Why Square Apertures Provide More Solder Paste than Circular Apertures


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)


Dr. Ron

Dispelling The ‘Five Ball Rule’

Michel writes:

Dr. Ron, when if comes to SMT printing of solder paste, why do some people use the five-ball rule for rectangular apertures and the eight-ball rule for circular apertures?


The “Five Ball Rule” is another metric that SMT assembly industry leaders believe, but it is difficult to find its origin. It states that when selecting a solder paste, five of the largest solder balls should be able to fit across the width of the smallest rectangular stencil aperture. See Figure 1a for a 0.2mm wide rectangular aperture.

Typically, the largest solder ball diameter is assumed at the 90th percentile. See Figure 2. So, in this example, a type 4 solder paste would fit the five ball rule as the largest solder ball is 0.038mm. Five times 0.038 is 0.190mm, just a little less than the aperture width of 0.2mm. It should be remembered that this is a “rule.” not a “law.” So let’s say you had 4.5 balls across the aperture with instead of 5, it would most likely be OK. 

Figure 1. A comparison of the Five and Eight Ball Rules

Figure 2. Solder Powder Sizes

A generation ago, the advent of circular apertures to support BGA and CSP packages necessitated a new “rule.” Figure 1b shows why the five-ball rule is inadequate for circular apertures. Although five type 3 solder balls fit along the 0.275 diameter, off the diameter, there is not enough room for many solder balls.  Hence, an insufficient amount of solder paste would be printed.

For the same aperture, if a type 4 paste is used, 7 or 8 solder balls span the diameter and the amount of paste printed would be much closer to the volume of the aperture.

For a little more on this topic, see a past post.


Dr. Ron

Alloy Metal Weight Fraction Calculation

Iasad writes,
“Dear Dr. Ron,

I see that you have developed software to calculate the density of an alloy if given the weight fractions of the constituent metals. Is it possible to find the weight fractions of the metals in an alloy given the alloy’s density? Thank you!”
Unfortunately, finding the weight fractions of the metals in an alloy from the alloy’s density can only be accomplished with a two metal alloy. First we must use the equation:

Equation 1

Where x is the weight fraction of metal A and the rhos are the associated densities.  All that has to be done is to solve for x.  The solution is worked out below in Figure 2, the final result is:

Equation 2

As an example, let’s say you have a gold-copper alloy with a density of 18.42 g/cc.  The density of gold (metal A) is 19.32 g/cc and that of copper (metal B) is 8.92 g/cc.   Substituting these values into equation 2 gives the weight fraction of gold as 0.958.  Hence the weight fraction of copper is 1-0.958 = 0.042.

I have developed an Excel-based software tool to perform these calculations. An image of it is shown in Figure 1.  If you would like a copy of this tool send me a note.


Figure 1.  A screen shot of the alloy metal weight fraction calculator.


Figure 2. The derivation of the weight fraction formula.


Dr. Ron

BGA Voiding in Electronics Assembly

Patty had to admit that the last few weeks were exciting.  Her talk to US Army Rangers and Navy Seals on critical thinking went really well.  Now, the local newspaper was asking her to comment on political polling in the current presidential primaries.  Patty was just finishing her response to the paper before a meeting with Pete to discuss the voiding presentation that they were working on for Mike Madigan.  Her response follows:

Dear Editor:

My favorite candidate was trailing in the polls by only 1% in my state, but on primary day he lost by 5%.  Why isn’t polling more accurate?




Dear Disappointed,

Pity the pollsters. They have to predict what will happen by sampling a manageable number of people, say 1,000. This situation creates several challenges. The first is that their sample should represent the population as a whole. This challenge is not easy. They need to assure that the 1,000 people represent the population of the entire state. If they get an inappropriate number of old, young, wealthy, lower income, educated, less educated, etc., in these 1,000 people then their prediction will be off. As an example, let’s say that 45% of a state’s residents have a bachelor’s degree or higher, yet their sample has 60% with a bachelor’s degree or more. This difference will likely make their sample non-representative of the population as a whole and will skew the results.

Let’s go back to your candidate, whom we will call candidate A. It ends up that candidate A was supported by only 47.5% of the total population and his opponent, candidate B, by 52.5%, giving the difference of 5% that you mentioned. Let’s assume that the pollsters establish a good sample of 1,000 people that is very close to representing the state as a whole. It is unreasonable to expect that the 1,000 people polled would exactly have 47.5% or 475 supporting candidate A, due to statistical variation.  To show the likelihood of a number different than 475, we have to use the binomial distribution as seen in  Figure 1 below. Note that there is about a 10% (0.1085 in the figure) chance that a population of 1,000 will have 495 or greater supporting candidate A. This uncertainty, added to the difficulty of establishing a perfect sample, makes polling error of 5% or so not uncommon.

Figure 1. Note that, even though 475/1,000 is the most likely, if the larger population has 47.5% supporting candidate A, there is a 10% chance a sample of 1,000 could have 495 or greater favoring candidate A.


Just as Patty finished her response, Pete came to her office door.

“Hey kiddo! Can we go over my thoughts on the voiding in BGA balls section on voiding for Mike Madigan?” Pete asked cheerfully.

“Sure. What do you have so far?” Patty asked.

“I’m focusing on the importance of the reflow profile.  Have you seen this graph,” Pete began.

Figure 2. The hot soak profile produces the fewest voids in CSP and BGA balls.

“Wow! That really shows the benefit of a hot soak profile over a cool soak profile. But, I am most surprised at how much benefit a hot soak profile has over a ramp-to-peak profile (RTP),” Patty commented.

“Isn’t the timing of the higher temperatures important, too?” Patty asked.

“My next point precisely. Look at this graph,” Pete said enthusiastically.

Figure 3.  The combination of the reflow profile and flux characteristics that produces outgassing before the solder becomes liquid (the red curve) will minimize voiding.

“The process engineer needs to assure that most of the flux is volatilized before the solder melts, as in the red curve, not as in the black curve where almost all of the flux is outgassing during the melting it the solder (Tm). This situation is assured by the correct combination of flux and reflow profile,” Pete said confidently.

“Anything else, Professor Pete?” Patty asked.

“It is really helpful to work with your solder paste supplier to obtain the red curve. They should be able to tell you what type of reflow profile and solder paste will most likely provide this kind of result,” Pete finished with a chuckle.

And he added drolly. “Right … Professor Pete.”

“Rob’s working on voiding on thermal pads for BTCs right?” Patty asked.

“Yep. He said he will be ready in two days,” Pete answered.

What will Robs plan be for minimizing voiding with BTCs?  Will Patty be happy with it?  Stayed tuned for the details.

Best Wishes,

Dr. Ron

Lead Free 2015

It is hard to believe that in July we will celebrate the 9th anniversary of the advent of RoHS. So the timing seemed right when I was recently asked to speak at the Boston SMTA Chapter on The Status of Lead-Free 2015: A Perspective.

An overview of the entire 75-minute presentation would be a bit long, so I am going to discuss three of the “questions” that I covered.

  1. Q: We are now almost nine years into RoHS’s ban on lead in solder. How has lead-free assembly worked out?

A: Something over $7 trillion of electronics have been produced since RoHS came into force, with no major reliability problems. One senior person, whose company has sold hundreds of millions of lead-free devices since 2001, reports no change in field reliability. The challenge that implementing lead-free assembly placed on the industry should not be minimized, however. Tens of billions of dollars were spent in the conversion. In addition, failure modes have occurred that were not common in tin-lead assembly, such as the head-in-pillow and graping defects. But assemblers have worked hard with their suppliers to make lead-free assembly close to a non-issue. Some people ask how I can say that lead-free assembly is close to a non-issue. My office is across the hall from some folks that purchase millions of dollars of electronics a year for Dartmouth. Several years ago, I asked them how they feel that electronics perform since the switch to lead-free. They answered by saying “What is lead-free?” If people that buy millions of dollars of electronics have not even heard of lead-free it can’t be a big issue.

  1. Q: In light of sourcing difficulties, is there an industry consensus regarding lead-free conversion for military, medical, aerospace etc. assemblers that will continue to be exempt?

A: The main issue is getting components with tin-lead leads, especially BGA balls. Many assemblers are reballing BGAs, which has become a mature technology, although with an added cost. As years go by and there becomes more confidence in medium to long term lead-free reliability, some exemptees may switch to lead-free. However, I think mission critical applications with 40-year reliability requirements must be extremely cautious to make the switch. There may be subtle reliability issues that may show up in 40 years, that are not found in accelerated testing. One concern is aging. Even at room temperature, solders are at over 50% of their melting temperature on the absolute scale (300K/573K = 0.52). So aging can occur at room temperature. Some research suggests that lead-free alloys may be more affected by aging than tin-lead alloys.

  1. Q: It has been said that you claim that lead-free assembly has some advantages. Can this be true?

A: Guilty as charged. Lead-free solder does not flow and spread as well as tin-lead solder. This property can result in poor hole fill in wave soldering and some other assembly challenges. However, this poor wetting and spreading means that pads can be spaced closer on a PWB without the concern of shorting as seen in the image below. Your mobile phone would likely be bigger if assembled with tin-lead solder.

Lead-free solder does not flow as well as tin-lead solder. Hence, closer pad spacings are possible.



Dr. Ron

Photo courtesy of Vahid Goudarzi.


Alloy Melting


Richard asks:

Dear Dr. Ron,

Recently we had a solderability problem with tin-finished component leads and SAC305 solder paste.  One of our engineers claimed that the problem was that the tin finish melts at too high a temperature (Tm= 232°C) for the SAC305 solder paste (Tm = 219°C) to melt it.

My understanding is that certainly above 232°C both will melt and form a good solder joint, but even if the temperature was less than 232°C, say 225°C, the tin would melt. Can you explain this phenomenon?


Thanks for this question, which can be interpreted two ways. The first would be that, in a reflow oven at temperatures above the melting point of both metals, the one with higher melting temperature prevents the metal with a lower melting temperature from melting it. This is not true, since both metals would come near to the temperature of the air in the reflow oven and melt.

The other perspective would be that the temperature in the reflow oven is above the melting temperature of SAC 305, but below that of tin. So, how can the tin melt?  To consider this situation let’s say the oven is at 228°C. Will the tin on the lead or pad finish melt? The answer is yes. But, let’s try to understand the phenomenon with gold and tin first.

Metals that have extreme melting point differences often dissolve in each other. As you stated, tin melts at 232°C, whereas gold melts at 1064°C.

This phase diagram can be found here.

Figure 1. The gold tin phase diagram

To make a gold-tin solder, all one has to do is have a bath of tin at some moderate temperature, say 350°C. Insert the gold and the gold will melt and flow into the molten tin. This is true even though the gold melts at 1064°C. This effect can be shown experimentally. A similar phenomenon exists with gold and mercury. Mercury reacts with gold at ambient temperatures. The phenomenon can be used to extract tiny gold particles from soil and is commonly used today in artisanal gold mining. Unfortunately this use of mercury is often toxic to the miners and pollutes the environment.

Considering electronics assembly solders again, let’s assume that some liquid tin-lead solder is heated to 200°C. See Figure 2a. As seen in this figure, a ball of tin at 25°C is held above the molten tin-lead solder. The ball of tin is immersed into the molten tin-lead solder in Figure 2b. The tin-lead solder forms a meniscus around the solid tin. Even at room temperature the tin atoms are vibrating, and as a result, some of these atoms on the tin ball will end up flowing into the tin-lead solder. This action will leave a vacancy in the tin ball that may be filled by a lead atom from the tin-lead solder. In the vicinity of the newly arrived lead atom, the melting temperature of this micro spot of tin-lead alloy will be lowered as tin-lead solder has a melting temperature below that of tin. This process will continue until all of the tin will intermix with the tin-lead solder and flow into it as seen in Figures 2c through 2f.

Figure 2a

Figure 2b

Figure 2c

Figure 2d


Figure 2e


Figure 2f


Dr. Ron

The Importance of Oxygen Barrier in Solder Pastes


Pity the solder scientists of the late 1970s and early 1980s. SMT was an emerging technology and the world wanted to buy solder paste. However, the only experience many solder scientists had was wave soldering. In wave soldering, the flux’s main job is to remove the oxides from the PWB pads and components. The solder is in a molten state and its oxidation is not a main concern. In the soldering process, the solder only touches the board for a few seconds and the board only experiences high temperatures during this brief period.

I imagine some early solder pastes consisted of solder powder with fluxes similar to those used in wave soldering. If so, they probably didn’t work too well. Consider the dramatic differences that solder paste experiences as compared to solder in wave soldering. The “flux” in solder paste has to remove oxides from the PWB pads, component leads and solder particles, but it also has to protect all of these surfaces from re-oxidation for several minutes in the reflow oven. To achieve this protection, the “flux” has to contain materials that act as oxygen barriers. The most common oxygen barrier materials used in no-clean solder pastes are rosins/resins. Rosins, or resins, which are modified or synthetic rosins, are generally medium- to high-molecular weight organic compounds of 80-90% abietic acid. They are typically found in coniferous trees. Rosins/resins are tacky in nature, they provide some fluxing activity, and provide the critical oxidation resistance during the reflow process.

The reason I wrote “flux” in quotation marks in the above paragraph is that what most people call the flux in solder paste is actually a complex combination of materials. These “fluxes” consist of:

  • Rosins/resins: for oxygen barrier and some fluxing activity
  • Rheological additives: to give the best printing properties. e.g. good response to pause, good transfer efficiency, excellent slump resistance, good tack, etc
  • Solvents: to dissolve the other materials
  • Activators: to perform the main fluxing action (removing oxides).

Because of these complexities, and the material’s multi-functionality, they are sometimes referred to as, “flux-vehicles.”

Modern solder pastes must have good oxygen barrier capability. In most reflow profiles, the solder paste is at temperatures above 150°C for several minutes. During this time an oxygen barrier is needed to protect both the solder particles and the surfaces of the pads and leads.

The graping defect. A common example of cases where the solder barrier was insufficient is seen in the graping defect, or its relative, the head-in-pillow defect. If you are experiencing one of these defects, a solder paste with better oxygen barrier properties is bound to help.


Dr. Ron

Density Calculation Still Raises Questions


It is hard to believe, but I have been blogging now for over 8 years. In all of that time, the most popular topic by far has been the calculation of alloy densities. Many people are troubled by the required equation. At first blush it doesn’t seem logical. I have derived it, but here is an effort to try and make it more intuitive.

A reader wrote the following, which inspired my explanation that follows.

Craig writes:

Why do we calculate the theoretical density of an alloy by using the formula of:
1/Da = x/D1 + y/D2 (binary) or
1/Da = x/D1 + y/D2 + z/D3 (ternary) as opposed to multiplying the individual metals’ densities by their percentages in the alloy and adding them together ? I thought the density of an alloy would be analogous to a weighted average of the densities of the metals in the alloy. Obviously, that is wrong, but I don’t understand why?

The reason you have to use the non-obvious formula is that density is given as mass/volume. So there are two quantities to be concerned with, and one is inverse (i.e. volume is 1/volume in the density formula). So you have to use the equations above, or you will be in error. I derived these equations in a previous post. However, many people do not find the result intuitive.

Here is, hopefully, a more intuitive example. Let’s say, for some strange reason, you were interested in inverse height. So let’s say John is 5 feet at his shoulders, or 1/5 inverse feet. His girlfriend Kathy is 5 feet tall, also 1/5 inverse feet. Kathy stands on John’s shoulders. How many inverse feet is their total height? The tendency would be to say they are 1/5 +1/5 = 2/5 inverse feet. But we know that when Kathy stands on John’s shoulders they span 10 feet, so they must be 1/10 inverse feet. So, to calculate their height in inverse feet, we need to use:

1/IFT = 1/John IF + 1/Kathy IF = 1/(1/5) +1/(1/5) = 10

IFT = 1/10

Note: (IFT = Inverse Feet Total)

The same is true for density, since density = mass/volume (inverse volume is like inverse feet).

Another engineering example is electrical resistivity and conductivity. If the resistance of one wire is 2 ohms, and another is 1 ohm, and we connect them in series, the total resistance is 3 ohms. However, if we consider conductivity, one wire has a conductivity of 0.5 mho and the other 1 mho. Again, to get the total conductivity we don’t add the conductivities, we must use:

1/Ctotal = 1/C1+1/C2 = 1/0.5 + 1/1 = 2+1 = 3. So Ctotal = 1/3 mho.

I hope this helps.


Dr. Ron


Is the PC Dead?


Let’s see how Patty is recovering from her conflict with Hal Lindsay.

Patty saw a link on one of the daily industry newsletters that piqued her interest. It was titled, “A New Alloy for Medical Electronics Applications.”

The paper talked about a new SAC (tin-silver-copper) lead-free solder alloy that contains a small amount of manganese. Apparently, the manganese modifies the metallurgical structure and enables the alloy to perform well in both drop shock and thermal cycle tests.

She finished the article and decided that she needed to look into this alloy more. It isn’t common for an alloy to perform well in both drop shock and thermal cycle. The title also reminded Patty how relieved she was that their St. Paul facility was getting ready for Medical RoHS on July 22, 2014.

She then checked her email and saw she had a note from ACME CEO Mike Madigan.


I just read an article about the death of the PC. PC sales are off by 10% per year. That means in 10 years they will be gone. PCs are 30% of our business. Develop a plan to modify our business so that we can replace this loss. Be ready in two weeks.


Patty had to chuckle at Mike Madigan. He seemed to totally lack social skills. She wondered how someone could get so far in the company without them. Anyway, she had a new assignment. Also, she quickly determined that a business decreasing 10% per year would still be at about 35% (0.9^10= 0.349) of its original size in 10 years.

Right off the bat, her sense was that the “Death of the PC” was exaggerated and misunderstood. She performed a search and read some more articles, finding several that talked about the surge of tablets like the iPad and the Kindle Fire as being behind the PC’s decline. She and Rob owned two iPads and one Kindle Fire. Were they part of the trend? She suspected that there was more to the story and thought that discussing this topic with The Professor would be interesting. She was visiting him at Ivy University later in the week.

The week went by quickly and, before Patty knew it, she was in her car with her husband Rob, driving the two hours from Exeter, NH, to Ivy U. Rob was getting his Ph.D. there, part-time, and made the trip several times a month. The Professor was his advisor.

Rob needed an hour with The Professor. During that time, Patty checked her email. After he and Rob were finished, The Professor said, “Let’s go the lunch at Simon Pearce and chat about your latest adventures.”

“That’s great! Simon Pearce is my favorite restaurant, Professor,” Patty responded. Simon Pearce Restaurant in Quechee, VT.

They both looked at Rob, but he had so much work to do that a slice of pizza from the student union was his destiny for lunch today.

As Patty and the Professor arrived, Simon Pearce was at its most beautiful. The sky was blue and the leaves were just starting to turn. Patty had alerted The Professor that she wanted to talk about the “Death of the PC,” so he was ready as they sat down.

“Patty, I have read all of the articles that you sent. They emphasize the impact on the PC by tablets and mobile phones, but they miss an important point,” The Professor began.

“What’s that?” Patty asked. “The effect of the constancy of ‘Memory Metrics,’” The Professor replied.

“What are ‘Memory Metrics?’” Patty asked.

“Let me explain with a story,” The Professor answered.

“When you were a toddler, back in 1986, my family got our first computer, an IBM PC XT. It cost $6,000 and had 0.512MB of RAM and 20MB on the hard drive. We had it for 3 years and needed a replacement. Can you guess why? The Professor asked. The IBM PC XT circa 1986.

“You ran out of memory,” Patty answered.

“Yes, the Kids where playing, ‘Where in the World is Carmen Sandiego,’ and some other games,” The Professor said.

“Then we had to get another computer in 1989 with 4 MB of RAM and a 200MB hard drive. We tripled the hard drive memory and still needed a new computer in 1995. This trend continued until about 5 years ago, but it then slowed down,” The Professor continued.

“Why did the trend slow down?” Patty asked.

“Before I answer that, we need to discuss ‘Memory Metrics,’” The Professor said. “How much memory does a decent photo require?”

“About 1MB,” Patty answered.

“A song?” The Professor quickly shot back.

“About 5MB,” was Patty’s fast reply.

The banter continued.

“A book?”

“About 1MB.”

“A full length movie?”

“About 5,000MB.”

The Professor was impressed.

“Wow, Patty! It’s amazing that you know all of the critical ‘Memory Metrics.’ Where did you learn them?” The Professor said.

“Rob went to lunch with the professor who teaches a class called Everyday Technology. They discussed these metrics. It helps me to estimate how much memory I need on my smartphone to store all of my photos, songs, and books,” Patty answered.

“So how many songs could you store on the hard drive of my first computer?” The Professor asked.

“Only 4 or only 20 photos or books,” Patty answered.

“So, we weren’t putting songs, books or photos on early computers. Even the $10,000 laptop that I needed for my research in 1996 only had a 1000 MB hard drive … only 200 songs,” said The Professor.

“I get it,” Patty responded, “Let me see if I can fill in the blanks.”

“Memory Metrics, are close to constant, with the exception of perhaps photos with more megapixels. Anyway, the memories on new computers have gotten so large that we don’t need to replace them as often, we just don’t fill the memories up any more. Unless we store movies,” Patty continued.

“Add in the advent of smartphones and tablets and see if you can make an argument for why PC sales are down,” The Professor said, beaming at his favorite student.

“Well, before smartphones and tablets you might get a new PC even if you didn’t need it. But, now you have two or three devices to buy, so you may put off a new PC for awhile to buy a new smartphone or tablet,” Patty responded.

“So, what will happen to smartphones and tablets?” The Professor asked.

“A similar thing. They are developing so many features, memory, and computing power, that their sales will slow done in a few years,” Patty thoughtfully said.

“So, the PC is not dying!” Patty exclaimed.

“But, what about the tablet forcing out the PC as a replacement?” Patty asked.

“Let’s ask my statistics class after lunch,” The Professor suggested.

Although Patty had been around campus off and on, she was immediately struck by how young everyone in the class looked. Just last night, Rob was teasing her about her first gray hair. That, no doubt, added to this effect. As the class of 60 students settled down, The Professor began to speak.

“Before we start the lecture, my colleague, Patty Coleman, would like to ask you a few questions,” The Professor said.

“The Professor calls me a colleague! Wow!” Patty thought.

So, Patty stood up and started speaking. “If you could only have one device, a PC or a tablet, who would choose the PC? “ Patty asked.

Everyone raised their hand.

“Why?” Patty asked.

A boy, who looked, to her, to be about 12 years old, answered, “Well, Dr. Coleman, it is very difficult to write a 10-page paper on a tablet, or perform Excel calculations, or prepare a PowerPoint file on a tablet.”

“How many of you own a tablet?” Patty asked.

Only about 20% of the students raised their hands.

“How about a smartphone?”

Everyone raised their hands.

“Any thoughts on why many people think the PC is dead and the tablet is the future?” Patty asked.

After a little murmuring a female student raised her hand. Patty was relieved that she looked to be about seventeen. Patty acknowledged her.

“Well, Professor Coleman, my family has two tablets at home. My 7- and 9-year old brother and sister play games on them all the time. My mom and dad use one as an e-reader, and they use one to control our TV. But, for any serious work, even they prefer a PC. Only my kid brother and sister would prefer a tablet over a PC. But, the most recent devices we bought were tablets. Our PCs just haven’t needed replacing,” the young women said.

“Why do so few of you have tablets at school?” Patty asked the group.

Again more murmuring ensued and, finally, a graduate student raised his hand. Patty acknowledged him.

“Professor Coleman, I find that the combination of a PC and a smartphone fills all of my needs. I can listen to music or read a book on my smartphone. I can’t not have a smartphone and function at today’s university. A tablet would be a luxury, I don’t feel I need or can afford,” he said.

With that Patty sat down. This information on tablets and the college scene was consistent with what Patty had read. The Professor’s lecture was on Bayes Theorem and Patty decided to stay. She chuckled that the students thought she had a Ph.D. and was a professor. “Yikes, does this mean they think I am old?” she thought.

After the lecture Patty thanked The Professor and went to find Rob. She now felt she understood why PC sales were down. Tablets had some effect, but the near stability of Memory Metrics and the tremendous computing power and increased memory of a modern PC simply meant people didn’t need to upgrade as often. She expected similar trends for smartphones and tablets. As she organized her backpack, she put her 6 month old laptop, with its 16,000MB of RAM and 1,000,000MB hard drive away.

The PC was alive and well.

Best wishes,

Dr. Ron

Solder Alloy Density


I have occasionally written on calculating solder alloy density, as there is surprisingly more interest than I thought there would be in this topic. Recently, it occurred to me that it might be beneficial to compare the calculated densities to actual densities of a few alloys to see how accurate the correct formula is (for the derivation of the correct formula see below). The formula assumes “perfect mixing” (i.e., no interactions between the alloy elements). The alloys we investigated were tin-bismuth-silver, tin-silver, tin and tin-bismuth.

To measure the density, I obtained a few alloys from Indium. My student, Evan Zeitchick, determined that a good technique to measure density is to machine the alloy into a rectangular parallelepiped (see photo), weigh it, and calculate its volume from its dimensions.  The results agree with the correct formula to about 1 to 2%. Some people would ask why there is any difference. The reason is that all alloys form different phases, and some form intermetallics. These phases and intermetallics would typically have different densities than that calculated for the alloy. I will have more detail on this work in a future post.

Here is a derivation of the correct density formula:

Many people incorrectly assume that if you have an alloy of x % tin by weight and y % silver, that the density of this alloy would be 0.x*Density tin +0.y*Density silver. This intuitive linear formula is incorrect however, as density has two units (mass and volume).

An easy way to understand the derivation of the correct formula (proposed by Indium  engineer Bob Jarrett) is to consider a 96% tin, 4 % silver example.

Let’s assume I have 1 g of this alloy, 0.96 g is tin and 0.04 g is silver.

The volume of the tin is 0.96 g/7.31g/cc = 0.131327cc

The volume of the silver is 0.04g/10.5g/cc = 0.00381cc

So 1 g of the alloy has a volume of 0.131327 + 0.00381 cc = 0.135137 cc

Hence it’s density is 1g/0.135137cc = 7.39989g/cc

Hence, the general formula is:

1/Da = x/D1 + y/D2 + z/D3

Da = density of final alloy

D1 = density of metal 1, x = mass fraction of metal 1

same for metals 2, 3

The formula continues for more than 3 metals.

I have developed an Excel spreadsheet that calculates density automatically. If anyone wants a copy, send me an email at [email protected]

Dr. Ron

P.S.: Interesting thought: About 165,000 tonnes of gold have been mined throughout history. If all of this gold was gathered into a cube it would only be about 21 meters on a side. At $1,550/oz, its value would be $8.5 trillion, quite a bit less than the almost $15 trillion debt of the US government. Yikes!