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# Installing a 240 Watt off-grid solar power system in Sweden

This summer I made some improvements to the off-grid solar power system I have at my summer cabin here in Sweden. It is an old farm house from the 19th century, which is located quite far out in the forest without any neighbours nearby. Until this day, the house has never had electricity installed. Living in a simple way without electricity and without running water definitely has its charm, as a way of escaping the stressful life of the big city. However, I thought it would be an improvement if I could make it possible to charge my computer, my phone and some further electronics – so I set up an off-grid PV solar power system! Since I’m also an electrical engineer, running this homepage where I size off-grid solar power systems, I thought it might be fun to show you how the installation was done. I will start by describing the different steps of the work and showing you some images. If you’re interested in the full circuit diagram, it can be found at the end of this article.

1. Sizing the off-grid solar power system

2. Mounting the solar panels

3. Connecting the cables to the solar panels

4. Parallell connecting the solar panels in the string combiner box

5. Connecting the MPPT solar charge controller

6. Connecting the battery

7. Connecting the inverter

8. Installing a cable to the house

9. Charge my computer

10. Chill out and enjoy the scenery

11. Full circuit block diagram

12. Calculations

#### 1. Sizing the system

I had decided to install an off-grid PV system with three 80 W solar panels, giving 240 W power in total. The three panels are connected in parallel together with one 72 Ah 12 V battery, which gives the battery energy in Watt-hours: 865 Wh (see calculation 1 at the end of this article). This should be enough to power some of my electronics for the bright part of the year here in the middle of Sweden (in the winter we see almost no sunlight here, but the summers are very bright).

#### 2. Mounting the solar panels

I started out with building a wooden frame which could hold up the three 80 W solar panels in a stable manner. I leaned this frame with the panels against the wall of an old barn.

#### 3. Connecting cables to the solar panels

The 80 W solar panels I am using have a voltage of around 20 V as they are connected to a charge controller, charging a battery. Since they are 80 W panels, they will give a maximum current of around 4 Amps (calculation 2) when the sun is shining really intensively in the summer here in Sweden. Based on this maximum current I chose cables with an area of 2.4 mm2, which is enough for running at least 17 amps continuously (source). I connected the solar panels in one end of the cables (see image 5) and in the other end I cut the cables in two halves. The cut-off cable parts I connected to the string combiner box and I then soldered on some banana plugs, so that I could connect the cable halves together (see image 6). I thought banana plugs would be a good way to quickly connect and disconnect the solar panels from the cables going to the PV junction box. This is because the standard MC4 PV panel connector can be quite a hassle to work with, in my opinon, if you want to be able to pick apart your connections easily.

#### 4.  Parallell connecting the solar panels in the string combiner box

The cables from the solar panels go into a homemade string combiner box. The purpose of this circuit is to make it possible to interconnect multiple solar panel strings in parallel. The string combiner circuit I’ve built follows the schematic in image 8. There is one power diode of (BYV29, 9 A current rating) in series with the positive voltage cable from each solar panel. The reason for this is that you want the total sum current from all the panels to go to the battery. Without the diodes, there would be a risk of having current going from one panel into another, if there is some small voltage difference between the panels for some reason. This voltage difference could for example occur if one panel is in strong sunlight and another one is slightly shaded.

#### 5. Connecting the MPPT solar charger controller

Out of the string combiner box, the current from the three panels is now flowing in one pair of cables instead of in three pairs. This total current should be fed into the battery. In order to control this charging current to the battery, a solar charger controller is used. I use a type of controller called an MPPT (maximum power point tracker – it finds the maximum power), which is a very efficient type of charge controller. At the input of the MPPT, where there is a depicting a solar panel, I connect the two cables coming from the string combiner box.

#### 6. Connecting the battery

I connect the battery to the MPPT, at the input with an image depicting a battery. When I look at the measurement of the current up in the left corner of the MPPT display, I can see that the current is 11 A. That gives a power of 220 W (calculation 3), which is almost the full 240 W power!

#### 7. Connecting the inverter

At the load output of the MPPT,  where there is a picture of a light bulb, you can connect an electrical load, i.e. the things you want to power with the electricity from the solar power system. In my case, I want to charge my computer, and it runs on a 230 V AC voltage. Therefore I have to connect an inverter, which converts the 12 V DC voltage from the battery into a 230 V AC voltage. The inverter I am using is a 300 W pure sine wave inverter. 300 W is enough for charging my computer and some other electronics (my computer’s charger is 45 W and the fast charger for my phone is 15 W).

#### 8. Installing a cable to the house

I want to charge my computer from my solar power system, but I normally don’t use the computer down by the wall of the barn. I want to be able to charge the computer up in my house in the kitchen. Therefore I connect a 50 metre long cable with two wires to the inverter. It is good that the inverter increases the voltage from 12 V to 230 V, because that makes the losses in the cable smaller. The conductor area of the cable is 0.75 mm2, which is not so much, but it is enough to transmit 300 W power over 50 metres. The cable area is enough for not being damaged by the current (calculation 4). Also, the voltage drop over the cable is only about 1% and not a problem (calculation 5).

#### 9. Charge my computer

It was nice to see that everything worked out perfectly! When the sun was shining, the solar cells supplied my electronics with power and charged the battery bank. When it was cloudy, the battery supplied the power instead. I measured the voltage up in the house, at the end of the cable coming from the inverter. The voltage there was 234 V, which is good, so I could just connect my computer and start charging.

#### 10. Chill out and enjoy the scenery

So now, except for having a beautiful house out in the wild nature, far from stress and intruding neighbours, I also have electricity. At least during spring, summer and autumn. Previously I had to either charge my phone in the car or bring around lots of small power banks. Now there is a more stable solution where I can charge my devices inside the house. Now it’s time to celebrate, grab a beer and a good book, play some music and go down to the still lake and take a swim and enjoy my new off-grid way of life!

#### 12. Calculations

1. Battery energy content: $E=Q \cdot V=72 |Ah| \cdot 12 |V| = 865 |Wh|$

2. SY-S80W solar panel rated power: $P_{MP} = V_{MP} \cdot I_{MP} = 18 |V| \cdot 4.46 |A| = 80.28 \approx 80 |W|$

3. Solar power system measured power: $P = V \cdot I = 20 |V| \cdot 11 |A| = 220 |W|$

4. A 0.75 mm2 cable can withstand 6 A of current or 1380 W continuously (source) and the maximum current for me at 300 W power and 230 V voltage is:

$P = V \cdot I \implies I = \frac{P}{V} = \frac{300}{230} = 1.3 |A| \label{eq:power_current}$

5. If I’m running electronics with 300 W power up in the house, the electrical load has an equivalent impedance of

$P = \frac{V^2}{R_{load}} \implies R_{load} = \frac{V^2}{P} = \frac{230^2}{300} \approx 176 |\Omega|$

The resistance of a 0.75 mm2 cable per kilometer is 23 ohm (source) and the resistance of my 50 m cable is then:

$R_{cable} = 23 \cdot \frac{50 |m|}{1000 |m|} = 1.15 |\Omega|$

The voltage in the kitchen at the end of the cable will then be:

$V_{kitchen} = V_{inverter} \cdot \frac{R_{load}}{R_{load}+R_{cable}} = 230 \cdot \frac{176}{176+1.15} \approx 228 |V|$

This is a voltage drop of less than 1 percent, so it is nothing to worry about.