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Forum Index : Electronics : Oh, no, MPPT again!
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| domwild Guru  Joined: 16/12/2005 Location: AustraliaPosts: 873 |
Friends, Gordon's cap solution for the F&P means, as far as I understand it, that MPPT has finally been buried for the F&P but may still be of interest for other and/or larger mills. Have just come across an interesting discussion about wind+MPPT on fieldlines. Although I do not have a black belt in electrickery, this subject has always intrigued me, despite the fact, that on smaller mills a gain of 10% or more may hardly be worthwhile. As the Picaxe can drive the PWM for a buck converter, I became even more intrigued. A contributor named David has come up with an extremely simple C-program to drive the PWM for the buck controller and he backs up his research with lots of maths. http://www.fieldlines.com/story/2010/2/14/221919/278 Taxation as a means of achieving prosperity is like a man standing inside a bucket trying to lift himself up. Winston Churchill |
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GWatPE Senior Member  Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Hi Dom, I have just skimmed through the otherpower thread, and there is merit in the discussion. There is a similar open loop solar MPPT made by AERL. I have one of these on my solar panels. These open cct the panel, and use the relationship with the open cct voltage and the MPP voltage to track the maximum panel power. The battery voltage is needed, as the pulse width is basically related to the voltage gain required to match the loaded windmill voltage to the battery voltage. The current that flows, is proportional to the how close the ratio is matched and the losses in the conversion. I see problems of matching with a battery system, as input capacitance affects the open loop response time, and capacitance is required as an interim storage after the rectifiers. I used a lookup table type boost maximizer, with battery voltage compensation with a micro controller, and in another system, a voltage controlled shunt that adjusted the current limiting of a PWM chip. I prefer a boost type controller, as the bulk of the power is passed directly by the main rectifiers, as the boost effectively turns OFF at maximum power, and the boost is in parallel with the main rectifiers. In contrast, The buck controller has to pass maximum power, at a minimum pulse width. This places maximum stress on the switching device, and freewheeling diode at maximum power, and the inductor has to pass all the power. The capacitor arrangements work on windmills up to 2-3kW, even low frequency units like AxFx mills. I suspect that a windmill commercial MPPT is still a while away, in a black box type automatic model. Gordon. become more energy aware |
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| domwild Guru Joined: 16/12/2005 Location: AustraliaPosts: 873 |
Gordon, Thanks for that. If I read his research correctly, he needs to unload the mill to read the battery voltage at rest (?). When I charge my battery on the tractor I notice a slow drop from the charging voltage of around 14V to the "at rest" voltage of 12.7V. The time of the drop, IMHO, is too long to unload the mill for and could lead to a run-away. Is he right or am I wrong? Taxation as a means of achieving prosperity is like a man standing inside a bucket trying to lift himself up. Winston Churchill |
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| GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Hi Dom, The battery voltage will be just the voltage it is at the time normally. The windmill will be unloaded, and the voltage measured. 60% of this will be a good loaded voltage. Calculate the ratio of this voltage to the battery voltage. Make the pulse width this ratio. Choose to slightly underload the windmill, as this allows the windmill to control the process. Measure and adapt at least at 1 second intervals. Large filter Capacitance on the MPPT input side, will slow down the unloaded mill rewsponse, and result in poor tracking of the wind energy. Gordon. PS: open loop solar MPPT, only sample every 45secs or so. At least the windmill has inertia, so the energy during the time required to sample is not lost, so a faster sample rate is workable. become more energy aware |
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| domwild Guru Joined: 16/12/2005 Location: AustraliaPosts: 873 |
Gordon, Thanks for that. This begs the question why there are so few cheap commercial MPPT units available, which combine solar/wind, as there are table methods (not generic, but biased for specific models) and other methods, like the ones you mentioned? Taxation as a means of achieving prosperity is like a man standing inside a bucket trying to lift himself up. Winston Churchill |
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| domwild Guru Joined: 16/12/2005 Location: AustraliaPosts: 873 |
Gordon, Another question, if I may? " ... capacitance is required after the rectifiers as an interim storage ...". Is this the capacitance, i.e., storage required to stop a possible run-away situation, once the mill is unloaded? If, as you are saying, the voltage is the instanteneous battery voltage, the inertia might just prohibit a sudden RPM increase? The voltage could be read in a few milliseconds. Is that too long? Taxation as a means of achieving prosperity is like a man standing inside a bucket trying to lift himself up. Winston Churchill |
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KarlJ Guru Joined: 19/05/2008 Location: AustraliaPosts: 1178 |
I would have thought a second or two wouldnt hurt, but of course as soon as she's unloaded not only will it pick up speed but unloaded the voltage will jump instantly (up to possibly double), this opens the door for nasty electric shocks and lots of smoke getting out of things sensitive to high voltage Luck favours the well prepared |
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| GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
windmill MPPT This is a boost type unit. Scalable design, and adaptable to buck operation. Gordon. become more energy aware |
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oztules Guru Joined: 26/07/2007 Location: AustraliaPosts: 1686 |
Gordon, I may have gotten this wrong somewhere, but pondering this [quote]The battery voltage will be just the voltage it is at the time normally. The windmill will be unloaded, and the voltage measured. 60% of this will be a good loaded voltage. Calculate the ratio of this voltage to the battery voltage. Make the pulse width this ratio. Choose to slightly underload the windmill, as this allows the windmill to control the process. Measure and adapt at least at 1 second intervals. Large filter Capacitance on the MPPT input side, will slow down the unloaded mill rewsponse, and result in poor tracking of the wind energy[/quote] As I see it, this is just a difficult way to measure current 
In reality, the voltage you measure unloaded is just the loaded battery voltage plus the voltage lost due to IxR losses of the stator. By knowing R, you can measure I and find the same result..... with out the unloading and bothering about capacitance skewing the measurement (or use a separate isolating diode to get over this bit anyway if you still want to do this)..... This begs the question of what are we really doing then? The ratios and massaging of the figures to get the pulse width, can be done with the current measurement alone.. for exactly the same result.....I think....? I mean: By knowing the current, you can get the voltage lost in the stator (IxR) add that to your loaded voltage, and get the same result as unloading it for a small time.... So ((60/100) x ((IxR) + loaded V))/ loaded V = pulse width ratio perhaps?(minus your offset to keep conservative)... That "((60/100) x ((IxR) + loaded V))" should represent 60% of your unloaded voltage... the rest you know. The 60% you mention will be a figure dependent on the R in your first iteration I suspect. A tight R will result in a different "best" percentage ... yes? Or did I miss something in the translation 
..............oztules Edit. I guess you need to add line R to stator R as well. Diode loss V should be a constant. This also means I would probably opt for an op-amp with controlled gain instead of a pic.... constant control. Same as Flux did with his booster as I recall. Village idiot...or... just another hack out of his depth |
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| domwild Guru Joined: 16/12/2005 Location: AustraliaPosts: 873 |
oztules, Thanks for that. Pls follow the given fieldline link as this David bases his research on reading the voltage and not amps. I cannot justify his reasoning of why this is so. Once you have read the two pages with its C program and maths it will become clearer why he does it. I hope you did get your supply ship back into service; The Australian had a story about the ferry problem a good few months back. Taxation as a means of achieving prosperity is like a man standing inside a bucket trying to lift himself up. Winston Churchill |
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| oztules Guru Joined: 26/07/2007 Location: AustraliaPosts: 1686 |
After seeing the reference to C programs and Maths... I deliberately didn't look at that link.... up until now I was happy in my ignorance 
Somehow, the edit turned up below here 
............oztules Village idiot...or... just another hack out of his depth |
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| oztules Guru Joined: 26/07/2007 Location: AustraliaPosts: 1686 |
Hmmm.... quietly curses Dom...... Ok, they are one and the same thing really. (I must say I needed to take off the shoes .....I ran out of fingers )
[quote] Meat of the Matter RPM = V * TSR / (pi * D) * 60 => V = RPM * pi * D / TSR / 60 Pp 1/2 * rho * V^3 * pi * D^2/4 * Cp Voc = Vrpm * RPM I = Pg / Vg = Pp / Voc (non-intuitive, but work it out) Vgl = I * Zg Vg Voc - Vgl Vll = I * Zl Vl = Vg - Vll For a buck converter, Vb = Vl * d => d = Vb / Vl For a boost converter, Vb Vl * 1 / (1 - d) => d = 1 - [1 / (Vb / Vl)] The important thing is to calculate Vb / Vl; then we can calculate d whether for a boost or buck converter. The following assumes a buck converter, but the method would work for either. Working backward, duty = Vb / Vl = Vb / (Vg - I * Zl) Vb / [Voc - I * (Zl + Zg)] = Vb / [Voc - Pp / Voc * (Zl + Zg)] = Vb / {Vrpm * RPM - [Pp / (Vrpm*RPM)] * (Zl+Zg)} = Vb / {Vrpm*RPM - [(1/2*rho*V^3*pi*D^2/4*Cp)(Vrpm*RPM)] * (Zl+Zg)} = Vb{Vrpm*RPM-[(1/2*rho*(RPM*pi*D/TSR/60)^3 pi*D^2/4*Cp) / (Vrpm*RPM)](Zl+Zg)}[/quote] So it appears that Vl is the thing for him to find..... Vl = line volts at converter (Vg - Vll) The sad thing is, Vl is only the voltage at the genny minus the losses in the cable in the line due to line impedance. So his "{Vrpm*RPM-[(1/2*rho*(RPM*pi*D/TSR/60)^3 pi*D^2/4*Cp) / (Vrpm*RPM)](Zl+Zg)} could be found in a heartbeat by measuring the I and multiplying by the Zl (line impedance... or R will do).... and subtracting from Vg?..... or if youre lazy like me...... you could just measure it perhaps???? This does not mean I agree with the hypothesis promulgated by either Dave or Gordon.... just different ways to implement their calcs..... The guts of it is this line : [quote]For a buck converter, Vb = Vl * d => d = Vb / Vl [/quote] Vb is nasty simple, Vl is easy to calculate (or measure directly) by a lot of means (see above). It makes sense that this d=Vb/Vl works to make life decent for the stator. Gordon uses stator loss, this fellow uses [Voc - I * (Zl + Zg)].... both current dependant. So guess how I would measure for it???... But I'm still to be convinced deep down that this will accurately reflect the best TSR outcome. Luckily, the curve is so forgiving, that close enough is good enough I suspect It is interesting he believes that CP is a constant.... this is the very parameter we are fighting to hold together. He has to measure V because he relies on Voc = Vrpm * RPM. In his case, he cant find RPM if he can't find V. Or put simply.... he could have (and should have) stopped here: duty = Vb / Vl Vb=measure battery voltage and Vl= measure converter input voltage. Should ba a damn simple program I should think. ...........oztules Village idiot...or... just another hack out of his depth |
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| GWatPE Senior Member Joined: 01/09/2006 Location: AustraliaPosts: 2127 |
Hi Karl, there is no chance with a boost cct for the condition you think will happen, to actually happen. Remember that there is always the normal rectifier bank in parallel, and also the flyback diode in the boost unit directs current to the load as well, even when the boost is turned OFF. The only way the voltage can rise too high, is if there is no load connected, and there is no overvoltage protection. Both these conditions would likely to be operator caused, if they occured, unless there was catastrophic failure with the power cable being cut ahead of any electrical protection, and then there would be nothing to let the smoke out of. Gordon. become more energy aware |
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