure.

Solution• The following information is required for a solution:• E = 2 x 106 Ib/in2 (epoxy fiberglass modulus of elasticity) • h = 0.080 in (PCB thickness)• μ = 0.12 (Poisson's ratio, dimensionless)• W = 1.2 Ib (weight)• a = 9.0 in (PCB length)• b = 7.0 in (PCB width)• g = 386 in/sec2 (acceleration of gravity)• Substitute in Equations 27.2 and 27.3 yields to

• Substitute in Equations 27.1 to get the resonant frequency of PCB.

in

Ibsec x10493.0

)7)(9)(386(

2.1

)(stiffnessin Ib6.86))12.0(1(12

)08.0)(10x2(

3

24-

2

336

hD

HZ2.68

)7(

1

)9(

1

10x493.0

6.86

2 224-

nf

Desired PCB resonant frequency for sinusoidal vibration

• When the component is mounted at the center of the PCB.

nt)displaceme PCB desired maximum(00022.0

LChr

BZ

Where:• B = length of PCB edge parallel to component, in• L = length of component body, in• h = height, or thickness of PCB, in• C = component type = 1.0 (for standard DIP or a standard pin grid array) = 1.26 (for a side-brazed DIP, hybrid, or pin grid array; two

parallel rows of wires extending from the bottom surface of the component)

= 2.25 (for a leadless ceramic chip carrier (LCCC))• r = relative position factor = 1.0 at center of PCB• = 0.5 at 1/4 point on X axis and 1/4 point on Y axis

22

Q8.98.9

n

in

f

G

f

GZ

Desired PCB resonant frequency for sinusoidal vibration

The maximum single-amplitude displacement expected at the center of the PCB during the resonant condition can be obtained by assuming the PCB acts like a single-degree-of-freedom system, as shown in the following equation:

Where:Q = transmissibility (Q) of the PCBG = Peak input acceleration level of vibration

Desired PCB resonant frequency for sinusoidal vibration

• The transmissibility (Q) of the PCB at its resonance can be approximated by the following relation:

• The minimum desired PCB resonant frequency that will provide a component fatigue life of about 10 million stress cycles can be obtained by combining Equations. 27.4. through 27.6. Yields to:

nfQ

frequency)resonant PCB desired mimimum(00022.0

8.93/2

B

LChrGf ind

Example • A 40 pin DIP (Dual inline package, electronic

equipment) with standard lead wires, 2.0 in length will be installed at the center of a 9.0 x 7.0 x 0.080 in plug-in PCB. The DIP will be mounted parallel to the 9 in edge. The assembly must be capable of passing a 5.0G peak sine vibration qualification test with resonant dwell conditions. Determine the minimum desired PCB resonant frequency for a 10 million cycle fatigue life, and the approximate fatigue life.

Solution

• B = 9.0 in (length of PCB parallel to component)• h =0.080 in (PCB thickness)• L = 2.0 in (length of a 40 pin DIP)• C =1.0 (constant for standard DIP geometry)• G = 5.0 (peak input acceleration level)• r = 1.0 (for component at the center of the PCB)• Substitute into Equation 27.7 for the desired PCB

frequency.

• The approximate fatigue life for 10 million cycles will be.

Hz6.198)0.9)(00022.0(

)0.2)(0.1)(08.0)(0.1)(0.5)(8.9(3/2

df

hr 14sec/hr))(3600cycles/sec(198.6

fail tocycles10x10Life

6

Miner's cumulative damage fatigue ratio

• Every time a structural element experiences a stress cycle, a small part of the fatigue life is used up. When all of the life is used up, the structure can be expected to fail. This simple theory is widely used to determine the approximate fatigue life of structures operating in environments that produce stress reversals. The damage that is accumulated is assumed to be linear, so the damage developed in several different environments can simply be added together to obtain the total damage to determine if the part will fail. This is known as Miner's rule, or Miner's cumulative damage ratio R, which is defined below.

• Where: • n = actual number of fatigue stress cycles accumulated at

stress levels 1, 2, 3...• N = number of fatigue stress cycles required to produce a

failure at stress levels 1, 2, 3…

1.............N

n

N

n

N

nR

3

3

2

2

1

1

Miner's cumulative damage fatigue ratio

• Different fatigue cycle ratios are often used for different applications, depending upon how the electronic product will be used. For commercial electronic systems that have no involvement with the public safety, an R value of 1.0 is suggested. Where the public safety is involved, as in an airplane, train, or automobile, then an R value of 0.7 is suggested. Where a critical life system, such as a space shuttle, is involved, a higher safety factor is recommended, so an R value of 0.3 is suggested.