Solar Panel Calculator

Proper solar system sizing depends on:

• your energy consumption
• available roof space
• local sunlight
• budget

Most homes need 5-10kW systems to offset electricity bills.

Optimize output by:

• choosing south-facing orientation
• proper tilt angle (roughly equal to latitude)
• minimizing shading
• keeping panels clean

Calculate daily solar output with the formula: Daily Energy (kWh) = System Size (kW) × Peak Sun Hours × System Efficiency.

For example, a 6 kW system in a location with 4.5 peak sun hours running at 80% efficiency produces 6 × 4.5 × 0.80 = 21.6 kWh per day.

The energy unit is the kilowatt-hour (kWh), defined by NIST as the energy delivered by one kilowatt of power over one hour, equal to 3.6 megajoules.

Peak sun hours are not clock daylight hours; they represent the number of hours the sun delivers 1000 W/m² of irradiance, the reference intensity used to rate panels.

To supply 1000 kWh per month, first find daily need: 1000 ÷ 30 ≈ 33.3 kWh/day. Divide by peak sun hours and efficiency to get required array power: 33.3 ÷ (4.5 × 0.80) ≈ 9.26 kW. With 400 W panels, panels needed = 9260 ÷ 400 ≈ 24 panels (round up).

Fewer panels are required in sunnier regions; a location with 6 peak sun hours needs about 33.3 ÷ (6 × 0.80) ≈ 6.94 kW, or roughly 18 panels.

Always size against your actual annual utility statements rather than a single month, since consumption and sunlight both vary seasonally.

Solar power is measured in watts (W) and kilowatts (kW), the SI unit of power defined by BIPM as one joule per second.

Energy over time is measured in kilowatt-hours (kWh); one kWh equals 3,600,000 joules (3.6 MJ) per NIST.

Panel ratings use watts-peak (Wp), the output under Standard Test Conditions of 1000 W/m² irradiance, 25°C cell temperature, and air mass 1.5.

Solar irradiance itself is expressed in watts per square metre (W/m²), and daily insolation in kWh/m²/day. The SI base for these derived units traces to the second, metre, and kilogram, as codified by the International System of Units (SI).

Panel efficiency is the fraction of incident sunlight a module converts to electricity; most residential silicon panels convert 19-22% of solar irradiance, according to figures published by the National Renewable Energy Laboratory.

This module efficiency is separate from system efficiency, which accounts for downstream losses:

• inverter conversion (2-4%)
• DC and AC wiring resistance
• module mismatch
• soiling
• temperature derating

HyperPhysics (Georgia State University) explains that photovoltaic conversion is limited by the semiconductor bandgap, which caps single-junction silicon near the Shockley-Queisser theoretical limit of about 33%.

In practice, combine these factors into a single system efficiency (typically 75-85%) that multiplies the ideal output in your calculation.

Photovoltaic panels generate electricity through the photovoltaic effect, first explained in principle by Einstein's work on the photoelectric effect.

As described by HyperPhysics, photons with energy above the silicon bandgap (about 1.1 electronvolts) excite electrons across the p-n junction, producing a direct current.

Khan Academy notes that the panel's electrical power follows P = V × I, where the current scales with light intensity and the voltage depends on the semiconductor. An inverter then converts this DC into grid-compatible AC.

Temperature raises electron thermal energy and reduces voltage, which is why hot panels lose roughly 0.3-0.5% of output per degree Celsius above the 25°C rating point.

Solar output calculations drive several real-world designs:

• Grid-tied systems size the array to offset annual consumption and feed surplus to the utility under net metering.
• Off-grid systems must oversize both the array and battery bank to survive low-sun days, sizing storage in kilowatt-hours to cover autonomy periods.
• Hybrid systems pair panels with lithium batteries and a smart inverter for backup power.
• Utility-scale solar farms apply the same output formula across thousands of panels, adding tracking mounts to raise effective peak sun hours.

Engineers also use the calculation for RV and marine power, water pumping, and telecom sites, where accurate sizing prevents both blackouts and wasted capital.

Estimate annual savings by multiplying yearly energy production by your electricity rate: Savings = Annual kWh × Rate. A system generating 8,000 kWh at $0.15/kWh saves $1,200 per year.

Simple payback period equals net system cost divided by annual savings; a $14,000 installation (after a 30% federal tax credit on a $20,000 gross cost) saving $1,200/year pays back in about 11.7 years.

Net metering, time-of-use rates, and rising utility prices shorten payback further.

Because panels degrade roughly 0.5% per year, most manufacturers warranty at least 80-85% of rated output at 25 years, meaning decades of continued savings after payback.

Location determines peak sun hours, the single largest driver of output. The Desert Southwest can exceed 6 peak sun hours daily, while the Pacific Northwest averages 3.5-4.

Orientation matters too: in the Northern Hemisphere, true-south-facing panels capture the most annual energy, per NREL guidance, while south orientation is optimal in the Southern Hemisphere.

Tilt angle roughly equal to your latitude maximizes year-round yield; steeper tilts favor winter production.

Shading is especially damaging because series-wired panels behave like a chain, so one shaded module can throttle a whole string.

Use local irradiance data from NREL's dataset rather than assumed values for accurate results.

• A frequent error is confusing peak sun hours with total daylight hours, which inflates estimated output by two- to threefold.
• Another is ignoring system efficiency and using nameplate wattage directly, overstating real production by 15-25%.
• Many people size against one summer month instead of annual consumption, leaving winter shortfalls.
• Forgetting temperature derating in hot climates and neglecting future load growth (an EV, heat pump, or addition) are common too.
• Finally, undersizing roof area (each 400 W panel needs roughly 1.8-2.0 m²) or overlooking shading and inverter clipping leads to disappointing results.

Always cross-check estimates against NREL's PVWatts and your real utility bills.