Transcript 21. Packaging of electronic equipments

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
expansion 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 Δt
Where:
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
PL3W
X
7.5EW I W
•
•
•
•
•
Where:
X = 0.000675 in (wire displacement in X-Y plane)
LW = 0.1 in (vertical and horizontal wire length)
•
•
EW = 6106 psi (modulus of elasticity, copper wire)
Substituting in Equation 25.2 yields to
IW =
d 4
64
=
 (0.032) 4
64
 0.051x10-6 in 4 (wire inertia)
7.5(0.000675)(16x106 )(0.051x106 )
P
 4.13 Ib
3
(0.1)
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.
Sb 
KMC
IW
Solution
•
•
Where:
K = stress concentration factor
= 1 here
C = Wire radius to neutral axis
= 0.032/2 = 0.016 in
Substituting in Equation 25.4 yields to
Sb 
•
(0.459)(0.
016)
2

155294
Ib/in
0.051 x10-6
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.
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.
Ss 
P
A
• Where: P= 4.13 Ib
A= 0.09 in x 0.032 in = 0.00288 in2
• Then the direct shear stress is
Ss 
4.13
 1434 Ib/in 2
0.00288
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.
PL3W
X
 R
7.5EW I W
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.
N1S1b  N 2Sb2
• 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
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 proportional to the temperature
change. The high and low temperatures to the neutral points
are:
95 - (-55)
 75 C
• Condition (A)
2
75 - (-25)
 50 C
• Condition (B)
2
• Using a linear ratio of the temperature change, the solder joint
shear stress for the 50°C temperature change will be:
50
SS 
(1434)  956 Ib/in 2
75
• By Figure26.4 the approximate fatigue life is.
Life = 1600 cycles to fail
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:
N1  N 2 (S2 / S1 ) 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:
N1  80000(200/ 956) 2.5  1601 cyclestofail
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 to
the flexing of the PCB at its resonant frequency
PCB resonant frequency
fn 
•
•
•
•
•
•
•
•

2
D1 1
 2  2  ExpectedPCB resonantfrequency
 a b 
Eh3
 flexuralstiffness
D


2
in
12(1   ) 

W

(mass per unit are)
gab
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
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
(2 x 106 )(0.08) 3 h 3
D
 86.6 Ib in (stiffness)
12(1  (0.12) 2 )
1.2
Ibsec2
-4

 0.493 x10
(386)(9)(7)
in 3
Substitute in Equations 27.1 to get the resonant frequency of PCB.
fn 

2
86.6
0.493 x 10-4
 68.2 HZ
 1
1
 2 
(7 ) 2
 (9)



Desired PCB resonant frequency for
sinusoidal vibration
• When the component is mounted at the center of the PCB.
Z
0.00022B
Chr L
(maximumdesired PCB displacement)
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
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:
9.8G 9.8Gin Q
Z 2 
f
f n2
Where:
Q = transmissibility (Q) of the PCB
G = 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:
Q  fn
• 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:
 9.8GinChr L 
fd  

 0.00022B 
2/3
(mimimumdesired PCB resonantfrequency)
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.
 (9.8)(5.0)(1.0)(0.08)(1.0)( 2.0 ) 
fd  

(
0
.
00022
)(
9
.
0
)


2/3
 198.6 Hz
• The approximate fatigue life for 10 million cycles will be.
10x106 cyclesto fail
Life 
 14 hr
(198.6cycles/sec)(3600sec/hr)
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.
n
n
n
R
•
•
•
1

2

3
 .............  1
N1 N 2 N 3
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…
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.