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    • CommentTimeFeb 23rd 2016 edited
    If you look at a data sheet for a module, you usually get two voltages stated.
    Open Circuit (Voc) and Maximum Power (Vmp)
    Regardless of the amount of light hitting the module, the current (A) will be zero until it is connected to a load (your water heater).
    Power is Voltage times Current.

    So taking a typical 250W module, the Voc may be 37.6V and the Vmp may be 30.9V.
    The current, when the load is perfectly matched (the resistances are the same), could be 8.1A
    So at maximum power (when the light levels are high enough) you get 30.9V x 8.1A = 250.29W
    If you get a lower light level, but keep the load the same, the voltage will stay the same, but the current will reduce, say to 4A
    30.9V x 4A = 123.6W

    Now if you reduce the load resistance by half, the supply voltage will increase, say to 34V.
    34V x 4A = 136W

    This implies that at different light levels you need to vary the load resistance, which varies the power available.
    Now it would be possible to design an electrical heating element that can cope with this, they already make variable resistors that do it. All that need to happen is that every few millimetres a tapping is taken off the resistive element and a bank of switches constantly varies where the current is introduced. This is possible but complicated and expensive.
    So the easy way is to have fewer option and accept that it is a compromise.

    This compromise really has to be set to the marginal times, not the modal times. There is plenty of sunlight in the summer, so you can afford to be less efficient.
    There is not enough in the winter, so why bother.
    This just leaves a 2 or 3 months to worry about. So optimise for then.
    But remember that you are limited by the amount of available power, which varies from 0 to 250W (per module). No amount of power point tracking is going to get you more, and most of the time, you will get a lot less.
    A quick look at my solar data shows that 55% of the daylight time, my levels are 100W.m^-2 or below. So a 250 PV module will at very best, only produce 31W, and probably less.

    I am not saying that it is not worth doing, just remember the limitations and work to them.
    • CommentAuthorEd Davies
    • CommentTimeFeb 23rd 2016 edited
    Posted By: SteamyTeaIf you get a lower light level, but keep the load the same, the voltage will stay the same, but the current will reduce, say to 4A
    If by “keep the load the same” you mean keep the load resistance the same then this is wrong - the voltage will decrease. Or are you one of these Ohm's law deniers?

    Posted By: SteamyTeaSo at maximum power (when the light levels are high enough) you get 30.9V x 8.1A = 250.29W
    So the resistance is 30.9 V/ 8.1 A~= 3.8 Ω.

    At a light level giving 4 A into this resistance the voltage would be 3.8 Ω × 4 A = 15.2 V and the power 15.2 V × 4 A = 60.8 W.

    I thought this was all sorted out on the first page of this thread.
    • CommentTimeFeb 23rd 2016
    Don't the internal resistance of the modules vary with the light levels? By how much I am not sure, would need to read that of an IV chart.
    Why you need to vary the load resistance constantly to get maximum power.
    I seem to remember that when we did this at University, we got some strange results because we could not control the solar power very well.
    • CommentTimeFeb 23rd 2016
    Posted By: Ed DaviesIf by “keep the load the same” you mean keep the loadresistancethe same then this is wrong - the voltage will decrease. Or are you one of these Ohm's law deniers?
    I think I explained that badly.
    • CommentTimeFeb 23rd 2016
    Right, I think this may help explain what I mean, or where I am totally wrong.
    It is the area between the curves, the IV and the PV that is important, and they are far from linear.
      Power and IR Curves JPEG.jpg
    • CommentAuthorEd Davies
    • CommentTimeFeb 23rd 2016 edited
    Posted By: SteamyTeaDon't the internal resistance of the modules vary with the light levels?
    Thinking about the internal resistance of the module is, I think, the source of your confusion.

    The modules are essentially constant-current sources where the current is set by the light level - more light gives more current, of course. As constant-current sources they have effectively infinite internal resistance. [¹]

    V = IR
    V + ΔV = (I + ΔI)R
    IR + ΔV = IR + ΔIR
    ΔV = ΔIR
    R = ΔV/ΔI

    but it's a constant current so ΔI = 0 so R tends to infinity and beyond.

    When a pure constant-current source is fed into a very high resistance the voltage goes way up. This doesn't happen in the real world because there are various limits to what voltage sources can produce. [²] With PV panels the main limit is that the P-N junctions in each of the cells gets to 0.6 or 0.7 volts or so and start to forward conduct effectively shorting out some of the current generated. That's why the current drops off so rapidly approaching and passing Vmp.

    So what you want to do is, for any light level, run the modules so that the output voltage is just about Vmp. That voltage does decrease a little bit as the light level goes down but not much. If you ran the modules all the time with a load resistance aiming to keep the voltage at, say, 0.9 × nominal bright-sunlight Vmp you wouldn't do too badly, I think.

    [¹] When I was a research student in the late 1970s I built prototype components of a bus-type network to work at roughly mega-bit per second rates over twisted pair cable. We (my supervisor and I) were well aware of coax Ethernet but didn't have access to anything of the sort ourselves (just read the Xerox PARC papers) and this was well before twisted-pair Ethernet came on the scene though there was some high speed point-to-point twisted pair stuff around including, IIRC, token ring. Anyway, to allow for transmission at times when there were already signals on the wire we used constant-current drivers to allow voltage addition. In a nod to Douglas Adams, whose Guide was fairly new then, I referred to them as infinite-impedance drives.

    [²] E.g., for not-so-well-designed bench power supplies electrolytic capacitors can get all explody: http://edavies.me.uk/2013/07/lifepo4-charge/
    • CommentTimeFeb 25th 2016
    Posted By: Ed Daviesbut it's a constant current so ΔI = 0 so R tends to infinity and beyond.
    It is that bit I have trouble with. I can't find any evidence from a quick search that PV modules (and the connects etc) have anything other than a relatively low resistance.
    • CommentAuthorEd Davies
    • CommentTimeFeb 25th 2016
    Indeed, they don't have much in the way of series resistance, as such. If you have an open-circuit panel and you connect a load across it which has a high-enough resistance that it causes somewhat less than Imp to flow then the voltage will only drop a bit so it looks like the series resistance is small.

    If, on the other hand, you connect a smaller-resistance load such that a simple Ohm's law calculation from the Voc would say that more than Isc would flow then the current will limit to somewhere between Imp and Isc and the voltage will drop significantly so it'll look like the series resistance of the module is high.

    If you're approximating MPP tracking (as we're discussing here) then you're better off approximating on the low-voltage/high-current side as your blue power curve shows (it drops off steeply once you go over Vmp) which means you're best operating in the regime where the apparent series resistance of the module is high.

    Still, I think that's a bad way of looking at it: a Thévenin-equivalent circuit (constant voltage source in series with a fixed resistor [¹]) can only approximate the behaviour of the module over a narrow voltage/current range.

    [¹] https://en.wikipedia.org/wiki/Th%C3%A9venin's_theorem

    Slightly less confusing would be a Norton-equivalent circuit (constant current source with a resistor in parallel [²]) but even that's not so helpful as the “resistance” (forward conduction of the P-N junction) is, as you rightly point out, very non-linear.

    [²] https://en.wikipedia.org/wiki/Norton's_theorem

    This whole thing of matching source and load resistance/impedance might well be a good approximation for nice linear devices (as you'd hope that, e.g., audio stuff is) but really messes you up if you think it'll work over a wide range with non-linear things.
    • CommentAuthorEd Davies
    • CommentTimeFeb 25th 2016 edited
    Posted By: SteamyTea: “There is not enough in the winter, so why bother.”

    A panel inclined at 60° with a good view of the southerly horizon in the south west of England generates for an average of about 1.27 hours per day equivalent says PVGIS (1.27 Wh/Wp) in January (the worst month) so to get the typically suggested 3 kWh/person/day you need 2.36 kW/person. Call it 10 255 W panels at £125 each [¹] so £1250 per person plus whatever's needed to get the electricity into the water.

    Say you set up a system for three people (e.g., two adults + one child) then in October you'll have spare (2.09 - 1.27) * 3 * 3 = 7.38 kWh/day and in November (1.56 - 1.27) * 9 = 2.61 kWh/day for an excess of 7.38*31 + 2.61* 30 = 307 kWh. With similar (actually slightly larger) excesses in the symmetric spring months (February and March) which are typically colder but sunnier you also get a substantial proportion (maybe a bit less than half) of the energy needed to heat your 100 m² Passivhaus.

    That's why one might bother.

    [¹] http://www.bimblesolar.com/255w-gb-sunsolar-b-grade
    I was thinking again about bridging electronically between the PV dc output and any normal AC resistive load (immersion, ufh, storage heater, towel rail...) by using a switch mode converter.

    This uses the PV input DC to 'charge up' an inductor. A semiconductor switch cycles on and off many times per second to partly discharge the inductor into the load. The output voltage can be continuously adjusted higher/lower than the input voltage, by adjusting the ratio of on:off time.

    In this way the input voltage can be set to whatever best suits the PV MPP, while the output voltage is continuously adjusted to dissipate exactly the right amount of power in the load to match all the available PV power when the sunshine varies.

    This would avoid needing multiple bespoke immersion elements, etc.

    The pulsed DC output is usually smoothed with a capacitor filter, but I guess it could be inverted to ac if the load preferred it (mechanical switch contacts etc)

    I guess many folks already thought of doing this, but I couldn't immediately spot one on the market. This principle is used in mppt PV battery chargers and PV diverters, so the parts must be available at the right power ratings.

    Has anyone found such a thing available? I don't have enough skills or fire insurance to build one myself - has anyone else tried?

    • CommentAuthorEd Davies
    • CommentTimeApr 11th 2016
    Indeed, using a buck or boost converter to go from some approximation of the PV Vmpp to a suitable current into the immersion would be ideal. There are plenty around which would do this from the electrical-power point of view but normally they control the output voltage or, in some cases, the output current (e.g, LED drivers or battery chargers in the constant-current phase of the charge). I don't know of any, though, that are set up to control the power transfer to get to a target input voltage.

    It's what the two devices I reference at the bottom of this blog post do: https://edavies.me.uk/2012/11/pv-immersion-gotchas/ . But they're quite expensive for the relatively low power they handle - probably OK in the contiguous US where the days are longer but here we need to make the maximum use of short bursts of sunshine (or reduced darkness) in winter so would probably need somewhat bigger panels.

    Perhaps the thing to do is to get an off-the-shelf programmable constant-output voltage or current device and add a separate controller to program the output to get the required input voltage.

    I'm not convinced you actually need to do the constant scanning and adjustment that MPPT controllers do. I think it might well be sufficient to just aim for a fixed voltage, about 0.9Vmpp.

    Still, if you're setting things up from scratch then multiple elements would likely work out cheaper.
    • CommentAuthorjohnuready
    • CommentTimeJan 24th 2017
    1 year on.... I joined the raspberry pi group in Cambridge and started the design of the software to interface between the panels and the immersion using the step up and down over three immersions in one unit.

    The immersion I used is a 36 Volt dc with 3 * 400W coils. The coil use compared to a built to order was much cheaper and I managed to get 6 steps/combinations out of the 3 No. coils.

    Have installed 4 panels on the roof wired in parallel.

    Switching between coils using solid state relays.

    I'm at the stage:

    Panels wired to 3 coils in parallel for max power transfer - just to test out

    Software getting there

    Solid State Relays ready to be wired

    Anybody any ides on why:

    I'm writing the software to keep track of the kWh generated but find that my numbers are not working out as I thought. The coils have a resistance of 3.5 ohms each therefore 3 in parallel are 1.17 ohms.

    I need an accurate resistance to achieve a kWh calculation that I can continually accumulate but my standalone monitoring hard ware on the 4 panels and 3 coils in parallel show

    21.73 volts
    41.79 A
    908 power
    Water temperature around immerion 38C

    That means the resistance of the coils or the combination of panels and coils (using V2/P) come out as 0.5 ohms.

    Any ideas why the shift in resistance?
    • CommentTimeJan 24th 2017 edited
    Posted By: johnuready3 in parallel are

    1/Rt = 1/R1 + 1/R2 + 1/R3

    Just realised that I made a mistake in the formula and I forgot to divide 1 by Rt (put right now)

    So your total resistance is right.
    • CommentAuthorjohnuready
    • CommentTimeJan 24th 2017
    I must be missing something,

    My coils are rated at 400w @ 36 volts therefore the Resistance is 3.5 ohms per coil

    Putting 3 No. coils in parallel drops the resistance to 1.7 ohms

    But my real life connection showing power and voltage works out at 0.5 ohms
    • CommentTimeJan 24th 2017
    Is it anything to do with the internal resistance of the PV modules.
    Ideally they should be equal to produce maximum power (I think).

    Ed will be along sometime, he understands all this a bit better.
    • CommentAuthorEd Davies
    • CommentTimeJan 24th 2017
    Hmm, odd. I get slightly lower resistance than you but not as low as 0.5 ohms.

    V = IR
    I = V/R
    P = IV = V²/R
    R = V²/P = 36² / 400 = 3.24 ohms per element so 1.08 ohms for the three in parallel.

    It's nothing to do with the elements having a positive temperature co-efficient and being run at somewhat less than full power, is it?
    • CommentAuthorjohnuready
    • CommentTimeJan 24th 2017
    It could be my Ebay (Volt, Current and kwh) reading device. The voltage is correct I',m reading on the hardware as i have checked that with a voltmeter but could I have shunt that's is not that accurate perhaps?
    • CommentAuthorEd Davies
    • CommentTimeJan 24th 2017
    You ought to be able to check the voltage across the shunt with a multimeter on, say, the 200 mV range. Shunt resistances are usually chosen for something like 50 mV across them at maximum current so you should at least be able to see if it's right to considerably better than a factor of two.

    What are the specs of your panels? Is a bit over 10 A per panel plausible in winter sunshine? Good if it is, usually they're about 6 to 8 amps in nominal conditions. How about trying with just one panel and a meter on the 10 amp range to check the short-circuit current (assuming the panel nominal Isc is greater than that)?
    • CommentAuthorjohnuready
    • CommentTimeJan 29th 2017
    After many days of limited sunshine at last I managed to look at the meter:


    Its a dual current 50 /100 amp set by a push button. My unit was set to 100 amps hence the double values. Now back on track, hopefully soon able to check my software to switch up and down through my coils.

    I currently have the new 1 Kw PV DC hard wired to the 1.17 ohm coil matrix ( 3 in parallel ) producing max about 850w
    • CommentAuthorEd Davies
    • CommentTimeJan 29th 2017
    Presumably the 100 A setting would be for use with a lower-resistance shunt?
    • CommentAuthorjohnuready
    • CommentTimeJan 29th 2017
    I think its all soft, the same shunt for 50 amp or 100 amp good value for money for a cheap accurate DC monitor. It needs a minimum of 6 volts to start readings.
    • CommentTimeJan 30th 2017 edited
    On 25.1.16
    Posted By: fostertomWhy can't an immersion element just accept whatever voltage is available (up to its rated max, 250v or whatever)?
    Posted By: Ed DaviesTom, see the second post in this thread - it's very inefficient at low light levels. (And a lot of the rest of the first page, for that matter.)
    Was the controversy there, between Ed and Sprocket (agreeing with me), really settled?

    It still doesn't feel right that half or more of the power in the circuit gets dissipated in the PV rather than in the immersion element, when the PV has very much higher resistance than the immersion.
    • CommentAuthorEd Davies
    • CommentTimeJan 30th 2017
    Posted By: fostertomWas the controversy there, between Ed and Sprocket (agreeing with me), really settled?
    Sprocket was “converted”:

    Posted By: Sprocket on Feb 24th 2013(I'm lagging behind a bit here)
    Doh! Of course. Thanks Ed & Jim for the patient explanations.
    • CommentAuthorjohnuready
    • CommentTimeJan 30th 2017 edited
    I am working on my immersion coils, 1.17 , 1.75, 3.5 , 5.25 , 7.0 and 10.5 ohms. Hopefully tracking the lower light levels for max power transfer to match the coil resistance. That should get closer to max power transfer from PV, I hope.
    • CommentTimeJan 30th 2017
    Posted By: Ed DaviesSprocket was “converted”
    Thanks Ed - but I couldn't see what he 'got', and I still don't. I read your blog and as presented it seems to make sense but ...

    "Say the current available drops to 5 A. With that going into the 10 Ω immersion the voltage will now be 50 V and the power being transferred into the water just 250 W.

    In other words, when the sunlight drops to half its brightest the heat getting into the water drops to a quarter. Though we've got nearly 500 W of power available from the panels (5 A times nearly 100 V) we'll actually only be getting just over half of that. The other half (another nearly 250 W) will be lost as heat in the panels."

    Maybe the last sentence - could there be another calc, matching the above one for the immersion element, but proving rather than deducing the heat dissipated in the PV?

    I mean, how can the PV be dissipating equal heat to the immersion element, or more, as the 2 principal items in a single circuit, if the PV has much higher resistance than the immersion? Heat generation equals low resistance (or is it resistivity) - a 'controlled short-circuit', doesn't it?

    The voltage drops redistribute themselves around the circuit, or something?
    • CommentTimeJan 30th 2017
    Solar cell efficiencies are only 15% or so, so the vast majority of the solar energy is dissipated as heat in the solar panel in any case. If nothing is connected to the solar panel then ALL the solar energy is dissipated as heat, and the same happens if the solar panel is short circuited. When there's anything other than the 'ideal' current for the sunlight intensity being drawn, the efficiency will be somewhat less than its maximum.

    Solar panels are [imperfect] diodes - active components - not simple passive resistors.

    Posted By: fostertommean, how can the PV be dissipating equal heat to the immersion element, or more, as the 2 principal items in a single circuit, if the PV has much higher resistance than the immersion?

    The solar panel and the immersion are in series, with the same current flowing through both. So if the solar panel is dissipating more power (as indeed it is), then its resistance must be greater than that of the immersion.

    Try http://www.pveducation.org/pvcdrom/modules/heat-generation-in-pv-modules and other pages on the site.
    • CommentAuthorEd Davies
    • CommentTimeJan 30th 2017 edited
    It all depends what you mean by the resistance of the PV panel.

    In the regime we're discussing where the load resistance is a lot less than that needed for maximum power transfer, so the panel is almost short circuited, the voltage is low so the forward bias on the diode forming the cell is small so it's hardly conducting and is almost irrelevant and the shunt resistance is also doing less than it might otherwise so the cell is more-or-less equivalent to a constant current source with a series resistor.


    That series resistor is, indeed, fairly small so is quite capable of generating the heat required.

    On the other hand, because it's mostly a constant-current source changing the external load resistance a small amount varies the output voltage nearly proportionally but doesn't change the current much so from that point of view the apparent resistance (perhaps you could abuse the word slightly to call it impedance) looks high.

    It's not a linear device with a simple notion of its unique resistance (even at a single point on the IV graph). Getting these two different views of the resistance/impedance of the PV panel mixed up is a serious source of confusion. Either you need to look at the equivalent circuit in detail (with the effect of both its resistors and the diode) or treat the whole thing as a black box and look at the IV curve of its output (which is what I prefer).

    Also, heat generation doesn't “equal” low resistance. Power is simply the product of voltage and current so you can have a lot of voltage and a small current through a high resistance and get things warm. However, with PV the voltage is limited by the forward conduction of the junction diode so, yes, then you need a relatively low resistance to get much useful power.
    • CommentTimeJan 30th 2017
    Posted By: djhIf nothing is connected to the solar panel then ALL the solar energy is dissipated as heat, and the same happens if the solar panel is short circuited
    Of course - that clarifies.

    Thanks Dave and Ed - am away now, hope to look again properly shortly.
    • CommentAuthorbxman
    • CommentTimeFeb 1st 2017 edited
    It seems that welshjim99 our OP has not been in touch for a while and we are unlikely to learn anything of value from his experience ..

    Fortunately our saviour the Sun is now a little higher in the sky and also put in an appearance for a while today.

    So I was able to take some measurements , from my recently installed system of 8 (JAM6 270) panels directly coupled to a night storage heater that it's self has a recorded resistance of 22 ohms

    When clouds cleared around mid day there was a current approaching 7.5 amps at around 210 volts
    as clouds limited its power both voltage and current fluctuated in unison from approximately 1.5 amps at 20 volts back up to the 7.5 at 200 odd..

    I can not see that a controller would make any significant difference to the amount of Heat that I harvest .

    I maybe wrong and will happily test any devise that anyone feels may work.
    IMO when the sun is not there there is nothing to harvest anyay about .
    • CommentAuthorEd Davies
    • CommentTimeFeb 1st 2017 edited
    Posted By: bxman…as clouds limited its power both voltage and current fluctuated in unison…
    Given that it was feeding into a constant resistance that would have to happen. It's called Ohm's Law.

    …from approximately 1.5 amps at 20 volts
    So 30 watts. But the Voc and Vmp of the panels will only be reduced a bit at low sun levels. You could probably have got something like 1.4 amps at 190 volts at the same level of sunlight so perhaps 260 watts or about 8 times more. That would need the resistance to increase to 135 ohms, of course.
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