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You are here: Home / Articles / Introduction to High-Temperature Life Testing

by Fred Schenkelberg 8 Comments

Introduction to High-Temperature Life Testing

Introduction to High-Temperature Life Testing

A common reliability test involves ‘baking’ a few units.

Various standards list temperature, duration, and sample size requirements.

When the units survive the test, meaning there are no failures, what does that mean?

How do you interpret a system or component life test using high temperature?

Do the results suggest your product is reliable? Maybe it is maybe it isn’t.

Let’s examine one way to design and interpret high-temperature testing.

What does high-temperature life testing test?

Let’s say we have a handheld consumer product and plan on testing 20 units in an oven at 60°C for a week or 168 hours.

Given we expect most units to (at most) experience 40°C and then relatively briefly, what does the week at temperature do to the product?

Let’s assume the test temperature of 60°C is well below any material softening or melting points.

We’re not close to any of the polymer, adhesive, ink, glass, film, ceramic, silicon, or coating material transition points. The unit has an electronic thermal protection feature within the ASIC and we’re 20°C below that trip point.

The exposure to increased temperature will do a few things to the devices that may be of interest.

The range of materials within the unit will change shape according to each specific material’s coefficient of thermal expansion (CTE).

Simultaneously the increased temperature will accelerate the many different possible chemical reactions within the materials.

By keeping the exposure below melting points (transition points) the types of failures that may occur at the elevated temperature may occur at use temperatures, just taking longer to occur.

Given the nature of the test, there is possibly only one thermal cycle involved.

Room temperature to 60°C and back at the end of the test.

Unless there is a major flaw in the design a single cycle will likely not reveal a failure due to the CTE effect.

The acceleration of chemical reactions may reveal issues.

The chemical reactions of interest include:

  • Diffusion
  • Chain scissioning
  • Electromigration
  • Corrosion

To name only a few. Within any consumer handheld product, there are thousands of potential chemical reactions.

Some are harmful to the durability of the product.

How to interpret the high-temperature life test?

If you know the specific chemical reaction that causes your device to fail, you are then able to use that knowledge to determine the activation energy of that reaction.

If not, you will likely have to assume a failure mechanism and it’s associated chemical reaction, thus the activation energy.

You may have heard of the Arrhenius rate reaction equation.

In reliability work, we use this equation to translate time to failure information at high temperature to the expected time to failure at use temperatures.

The equation for the acceleration factor (AF) based on the Arrhenius equation is:

$$ \displaystyle\large AF=\frac{{{t}_{u}}}{{{t}_{t}}}=\exp \left[ \frac{{{E}_{a}}}{k}\left( \frac{1}{{{T}_{u}}}-\frac{1}{{{T}_{t}}} \right) \right]$$

Ea is the activation energy, in electron volts

k is Boltzmann’s constant (8.617 x 10-5 eV/K)

T is the reaction temperature in Kelvin, K

The activation energy term defines the relationship between the test and use temperatures. Examine this equation.

A low activation energy means there is not much difference in the reaction rate is little changed with different temperatures.

A high value suggests a dramatic change in reaction rate with a change in temperature.

If you know the activation energy, you can estimate the acceleration factor.

Then multiply the AF times the test duration, in our example 168 hrs, to find the estimated equivalent time of normal use.

Here is a table of semiconductor device failure mechanism activation energy values.

Failure Mechanism Activation Energy, eV
Mital migration 1.8
Charge injection (slow trapping) 1.3
Ionic contamination 1.0 – 1.1
Gold-aluminum intermetallic growth 1.0
Corrosion in humidity 0.8 – 1.0
Electromigration in aluminum 0.5
Electromigration of silicon in aluminum 0.9
Time dependent dielectric breakdown 0.3 – 0.6
Electrolytic corrosion 0.3  – 0.6
Hot carrier trapping -0.1
Surface charge accumulation 1.0
Gate oxide defects 0.3
Silicon junction defects 0.8
Metallization defects 0.5
Charge loss 0.6
Dark line and dark spot defects in laser diodes 0.6

Table details from Condra, Lloyd W. 2001. Reliability Improvement with Design of Experiments. New York: Marcel Dekker. p 243.

If you do not have the activation energy, which value should you use?

The right answer is to find out which one to use before just picking a value. All of these chemical reactions are occurring in your device at their specific reaction rates.

The higher temperatures will excite the higher activation energy mechanisms faster than low activation energy mechanisms.

The issue is the lower activation energy items may be the one to fail first at normal temperatures and not reveal themselves in a test very quickly.

Therefore, if you select a higher activation energy value, say 0.9 then you will over estimate the time to failure at use temperatures.

You’ll be surprised at failures that occur due to the lower activation energy.

If you select a lower activation energy value, the acceleration factor is lower thus a 168-hour test may only simulate 500 hours at use conditions.

How to determine the right activation energy to use

The first thing to do is run the test till failure. Get a failure and determine the failure mechanism involved.

From there a literature search may reveal an appropriate activation energy to use for your testing.

Better if you have multiple failures and all are from the same mechanism.

In conjunction with the test to failure recommendation, also conduct risk analysis (hazard analysis or FMEA, for example) and determine what is likely to fail at use conditions.

The table above is only for silicon related mechanisms, there are more mechanisms and associated activation energy values to consider.

Another approach is to conduct multiple stress levels (different high temperatures) and run till you have sufficient failures at each temperature to estimate the time to failure distribution.

Let’s assume you only experience one failure mechanism otherwise the analysis becomes complex.

The idea is to determine the relationship between temperature and time to failure.

Using the Arrhenius model you are empirically estimating the activation energy. Also, double check with a chemist is the estimated values make sense for the specific chemical reaction involved.

Summary

168 hours of testing at 60°C will tell you how your product works at 60°C for a week.

It might reveal information about your product for a month or many years, depending on what failure mechanisms limit the use condition lifetime.

Take care when assuming an activation energy, it has a significant impact on how you interpret your test results.

How do you interpret your high-temperature life testing? Leave a comment below.


Related:

Basic Approaches to Life Testing (article)

Life Testing Starting Point (article)

When to Stop Testing (article)

 

Filed Under: Articles, CRE Preparation Notes, Reliability Testing Tagged With: Life testing and accelerated life testing (ALT), Physics of Failure (PoF)

About Fred Schenkelberg

I am the reliability expert at FMS Reliability, a reliability engineering and management consulting firm I founded in 2004. I left Hewlett Packard (HP)’s Reliability Team, where I helped create a culture of reliability across the corporation, to assist other organizations.

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Comments

  1. Hilaire Perera says

    November 21, 2016 at 5:30 PM

    Fred,

    A single, best-estimate activation energy value facilitates accurate estimation of the acceleration factor for the device failure rate estimation in the system application. JEP122E: Failure Mechanisms and Models for Semiconductor Devices provide a good selection of activation energies

    Reply
  2. Bill Meeker says

    November 27, 2016 at 4:41 PM

    Fred:

    I really enjoyed this article. Is there some way to get the Arrhenius equation to look better? Perhaps inserting an image of the equation produced elsewhere?

    Bill

    Reply
    • Fred Schenkelberg says

      November 27, 2016 at 8:54 PM

      Hi Bill, Thanks for the kind words and note on the equation. Yeah, unrendered Latex is not very pretty… forgot to properly close the equation and it should look better now. cheers, Fred

      Reply
  3. Nigel in Santa Barbara says

    February 28, 2018 at 4:19 PM

    “A low activation energy means there is not much difference in the reaction rate with different temperatures.”

    I think you meant to say: “A low acceleration factor means there is not much difference…”

    Reply
    • Fred Schenkelberg says

      March 1, 2018 at 6:23 AM

      Hi Nigel, I did want to mention the activation energy. With a low activation energy the change in the reaction rate with a change in temperature is small, thus not highly dependant on temperature. Likewise, if the reaction has little change with a change in temperature the acceleration factor will also be slight. Thanks for the careful read and comment. I edited the piece to hopefully be clearer. cheers, Fred

      Reply
  4. valeria says

    March 2, 2020 at 11:58 PM

    Hello,

    I have a question regarding the acceleration factor.

    We know that the lifetime (TF) of a component can be expressed as:

    (1) TF=A0exp(Ea/(kT))

    where A0 is a constant (component dependent), Ea is the activation energy, k is the Boltzmann constant and T is the temperature.

    For a given temperature, the larger is Ea the larger is TF.

    If we test the same component at different temperature (T1 and T2), we can define an acceleration factor AF as:

    (2) AF=TF1/TF2 = exp((Ea/k)*(1/T1 – 1/T2))

    so we can write:

    (3) TF1=TF2*AF=TF2*exp((Ea/k)*(1/T1 – 1/T2))

    – If T1>T2:
    AF>1 so if we vary Ea we find that the larger is Ea the longer is TF. This is in agreement with (1)

    – If T1<T2:
    AF<1 so if we vary Ea we find that the larger is Ea the shorter is TF. This is NOT in
    agreement with (1).

    Can someone explain me why?

    Thanks a lot and I hope I was clear enough

    Reply
    • valeria says

      March 3, 2020 at 4:35 AM

      just to explain better my issue:

      Let’ s say that I do a test at temperature T1 and I find the value of TF1.
      I want to know now the TF2 associated to T2, I apply formula (3). As I have some incertitude on the exact value of Ea I vary its value in a certain range.

      case #1: T1>T2 –>I find that the smaller Ea will give me a smaller TTF2
      case #2: T1I find that the smaller Ea will give me a larger TTF2

      Is this normal or there is something wrong with this reasoning? I find case#1 in agreement with my understanding of the meaning of activation energy. On the contrary I do not understand the meaning of case#2: How is it possible that a smaller Ea gives me a longer TF? This is not in agreement with formula (1).

      Thanks again

      Reply
    • Fred Schenkelberg says

      March 3, 2020 at 11:53 AM

      Hi Valeria,

      It might just be the simple math at play here. The formula is interested in the change in temperature – and you should take care with keeping track of the sign of the term. Different authors order the 1/t1 and 1/t2 differently and include or not a minus sign in front of the Ea term.

      Keep in mind that Ea is the activation energy and represents the rate of reaction for a chemical process – in some cases, it is simply treated as a fitted variable. In general, the acceleration factor equation is only interested in the change in time to failure difference when applying higher temperatures than used in normal applications.

      I played with my excel model of the equation and you are right if I only change the temperatures and raise or lower the activation energy term, I see the behavior you describe. I suspect the equation really only works or is valid when the T1 and T2 are set with the higher temp in the first position then subtracting the lower temperature… both inverted and in units of Kelvin, as well.

      not sure that explains what is going on – a bit of exploration using a spreadsheet may help you determine what is happening mathematically. Yet, I do not believe the formula is useful unless setting up the ratio of the two Arrehenisu equations with higher stress on top of lower stress to estimate the acceleration factor.

      Cheers,

      Fred

      Reply

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CRE Preparation Notes

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