Sizing a photovoltaic electric system for your home and selecting the right one may be challenging. One of the main questions you will face is “How many solar panels do I need to power the house?” Well, as you probably expected, there is no simple answer to this question. It all depends on many factors. You would need to decide how much solar electricity you want to generate from your PV array, estimate the average amount of sunshine in your area, decide where your PV array will be installed, and calculate the amount of energy the panel you select would produce in your location. So, let's discuss these questions step by step.


First you need to determine an average amount of energy your home uses daily. If you are on grid, this amount may be indicated in your electric utility bill. If it is not, you may gather your bills over entire year, add all monthly kilowatt-hours and divide the result by 365. Just for reference, according to EIA, in 2010 the yearly average electricity consumption for a residential customer in US was 11,496 kWh. This is about 31 kWh per day. In general, it may not be economically feasible to power your entire house from sunlight unless it consumes very small amount of electricity. So, the second step is to decide what percentage of your electrical needs you want to offset with solar energy. Many guides and tutorials recommend 50%, but this is pretty much an arbitrary number.

Desired daily generation (kWh)
Panel's rated wattage
Daily insolation (kWh/sq.m)
Inverter efficiency (%)
Required number of panels


Once you decide on the amount of electricity you want to generate from sunlight, you can estimate the required size of your system. This is often a confusing part of the calculation. Indeed, the amount of electricity is measured in kilowatt-hours, while solar generators and PV panels are rated in watts. People are often wondering how to convert kWh to watts? This calculator would just do it for you, but I'll still provide a simple explanation for those who would like to understand the details. Let's first revisit the difference between watt and kilowatt-hour. Watt is a unit of power, which is energy rate per unit time. A kilowatt-hour is a unit of energy corresponding to the flow of power of one kilowatt (1000 watt) for one hour. So, basically, to “convert” kWh to watts you need to divide kWh by the time and multiply by 1000. Another fact to remember is the characteristics of photovoltaic systems are usually provided for so-called Standard Test Conditions (STC). STC are close to the conditions on a bright noon with the panel tilted so that it is exactly perpendicular to the sunrays. Accordingly, your panel may provide its rated power only for a short period of time at noon. As the sun moves in the sky, the amount of sunlight that falls onto the cells obviously varies. Since we are trying to size a PV system that would deliver a required amount of kWh per day, we don't care much about peak power. What we need to find is the net amount of radiated energy your system will receive over an entire day. This value is called insolation. Its average amount depends on your latitude, on orientation of your PV array surface and on it's the angle to horizon. In most places of U.S., the annual average insolation on flat-plate PV collectors tilted south at latitude is from 4 to 7 kWh/sq.m per day. You can quickly find an approximate amount of kWh for your geographical area from the solar radiation resource map. If you want more accurate data, you can determine an exact irradiance for your address and your collectors orientations from PVWATTS tool. The sunlight intensity of 1 kW/sq.m is sometimes called “standard sun”- this is peak of radiated solar power that you can get at a bright noon with optimum tilt. So, the insolation numerically equals to the amount of standard sun-hours a day. Once you know this value, you can calculate the required size of your solar generator. Suppose you use 30 kWh a day and you want your generator to offset 50% of your electric usage, or 15 kWh. An important fact to remember is the wattage rating of a photovoltaic system tells you how much DC power the PV array can produce under sunlight of 1kW/sq.m. If you determined that in your area at a given tilt you get for example, 5 kWh/sq.m daily average, and if there were no losses, you would need a system rated at 15/5=3 kW. Obviously, the actual wattage has to be higher in order to compensate for losses in the inverter and wiring. The efficiency of most grid-connected inverters is from 90 to 97%- you can find the actual number in the inverter's data sheet. Typical wiring looses are 3-5%. Our solar panel calculator assumes wiring losses of 4% and rounds the result. For example, if the inverter that you are considering has efficiency of 95%, you would need a system with 15/51.04/0.95=3.3 kW rated DC output.


Once you found the desired net wattage of the PV array, and selected a particular type, it is easy to determine the required amount of the panels. You just need to divide the desired net DC wattage by the rating of a single panel and round the result. This is basically what our tool does. In the above example, if you choose for example a 300W model, you would need 3300/300=11 modules. If during daylight hours, i.e. when PV cells generates electricity, your home will consume less energy than you produce, the excess will be fed back into the grid. Alternatively, if you store energy in batteries you can use it gradually over entire day.