Harmonic Elimination PWM Comparison and Uses

I've been doing a series about how to calculate the switching times of harmonic elimination PWM waveforms and I thought it was time to compare HEPWM to another method and look at how it can be used.  To catch up on the theory so far, have a look at the rest of the series.

Fourier Series of a Quarter-Wave Symmetric Pulsed Waveform 

Fourier Series of Harmonic Elimination PWM Waveforms

Octave Code For Generating Harmonic Elimination PWM Waveforms

All the code for this post can be found here.

A simple well known method for generating PWM waveforms is to compare a sawtooth wave to the desired signal and set the value of the PWM waveform high whenever the signal is higher than the sawtooth wave.  It gives reasonably good results, but I'd like to know how it compares to the HEPWM method when it comes to controlling harmonics.  When you're switching power loads there will be switching losses.  Ideally you want to minimise the amount of switching you do, but it comes with the price of not being able to control harmonic distortion as well.  It's one of those engineering trade-offs you have to make based on your design.  So to be fair and compare apples with apples, both the PWM and HEPWM waveforms below each have 40 switching transitions per cycle.  This means that switching losses should be equal and we can compare them using other metrics.

A half cycle of the PWM waveform is generated as mentioned above.  This and an inverted copy of it are concatenated to produce a full wave.

Although the generated waveform is inverted (my bad) it's hard to tell it apart from the HEPWM wave below.  Upon closer inspection you can see that the switching times are different, but on first glance everything looks the same.

Below is the real test.  The HEPWM waveform knocks out the harmonics right up to the 20th harmonic, whereas the basic PWM signal only kills the harmonics up to the 10th before they start creeping up.  This may be fine for what you're doing, but if you absolutely need to control certain harmonics, HEPWM is the way to go.  It might mean that your output filter is cheaper, or lighter, or takes up less space.  If you need to control a specific harmonic, HEPWM can do it.  It may be an issue with EMC you have to take care of.  If for some reason the geometry of the enclosure you're using is letting a certain frequency through, taking care of that in software rather than adding more shielding is going to save yourself a headache.

HEPWM also allows you to easily control the magnitude of the output while controlling harmonics.  If you're making a power inverter, ideally you want to run at nearly full magnitude to get the most out of your design, but you can still trim the output if you need to.  The graph below shows how you can pre-compute switching time for different magnitudes.  While running, the magnitude can be changed by selecting the set of switching times for the desired output magnitude.

HEPWM has some disadvantages, it's only useful in situations where you can pre-calculate the waveform.  It's not going to work with a signal like audio.  It's suited to power inverters and other niche applications where you want to reduce harmonics of a simple waveform like 50/60 Hz mains power.  It may not be for everything, but when the situation does call for it you've now got the right tool for the job.

To learn more I encourage you to read this thesis by Yu Yang.  It gives an overview of some other switching techniques and goes into a little more depth on some of the details.

Evaluation of New HEPWM Patterns in HVDC Applications - Bad Link

Cached backup I have of the above file.

Probability of Collecting a Full Set

The frenzy surrounding the Aussie Animal cards promotion got me thinking.  How many randomly collected cards would you need to collect before you had the whole set?  More specifically, if you were randomly given n cards what's the probability of having the full set.  I'm going to use the 108 card Aussie Animal set as an example.

All the files associated with this post can be found here.

When I initially came up with the idea for this post I thought it would be easy.  Turns out I was wrong.  A couple of quick simulations showed my understanding of the problem wasn't right, but after a little research I was back on track.  It did take me 2 days though.

The problem is mathematically the same as a well known problem called The Coupon Collector's Problem.  In this context the problem can be restated in the following way:

"Suppose that there are k different coupons, equally likely, from which coupons are being collected with replacement.  What's the probability that after n sample trials the complete set of k coupons is collected."

The formula to calculate this is deceptively simple.

There are k^n different ways to select the coupons. The next step is to figure out how many of those result in a complete set being collected.  I'm still getting my head around this, but it comes down to something called Stirling numbers of the second kind that equal the number of ways to partition a set of n objects into k non-empty subsets.  This suits the problem.  The collected coupons need to partitioned into k sets and none of these sets can be empty, i.e. the coupon hasn't been collected.  This number is represented as S(n,k).  This number has to multiplied by k!, as there are this many ways to arrange the sets.

This is where things got hard, I tried GNU Octave to calculate the results but as there was no built in function I tried to roll my own. I used an explicit formula for S(n,k) but that required numbers like 1000! to be calculated.  A number this large is just too big for Octave to handle.  I came up with a way to do all the calculations using the log of the number, but there was a precision problem for low values of n. Although with more time I could have got it working, I scrapped it and moved to Maxima, a computer algebra system that uses arbitrary precision (should have started there).  A couple of lines of code later I had my result.  The data was imported into Libre Calc to graph the solution.

Back to the Aussie Animal cards.  I've chosen a couple of data points to illustrate the how hard it can be to collect the set.

n = 200,    P = 8.99E-11

n = 400,    P = 0.0620

n = 600,    P = 0.662

n = 800,    P = 0.938

n = 1000,    P = 0.990

After collecting 600 cards you still only have a 66.2% chance that you've collected a set.

The expected value of how many cards you need to collect before you have a set is given by the formula nH(n), where H(n) is the harmonic number of n.  In our case this is equivalent to 108*H(108) = 568.5  To double check this, the Probability Density Function of the above data was generated to manually calculate the Expected number.  I get 557.  Given it was done in a spreadsheet with low precision data only up to n=1000, I'm pretty happy they agree that much.

This shows that without trading cards with other people it can be difficult to collect an entire set.  It assumes that all cards are equally likely and independent.  This may not be the case but it's not unreasonable to assume it is.

I'm glad I tackled this problem, learning about Stirling numbers made it worthwhile.  I now have another mathematical tool under my belt.  Would've been nice to have heard about them in statistics class, but there's only so much you can cover.

ALDI LED Fluorescent Tube Replacements

Warning - This article describes equipment and circuits that operate at high voltages.  Don't attempt to repair any high voltage circuits if you're not trained to safely work with electricity.  You may be seriously injured or even killed.  For further information read the blogs Terms Of Use.

ALDI recently had a "sale" on LED lighting tubes that fit into standard fluorescent lighting sockets.  I'd been wanting to try LED fluoro replacements for a while but was unsure of the quality of different products, so I decided to just dive in and find out the pros and cons first hand by buying 4, 1700 lumen, 18 Watt, 1200 mm modules.  To make things a little more interesting I took some power measurements before and after performing a few modifications to the existing fluorescent batten to save more power.

LED Light Tube & Replacement Starter

They come with the standard European Union Energy Label on the box to indicate the energy efficiency of the module.  The exact definition of the A+ rating is a little hard to find, but seems to mean that for a non directional light, it has an Energy Efficiency Index (EEI) of between 0.11 and 0.17.  The EEI has a bit of a convoluted definition but at its core it's a measure of how much power is used to generate an amount of light, with smaller numbers being better.

European Union Energy Label

For those interested the electrical specs are on the side as well.  The lights have a neutral white colour temperature of 4000K, and a colour rendering index of above 80.  They're supposed to survive 15,000 on off cycles and have a life of 30,000 hours.  Those figures are a little suspicious.  I have a feeling that although the individual LEDs may last that long, the electronics driving them will fail long before that.

LED Light Specifications

The installation process was simple.  The old starter and tube needs to be removed and replaced with the new ones.

LED Installation Instructions

The replacement starter is really simple, it's just a short circuit.

Shorted Starter

To understand the purpose of the shorted starter it may help to gain an understanding of how a fluorescent tube normally operates.  This post from a while back should be a good starting point.  Using the diagram from that blog the short makes sense.

In the normal operation of a fluorescent tube, a series path between active and neutral is created at start up via the two end filaments (F), the starter (C), and the ballast (G).  A LED tube replacement simply has one end shorted instead of a filament (let's say the right one for arguments sake) and draws all its power from the other end (where the left filament would normally be).  For this to occur, the starter needs to be replaced with a shorted link.  The power will then flow from the input, through the ballast, through the shorted filament on the right, through the shorted starter and into the power supply for the LEDs on the left.

Fluorescent Tube Operation Schematic

The below EEVblog video featuring Doug Ford has a really good explanation of the different ways LED replacements can be wired.  Skip to 12:30 for the explanation, but the whole video is worth a watch.

In the video it was mentioned that you may be able to save some power by removing the ballast and capacitor, and I wanted to test this out.  Using a fluoro batten and some basic measurement equipment I set up a small test jig.

Fluoro Batten

The batten is connected to an earthed power lead and allows easy access to internal components for testing.

Fluoro Batten Components

The ballast or inductor is used in a fluorescent light for two purposes, it helps generate a striking voltage to start the lamp, and when in operation it regulates the current flow through the tube.  The LED module works with it in place but doesn't need it to work.  This particular one is an old magnetic core variety.

Ballast / Inductor

The starter will be replaced with the shorted version.

Starter

Although the inductor regulates the current through the tube it introduces a lagging current.  This causes a higher current flow than necessary in the supply network and is counteracted by placing a capacitor across the supply to the batten.  As the ballast will be removed and the load is close to linear, this capacitor can be removed.

Power Factor Correction Capacitor

To measure the current, voltage, and power of the load, I'm using one of the power meters that the state government were selling.  They aren't super accurate, but it will allow me to get a ballpark estimate of the parameters I'm trying to find.  Another advantage is that I don't have to mess around with 230 Volts.  I set up my experiment, take a step back, plug it in, record the data and then unplug everything.  No points for being a hero around mains power.

Power Meter

Test 1 - No Light Installed, Ballast and Capacitor Installed

Measured
Power 0.3 Watt
Voltage 245 Volt
Current 0.263 Amp

Derived
Phase Angle 89.7°
Real Load = 4 Ω
Reactive Load Magnitude = 931 Ω

In this test the only thing connected across the supply voltage is the Power Factor Reduction Capacitor.  This accounts for the reactive load.  The reactive load can be used to calculate the size of the capacitor.  Using the equation X = 1/jωC the capacitance is be calculated to be 3.4 uF.  This agrees with the 3 uF value printed on the side.

Test 2 - Fluorescent Tube, Ballast and Capacitor Installed

Measured
Power 40.9 Watt
Voltage 245 Volt
Current 0.2 Amp

Derived
Phase Angle 33.4°
Real Load = 1022 Ω
Reactive Load Magnitude = 674 Ω

When in operation the load is still reactive but it's not too bad.  Due to the limited measurements taken, it's unknown if the load is inductive or capacitive.

Test 3 - LED Light, Ballast and Capacitor Installed

Measured
Power 19 Watt
Voltage 245 Volt
Current 0.283 Amp

Derived
Phase Angle 74.1°
Real Load = 237 Ω
Reactive Load Magnitude = 832 Ω

After the LED tube was installed, real power use dropped to 19 Watts.  The phase angle is still high and the current is quite large.  Now let's see what removing the capacitor and ballast does.

Test 4 - LED Light and Capacitor Installed, Ballast Removed.

Measured
Power 18.3 Watt
Voltage 245 Volt
Current 0.296 Amp

Derived
Phase Angle 75.3°
Real Load = 208 Ω
Reactive Load Magnitude = 801 Ω

The phase angle and current are both still high.  The real power did go down if you believe the power meter, but only by a small amount.

Test 5 - LED Light Installed, Ballast and Capacitor Removed.

Measured
Power 17.9 Watt
Voltage 245 Volt
Current 0.087 Amp

Derived
Phase Angle 32.88°
Real Load = 2364 Ω
Reactive Load Magnitude = 1528 Ω

The phase angle and current have dropped significantly. The readings don't make a lot of sense compared to the test 5, but you have to remember that the load numbers given are for series loads.  If you work out the equivalent parallel load things are a bit clearer.

The Conclusion

I've been living with the lights for about a week now and I'm quite impressed.  The amount of light they give out is equivalent.  I don't have any test gear to measure this, but they seem very similar to the fluorescent tubes they replaced.  The eye has a remarkable ability to compensate for different lighting conditions so it's hard to tell, but I think they're close.  When choosing lighting the main number you use to compare lamps is luminous efficacy.  These come in at 94 lumens per Watt, a high number, and although there are fluorescent tubes out there with better ratings they aren't directional.  If you have a fluorescent tube in a trough with a good reflector to redirect the light that goes towards the ceiling, they might be a better option.  In my case however any light that a tube emits upwards is pretty much lost, this makes the directionality of the LED replacement tubes more attractive.

As for removing the ballast and capacitor from the batten to save power, that's up to you.  You might save a couple of Watts, and when your only using 18 Watts that's significant, but it does require you to know exactly what you're doing.  You don't want to mess around with mains power.  Personally, I left them in.  If I ever need to use a fluoro tube in the fitting the change over is simple.  I'm happy enough cutting around 17 Watts from the lighting load.  Another positive is that they don't seem to attract moths.  Which is bad news for the gecko that used to hang around my window eating moths, he was entertaining to watch.

For a bit more information I reccomend the following two EEVblog videos about these exact tubes.