Grüezi Youtubeers. There's the guy with the Swiss accent. This time with a new episode about sensors and microprocessors. We all have our gadgets, and they all need them some electricity. We usually use batteries or a power supply. I'll start small today a project to use solar energy to power our facilities throughout the year. If we want to do so, we need to answer the following questions: First, what size must the solar panel be to power our device Second, what size battery we need to power at a time when the sun is not shining? These questions lead to another series of questions: How much energy can we get in one year? How much energy does our device consume during year? How is this energy distributed during day and throughout the year? How long do we want the power supply to last with limited light or without sunlight? Many questions. So let's get started! But wait: I have to warn you! It will not be easy. But save the planet with green energy it will never be easy! We will use a simple design: solar panel, "Charging unit", battery, and our well-known ESP8266 How much energy can we get in one year? It depends mainly on three factors: The place where your solar panel is located and its direction to the sun The size of the efficiency panel will transfer your circuit solar energy to your ESP The "location" question can be answered looking at this map.
To find out more, you can go to "solargis.com" I live near Basel and get about 1200 kWh / m2 of solar energy per year. What this means in relation to running the ESP8266 without putting it to deep sleep, which uses about 100 mA at 3.3 volts, which corresponds to 0.33W? The year has 8760 hours. If we divide annual energy consumption this time period, we get Watts: 1200 kWh / 8760 h equals 137Watt / m2. By this calculation we obtain total sun radiation per year. Unfortunately, solar cells only have an efficiency of approximately 15%. So we get about 20 W / m2.
This value is further reduced by the efficiency of our charger and the loss during battery charging. lets go assume we are losing another 33%. Then we get only 14 W / m2 of usable energy or 1.4 mW / cm2, because 1m2 = 10'000cm2. With these two values, we can calculate required panel size: 0.33 / 14 = 236 cm2, which is about 15 x 15 cm. So this panel should be sufficient to power our ESP8266 all year round. Great! But let's quickly calculate everything from the other side: The supplier of this panel states that delivers 4.5 watts. And only our ESP needs 0.33 watts. That is a big difference. So, do you know where I made a mistake in my calculation? Didn't find a mistake? You're right: There's no mistake (at least I hope not), just another problem: The sun does not shine all the time. It fluctuates during each day, and also during during the month.
The specifications of the panel show us only top performance, and moreover not in Switzerland, but somewhere completely different with lots of sunshine! Maybe this solar panel performance is still a bit exaggerated, as usual in the specification for Aliexpress … We must therefore continue our calculations. But since it's boring, and the sun is shining outside, let's do some tests first: I bought a few small solar panels and a bigger one and I would like to do some tests now.
This is a simple test: place the solar panel on the sun and attach it to my new one electronic load. Electronic load is a simple device: behaves like a variable resistor plus a volt and ampere meter. The only difference is that the electronic load automatically sets the resistance to either to constant current, constant voltage, or constant power input. At the same time automatically calculates and displays power consumption, which is useful for these experiments. Shooting today is not easy, but I hope you see the values. I'll start without load and measure the "open voltage" from the 6.5 Volt solar cell. If they start subscribing current, we see that the power input increases while the voltage drops only slightly. Suddenly, the voltage drops and we lose most of the power. If I try to repeat it again in smaller steps, we see that we can draw a maximum current of about 550 mA and get 2.8 watts at the output.
At Consumption of greater voltage will dramatically reduce voltage. Why is that so? This is characteristic of solar panels. Here is the result of my measurements 16x16cm panel, here is a theoretical curve. Both curves are quite identical. And here you see the term MPP, or maximum power point. In order to get maximum performance from the panel, we must always approach this point. Unfortunately, this point is changing when the lighting conditions of the panel change. As a result of this change, we need to find this point again. There are specials devices that do exactly this. They are called MPPT or "maximum power finder". I would like to look at this topic in a future video. You can buy monocrystalline or polycrystalline articles or panels. Monocrystalline silicon is used for most of our electronic chips and panels of this material have theoretically higher efficiency, which means they should produce more electricity with a defined light intensity. They should also be more expensive than polycrystalline modules. In fact, the differences are small and we shouldn't to deal with them too much.
You can easily see the difference between mono or poly modules, as they are usually called: Mono Modules are darker, almost black and POLYS are gray. Results on a sunny Sunday afternoon see this table. Please note that the results are not entirely meaningful given the fact that that even the smallest clouds can have an effect, and above all that I made the measurements in series, not in parallel. In the middle of the measurement, I had to take a beer break because the weather was really hot… We see that I have a power input of cm2 between 5 and 10 mW. However, we must also take into account that not the entire area of the panel is used to convert light into electricity. There are also areas for linking individual cells because there is only one article produces about 0.5 volts. So at this point, we know how much energy we can get throughout the year, and also during a sunny day.
So let's find out the actual panel size and battery size large enough to power our ESP for safe power supply throughout the year. Here in Basel, we get 2.6 times less solar radiation in December than in July. The sun also disappears every night for a few hours. Especially in winter, we experience bad weather and sometimes we do not see the sun all days. This creates three more problems for our project: We need to make sure that our facility survives all night in the winter. At the same time, we must be sure that our facility will survive the period bad weather without the sun and further that our facility will survive the whole December. Of course, these problems vary in different ways localities. This is also the reason why I show individual equations and sources of my data. With this you would they should be able to create their own calculations. The first problem can be solved with batteries, which is charged during the day and discharged during the night. Let's count quickly battery size for the shortest day in December. The length of the day is about 8.5 hours plus night, therefore 15.5h. So our battery must be: 15.5hx 0.1 A = 1.6 Ah.
That is less than capacity than 18650 articles. Now the second problem: if we assume bad weather without sun for 2 weeks, you will need a larger battery: 14 daysx24hx0.1A = 34 Ah. In this case we will need about 14 18650 cells connected in parallel. If this is the worst case, we know now battery size. Another thing is to calculate the size of the solar panel. We can assume that bad weather conditions are included in our average values for a certain locality.
So we can design our solar panel for the worst month of the year. Let's take that our 1.62 kWh / m2 per day of average solar energy in December, divide it by 24 hours, and multiply it 10%, thus obtaining the necessary electricity. This is 6.7W / m2 Because we need 0.33W, we need a 448 cm2 solar panel to get enough power to power it our facility for the entire month of December. These values are for the average year. In recent years, however, we have they never had an average year. So maybe we should add a little and we come to the size of the panel 25x25cm, which is equal to 625 cm2. Quite big! To sum it all up, we can do the same calculation for Dubai, where it is quite hot in the summer. First, we are looking for sunlight per m2 for the worst month of the year. That is 3.68 kWh / m2 / day.
Subsequently, we divide this value by the number 242, and divide it again by the consumption of our equipment This gives us the size of our panel 197 cm2. The battery size may be smaller as you don't have to predict 14 consecutive days of bad weather. Suppose 5 days. Day is 10.5 hours long, so the night lasts 13.5 hours. The battery size is therefore only 12 Ah, which is about 5 18650 cells. Some of you may remember my videos about sleep modes.
If we are able to reduce the electricity consumption in our equipment by a factor 10, the size of our battery will be reduced to one 18650 cell panel in the case of Switzerland. At the same time our solar panel will be the size of 10×10 cm. If we can reduce energy consumption even more, for example, by using LoRa instead of Wi-Fi, the battery and panel size will be further reduced Great! Today we calculated the size of the solar panel and batteries for year-round use. In one of the other videos, we should focus to the charger between the solar panel and the ESP. This equipment must meet certain needs: First, to find and maintain the MPP in all lighting conditions; to get the maximum power from the solar panel Second, make sure we have a constant voltage of 3.3V for ESP Third, turn off the battery charge if it gets to 4.2V.
This is especially important because we need to design a solar panel for the worst month of the year. In all other months, we will have too much energy Fourth, you need to protect the battery from too low voltage, Fifth, signal low voltage on the ESP with respect the possibility of reacting accordingly. E.g. Send a message I hope this video was useful to you, or at least interesting. If so, press the LIKE button. Hello..