Following on from the last post, I wanted to start with a picture again. Who said it was a dog’s life being an engineer?

A young mate of mine, after browsing this blog, suggested that I should make it a movie blog like that of Dave Jones at http://www.eevblog.com/
To be honest, I am not sure how I was supposed to take this. The subject of Dave Jones is too big to deal with in just one post, so I will just report on my latest squint at eevblog here.

The latest posting on eevblog at the time of writing is EEVblog #533 – LED Fluoro Tube Teardown He might have moved on by the time you read this, but you can work your way back through his postings to see this one. When I visited the site, it was sporting an advertisement. I suppose that is how Dave makes a quid out of his eevblog site. (He says it is his full time job.) This particular ad had a picture of a woman with a look on her face like a rabbit that is frozen in the spot light whist the hunter cocks (bad choice of word, perhaps) his gun. Underneath it says in big bold lettering with a blue background “Date Mature Women!” I don’t think that the woman in the picture could be very mature, or she would not have posed for that photo. I think that in this instance, the word “mature” is a code word for the 20 year old reader, and it means “older than you are”.

In the posting, Dave pulls apart a device that incorporates some electronics and a row of LEDs. The whole thing has the physical form of a fluorescent tube, and is designed to plug into a fluorescent tube fitting.

Dave determines that the electronics consists of a bridge rectifier and res caps up front – a cheap ac to dc conversion which usually comes at the price of low power factor. It seems that we have to be assuming that this device will fit in a light fitting that is equipped with a traditional ballast inductor. I suspect that it would not work in an “electronic ballast” fitting. The power factor would probably not be as bad as it looks on first blush, because the ballast makes this a choke input filter, and together with the power factor correction capacitor, it might come out quite well. This is an interesting situation! It is part of Dave’s style that he skates right past interesting points such as this.

What is the inductance of a fluoro tube ballast? Dave doesn’t say, and I could not find the answer anywhere. If you know where this information is to be found, please let me know. I somehow had the idea that the “on” voltage of a tube is about 70 volts. If this is the case then a 37 watt tube might be modelled like this:

240 volt RMS mains is modelled by a sine generator with an amplitude of 340 volts (root 2 times 240)

Of course the real tube will conduct in both directions, but I have modelled it as a bridge rectifier and then a constant voltage source. As the voltage is constant, the power dissipated in the tube is 70 times the current, or 700 times the volt drop in a 0.1 ohm resistor. R2 and C1 give me an average power signal. I have just varied the value of the inductor until I get 37 watts dissipation in the tube. This is a crude model, but it gives me a figure for the inductance to work with – 1200 mH.

The LED light doover that Dave is playing with is advertised as having 18 watts dissipation. There will be two ways of drawing less power from that inductor. One will be with a lower impedance circuit, with a lower voltage and hence less power, and the other with a higher impedance circuit with a lower current and less power. This is the same as saying that the low frequency volts vs. current characteristic of the inductive source will cross the 18 watts hyperbola in two places. Here is a plot of output power vs the voltage at which the energy is taken from the rectifier.

Note that the vertical scale has a “V” after the number. This is volts as far as LTspice is concerned, but my model makes this voltage represent power. The unit is Watts. The horizontal axis is Load Voltage. Again, I varied this with time in my model, and LTspice has an “s” for seconds after the numbers. The units are volts.

If the supply were DC, this would be a parabola, but it is distorted as when the voltage rises, the proportion of the time that the bridge diodes are conducting is reduced. The maximum power point is where the voltage is about half the mains voltage as we would expect. Eighteen watts is drawn when the voltage at the rectifier output is either 32.69 volts or 262.55 volts. What’s the bet that the lower voltage is chosen! Either way, the light-fitting inductor is a critical component in the circuit. In a different posting, a mate of Dave Jones’ tells us that the inductor can be eliminated (shorted) to reduce losses. I DON’T THINK SO!
1. It looks to me that the impedance of the ballast is a key component and that it works in conjunction with a shunt regulator to determine the 18 watts dissipation.
2. Even if point 1. (above) were wrong, an ordinary switch mechanism has trouble working a room full of fluoro tube fitting power factor correction capacitors. It would vapourize if closed at mains peak with a load of many paralleled rectifier/res-cap combinations!

After the reservoir caps, there is a regulator, that Dave identifies as a buck regulator. On his video, Dave applies a current clamp to the LED array load and shows us the waveform.

He provides commentary to go with this waveform. He says “You can see the current ramp here, with a whole lot of ringing at the bottom of it.” Hmmmmm! I am not sure that I reckon that Dave’s comments are adequate in the face of the interesting details we see on his oscilloscope.

If Dave is right, and this is a buck regulator, then it is running in discontinuous inductor current mode. Discontinuous current can sometimes be just a waste of time – the circuit is not transferring energy when the current is not flowing. However in a lighting application this might not be so. Some say that the human eye responds to the peak brightness in a varying waveform. If we create subjective brightness at the peaks, then perhaps it is just a waste of energy to maintain brightness between peaks. This is an interesting subject that I will save for another day.

The thing about inductors (when not saturating) the rate of change of current is proportional to the voltage across the inductor. In a buck regulator, this means that the rate of rise of inductor current is proportional to the volt difference between the input voltage and the output voltage. When the switch is off, and the freewheeling diode is conducting, the rate of fall of inductor current is proportional to the output voltage with the same constant of proportionality.

This is the Buck Regulator Topology

This is Buck!

We can obtain good estimates of the rates of change of current graphically from the oscilloscope picture.

The green line (Switch ON) rises 4 divisions in three time divisions.
The red line (freewheeling diode conducting) falls 5 divisions in half a division of time. Dave tells us that there are four strings of 24 LEDs. These LEDs have a forward volt drop of about 3 volt. That gives us 72 volts supply to the LEDs. A little algebra with the rates of change of voltage gives us:
Input voltage to buck regulator = 82.26. Of course this input volts to the buck regulator is the same as the output volts of the bridge rectifier that we were considering earlier. But this is the wrong voltage, isn’t it?

I cannot be confident that I have the inductance value of the ballast any closer than within an octave.

If the inductor is 600mH, then the low voltage for 18 Watts is 16.54 volts
If the inductor is 2400mH then the low voltage for 18Watts is 66.97 volts
If the inductor is 3000mH then the low voltage for 18 watts is 86 volts.

Maybe that is it. Dave Jone’s observation about the LED strings might be perfectly correct, and the ballast is actually about 3H.

There is more mystery yet. What is that ringing on the load current waveform when the inductor current ceases? I can simulate it. Compare these:

Not a bad match. What worries me about it is that the damped ring cannot be sustained without a path going somewhere from the switch end of the inductor. When that ring is occurring, both the switch and the freewheeling diode are not conducting. To get the simulated waveform, I had to place 12nF across the switch, a most unlikely circuit feature.

Dave Jones has teased us here with more questions than he had answered. I see the buck converter working as a shunt regulator in conjunction with the high input impedance provided by the ballast. There was no hint of this in his analysis. He is pitching at a viewership that is quite different from this blog’s readership. I hope that neophytes don’t mistake his superficiality for the real thing.