DC Voltage Drop

P

PV Noob

Member
Location
Midwest
Occupation
Designer
Sorry if this has been covered before but I am trying to better understand DC voltage drop. When I took my SEI online classes, they said to keep DC VD to 2% or less, but since then I've seen numerous people say VD on the DC side is not an issue. Lots of people say that if the voltage drop is not too much to take you out of the lower end of the MPPT then you are fine.

So my question is, is DC VD just a matter of less overall energy from the lost voltage as opposed to being something operational (ie. something doesnt work and the array/inverter is not functioning at all).
 
My understanding is that DC voltage drop is simply a matter of energy loss, and not a functional issue unless the voltage gets so low it causes something not to function.

Thought about that way, you don't want to aim for a VD of 2% or less, but rather you want the VD that gives you the best overall production per $ of installed system cost.
 
PV Noob said:
So my question is, is DC VD just a matter of less overall energy from the lost voltage as opposed to being something operational (ie. something doesnt work and the array/inverter is not functioning at all).
Click to expand...
Those factors are not in opposition. Vd in the DC conductors is power production loss to heat, and in extreme or borderline cases it could possibly interfere with the operation of the inverter.
 
If it isn't going to push the voltage below the inverters operating window, then it comes down to the economics. Is the cost of upsizing the wires more or less than the price of the energy output gained at the inverter? You have to crunch the numbers.

Years ago there was an article in Solar Pro magazine with a bunch of math arguing that upsizing the wires didn't pay for itself on utility scale solar. Prices may or may not have changed to where that argument is no longer valid. YMMV.
 
jaggedben said:
If it isn't going to push the voltage below the inverters operating window, then it comes down to the economics. Is the cost of upsizing the wires more or less than the price of the energy output gained at the inverter? You have to crunch the numbers.

Years ago there was an article in Solar Pro magazine with a bunch of math arguing that upsizing the wires didn't pay for itself on utility scale solar. Prices may or may not have changed to where that argument is no longer valid. YMMV.
Click to expand...
Also there is no cheap way to connect a PV system to a service that is (for example) ~1500 feet away. I did it once on the AC side with two transformers 240V to 2000V to 240V but the cost of the transformers and their inherent losses made it pretty much a wash.
 
What is the highest DC voltage that can be used in a residential setting, both as a practical matter, eg. available inverters and code wise?
 
winnie said:
What is the highest DC voltage that can be used in a residential setting, both as a practical matter, eg. available inverters and code wise?
Click to expand...
600V for one or two family dwelling rooftops and 1000V for everything else. There has been talk about raising it to 1500V for ground mounted systems but I don't know if there has been any movement on it in the US.
 
  • Like
Reactions: Zee
I think technically the NEC would allow greater than 1000V for a ground mount away from buildings, even if the property was otherwise residential. But between equipment availability and more stringent requirements I don't think it would ever make sense. In my mind over 1000V PV is for fenced off utility/industrial installations.
 
Side notes:
It seems V drop is likely to be more of an "operational" issue with AC rather than DC given the wide voltage window of inverters vs the narrower tolerance of AC grids.
Not to mention, V can usually drop even below the MPPT window before shutting down the inverter, as most inverter's "operating" V window is often much wider than the MPPT range.

Also it is useful to run the V drop calcs using Imp (Ioperating) and not Isc as that is what is closer to what the panels produce. An argument can be made that even Imp is quite high.....it is only present at noon on June 22nd (roughly!).

On the other hand the lower Vmp is also more realistic (which makes the Vdrop worse) and not the higher Voc (which would help calcs).

But I digress.
 
Zee said:
An argument can be made that even Imp is quite high.....it is only present at noon on June 22nd (roughly!).
Click to expand...
. . . if your array is pointed due south with an elevation angle equal to your latitude minus 23.5 degrees (for locations north of the tropic of cancer).

Cheers, Wayne
 
Another thing is that since vdrop in PV DC circuits is primarily a function of current, it will be less severe when insolation is low and more severe when insolation maxes out the inverter. So you'd really need to integrate a model function of production to get an accurate estimate of losses, rather than just use a max current. ('Integrate' mathematically, i.e. do calculus, to be clear.) I think fancier PV modeling software like PVSyst may do this. I'm pretty sure PVWatts doesn't.
 
jaggedben said:
Irradiance can reach or exceed 1000W/m^2 under other conditions, it's just pretty rare.
Click to expand...
Sure. Just wanted to clarify the relationship between date and when the sun is perfectly normal to a given solar array. If I understand correctly, it happens 0, 1, or 2 times per year, depending on the array orientation/elevation and the latitude. 1 being the exception (codimension 1 set of configurations) which occurs on the boundary between 0 and 2. [And ignoring corrections for the fact that the solar year is not an integer number of days.]

Of course, the correction for the direct (non diffuse) irradiance is cosine of the angle from normal, so within 8 degrees of normal is at least 99% good as exactly normal.

Cheers, Wayne
 
I have attached my calculations. If I am understanding it correctly, I am losing ~50V per string which will result in a power loss and therefore less energy produced by the inverter, but I am still well within the MPPT voltage. I could increase my conductor size to #8's and produce more power but obviously with the cost of larger wires. Energy is so cheap here I think the loss of KWh is a more cost-effective solution than increasing wire size. Can someone smarter than me let me know if I'm missing/not understand anything? Thanks!
 

Attachments

  • DC TABLE.pdf
    32.2 KB · Views: 4
PV Noob said:
I have attached my calculations.
Click to expand...
Based on your Imp (13.27A), your calculated voltage drop of 49V over 1500' one-way of #10 copper implies a DC resistance of 1.23 ohms/kft. NEC Chapter 9 Table 8 gives a DC resistance at 75C of 1.26 ohms/kft for solid and 1.29 ohms/kft for stranded. [Stranded is higher as the outer strands of a foot of stranded wire are in fact longer than a foot because they spiral around the inner strand.]

So presumably you've used a different source for the DC resistance, which must be based on a lower temperature than 75C? Determining the actual operating temperature is beyond what I'm familiar with, and the difference is small enough not to really matter.

Anyway, that is the worst case scenario for when the array is fully illuminated and producing maximum power at STC, presumably. So if you want to evaluate the economic cost of the voltage drop, you really need a model for the actual current, and how that changes over the course of the day and year.

PVWatts will give you an hour by hour spreadsheet of various values based on the parameters specified. In particular, calling the last two columns exported X and Y, they are X = DC Array Watts and Y = AC System Output Watts. So you can get at least an approximate answer by running it with the proper parameters for your array and then doing a few simple operations on the spreadsheet. Namely, if you assume the string voltage at the array is always Vmp, then:

I = X / Vmp
Vdrop = I * R
Vinverter = Vmp - Vdrop
Pac = Y * Vinverter / Vmp

Do this for each row of the spreadsheet, add up the Pac values for each hour of the year, and compare that to the sum of Y over the year to get an approximation of the production loss due to voltage drop.

I think this will be a slight underestimate of the production loss, as when the DC Array Watts is below peak, the actual string voltage will be somewhat lower than Vmp. I'm not so familiar with PVWatt's methodology; it may have a model for string voltage, and if it is simple enough and explained in the documentation, you could use that model to more accurately backcalculate string current from the values the model exposes in the spreadsheet.

Cheers, Wayne
 
Last edited:
solarmatt said:
I recently enjoyed reading this article from Ryan Mayfield's consulting group. They used Helioscope for their modeling.
Click to expand...

Wow, that is a great article, and it makes a critical point: voltage drop when your system is clipping is 'free'.

Solar panels have gotten so cheap that the optimal system (in terms of electricity produced per $ of investment) generally has more PV capacity than inverter capacity. This means that during the peak of the day, electricity that the solar panels _could_ produce is going unused. It really doesn't matter if some of this excess energy is going to heating up wires (voltage drop) or letting the panels get a touch warmer. The output from the inverter is maxed out.

-Jonathan
 
winnie said:
Wow, that is a great article, and it makes a critical point: voltage drop when your system is clipping is 'free'.

Solar panels have gotten so cheap that the optimal system (in terms of electricity produced per $ of investment) generally has more PV capacity than inverter capacity. This means that during the peak of the day, electricity that the solar panels _could_ produce is going unused. It really doesn't matter if some of this excess energy is going to heating up wires (voltage drop) or letting the panels get a touch warmer. The output from the inverter is maxed out.
Click to expand...
That has been the case for as long as I have been designing PV systems, ~15 years. Usually a little clipping in the middle of the day is made up for or more by output on the shoulders of the time vs. power curve.
 
Have you given any thought to electrofelon's suggestion in your other post to use #6 AL XHHW instead of #10 CU XHHW? I just checked our distributor's prices and the #6 AL is $.25/ft vs. $.44/ft for the #10 CU. And #4 AL XHHW is still less than #10 CU at $.31/ft. Of course the conduits will be bigger, and the pull will be a lot harder. The #6 seems like a good option at 57% of the wire cost and 65% of the voltage drop.
 
  • Like
Reactions: Zee
You must log in or register to reply here.
Top