New Excel Software Tools to Practice for SMTA Certification #2: Reflow Profiling

Folks,

In my last post, I shared about an Excelbased software tool called Line Balancer to help candidates for SMTA Certification prepare for the line balancing part of the program. They can use Line Balancer to check the correctness of practice line balancing problems. This post will discuss another Excel-based software tool, Reflow Profiler, to help candidates prep for the reflow profiling part of the certification.

Typically, the reflow profiling goal is to determine if the reflow profile matches the requirements of the solder paste specification.

As an example, let’s consider a reflow profile as shown in Figure 1. The solder paste specification is shown in Figure 2. We will first solve the problem by hand and then use the software.

Figure 1. A ramp-to-peak reflow profile.
Figure 2. The solder paste reflow specification

The first task is to determine if the ramp-to-peak rate matches the solder paste specification outlined in red in the specification shown in Figure 3. By measuring the change in temperature in Figure 4 from point A to B and dividing it by the change in time from those points, we see in Figure 4 that the ramp-to peak-rate is 0.857°C/sec., and is within the recommended specifications 0.5 to 1.0°C/sec.

Figure 3. The solder paste specification with the ramp-to-peak rate highlighted.
Figure 4. The reflow profile with the ramp-to-peak rate calculated.

Figure 5 shows the solder paste specification with the time above liquidus (TAL) with the peak temperature highlighted. While Figure 6 shows the reflow profile, where the TAL is measured as 60 seconds and the peak temperature at 240°C, both are consistent with the recommended values.

Figure 5. The solder paste specification with the TAL and peak temperature highlighted.
Figure 6. The reflow profile with the TAL and peak temperature identified.

Lastly, Figure 7 shows the solder paste specification with the cooling ramp rate highlighted and Figure 8 shows the reflow profile with the cooling rate calculated as -2.8°C/s, again within the specification.

Figure 7. The solder paste specification with the cooling rate highlighted.
Figure 8. The reflow profile with the cooling rate calculated.
Figure 9 shows all of the calculations performed and matched to the specification with Reflow Profiler.

If you are interested in a copy of Reflow Profiler send me an email at [email protected]

Cheers,
Dr. Ron

New Excel Software Tools to Practice for SMTA Certification #1: Line Balancing

I recently developed some Excel-based software to help those who are planning to take the SMTA certification exam to practice. 

In this post, I will discuss the tool that performs line balancing. In a typical SMT assembly line, the placement machines are the “gate” in the cycle time of the line. To assure that their cycle time is the lowest, the placement machines must be time balanced. For example, suppose a simple SMT assembly line has one chip shooter and one flexible placer. Let’s say that the chip shooter takes longer to place all of the chips than the flexible placer takes to assemble the simple and complex integrated circuits. So, in this case, chips should be removed from the chip shooter and be placed on the flexible placer. But how many should be moved to the flexible placer? Determining the number requires algebra, and to understand how to do it, we need a numeric example.

Let’s do an example. In an assembly line, the “gate” in the cycle time is component placement.

  • The chipshooter (CS) places passives at 60,000/hr and Simple ICs (SICs) at 4,000/hr
  • The flexible placer (FP) places complex ICs (CICs) at 3,000/hr and SICs and passives at 8,000/hr
  • The bill of material (BOM) is 354 passives, 12 SICs, and 4 CICs
  • If the FP takes less time to place the CICs and SICs than the CS takes to place all of the passives, then move some of the passives to the FP to time balance the line
  • Let’s check the situation that the FP takes: (4 CICs / 3,000/hr) + (12 SICs / 8,000/hr) = 0.001333 + 0.0015 = 0.002833hrs
  • The CS takes: 354/60,000/hr = 0.0059hrs
  • So, move chips to the FP—but how many? Let’s call the number x. The times should be equal, so:
    • 0.002833+x/8000 = (354-x)/60,000.  Solve for x to time balance the line.
    • 0.002833+x/8000 = (354-x)/60,000, multiply each side by 60,000
    • (60,000*0.002833) + (60,000x/8000) = 354 – x
    • 170 + (60/8)x = 354 – x,  gather x terms
    • 170 + ((60/8) +1)x = 354, gather numbers
    • (68/8)x = 354-170 = 184, solve for x
    • x = (8/68)*184 = 21.65 or 22 passive moved to FP

Let’s see if the times on each machine are the same.

  • CTCS= 332 passives/60,000 passives/hr =0.005533 hrs or 19.92 secs

Is the FP the same?

  • CTFP = 0.002833 + 22/8,000 = 0.005583 hrs or 20.10 secs

Why the difference?

The times can’t be exactly the same as we rounded the number of passives moved to the FP.

Figure 1. The “Line Balancer” answer to the problem

Figure 1 shows the calculations from the Excel® software tool I developed called “Line Balancer.”  Note the answers are the same. If you would like a copy, send me an email at [email protected].

Here is a problem for you to solve:

  • The chip shooter (CS) places passives at 50,000/hr and Simple ICs (SICs) at 3,000/hr
  • The flexible placer (FP) places Complex ICs (CICs) at 4,000/hr and SICs and passives at 7,000/hr
  • The bill of material (BOM) is 390 passives, 14 SICs, and 6 CICs

How many components need to be moved, and to which placement machine? What is the cycle time?

To the first person that sends me the answer, I will send them a Dartmouth hat.

Using FMEA to Determine Your Tin Whisker Mitigation Strategy

In our last post, we discussed techniques to mitigate tin whiskers (TW). To help determine what your tin whisker mitigation strategy should be, consider using failure modes and effects analysis (FMEA). The central metric of FMEA is the risk priority number (RPN). For tin whiskers, the RPN is equal to the product of: (1) the probability of tin whiskers (P); (2) the severity, if a tin whisker exists (S); and (3) how hard it is to detect a tin whisker (D).  In equation form:

RPN = P*S*D

As a first example, consider a consumer product, like a mobile phone with a life of 5 years. With mitigation, on a scale of 1 to 10, P might be 2. For S, we might rate it at a 3, as a failure in the device is unlikely to cause severe harm to anyone. Detection (D) is a problem because the tin whiskers that form later cannot be detected during manufacturing; hence, we would have to rate D as a 10. So, the RPN is: 2*3*10 = 60, which is not too high. Therefore, with P and S at relatively low values, a tin whisker mitigation strategy would likely be successful for any consumer product. It should be pointed out that determining the RPN numbers would almost certainly require supporting data, brainstorming sessions, and a buy-in from the entire product team. The team would also have to determine any appropriate mitigation strategy such as avoiding bright tin coatings on component leads and perhaps using a flash of nickel between the copper and the tin (Figure 1).

Figure 1. In  mission-critical products, coatings may be required. It is almost impossible for a TW to penetrate both layers of coating as shown above.

Now consider a mission-critical product, such as certain types of military equipment. If we assume that the electronics have a service life of 40 years and that a failure could cause bodily harm or death, we could likely end up with a consensus that RPN = 10*10*10 =1000, the highest RPN possible. This situation would demand that special tactics be used to address the tin whisker risk. These tactics were discussed in my paper and presentation given at SMTA Pan Pacific 2019.

Cheers,

Dr. Ron

Tin Whiskers IV: Mitigation

Folks,

In the last post on tin whiskers, we discussed detection. In this post, we will cover mitigation. Since compressive stresses are a primary cause of tin whiskers, minimizing these stresses will help to mitigate tin whisker formation. There are several approaches to accomplish this compressive stress reduction. The first is to establish a process that produces a matte finish as opposed to a bright tin finish. Experience has shown that a satin or matte tin finish, which has larger grain sizes, has lower internal compressive stresses than a bright tin finish. Studies have shown that avoiding a bright tin finish alone can reduce tin whisker formation by more than a factor of ten. Thicker tin layers will often reduce compressive stresses.

Since a major source of the compressive stresses in tin is due to copper diffusion into the tin, minimizing this diffusion will significantly reduce tin whisker formation. One proven approach to minimizing copper diffusion is to have a flash of nickel between the copper and the tin. Since nickel does not readily diffuse into the tin after initial intermetallic formation, tin whisker formation can be all but eliminated in many cases.

Adding bismuth to the tin, in small amounts, can also reduce tin whisker formation. The bismuth solid solution strengthens the tin. This strengthening will often reduce tin whisker formation.

Another mitigation approach is the use of coatings. Acrylics, epoxies, urethanes, alkali silicate glasses and parylene C have been used. Parylene C appears to be the most promising.

Often a tin whisker will penetrate the coating as seen in Figure 1. However, to be a reliability risk, it must penetrate a second coating.

Figure 1. A tin whisker about to penetrate a polymer coating. Source: Dr. Chris Hunt, NPL.

This situation is almost impossible as the tin whisker is fragile and will bend as it tries to penetrate the second layer of coating. See Figure 2. So, coatings can be a very effective tin whisker mitigation approach.

Figure 2. To be a reliability concern, a tin whisker must penetrate two protective coatings.

The next and last tin whisker post will be on using FMEA (failure modes and effects analysis) to develop a tin whisker reduction strategy.

Cheers,

Dr. Ron

Tin Whiskers 101: Part III: Detection

Folks,

One of the great challenges of tin whiskers is detecting them. When one considers that their median thickness is in the 3 to 5 micron range (a human hair is about 75 microns,) they can be hard to see with direct lighting. Right angle lighting facilitates visual detection. See Figure 1. In this figure, Panashchenko shows that with direct light (left image), it is impossible to see the tin whisker, however with right angle light the tin whisker jumps out.

Figure 1.* It is not possible to see the tin whisker with direct lighting as in the left image. However, in the right image, right angle lighting makes it easy to see the tin whisker.

In her excellent presentation, “The Art of Metal Whisker Detection: A Practical Guide for Electronics Professionals,” Panashchenko offers these tips for identifying tin whiskers with a stereo optical microscope:

  • Use a 3x to 100x stereo microscope
  • Start with low magnification and work up to high magnification
  • Have the ability to tilt the sample in 3 axes
  • Use a flexible lamp that allows multiple angles of illumination, do not use a ring light
  • Use a LED or fiber optic lighting, not incandescent lights which can cause shadowing
  • Vary the brightness of the light source

The most important tip is to vary the angle of lighting while varying the magnification. Thus, analyzing a sample should take several minutes, at least. However, even the most thorough inspection may miss some tin whiskers. 

In the next post, I will discuss mitigation techniques.

Cheers,

Dr. Ron

*The image is from Lyudmila Panashchenko, “The Art of Metal Whisker Detection: A Practical Guide for Electronics Professionals,” IPC Tin Whisker Conference, April 2012.

Tin Whiskers 101: Part 2: What Causes Them

Folks,

Continuing our series on tin whiskers. In the last post we discussed what they are. in this post we will discuss what causes them.

Tin whiskers are primarily caused by compressive stresses in tin. The most common cause of the stresses is copper diffusion into the tin as seen in Figure 1a. Such diffusion is common when tin is plated, melted or evaporated on copper. Copper preferentially diffuses into tin exacerbating tin whisker production.

Figure 1. Some causes of tin whiskers

 Another cause of tin whiskers can occur when the tin is plated, melted or evaporated on a material that has a lower coefficient of expansion than the tin, such as alloy 42 or ceramic. When temperature increases, the tin is constrained by the lower coefficient of expansion material. This constraint causes compressive stresses in the tin that can result in tin whiskers. See Figure 1b.

Less common causes are corrosion, as seen in Figure 1c and mechanical stresses as seen in Figure 1d.

Since copper diffusion is one of the most likely causes of tin whiskers, this mechanism deserves elaboration. The left image in Figure 2 depicts the mechanism of copper diffusion into tin. The mechanism is so strong that the diffusion of the copper often leaves voids in the copper. Such voids are called Kirkendall voids. The right image in Figure 2 is an x-ray map of copper (green) diffusing into the tin (black).

Figure 2. Copper diffusing into tin.

Clearly, one way to minimize this type of tin whisker growth is to prevent copper diffusing into tin. In a future post, we will discuss this and other tin whisker mitigation techniques. 

Cheers,

Dr. Ron

Best Wishes,

Tin Whiskers 101: What Are They?

Folks,

Tin whiskers are very fine filaments or whiskers of tin that form out of the surface of the tin. See Figure 1. They are the result of stress release in the tin. Tin whiskers are a phenomenon that is surprising when first encountered, as their formation just doesn’t seem intuitive.

Figure 1. Note how thin a tin whisker can be compared to a human hair. The image is from the NASA Tin Whisker Website

They are a concern, as they can cause electrical short circuits or intermittent short circuits as a fusible link. Lead in tin-lead solder greatly suppresses tin whisker growth. Therefore, with the advent of lead-free solders there is a justifiable concern for decreasing reliability due to tin whisker growth in electronics.

Tin whiskers can vary in length and width, as is seen in Figure 2. Note that although only about 10% are as long a 1000 microns (1mm). That length and occurrence rate is such as to cause many reliability concerns.

Figure 2. The length and width of some tin whiskers. The source is also the NASA Tin Whisker Website.

Over the following weeks I plan to post how tin whiskers form and strategies to alleviate them. Most of the information I will post comes from a paper I presented with Annaka Balch at the SMTA PanPac 2019.

NASA has an excellent website that provides much information about tin whiskers and is a source for historic critical failures caused by tin whiskers.

Cheers,

Dr. Ron

Solution to Moore’s Law

Folks,

In a recent post, I discussed Moore’s Law. I challenged readers to solve for “a” and “b” from the equation a*2^(b*(year-1970)) from the graph in Figure 1.

Figure 1. Moore’s Law: Note that in 1982, ICs had about 100,000 transistors, whereas in 2016, they had about 10^10.

Moore’s Law posits that the number of transistors doubles every two years. If so, “b” should be 0.5. It ends up that “b”, from the solution in Figure 2, is 0.4885, so a double occurs about 1/0.4885 =2.047 years, but this number is really close to two years. The solution follows:

Figure 2. The Solution of a and b in the Moore’s law equation. 

BTW, congrats to Indium Corporation’s Dr. Huaguang Wang as he got a close solution.

Cheers,

Dr. Ron

Checking Moore’s Law

Folks,

Moore’s Law was developed by Gordon Moore in 1965. It predicted that the number of transistors in integrate circuits would double approximately every two years. Surprisingly, it has held true up to today. Figure 1 shows some of the integrated circuit transistor counts as a function of time. The red line is a good fit.

Figure 1. A plot of transistor count in selected ICs as a function of the year.

A reasonable equation for the red line is Transistor Count = a*2^(b*(year-1970)). What should “b” be if the count doubles every two years? To the first person that can solve for “a” and “b” using the red line and the equation above, we will send a Dartmouth sweatshirt.

Cheers,

Dr. Ron

Tin: The Foundation Metal of Soldering

Folks,

The vast majority of solders used in electronic assembly have, as their base metal, tin. There are some specialty gold solders, like gold-copper or gold-indium, indium based solders, and a few others that do not contain tin. Although these solders have important applications, the sheer volume of tin-based solders is overwhelming in comparison.

Tin was a metal known to the ancients, and it led them out of the Copper Age into the Bronze Age. Ten to twelve percent tin in copper yields bronze, which is much stronger than copper (see Figure 1) and has the added benefit of melting at about 950°C vs. copper’s 1085°C.

Figure 1. The addition of alloying elements, such as tin and zinc solid solution, strengthen copper. Note that about 8% tin in copper increases the copper yield strength by two and one half times. The solid solution effect also lowers the melting temperature. Find the image source here.

This difference in temperature is significant in that with primitive heating technology, 1085°C is hard to achieve. In addition, since bronze freezes at a lower temperature, it fills molds much better. This property enabled the casting of much more complex shaped objects. See Figure 2. All of these benefits resulted in a dramatically increasing demand for tin. This demand established much more sophisticated trade routes for tin and its most common ore, cassiterite; this enhanced overall trade and accelerated the spread of civilization and learning.

Figure 2. The addition of tin to copper created bronze, which is much harder and also easier to cast than copper. This castability enabled complex designs like this dirk. Image: Wikipedia

Back to solder. Soldering is a technology that has existed almost as long as the copper age. It is thought to have originated in Mesopotamia as long ago as 4000BC. Soldering was used for joining and making jewelry, cooking tools, and stained glass. Today, in addition to these applications, plumbing, musical instrument repair, and plated metal are common uses. However, electronics assembly is the largest user of tin-based solder by far. See Figure 3.

Figure 3. More than 50% of tin is used in solder. Source: Wikipedia 

One of the greatest benefits of solder is its reworkability. This property enables rework of electronics assemblies, plumbing, jewelry, and musical instruments. Without the ability to rework electronics, the industry would struggle to be profitable. Another benefit, of course, is the miracle of soldering I discussed in another post.

So, the next time you stare at your smartphone, tablet, TV, etc., remember tin-based solder and soldering are fundamental to its existence.  

Cheers,

Dr. Ron