Current method for Photovoltaic Calculations
6 Steps
Hi everyone!

This is my firs attempt to make an Instructable, being a huge fan of this site, I am pretty excited.

I am from Argentina, so it is probable that some misspellings and grammatical errors pop up here an there, I apologize in advance for them.

Current Method for Stand Alone Photovoltaic Systems Sizing
This particular method is known as "Current method" since it uses the currents (in Amps, Amps Hour / Day) to select the components.

The components:
1) The modules
In Argentina we use this scale: cell → Module → PanelThe PANEL is an array of several modules, plugged in series or parallels to achieve the right configuration. In this example we will have everything in 12 V systems, so 32-cells modules are going to be OK.
The MODULE is composed of several cells in series and parallel to achieve the right current and voltage.The CELL is the smaller component. Is made of silicon, with the electric contacts in front and rear to pick up some electrons, running around the silicon when some photons hits them.
2) The charge regulator
This is a critic component, since it keeps your batteries fresh and healthier. A bad chosed charge regulator will end up in an early death of the batteries.
3) The battery bank
Deep cycle solar batteries aren't cheap, nor light weight. So choosing the right size of that battery bank, suitable for your budget and energy needs is going to be criticall.

Hope someone finds this text helpful!
Lets begin!
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## Step 1: Basic Knowledge

You will need to know here that some skills on the electric field are required. Not much, but some really helps.
One thing you will found very often is:
Power = Current x Voltage
Power in Wats [W]
Current in Amps [A]
Voltage in Volts [V]

We are going to need an “Isolation” or “Irradiance” chart of the area where we are located, to calculate the average monthly energy that reaches our module every day.

In Argentina we have some charts, and tables with measured Irradiance at several angles. This is really helpful. But in fact, we use the “Peak Sun Hours” wich means the equivalent in hours at 1000 W m^2 / day (a constant known as 1 SUN).

You probably found your Irradiance charts or tables in Kw (Kilowatts = 1000 Wats) or MJ (Mega Joules).
If you have them in Kilowats, you are lucky, so there is no need to make any change or calculation. But if you have them in Megajoules, you will have to divide that number by 3,6 (we use comma as decimal separator. Take notice of this) to convert that unit to Kw.

So lets recap:
W = A x V
MJ / 3,6 = Peak Hours (HSP) in further formulas referred as "H"

The full formula looks like this:
1 HSP = [(1000 W x 1 h ) / m2] x [3600 s / 1 h] x [1 J/s / 1 W] = 3,6 MJ/m2
Ajiero says: Feb 9, 2012. 2:16 PM
(removed by author or community request)
Supernatural (author) says: Feb 17, 2012. 6:06 AM
Dear Ikenna:

You are asking for a full time job here!

if you follow the instructable, and build a spreadsheet with the step-by-step formula you will reach your goal, but I can't do all this job for you.

If this instructable is too simple, you could check the other I have put (for solar-eolic systems).

And I have to say thar RETScreen is free and has it's own course on-line, and you could run it for free.

Sorry I can't help you more, but is too much work and I don't have the time.

Good luck,
Ajiero says: Feb 17, 2012. 9:24 AM
(removed by author or community request)
Supernatural (author) says: Feb 17, 2012. 12:29 PM
I am glad you found the Instructable usefull.

If you check my site, you could see the actual spreadsheet (in spanish) with the formula.

http://www.pablomaril.com.ar

Also, in my site I have put a library on-line, is there. Just choose the logo on the right "pablomaril" and below "instalaciones y renovables".

If you have some particular question, write me an e-mail using the direct message feature of Instructables.

Good luck.
tbone56 says: Jul 8, 2011. 7:06 AM
Good job. You put it all together. Just what I needed.
Thanks.
Supernatural (author) says: Jul 8, 2011. 8:33 AM
Thanks a lot. I am glad you like it.

I am putting another method toghether. Wich is a power balance method. In a couple of weeks it will be up.

Thanks again.
Eliseosebas says: Feb 5, 2012. 8:05 PM
Dear Supernatural. Today is 5th february 2012. Nobody know your method. Have wrote on other paper?
Supernatural (author) says: Feb 6, 2012. 4:47 AM
Estimado Eliseo Sebastián, no entiendo su pregunta. El otro método lo publiqué hace ya un tiempo en otro instructable.

http://www.instructables.com/id/Hybrid-Design-PhotovoltaicEolic-for-rural-commun/

Fue publicado en noviembre de 2011.
ibarnett52 says: Aug 8, 2011. 9:15 PM
GREAT INSTRUTABLE
NightLord says: Jun 22, 2011. 8:23 AM
"MJ / 3,6 = Kw"

Noupt, actually 3,6MJ = 1kWh (not kW!).
Supernatural (author) says: Jun 22, 2011. 10:09 AM
Yes, you whre right, I am cheking it now, and putting the full formula. But for "easy to solve" ways, if you have your mesurements in MJ and divide that by 3,6 you get the Peak Hours you need.

I am putting the correct formula:
So lets recap:
W = A x V
MJ / 3,6 = Peak Hours (HSP) in further formulas referred as "H"

The full formula looks like this:
1 HSP = [(1000 W x 1 h ) / m2] x [3600 s / 1 h] x [1 J/s / 1 W] = 3,6 MJ/m2

Thanks for your advice, but remember, I sometimes forget on porpouse of the units just to give to the students an easy way to make the calculations.

And dometimes I make mistakes, like you point to me here. Once again, thanks for the advice. I am always looking to improve.
NightLord says: Jun 22, 2011. 10:21 PM
Sorry for such "agressive" comment, your article is still great and I knoy you've put a lot of work into it. I wasn't in a really good mood when I wrote it ;).

I just really don't like confusion between kW and kWh, because this is really common and nobody understands even so simply physics as difference between energy and power, and I didn't really check your formula in details, to see what you were really calculating. Sorry again.
Supernatural (author) says: Jun 23, 2011. 3:45 AM
Dear NightLord, don't worry.

I still think your comment where right, and help me to improve my instructable.

I know, sometimes energy and power are confusing, and I try to keep it simple, that is wy I make some mistakes.

Have a good day.

Pablo
nanosec12 says: Jun 22, 2011. 4:26 AM
Supernatural,

A couple of suggestions, since this is your first instructable:

1. Your English, while not perfect, is good enough to convey your message and is much better than most of our spanish. Muy Bueno.
2. I think it would help if you could add an english translation to all of the photo captions, either yourself or with help from others. Some of the captions are already written with both english and spanish, but a few are not.

Thank you for your effort, with this instructable and with your english write up.
Supernatural (author) says: Jun 22, 2011. 5:42 AM
Hi nanosec12.

Thanks a lot for your words. I have put all the translations in the pictures, that was a rookie mistake, since I didn't realize that the pictures are the same on all the instructables I use them. So I chanhe the captions for my spanish version, thinking that a copy of the english version is what I already have.

I have cheked that, and think is better now.

I hope this won't happend again.

Saludos,
kelseymh says: May 16, 2011. 3:51 PM
This is a nice writeup, with decent guidance for the beginner on how to deal with medium-scale solar installations. I do have some minor comments on your units, though.

Joules are a unit of energy; watts are a unit of power, which is energy divided by time. When you write, "MJ/3,6 = kW", you really should write, "MJ / 3.6 [1000 s] = kW", since you have to divide joules by seconds to get watts.

I'm confused by your irradiance explanation. "1 kW m2/day" appears to have too many units of time in the denominator.

The SI symbol for "kilo" is "k" in lower case, not "K".

I'm a little confused by your Ohm's law calculation. You write "V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on Ω/meters)". Obviously length × resistivity is just resistance, so you get V = IR. But where does the extra factor of two arise? You don't need to deal with the return cable from the load back to the solar panel.
Supernatural (author) says: May 16, 2011. 7:11 PM
Thanks a lot for your comment. It took me a lot of hard work, since my language is spanish, and its an efford to pull this of in English.

We all agree, Joules are energy, Watts are power.
I just go right for the simplified formula, without dealing with units.
Here you have the right formula, in spanish, sorry, couldn't find it in English:
http://es.wikipedia.org/wiki/Hora_solar_pico
Since I teach to students with basic knowledge on mathematics, sometimes I have to forget all the units, and put just the real deal. You are right about it, but the simplified formula is a choice here, not a mistake.

The irradiance is measured in our maps like this: MJ m2/day
This is, a device that measures how much power strikes the cell (The megajoules) on a particular surface area (the square meter) during an entire day. I made a mistake by converting that to kW.

I have to say here, that :
V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on Ω/meters)
It's OK, since you are going to take each meter of cable as a resistance, you will have to multiply that by the number of meters the cable has, here is Ω/meter multiplied by the meters of lenght. Is not resistivity, is resistances in series, one for each meter. And regarding the cable back, it has to be taken into account, since you can't have a circuit with just one cable. You need two and a load. And if you don't have a circuit, you don't have current. I can't translate the Kirschoff' Laws, but there you will find that if you have a current comming in, you must have a current going out. Is really hard for me to do this in english, I am so sorry if it sounds too bad.

I hope I haven't make a mess here with my explanations.
Thanks a lot for posting your comment.
Saludos, Pablo.
kelseymh says: May 16, 2011. 8:55 PM
Muchas gracias, Pablo. Mi lengua primera esta ingles, y mi espanol esta pobre.

I am very grateful that you will take the time to work through the English for me, and I am sorry that I don't know enough Spanish to properly return the favor.

Have you communicated with any of our other Spanish-speaking members? A few of the ones I know are Rimar2000, ricardoruizo, M.C.Langer, but there are many others. There's also a discussion group where members interested in multilingual or non-English I'bles have some topics going.

I am a physicist, so for me the units are very important. I appreciate wanting to keep things simple (and thank you for clarifying that you made deliberate decisions, not mistakes! :-), but my own training and teaching of students suggests that with proper units, you can tell quickly if you have the wrong formula (the units don't work out correctly).

I still think the factor of two in your resistance calculation is wrong. Here's why: What you're computing is the voltage drop from the output of your solar panel to your load. That's V = I×L×r (here r = resistivity, ohm/m). The load itself will introduce some additional voltage drop due to its own resistance (or impedance). The return path is necessarily dropping all the way to ground (by construction), so the voltage drop due to the return cable is irrelevant.

You can also think of it this way. You could set up your system with just an outgoing cable! Connect the negative side of the solar panel to a true earth ground, run a single wire from the panel to your load, and connect your load to a true earth ground at its location. The current flow will be earth -> panel -> load -> earth, with the two (possibly distant) earth locations connected implicitly by dirt, ground water, whatever.
scraptopower says: Jun 22, 2011. 3:55 AM
I'm not sure about the voltage drop point of view, but the return leg resistance is just as important to keep low as the other side of the circuit as all of the current charging the battery must flow through it.

I don't think your earth example would work as there would be too much resistance from the ground/earth/soil water, so almost no current would flow.
Supernatural (author) says: May 17, 2011. 5:21 AM
Hi again! I haven't comunicated with the spanish members yet, this is my first instructable, and I think my next will be this same subject, but in spanish.

I understand what you say about units, and for people with some degree it's easier. I teach in vocational training Centers, where you teach people that sometimes don't have more than elementary school, and in a small course with only 4 months. Sometimes, it's easier to show them without units, in practical matters. I mean, "hands on the job". I have show them the right formula, the one I put you in the earlier post, and they all look at me like I was going mad... so I drop it and go for the straight-forward-system.

With the two factor, it's ok what you say, I need the voltage drop on the load side. I think I am not too clear with resistance and resistivity. Here Resistivity is ρ (Rho) wich stand on the copper like this: 1/57 value in [Ω mm2 / m], and is related to the section of the cable, while resistance is the equivalent value of an electrical resistance on each meter of lenght.

If you build the equivalent circuit, it will be the Generator (module or panel) with a resistance which value will be the one given for your cable + (plus becouse is in series) the internal resistance of the load, and here we don't use a particular value for that + the resistance value of the return cable. If you use earth, the last one will differ from the first one, but if you use the same cable, you have 2 resistances with the same value.

As you say in the last paragraph, you could set up the system as you stand, and the earth will be the second conductor. I think I should say conductor instead of just cable, but conductor is som how a difficult term to use, I don't know why, but people when you say "conductor" they say "you mean the cable?"

The: earth -> panel -> load -> cable
Is like this: panel ->conductor -> load -> conductor.
So you have: 2 conductors.

Look at here:
http://www.mikeholt.com/technical.php?id=technicalvoltagedrop1
Near the bottom you wil found this sentence:

VD = 2 wires x 12.9 ohms x 44 amperes x 160 feet/26,240 Circular Mils

That is what I am using, a similar actually formula.

Here it is again:
http://www.elec-toolbox.com/Formulas/Useful/formulas.htm

Nice to have this talk! I am practicyng with my English ; )
cmac91000 says: Jun 22, 2011. 2:43 AM
Hi, I am currently investigating how to install Solar power in my home to supplement the mains supply. If you are a complete novice then this can be quite confusing.

I would like to say that this instructable is probably the easiest one I have come across to date. I congratulate you on your easy to understand expainations.

Thanks for your help, it has given me the confidence to go ahead with my project.
Supernatural (author) says: Jun 22, 2011. 3:51 AM
Hi!, thanks a lot for your encouraging words!

Has been quiate a ride to make this in English, you know. I can manage reading and seeing movies in plain English (without subtitles) but writing a wole instructable was a challenge.

I am really glad you found it usefull.

I am working on another one, with other calculation system known as "Energy Balance" and used in a paper of the Universidad del Nordeste. Is good for hybrid systems too.

Also, I have some nice pictures of the course I am running here in argentina. You could take a look here: