Current Method for Photovoltaic Calculations




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!

Step 1: Basic Knowledge

Ok, let's start with the 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

Step 2: Energy Needs

Here we will found our energy needs, or “demand”.

This is nothing else that our hours of use on any appliance on a day bassis. This means that you have to go and figure out how many hours do you use some appliances during a day.

I have to say, this method is really good for small scale installations, but if you have to make a whole house installation... well you better talk to a local spetialist.

To make this easier, we will use a table.
You will have to note like this:
Room or ambient | Appliance | Power [W] | Quantity | Hours of Use [h] | [W . h / day]
Bedroom                | Light          | 9 W             | 2             | 3 h                         | 54 Wh/day
Kitchen                   | Light          | 18 W           | 1             | 5 h                         | 90 Wh/day
Living-room           | Light          | 18 W           | 1             | 3 h                         | 54 Wh/day

So we have here some subtotals, on how much energy in watts are we going to need on each room for light.

Notice that I am using small appliances and a few hours for each, just to make things easy to follow. Once you have some practice with this, you could add as many appliances as you want to.

So here we are going to take this numbers and make 2 more calculations on this subject.

1) The “maximum simultaneous demand” or MSD (in spanish is DPMS).
2) The total daily energy demand, which we will cal ED.

For the MSD you have to take the “Power [W]” column and multiply this by the “Quantity” column, and make a sum on each result.In this example:
Bedroom        | Light | 9 W x 2 = 18 W
Kitchen           | Light | 18 W x 1 = 18 W
Living-room   | Light | 18 W x 1 = 18 W
                                               SUM = 54 W

So here you have the maximum energy you will have to manage on you cables, and with your regulator.

Since we use Amps on each case to run some calculations, you will have to divide this power (54 W) by the system voltage (12 V).
A = W/V
A = 54 W / 12 V = > A = 4,5 A

NOTE: We are using small appliances, and small energy needs, so 12 V will be more than right. If you have heavier appliances, and a lot of power, you are probably going to need a higher voltage system and an inverter. Those are not mentioned here, and therefor is something you sould probably consult with a professional on your area, with knowledge in the electric code.

The total daily energy demand is the sum of each Wh/day on the last column

Room or ambient | Appliance | Power [W] | Quantity | Hours of Use [h] | [W . h / day]
Bedroom                | Light          | 9 W             | 2             | 3 h                         | 54 Wh/day
Kitchen                    | Light          | 18 W          | 1             | 5 h                         | 90 Wh/day
Living-room            | Light          | 18 W          | 1             | 3 h                         | 54 Wh/day
                                  | MSD          |                 54 W           | ED                         | 198 Wh/day

As we stated earlier, we use currents, so you have to take this Wh/day, and transform them in Ah/day. Pice of cake...
198 Wh/day / 12 V = 16,6 Ah/day.

So, you have here some interesting values.

You know that you charge controller is going to handle, on the demand side, at least 4,5 A. And on a standard day, you will consume 16,6 Ah/day. This is called our “Daily current needs” or Iday. “I” stand for current.

Those are our basic numbers to continue to the next step: Calculate your energy resource

Step 3: Resource

 Lets go now with our resource.

Here you will have to do some research, since I can only provide some data from my country and city as an example, but you could found that data on the WWW.

Ask around, and it will reveal to you.

As a first impression I have to say that is quite easy to acquire this data.

One of the things with solar modules is that they work it's best when they are at 90º to the sun rays. So, if you could afford a solar tracker, or you are skilled enough to make one, good for you, this is almost solved.

But there is some things to take into account:
In southern hemisphere, your panel should be facing south.
In northern hemisphere, it should be facing south.
This is you ORIENTATION.

Now, the next thing is INCLINATION.

To choose the best inclination for your modules, if they are going to be stuck in one position is the one according to this table:

Your Latitude | Angle of the panel
                          | In Winter | In Summer
0º to 5º             | 15º             | 15º
15º to 25º        | Your Lat.   | Your Lat.
25º to 30º        | Lat + 5º     | Lat. - 5º
30º to 35º        | Lat + 10º   | Lat. - 10º
35º to 40º        | Lat + 15º   | Lat. - 15º
40º <                | Lat + 20º   | Lat. - 20º

As a practical note, I have to say that here, with a latitude of 34º we use a plain fix, Lat + 10º, which compromises a small amount in the summer, when we need less energy, and gives some advantage in the winter.

So you have two mayor problems solved: where to face them, and in which angle.

Next is our Kilowatt or Megajoule thing.

Here we have (take a look at the picture) a table with the average monthly energy on an angle of Lat+10º in MJ m^2 /day: 11,2

So we took this 11,2 and divide it by 3,6. This is 3,11 “equivalent hours” or “peak sun hours”. We are going to call them just “H” for HOURS.

This means that, if the sun where a full 1000 W lamp facing our modules, it will shine right above them for 3,11 hours straight. So if you take a look at the specifications of the modules, you will notice that they say: 80 W @ 1000 W m^2 – 25º C – 1,5 MA
That means: this particular module will produce 80 W when the sun gives 1000 W on each square meter at 25º Celsius and 1,5 mass of air.

But this is something you could investigate in further readings. For now, let's continue with the next step: Choose our module 

Step 4: Generator

Ok, next thing: chose the right generator.

We have to use here some of the data of our energy needs.

The formula for this method is:(1,2 x Iday) / (H x Im)
Where:1,2 is a factor, which includes aging of the module, dust, and several losses.
Iday is as we have seen, the energy need in Amps.
H is the hours, we have seen this on the “Resource”.
And Im is our new inquiry. Whis is nothing more than the current of our chosen module at standard test conditions (STC = 1000 W m^2 – 25º C, and so on).

So, here we could choose our desired module, and see what happens.
(1,2 x 16,5 Ah/day) / (3,11 H x 3,45 A) = 1,85
This means we are going to need 1,85 modules. Since this is a simplified calculation, we are just taking modules in parallels.

We are going to need 2 KS60T modules in parallel.

TIP: If the result is near 1,3 just keep 1. But if its more than 1,3 it is recommended to chose 2.

Pretty easy, right? 

Step 5: Battery Bank

The battery bank is related to our cloudy days.

So you are going to need more research. Look for data on your weather service on a year basis. Usually 3 to 5 days are good numbers.

The formula here is like this:
Bb = (Iday x Cdays) / DoDWhere Bb is, battery bank
Iday, is as we have seen, again, our energy need.Cdays is our cloudy days.
And DoD is or “Depth of Discharge”
Here we have to talk to the battery manufacturer. Since each technology and kind of battery has an unique DoD.

We are talking here of 12 V batteries, to simplify. But you could form your bank from single-cell batteries, that are very powerfull.

For deep cycle, lead-acid batteries, a DoD of 0,7 is quite right. AGM is 0,5. look for specifications on the web.

Bb = (16,5 Ah/day x 5 days) / 0,7 = > Bb = 117,86 Ah

I choose a 110 Ah /12 V AUTOBAT SOLAR Battery. 

TIP: Remember, serial connections increments voltage, but maintains same current.Parallel connections maintains same voltage, but increments the current.

So, I could use two 12 V / 60 Ah batteries in parallel, or two 6 V / 110 Ah in series, and we always have the same energy. 12 V / 120 Ah in the first case, and 12 V / 110 Ah in the second.

Step 6: Choose Your Regulator

The charge regulator has to manage all the currents involved on the circuit, and I would recommend to pick up one with at least:
• Reverse polarity protection.
• LVD, which stands for Low Voltage Disconnect. This will cut your energy on the load side if the battery is too low, and wont let you reconnect until is half charged.
• Battery type selection: Gel, AGM, sealed or classic lead acid.
• LEDs to check the state of charge.

Those are really basic features, and you have lots of regulators, with displays, remote controller and monitoring, USB connections.

Just don't get too picky, and choose one reliable, proved on the field.

A small advice on connections:
The Universal technical standard recommends some cable diameters and voltage drops.

• For the Panel-to-Regulator length: 2,5 mm^2 (AWG 10) and 3% voltage drop.
• For the Regulator-to-Battery length: 4 mm^2 (AWG 6) and 1% voltage drop.
• For the Regulator-to-Load length: Just 5% voltage drop, and choose the wire that don't let the voltage go below that.

How to calculate voltage drop?
Well, if your system is 12 V, this is your 100%.
For 3%: (3 x 12 V) / 100 = 0,36 V
For 1%: (1 x 12 V) / 100 = 0,12 V
For 5%: (5 x 12 V) / 100 = 0,60 V

Now, you have to use this formula to test if your cable run is not too long or too thin.
V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on Ω/meters)

You probably have the resistance in Ω/Km. So divide that number by 1000.

So in each cable run, you have to make that calculation and check that the result is less than the spected V (1, 3 or 5 %).

For a run from panel to regulator of 4 meters, with 6,9 Amps (two 3,5 amps modules in parallel) and you whant a 0,36 V voltage drop tops, goes like this:
V = 2 x 4 m x 6,9 A x 0,0095 Ω/m = 0,52 V

As you can see, is above 0,36 V, so you will have to go to a thicker cable, since that is the resistance of 2,5 mm cable. Probably with 4mm will works.

So, protections now:
You will have to keep an interrupter, dual pole, that cuts the connection from the modules to the regulator, so if there is a electric storm, or you have to do some mantenaince, you could cut there.
A fuse must be installed on the positive of the battery. You have to use a DC fuse. So choose it right. Remember that batteries could push a lot of power if a short circuit appears.
Also, you have to put a surge protection, or fuses on the load side, to disconnect and protection to short circuits.

So, this is a resume of all you need to calculate all the materials needed, with one method.
Hope this will be found useful for someone.



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    20 Discussions


    7 years ago on Introduction

    I am glad you found the Instructable usefull.

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

    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.


    7 years ago on Introduction

    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,


    Reply 7 years ago on Step 6

    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.


    Reply 7 years ago on Step 6

    Dear Supernatural. Today is 5th february 2012. Nobody know your method. Have wrote on other paper?
    Please let us know your method!


    Reply 7 years ago on Step 6

    Estimado Eliseo Sebastián, no entiendo su pregunta. El otro método lo publiqué hace ya un tiempo en otro instructable.

    Fue publicado en noviembre de 2011.


    Reply 7 years ago on Introduction

    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.


    Reply 7 years ago on Introduction

    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.


    Reply 7 years ago on Introduction

    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.



    7 years ago on Introduction


    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.

    1 reply

    Reply 7 years ago on Introduction

    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.



    7 years ago on Introduction

    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.

    4 replies

    Reply 7 years ago on Introduction

    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.

    I will try to answer your comments in order:
    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:
    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.


    Reply 7 years ago on Introduction

    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.


    Reply 7 years ago on Introduction

    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.


    Reply 7 years ago on Introduction

    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:
    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:

    Nice to have this talk! I am practicyng with my English ; )


    7 years ago on Introduction

    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.

    1 reply

    Reply 7 years ago on Introduction

    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:
    And, if you run over my site:
    You could found a link to a bunck of documents I have collected over the years on renewables. Many are in spanish, but most of them are in English.

    Once again, thanks a lot for your words, and good luck on your project, I hope it works well.

    Don't forget to make an "Instructable" with your results. Even if you have only pictures. That way we help the growing community of renewable users.