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Solar Power Simple Algorithm
- May 08, 2018 -

The solar AC power generation system is composed of a solar panel, a charging controller, an inverter, and a battery. The solar DC power generation system does not include an inverter. In order for the solar power system to provide sufficient power for the load, it is necessary to properly select each component according to the power of the electrical appliance. Solar power system design needs to consider the following factors:


Q1. Where is the solar power system used? How is the solar radiation in the area?

Q2. How much load power does the system have?

Q3. What is the output voltage of the system, DC or AC?

Q4. How many hours does the system need to work every day?

Q5. If there is rainy weather without sunlight, how many days does the system need to supply power continuously?

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The following uses (load) 100W output power, taking 6 hours per day as an example, to introduce the calculation method:

1. First calculate the number of watts per day consumed (including inverter losses):

If the conversion efficiency of the inverter is 90%, when the output power is 100W, the actual output power should be 100W/90%=111W; if it is used every day for 6 hours, the power consumption is 111W*6 hours. 666Wh, which is 0.666 kWh.

2. Calculate solar panels:

According to the daily effective sunshine time of 5 hours, taking into account the charging efficiency and the loss of the charging process, the output power of the solar panel should be 666Wh ÷ 5h ÷ 70% = 190W. 70% of this is the actual power used by the solar panel during the charging process.

3.

180W module daily power generation

180×0.7×5=567WH=0.63 degrees

1MW daily electricity generation capacity=1000000×0.7×5=3500,000=3500 degrees


Example 2: An 10w lamp, lighting 6 hours a day, 3 rainy days, how to calculate the solar panel wp? and 12V battery ah?

Daily electricity consumption: 10W X 6H = 60WH,

Calculate solar panels:

Assume that the average peak sunshine hours for your installation point is 4 hours.

Then: 60WH/4 hours, = 15WP solar panels.

Recalculate the charge and discharge losses, and add solar panels every day:

15WP/0.6= 25WP,

That is a 25W solar panel is enough.

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Recalculate the battery.

60WH/12V=5AH.

Use 12V5AH power every day.

Three days is 12V15AH.

The battery configuration needs to be designed so that the daily electricity consumption does not exceed 20%, or the electricity consumption does not exceed 50% in continuous rainy days. In order to achieve the longest battery life requirements.

In this way we conclude that the battery of this system is sufficient for 26AH-30AH.


Example 3: How many watts of solar panels are required to fill a 12V 45A battery with 6 hours?

12V45A battery is 648Wh (?) 6 hours to be full, then the solar panel theoretically only 108W but actually because of the sunshine intensity temperature PV controller efficiency and other factors such as the overall efficiency of the 108W battery panel 6 hours is less than 12V45 The overall efficiency of the battery will be calculated as 0.8 you need to choose 135 watts of solar modules. By the way, the best charging current for lead-acid batteries is 1/10. The battery capacity current is 4.5A. Excessive charge current will speed up the battery plate. Sulfidation affects battery life.

The simplest calculation method:

Battery: 12V×45A=540WH

Solar panel power = 540/6/0.8 (loss) = 112.5W


Example 4: Will the two 20-watt (36-piece) solar panels charge the 12-volt 17-amp battery for several hours? An ordinary 12v4AH battery, it takes a few hours to charge it with the two solar panels?

The 1.20W solar panel operating voltage is generally 17.2V and the current is 1.15A. If the board quality is good, the measured current is generally 1.1A (I tested it).

2. Suppose you say that the 6-hour light is from noon to afternoon, then you can calculate 4 hours full power, that is to say two 20W boards can generate 2*1.1*4=8.8A per day.

3. Such a 17AH battery can be filled in 2 days; 4AH is almost as long as 2 hours.

Or the total number of solar panels is 20+5=25W

The total number of battery watt is 12v*17A=204w

Filling time is 204/25=8 hours


4A battery:

4A *12=48w

48w / 25w = 1.92 hours

Or because the relationship between sunshine intensity and battery capacity is not accurate, actuarial calculation is not necessary.

Solar battery current: 20/12=1.7A

Charging time 1:17/1.7*1.5 Charging constant = 15 hours,

Charging time 2: 4/1.7*1.5 charging constant = 3.5 hours,

In fact, you can put two batteries and two solar panels together.

Charging time 3: (17AH + 4AH) / (1.7 * 2 blocks) * 1.5 charging constant = 9 hours,

If you have good daylight in your area, you can charge it for almost two days.

Nothing to pay attention to charging, if you have a multimeter, then charge the battery at both ends of the test, not more than 14V on the line. Remember to discharge less than 10.5V. Overcharge and overdischarge will ignite battery life.



Example 5 Assume that for two consecutive rainy days, the load power is 40 W. The lighting time is 8 hours per day. To reach the above lighting time, how many watts of solar panels need to be configured and how many watts of batteries?

The simplest algorithm is four times.

That is, the load power * 4 times, need 160W solar panels.



If you want to be precise, it is as follows:

The load power is 40W.

40W * 8 hours / ceiling * = 320WH / 12V (battery voltage) == 27AH.

Use 12V27AH every day,

The battery is best kept within 30% of the daily discharge. So we need a battery that is easily 90AH12V. In this case, we can only choose 100AH, because 90AH battery is difficult to buy, solar battery. 40W*8 Hour = 320WH.

320WH removes 20% of losses in the circuit and storage process, and it requires 400WH per day.

If the standard time of sunshine is 4 hours per day at the time, the calculation is as follows:

400WH / 4 hours = 100W.



Example 6 load 2 50w load input voltage 24v 3 consecutive rainy days 8 hours per day

Request the required system solar panel and battery calculation

1. Solar panel 2*50W*8H/0.6/4H=340W (total power consumption/system utilization factor/effective sunshine hours)

2. Battery 2*50/24*8*(3+1)/0.7=200AH (total current* self-holding time/remainder coefficient)

(Solar panel power = load power * operating time / loss 0.6 / average effective light)

(battery capacity = load power * operating time * continuous rainy weather / battery voltage / charge and discharge coefficient)

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Calculated in terms of solar radiation

Annual power generation (EP) = PAS * HA * K * 365 (days)

PAS: Solar Cell Column Capacity

HA: Cumulative solar radiation (kWh/m2 * day) for setting place and setting conditions

K: sum design factor (0.65 to 0.8≒0.7 degree)

Calculated with system utilization

Annual power generation = generation of solar array template * system utilization * 8760 (hours)

System utilization = 0.1 ~ 0.15 ≒ 0.12 degree




Total hours in a year = 24 (hours) * 365 (days) = 8760 hours.



Household electricity can be completely replaced by solar power generation. It will become fashionable in today's environmental protection. We can recommend the best solution for you based on the number of household appliances you use and the geographical location of your location.

Although the solar power system has the advantages of safety, environmental protection and no pollution, but its cost is quite high, so it is generally recommended only for lighting use.

About the approximate cost calculation, can be calculated according to the following simple method to see how to arrange the scale of solar power generation.

1. Calculate the total power consumption per day. The average household should use electricity from 5 to 10 degrees a day. You can use the total monthly electricity price in addition to the unit price of electricity and then the number of days.

2. It can simply apply the formula 5000W (assuming 5 degrees per day electricity)/5 hours (average daily effective illumination time, different in different regions) / 0.7 (the actual efficiency of solar panels) / 0.9 (various losses) = 1600W, then Also add 5% margin, which is almost 1700W.

3. The above figure is the power of the system. At present, the average unit price of the system is 60 yuan/W (including all materials and what is installed). Then the total investment is 1700X60=102000, more than 100,000 yuan. Electricity prices in most regions are now calculated at 0.6 yuan, 102000/0.6=170000 degrees, and 5 kWh per day, which can be used for 90 years.

4. From the above perspective, it is unrealistic for domestic household electricity to rely entirely on solar energy. Foreign countries have developed very well because of state subsidies. We must also have subsidy, and the cost must be substantially reduced. Solar power can truly enter the homes of the people.


The AC power generation system can consist of solar panels, batteries, controllers and inverters. When there is sunlight during the daytime, the battery can be equipped with a controller to charge the battery. At night, the battery is used to power the appliance.

In this case, it is recommended to use a 80W battery board, a 12V20AH battery (locally purchased), 12V5A controller and 300W inverter. In the case of full power, four 20W lamps can be used for 5 hours or more, and most people can use it. If not enough, one or more panels can be added.

This small system is suitable for areas that lack electricity or electricity, such as forest areas, mountain areas, or field work (beekeeping). The cost is not high, it is convenient to carry, you can adjust the system according to your needs, and it can fully meet your daily electricity needs.