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.

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.

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/

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.

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?

https://en.m.wikipedia.org/wiki/Buck–boost_converter

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?

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

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?

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)?

Presumably the 100 A setting would be for use with a lower-resistance shunt?

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.

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.

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.

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.

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.