Recently, I posted a derivation of the equations to determine the mass fractions of two metals in a binary alloy. I thought it may be helpful to develop an Excel software tool to perform these calculations.

To use the tool, you enter the densities of the two metals and the density of the alloy in the blue cells as seen in Figure 1 below. The calculated mass fraction of each metal is shown in the gray cells.

Figure 1. The data entry for the mass fraction calculator. The densities are entered into the blue cells and the mass fractions are calculated and shown in the gray cells.

As an example, let’s assume you purchase some 14 karat gold. Unfortunately, to your eye it looks more like 10 karat gold, so you want to check it out. As a reminder, when gold is expressed in karats, the alloying metal is copper. First you need to measure the density of the gold alloy. An easy way to do this is the wet gold technique as discussed in a past blog post. From using this technique, you determine that the density of the alloy is 11.53 g/cc. The density of gold is 19.3 g/cc and that of copper is 8.96 g/cc. You will recall that 14 karat gold is (14/24) gold or a mass fraction of 0.5833.

The weight fraction of gold is shown to be 0.4167 or 10/24, as shown in Figure 1, indicating that the gold is 10 karat, not 14 karat.

I was surprised to see the wrong formula for metal alloy density calculations on YouTube. The wrong (Eq. 1) and correct (Eq. 2) formulas are shown in Figure 1. In many cases, Equation 1 will give an answer only a few percent off. However, in some cases it can be off by more than a factor of 1,000 as seen in a past blog post. This blog post also gives the derivation of Equation 2.

The YouTube video mentioned above did suggest one interesting task—determining the metal mass fractions of a two-metal alloy, while only knowing the alloy density. Of course, we know the densities of the two metals. The solution to this problem is seen in Figure 2.

To check the result, assume we have a tin-lead metal alloy. The alloy density is 8.4 g/cc. Tin has a density of 7.29 g/cc and lead 11.34 g/cc. By plugging these numbers into the solution for x (tin), we get 63% and y = 37%. Hence, this alloy is the tin-eutectic alloy.

This technique can solve only two-metal alloy system mass fractions.

In my lastpost, I shared about an Excel^{–}based 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.

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 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.

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.

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

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.

CT_{CS}= 332 passives/60,000 passives/hr =0.005533 hrs or 19.92 secs

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

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.

I am heartbroken to share the news that Irene Sterian passed away on May 8. Irene was director, technology and innovation development at Celestica, and before that held engineering roles at IBM.

But she is better known as the always cheerful mentor to younger engineers through SMTA, the REMAP technology accelerator she founded, and before that, NGen Canada, Canada’s Advanced Manufacturing Supercluster.

She is survived by her husband, three children, brother and mother, not to mention legions of colleagues and friends.

For details on how to remember her, please click here.

When comparing the volume of solder paste provided by a circular versus square aperture, consider that if the side of the square is D and the diameter of the circle is also D, the square has greater than 25% more area. (i.e., (1-0.785)/0.785 = 0.274). See Figure 1.

Figure 1. Square vs. circle areas.

However, the greater area of a square is not the only reason square apertures deposit more solder paste. The curving of the circular aperture enables more surface of the stencil to contact more of the solder particle’s area. See Figure 2. So, the solder particles will adhere to a cicular aperture more readily and not adhere to the pad, resulting in a smaller solder paste deposit.

Figure 2. The curving of a circular aperture results in more contact area with solder particles than a square aperture

These two effects can result in dramatically different soldering results, as seen in Figure 3. Using the square aperture provides so much more solder paste; when compared to what a circular aperture provides, it is stunning in the soldering result.

Figure 3. Circular aperture/pad (left) and square aperture/pad (right), using the same Type 3 powder size, area ratio, flux chemistry (no-clean), and reflow profile (RTP)

John Foster felt very fortunate. Not only did he get his undergraduate degree summa cum laude, but was now a graduate student at Ivy University under the tutelage of the famous Professor Patty Coleman. While contemplating these pleasant thoughts, he was working on his homework in advanced statistics when Professor Coleman walked up to his desk.

“Hey, John, I have a little assignment for you. Mike Madigan, CEO of ACME, has a vendor that is guaranteeing zero defects in lots of diodes that ACME buys, yet when ACME gets the lots they find a defect rate of around 1% or more. Can you contact ACME’s quality engineer, Frank Ianonne, to see how you can help?” Patty asked. “We covered this topic in the intro stats class you took last term,” she finished.

“Sure, glad to help,” replied John.

“Thanks, I’m going to SMTA’s PanPac for the first time and have a lot going on there,” Patty said thankfully.

“Wow,” John thought, “the pressure is on.”

John contacted Frank and learned that the vendor’s sales engineer, Mike Gladstone, said that they sample 20 diodes from each 10,000-part lot. If they find no defects in in the sample of 20, they claim they can say that there are 0 defects with 95% confidence, since 19 out of 20 is 95% and they found no defects.

“Yikes,” John thought, “this can’t be right.”

He thought about this and finally came up with what he confidently felt was the answer, especially after looking at his notes from the class Professor Coleman mentioned. He contacted Frank and they set up Zoom call with Mike to discuss the issue.

On the Zoom call after introductions, Frank asked Mike how they determine that a lot has zero defects.

“I’m glad to have the opportunity to explain this to youse guys,” said Mike.

It seemed to John that his tone was arrogant.

Mike continued, “Well, you will agree that 19 out of 20 is 95%, right?”

“Yes,” responded Frank and John.

“So, if we don’t get no defects in 20 samples, we got zero defects in the lot with 95% confidence. If we had one defect in the 20 samples, we couldn’t claim to have no zero defects in the lot,” Mike said.

“Mike, look at the image I took of one defect (a red bead) out of 2000 beads.” (See Figure 1.) “If I selected 20 beads on the left side of the container, how would I know that the defect rate is 0.0005 (1 in 2000)?” asked John.

Figure 1. The red bead is one “defect” out of 2000.

There was long silence.

“Mike what is your answer?” asked Frank.

Still no answer.

“The answer is that the only way you can assure zero defects is to evaluate all of the samples,” said John.

“You’re just confusing the issue with that there photo,” Mike spit out.

“Seems quite clear to me,” said Frank.

“You Ivy League types is all the same. You confuse the issue with mumbo jumbo when any dufuss can see I’m right,” Mike screamed.

Some profanity followed and Frank cut Mike’s Zoom feed.

“I see your point John,” Frank said. “but, can you give me some math to back it up?”

“Sure,” John replied.

“Let’s consider a case where the defect rate is not zero, but quite low, say 1 in 10,000 in a very large population. When we select the first sample, the likelihood of it being good is 0.9999 (10,000-1)/10000). What is the likelihood that the second one will be good?” John asked.

“Ah, let’s see…0.9999, right?” Frank answered.

“But what is the likelihood of both events?” John asked.

“Wait, I remember from a statistics class I took a few years ago, it’s 0.9999 x 0.9999,” Frank said triumphantly.

“And the likelihood of three in a row being good?” John asked again.

“0.9999^{3},” Frank answered confidently.

“So let’s say we sample so many times, let’s call it n times, that 0.9999^{n} = 0.05. What does this tell us?” asked John.

“Hmm …., ” replied Frank.

“Well, how likely is this to happen if the defect rate is 1 in 10,000?” John asked.

“Wait, I see, it would only happen 0.05 or 5% of the time.” Frank responded excitedly.

“So, let’s say we didn’t know the defect rate, what could we say if we sampled n and got no defects?” queried John.

Frank was stumped.

“I’ll tell you what, think about it and we will get back tomorrow. It’s already almost 6PM. Oh, and see if you can calculate what n is. Let’s Zoom at 10AM,” John proposed.

Time flew quickly and John and Frank were Zooming again.

“John, you just about killed me, I had trouble sleeping, but I think I have it after reviewing my stats book and doing some Youtubing,” Frank began.

“Well, if we didn’t know the defect rate and wanted to see if it was at least as good as 1 in 10,000 and we sampled n such that 0.9999^{n} = 0.05, we could say with 0.95 (1 – 0.05) confidence that the defect rate was 1 in 10,000 or less,” Frank said triumphantly.

“Precisely,” John exclaimed.

“But what is n?” John asked.

“That’s where I am stuck. We have the equation 0.9999^{n }= 0.05, but I can’t solve for n,” Frank said dejectedly.

“Hint: Logarithms,” John replied.

“That’s it, I got it,” said Frank enthusiastically.

Frank worked for a few minutes with a calculator, and came up with the solution in Figure 2.

Figure 2. The defects calculation.

“So, to show with 95% confidence that the defect rate is 1 in 10,000 or less, we would have to sample almost 30,000 components and find no defects,” Frank exclaimed.

“By looking at the equation, you can see if the defect rate was zero, 0.9999 would be replaced by 1 and the log of 1 is 0 so you would need an infinite sample,” said John.

“So, the only way to show 0 defects is to sample all of the components,” Frank said.

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.

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.

The Printed Circuit Engineering Association (PCEA) today announced it has closed the acquisition of the functional assets of UP Media Group Inc., including its industry leading publications and trade shows.

The deal, which was announced during the PCB West conference and exhibition last October, includes the annual PCB West and PCB East trade shows; PRINTED CIRCUIT DESIGN & FAB (PCD&F) and CIRCUITS ASSEMBLY magazine; the PCB UPdate digital newsletter; the PCB Chat podcast series; the PCB2Day workshops and webinars; and Printed Circuit University, the dedicated online training platform.

Printed Circuit Engineering Association (PCEA) (pcea.net) is a nonprofit association that promotes printed circuit engineering as a profession and encourages, facilitates, and promotes the exchange of information and integration of new design concepts through communications, seminars, workshops, and professional certification through a network of local and regional PCEA-affiliated groups. PCEA serves the global PCB community through print, digital and online products, as well as live and virtual events. Membership is free to individuals in the electronics industry.