conversion chart for classic 200

[eajonesyk2]

Hello,

I was informed by a member over at the NAWS forum that the Charge Controller companies have conversion data available that would allow me to figure out conversion losses with a particular charge controller. I am looking at the classic 200. If my array was wired at about 108 volts (6 145 watt panels in series, 18.7 vmp)  and fed in to a 12 volt battery bank (12 volt system voltage) via the classic 200, what kind of losses will I see from the down conversion process in the charge controller?  Is there a chart or calculator I could use to find this answer?

[Vic]

Hi ea..

Believe that the PV modules are DM Solar 145s:
Vmp 18.7,  Imp 7.75,  Voc 22.3

Recall that you are at 5200 ft elevation,  but do not know the lowest recorded temps for that location.

With 6 of these PVs in series,  you certainly in the MN Cl 200 range.  Know that you have a fairly long run from the PV to the rest of the electronics.  It is possible to locate the CC,  batts and inverter near the PVs and run the AC output of the inverter to the point of use.  Believe that you are concerned about cost of wire vs all else.

Believe that the Efficiency Chart is not yet available.  It is correct that converting from 108 ish string V to 12 v batt,  is not as efficient as converting this V to a 48 V bank.   It is known that this  chart data is difficult to collect,  and often some other mfgs are very optimistic with their "data".  Furthermore,  there appears to be NO industry standard for the collection of this data.

But,   if the string V is 108ish V,  given your location,  you really have very few alternatives to the Classic 200.

Little time now,   more later,   Vic

[eajonesyk2]

hi Vic,

Yes they are the dmsolar 145s at 5200 ft ele. Cariboocoot suggested that between the conversion loss and v-drop that I might lose 6% of max power (not sure I know what that means yet). I wanted to run 10 awg wire to save money and thought that the higher the voltage from the array the smaller the wire I could get away with. According to a v-drop calculator the 100ft run with 10 awg at 108 volts would be less than 2%. So the remaining question would be how much power would be lost in the controller to convert the 108 volts to 12 volts for the batteries? Searched high and low but can't find that info anywhere.

Also when I use a v-drop calculator should I use the one way distance of the wire 100ft or the loop distance 200ft?

[eajonesyk2]

hi again,

I forgot to mention that yes I had the idea of locating the controller, batteries and inverter below the array, possibly in a tiny shed of sorts. I didn't get any feedback from the naws group on this subject so I thought it might not be a good idea. One problem I could foresee is the cold temps and keeping the batteries at a reasonable temp in the winter. Much of the winter is below freezing day and night. Security is also a big concern.

[Vic]

Hi eajones,

Sorry,   just scanned the NAZW&S post,  forgot that this remoteing of the electronics had been mentioned.  If you can keep the batteries fairly fully charged in the Winter,  they should do well in cold.  They really hate the heat,   tho.

Regarding the Volt Drop calcculator,  normally,  it is speced right up front what length to use,  RT or One Way.

Believe that niel,  a member here,  and there,  has a link to his Drop Calc program.

It is quite possible that the conversion losses added by using higher string V,  12 V batt,  and the Classic 200 will add one or perhaps even 2% additional conversion loss.  I am no expert on this.

There is a large Solar Trade Show in FL,  and believe that a number of key MN employees were there,
EDIT:  The show ends on 9/13 ...     Perhaps,  in a day or so,  some of the MN wizards will respond.

Because the Classic is such a new product,  it benefits from a lot of experience that the founders of MN have gained (they also founded OB Power,  and were at Trace/Xantrex),  So conversion loss is one factor,   but  believe that the Classic is very good at maximizing the power extracted from the PV (or Wind,  Hydro,  etc).  So it is not only conversion efficiency.

EDIT: Efficiency Curves have been a topic here in the past,  here is a link to one:
http://midnitesolar.com/smf_forum/index.php?topic=91.msg589#msg589

If you search this site  for " efficienty curves " (no quotes)  you should find more.

More later,  Vic

[eajonesyk2]

Here are my latest thoughts on the set up. According to the voltage drop calculator if I use the 10 awg wire it will only cause a 1.6% voltage drop with 6 panels in series. So even if I lost 2% more in the controller conversion, I am looking at less than 4% power loss. Does that sound about right?

So the original Cariboocoot approved setup was:

4 dm solar 145 watt panels 18.7 vmp 7.75 amps  $640 (already bought these)
Tristar 45 mppt controller $400 plus another $125 for display, battery temp sensor and shipping
8 awg wire  250 foot roll $450 plus shipping?
Fed in to 12 volt system (battery bank)

This would give me about 31 nameplate amps out of the contoller for $1615 or $52 bucks an amp

Scenario two:

6 dmsolar 145 watt panels $960
Midnite Classic 200 $610 with free shipping
10awg wire 250ft roll $240
Fed in to 12 volt system

This would give me about 44 (46 minus 4%) nameplate amps out of the controller for $1810 or about $41 bucks an amp.

I think the Classic controller would also give me more flexibility for future expansion although I don't think I would ever need more than six panels. I could always upgrade the battery system voltage to 48 volt with some new batteries and a inverter purchase. But the panels are so cheap right now that it seems to be the cheapest upgrade to do.

[eajonesyk2]

Also thanks Vic for that link to the post regarding efficiency curves. As bob stated in the post some of the efficiency is lost regardless of the array voltage due to "idle losses", etc. But that was interesting about the losses due to higher voltages causing the unit to run warmer. Good catch finding that. Since I would be running a fair amount of wattage, my guess is I would not notice the losses from running the higher voltage into the 12 volt system.

[vtmaps]

According to the voltage drop calculator if I use the 10 awg wire it will only cause a 1.6% voltage drop with 6 panels in series. So even if I lost 2% more in the controller conversion, I am looking at less than 4% power loss. Does that sound about right?

You are confusing voltage drop with power loss.  With solar panels the voltage drop is proportional to the power being produced by the panels.   The power loss is proportional to the square of the power being produced.

That is why in many/most situations it is better to use the lower voltage array configuration... only at the highest power are the line losses greater than the controller losses. 

The idea of a power shed is good.  If you do that, be sure to run two conduits... one for AC power and a separate conduit for control and monitoring.   Another advantage of a separate power shed (even if it is right next to your house) is that you won't have to hear the noise (hum, fans, clicking relays) of your electronics.

--vtMaps

[niel]

vtmaps,
although it is somewhat small in a well designed system, a v drop does pose a power loss. when one is putting out the high power via higher current (which you would want) the loss increases just as you state and it is less when the current is smaller. make no mistake about it that it is a power loss as whatever the v drop is multiplied by the current passing at that time is the power lost at that time. it is fluid just as you say in that it changes with the output of the pv(s). if the wire was not present and the pv output went straight to the load more power would reach the load to be available for it as the wire is a resistor even though it is a smaller one.

[eajonesyk2]

Hi vtmaps and niel,

Thanks for adding to the discussion. I am still having a hard time understanding the differrent terms and concepts with all this solar stuff. I think I understand the voltage drop that occurs in wiring and have done the calculations on running 10 awg from the array to the controller 100 feet away (200ft of wire). With the six dm solar panels wired in series the voltage drop was under 2%. So i am trying to get a number (%) of the loss that is going to occur in the controller? I have exchanged a few emails with Tom C at Midnite and I think he may be getting back to me with a number I can work with.

A gent over at the NAWS forum suggested 6% but I think this was including the voltage loss from the wire. I have read other post on this forum suggesting the Classic controller losses are greater than 6% by itself.

If I can get a somewhat firm number on the controller losses from running a high voltage array into a 12 volt system then I will be able to make decisions on how to proceed.

[eajonesyk2]

Ok looking back at the posting Vic linked to. boB provided this answer for somebody who had an array running 140vmp into a 48 volt system. He suggested 96 to 97% efficiency which is super in my book.

"Hi Stephen dv...    No efficiency "curves" quite yet, but depending on how much power you are putting out at those voltages,
the efficiency will be around 96% to 97% or so.  I am stating sort of on the low side here (as far as I remember), as I have seen higher but do not want to OVER state.
At low output powers, say, less than a couple hundred Watts, you might get below 90%, but at higher powers, you will be in the high 90's percentage wise.

I am also assuming a Classic 200 for this usage where the MPPV can be around 160 Volts with a 200V Voc

At low output powers, the idle losses become more important and eat into the efficiency numbers.
Also, the higher voltages eat into efficiency somewhat because the controller  will run warmer.

I hope to get efficiency curves for many different powers and voltages real soon.

boB"[/i]


So could I expect similar efficiency for my planned system of six 145 watt panels wired in series (18.7 vmp) going in to a 12 volt system?

[eajonesyk2]

wrong voltage drop calculation.

I think my v-drop calculation was off because I didn't use the total length of circuit (200ft instead of 100ft). so using 10 awg will yield a voltage drop of just over 3%. I would be happy with that if the controller efficiency is still decent converting the 108 or so volts down to the 12 system voltage.

[vtmaps]

I am still having a hard time understanding the differrent terms and concepts with all this solar stuff.

With the six dm solar panels wired in series the voltage drop was under 2%. So i am trying to get a number (%) of the loss that is going to occur in the controller?

Solar panels essentially provide full voltage from dim light to full sunlight.  The current is what varies with insolation.   

Therefore to calculate the voltage drop in the cable use ohm's law.   Your cable has a resistance of 0.2038 ohms.  With 6 panels in series (Vmp= 112.2) the current at full power is 7.75 amps.  Thus the voltage drop is 7.75 times 0.2038 = 1.579 volts.  That voltage drop is 1.41% of your Vmp. 

The power that lost in the cable is the current times the voltage drop = 7.75 times 1.579 = 12.241 watts.  That is 1.41% of your 870 watts.

If you drop to half power use the above formulas, but that 7.75 amps is now 3.875 amps:

thus, the voltage drop is now 0.790 volts which is a 0.70% voltage drop.  The power lost is 3.06 watts which is 0.70% of 435 watts.

When you are producing only a few hundred watts of power, the efficiency curves for the classic may be in the low 90's.  Lets say 94% efficiency.  That means 6% power loss.  Even when it is running at 97% efficiency, that's a 3% power loss.   How do these losses compare to you cable losses?

If you calculate the power losses for two strings of three panels (Vmp=56.1  Imp=15.5) you can see what happens to the cable losses.  Unfortunately, I don't have the efficiency curves for the classic, so I don't know what will happen in the controller. 

I am familiar with the Outback controller efficiency curves.  When I compare configurations on that controller, I am surprised at how often the lower voltage array configurations win out over the higher voltage configurations.

Bottom line: I think that 200 ft of 10 gauge cable is a bad situation, and I suspect that putting 6 panels in series is making the best of that bad situation.  A better situation would be heavier cable and 2 strings of three panels in series.

--vtMaps

[eajonesyk2]

Thanks vtMaps,

Very good info thank you. Now i can skip the calculators and do the wire loss math myself. As a matter of fact I did the math on the six panel system using the 8 awg that was recommended by cariboocoot on the naws forum. If I did the math right, my wire loss would be about 7 watts instead of of the 12.241 you came up with using the 10 awg.

Now you mentioned the 10 awg as being a "bad situation" but if our math is correct I would only be saving about 5 watts or so of power by going with 8 awg over 10 awg. Granted percentage wise this a 45% less loss in power but it is only 5 or so watts. The additional cost of 8 awg wire over 10 awg wire is over $200. So that works out to about $40 a watt. Does this sound about right?

I originally wanted a system that could generate 1 kwh of useable power a day during the warmer months and less in the colder months so I was ok with the 4 dmsolar 145 watt panels I already have. A morningstar tristar 45 mppt was recommended but would not have allowed any more panels to be added in series.

I understand a higher system voltage will allow the controller to work more efficiently but it would require new batteries and a new inverter and I am not sure the cost would justify the benefit. That is why I was trying to find out what the losses would be in the controller. I think you are probably close with your 94% estimate.

So lets use the numbers we have so far and look at the potential power from the system. I will start off with 6 dmsolar 145 watt panels wired in series for 870 nameplate watts.  Let's say actual watts may be closer to 670. Running 10 awg wire from array to Classic 200 controller will rob about 13 watts so I am down to 657 watts. Using a 94% efficiency rate for the controller will bring my power down to 617 watts. Let's say the xantrex prosine inverter acheives a 80% efficiency rate. That will leave me with very close to 500 watts of useable power per hour of full sun. In closing two hours of full sun a day would give me my 1kwh of power needed.

Dose this sound reasonable? 

[vtmaps]

Now you mentioned the 10 awg as being a "bad situation" but if our math is correct I would only be saving about 5 watts or so of power by going with 8 awg over 10 awg. Granted percentage wise this a 45% less loss in power but it is only 5 or so watts. The additional cost of 8 awg wire over 10 awg wire is over $200. So that works out to about $40 a watt. Does this sound about right?

yes, but you are comparing 10 gauge to 8 gauge, but only looking at 6 panels in series.  What happens if you make two parallel strings of three panels in series with 8 gauge wire?  Either of us can calculate the cable loss, but neither of us knows what the controller will do.  The only thing for sure is that the controller will run cooler and more efficiently.   btw, two strings through 8 gauge wire will lose 30.8 watts at full power.  That's about 18 watts more than one string through 10 gauge.  Will the increased efficiency of the controller make up for that?  (I don't know)  Remember, these numbers are at full power.  The cable losses will be much lower at more typical power levels but the efficiency gains in the controller will persist at the more typical power levels.

When I said "bad situation" I meant that trying to push 800 watts through 200 ft of 10 gauge wire is a bad situation.  Wiring your 6 panels in series is probably the best way to deal with 10 gauge wire.  Better would be to use heavier wire and then you would not have to wire 6 panels in series.

--vtMaps