Math Encounters Blog

Quote of the Day

I am not a product of my circumstances. I am a product of my decisions.

— Stephen Covey

Introduction

I was reviewing a test report today for an optical receiver with an integrated power measurement sensor. This sensor is not particularly accurate -- its accuracy was specified as within ± 3 dB of true. This is not good at all. As I looked at our test data, I immediately noticed that I could calibrate the sensor to get a much more accurate result. But as I continued to think about the problem, I decided not to calibrate the sensor. The decision was driven by the cost of calibration versus the value of a more accurate result to the customer. Let's look at how I made this decision -- no magic here -- just a common engineering tradeoff.

Background

Some Definitions

The Wikipedia has reasonable definitions for sensors and transducers.

Sensor

A sensor is a converter that measures a physical quantity and converts it into a signal which can be read by an observer or by an instrument.

Transducer

A transducer is a device that converts a signal in one form of energy to another form of energy. Energy types include (but are not limited to) electrical, mechanical, electromagnetic (including light), chemical, acoustic and thermal energy. While the term transducer commonly implies the use of a sensor/detector, any device which converts energy can be considered a transducer. Thus, a motor is a transducer.

I should also spend a bit of time discussion my views on calibration versus compensation.

Calibration

According to the Wikipedia, calibration is a comparison between measurements – one of known magnitude or correctness made or set with one device and another measurement made in as similar a way as possible with a second device. I view calibration as a measurement process designed to generate mathematical parameters needed to adjust a transducer's real output closer to some idealized value.

Compensation

According to the Wikipedia, compensation is a plan to mitigate the effect of expected side-effects or imperfections. I view compensation as the procedure that uses calibration data to "correct" the output of a transducer.

Sensor Errors

By the definitions given above, a sensor converts a physical quantity that cannot be directly measured into a form that can be directly measured. Unfortunately, all conversions are at some level imperfect. As long as the conversion errors do not vary with time, we can use simple table or polynomial-based error compensation approaches.

The degree of error compensation you require often drives the compensation approach you use. I remember talking to an engineer who was trying to measure temperature to a tiny fraction of a degree. His solved his problem using very  time-consuming, high-accuracy temperature measurements and an assortment of piecewise, seventh-order polynomials to compensate for the errors in his sensors. My accuracy requirements are not so severe. I would like to be able to measure optical power from -10 dBm to -30 dBm with an accuracy of ±0.5 dBm.

In the case I will be discussing here, I have a sensor with a digital interface (i.e. there is a processor with an I²C interface embedded in the sensor) that will tell me the optical power present on a fiber. Ideally, if I have -20 dBm of optical fiber on the fiber, my sensor will tell me that I have -20 dBm on the fiber. If my sensor with digital interface were ideal, I would have an optical power measured versus actual optical power that is linear  with slope 1 and intercept 0. Let's call the graph of optical power measured versus actual optical power my sensor's response curve.

Unfortunately, the digital value returned by the sensor and its interface are only specified by the vendor as accurate within ±3 dB. However, the sensor response is fairly linear -- the only problem is that the sensor plot has a slope that is not 1 and an intercept that is not 0 . This means that we can model the error using two parameters:

• gain error

  This term represents the slope error (i.e. deviation from 1)  in my sensor's response curve.

• bias error

  This term represents the intercept error (i.e. deviation from 0) of my sensor's response curve.

Role of Calibration

When a sensor has a processor embedded with it, you could calibrate the sensor to minimize the reported error. In my situation here, the sensor appears to have a linear characteristic that could be corrected well by a simple linear transform. However, the sensor vendor chose not do that. Why?

I am pretty sure I know why -- they did not want to spend the money to calibrate the sensor. Calibration of an analog sensor can be costly because:

• You need certified source (i.e. optical power in this case) to stimulate the sensor.
• You need to re-qualify your test fixture periodically.
• It is expensive to build a repeatable, certifiable test fixture.
• You need to pay for test time.

However, you only need to pay this money if you want the accuracy. In my case, this sensor vendor made their sensor less expensive by not calibrating it. If I want more accuracy, I would need to calibrate it myself. Now I need to decide if the cost of calibration is worth it.

Analysis

I spent some time looking at the sensor. Here are my results.

Raw Sensor Data

Figure 1 shows the actual sensor response (i.e. power in versus measured power). Observe that the characteristic is quite linear over the -10 dBm to -30 dBm range that I am interested in. However, the line does not have a slope of 1 nor an intercept of 0.

Figure 1: Raw Sensor Data Versus Input Data.

Compensated Sensor Data

Figure 2 shows the compensation approach. The compensation is based on the best-fit slope and intercept values measured in Figure 1.

Figure 2: Simple Compensation Approach Using Linear Model (Gain and Bias Error).

Figure 3 shows my compensated sensor reading versus the actual optical power. Simple linear compensation looks very effective.

Figure 3: Compensated Sensor Data.

Conclusion

By spending money on calibration, I can reduce my optical power error from ±3 dBm to ±0.5 dBm. The cost was a ~1 minute of tester time, which means ~$2 per unit. While I prefer to have more accuracy, does it really matter to a customer? How will they use this information?

• They will use the power data to ensure that their received optical power is within the dynamic range of their receivers

  Most customers (~95%) run their networks in the middle of the dynamic range of the system. Since we have 15 dB of dynamic range, ±3 dBm of inaccuracy is not important to most customers. Do we want to burden everyone with the expense of performance needed by a few -- probably not.

• They will use the power data to determine if any of their optical components are aging.

  In this case, accuracy is not so important as precision. The customers will simply track how their optical power levels move with time. As long as the measurements are repeatable, they really do not need to be that accurate.

So I have decided not to calibrate the sensor. The uncalibrated sensor is accurate enough for 95% of the customer base. For that 5% that need an accurate sensor, we will need to provide them with an off-the-shelf test equipment option. However, the bulk of our customers will not need to pay for the extra accuracy.

Quote of the Day

The creation of a thousand forests is in one acorn.

— Ralph Waldo Emerson

Introduction

Figure 1: Apex SA57 Evaluation Printed Circuit Board.

I am doing some self-training on electronic thermal analysis. As part of my self-training ritual, I often work through vendor application notes. This post will show how I work through an application note using Mathcad. I chose the Apex SA57 evaluation board (Figure 1) because Apex had the most complete worked example that I have seen.

You might think that there is nothing special about working through someone else's calculations, but this particular exercise shows how things do not always go smoothly. I frequently find errors in vendor application notes. This application note was no exception, and I worked with the vendor to correct it. They appreciated the in-depth review and are now in the process of correcting their documentation.

For those interested in the fine details, my Mathcad source and its PDF are here.

Background

Goals

I am always looking at new electronic parts. This blog post is about working through an application note for estimating the thermal resistance of a power Integrated Circuit (IC) on a two-layer Printed Circuit Board (PCB). Since I am just learning about thermal analysis, this will be a learning experience for me.

As one of my physics professors used to tell me, "You cannot say you understand the material until you can work the problems." To make sure that I can correctly apply new knowledge, I work through some of the application examples that the electronic part vendors put together. If I can duplicate their results, then I feel that I am ready to begin working on my problems.

Approach

As part of my self-education, I need to organize my work. There are three tools I use in this effort:

• Mathcad

  All of my learning involves mathematics. I NEVER do mathematics manually. I always create Mathcad worksheets to document my calculation efforts.

• Note Taking Tool

  I use a tool called Whizfolders. I also know folks who use OneNotes. It is a matter of personal preference. Religiously using a note-taking tool ensures that you never lose information. It also greatly speeds your access to information years later.

• Excel

  I use Excel for routine data management. I like its editor and I use it to beat data into forms convenient for analysis.

With my tools in place, it is now time time to dig into the application note. Here is a link to the original application note I found on the web. Because things move around on the web, I have included a copy on my personal web site. As I mentioned, I found some issues and the vendor gladly agreed to correct their note. Their corrected note has not been issued yet, so I have included a corrected version of the note here. My corrections are in red and they have been confirmed by the part vendor, Apex Microtechnology.

You may have noticed that I am detail-oriented – I always tell people that my career has been a celebration of detail …

Key Technical Points

Basic Thermal Resistance Equation

Equation 1 shows the formula for thermal resistance that I will be applying repeatedly in the following analysis.

Eq. 1

where

• λ_X is the thermal conductivity of the material that makes up object X (Reference).
• ρ_X is the thermal resistivity of the material that makes up object X .
• l_X is the length of object X along the path of conduction.
• A_X is the area of object X along the path of conduction.
• R_X is thermal resistance of object X.

Overall Thermal Model

Figure 2 shows a cross-section of the power IC I am looking at using and how it is mounted to a PCB.

Figure 2: Apex Power IC Mounting on a PCB.

The power from the integrated circuit moves from the silicon to the bottom of the circuit board. We assume the bottom of the PCB is an isothermal (i.e. constant temperature) region – on the real card, this means that the bottom-side heat sink (Figure 1) couples the bottom of the card to the ambient air perfectly (i.e. no thermal resistance). Obviously, this is an idealization but one that a more sophisticated model can easily include.

Figure 3 shows how they model the configuration of Figure 1 using thermal resistances.

Figure 3: Thermal Model Used in the Apex Application Note.

Thickness of Copper Planes

Figure 4 shows how we can compute the thickness of the copper foil used in the PCB to construct the copper traces, regions, and layers. The thickness of copper foil is usually specified in terms of the weight of one square foot of the copper foil.

Figure 4: Formula for Thickness of Copper Foil Based on Mass.

The application note computes each of the individual thermal resistances and a total thermal resistance. I will duplicate this work.

Analysis

Key Thermal Constants

Figure 5 shows some of the constants used in the analysis to follow.

Figure 5: Key Thermal Constants.

Thermal Resistance of the Silicon Die

The transistors on the IC dissipate power. Integrated circuits place their power dissipating elements on one side of a slab of silicon. This power must pass through the silicon to an aluminum slug, which conducts the heat from the silicon to the top of the PCB. Figure 6 shows the calculations.

Figure 6: Thermal Resistance of the Silicon Die.

Thermal Resistance of the IC's Heat Slug

Figure 7 shows the calculations for the thermal resistance of the heat slug.

Figure 7: Thermal Resistance of the Heat Slug.

Thermal Resistance of the Copper Pad on Top PCB Layer

Figure 8 shows the calculations of the copper pad on the top of the PCB. The IC's heat slug sits on this copper pad.

Figure 8: Thermal Resistance of the Copper Pad on Top of the PCB.

Thermal Resistance of the PCB

The PCB thermal model is a composite of vias, laminate, and planes. I break this modeling down in the following subsections.

Thermal Resistance of the FR4 PCB Material

Figure 9 shows how I calculated the thermal resistance of the FR4 PCB material.

Figure 9: Thermal Resistance of FR4 Material in the PCB.

Thermal Resistance of a 20 mil Via

Figure 10 shows the calculations for the thermal resistance of a 20 mil via.

Figure 10: Formula for the Thermal Resistance of a Via.

Thermal Resistance of the PCB with 24 Vias

Figure 11 show the calculations for the thermal resistance of the overall PCB (i.e. FR4 material thermally in parallel with 24 vias). Observe that the vias have a combined thermal resistance of ~2.6 Ω, which is modeled as a parallel resistance to laminate's thermal resistance of ~58 Ω. Clearly, most of the heat moves through the vias.

Figure 11: Thermal Resistance of the Overall PCB.

Thermal Resistance of the PCB Bottom Copper Layer

Figure 12 shows how the thermal layer of the PCB's bottom copper layer is computed.

Figure 12: Thermal Resistance of the Bottom Copper Layer on the PCB.

Overall Thermal Resistance Calculation

Figure 13 shows the overall thermal resistance calculation. Observe that nearly 85% of the thermal resistance is in the PCB – without vias, this percentage and R_Total would have been much larger.

Figure 13: Total Thermal Resistance Calculation.

Conclusion

This was a good computation exercise. Let's review the results:

• This exercise presented a good example of a power-focused, analog design.
  • This example used a heat sink – most digital designs do not include heat sinks.
  • In applications without fans, getting the heat to the air is a big challenge.
• All the calculations were based on Equation 1, which is analogous to the formula for the resistance of an electrical conductor.
• The thermal resistance of the PCB dominates the overall thermal resistance
  • Adding more copper to the PCB can help (e.g. 2 oz Cu planes), but this will make the PCB more difficult to solder and rework.
  • The role of the via to the overall PCB thermal resistance is critical.
  • You need to place as many vias as you can between the top and bottom layers.
  • For high -power circuit boards, you may wish to use other laminate materials – their thermal resistance varies quite a bit
    • 
    • 
  • FR4 is the cheapest laminate you can find. Any others will cost more.

Save

Save

Save

Save

Save

Save

Save