Solar Technician – Fundamentals
Topic: Key Terms
Purpose:
This study guide gives you clear, concise explanations of the most important terms you’ll encounter in a solar‑technician training program.
Use it for quick revision, exam prep, or to refresh your knowledge before a job site visit.
1. Simple Explanation
Solar technology is built around a few core concepts:
| Concept | What it means in plain English |
|---|---|
| PV (Photovoltaic) | The science of turning sunlight into electricity. |
| DC vs. AC | DC is the direct current produced by panels; AC is the alternating current used in homes and grids. |
| Inverter | The device that converts DC from panels to usable AC. |
| Array / String | A collection of panels wired together (normally in series or parallel). |
| MPPT | “Maximum Power Point Tracking” – a smart feature in inverters that extracts the most power from the panels. |
2. Key Terms & Definitions
| Term | Definition |
|---|---|
| Solar Module / Panel | A rectangular assembly of many solar cells that convert light to DC electricity. |
| Photovoltaic Cell | The basic unit (often silicon) that generates electricity when hit by photons. |
| DC (Direct Current) | Electricity that flows in one direction; produced by panels. |
| AC (Alternating Current) | Electricity that periodically reverses direction; used by most appliances and the grid. |
| Inverter | Device that converts DC to AC; can be string, central, or micro‑inverter. |
| String | A group of panels wired in series (or parallel) that make up a portion of an array. |
| Array | The complete set of panels on a roof or ground‑mount system. |
| MPPT (Maximum Power Point Tracking) | A control algorithm that adjusts the inverter’s input to harvest the most power from a panel under given conditions. |
| kWp (Kilowatt-peak) | The maximum power a panel or array can produce under standard test conditions (STC). |
| kWh (Kilowatt‑hour) | A unit of energy; the energy a 1 kW system would produce in one hour. |
| Irradiance | Solar power per unit area, measured in watts per square meter (W/m²). |
| Albedo | Reflective property of a surface; higher albedo means more light reflected onto panels. |
| Tilt | Angle of the panel relative to the horizontal plane. |
| Azimuth | Compass direction a panel faces (0° = true North, 180° = South in the Northern Hemisphere). |
| Shading Loss | Reduction in power output caused by partial shade on a panel. |
| Panel Efficiency | Ratio of electrical output to incident sunlight (e.g., 18 %). |
| BIPV (Building‑Integrated Photovoltaics) | Solar panels integrated into building components (roof tiles, façade). |
| Ground‑Mount | Panels mounted on the ground, usually on racks or poles. |
| DC Load | Device that directly consumes DC (rare in residential systems). |
| Site Assessment | The process of evaluating a location’s suitability for a solar installation. |
| Net Metering | Billing arrangement that credits solar producers for the electricity they send to the grid. |
| Performance Ratio (PR) | Real output ÷ theoretical output, expressed as a percentage. |
| Degradation Rate | Annual loss of panel performance, usually 0.5 %–1 % per year. |
| SIP (Solar Installation Program) | Training or certification program for solar installers. |
Tip: Memorise the acronyms by writing them in a flashcard app and testing yourself twice a week.
3. Worked Example: Power Output Calculation
Problem
A 5 kWp solar array is installed on a roof with a tilt of 35°. During a sunny day, the average irradiance is 800 W/m². The panels have an efficiency of 18 %. What is the expected DC power output?
Solution Steps
-
Determine the effective irradiance on the panel surface.
Since the panels are tilted, the irradiance is less than the horizontal value.
[ I_{\text{effective}} = I_{\text{horizontal}} \times \cos(\theta) ] where ( \theta ) is the angle between sun rays and panel normal.
Assume the sun is directly overhead (simplification).
[ I_{\text{effective}} = 800 \times \cos(35^\circ) \approx 800 \times 0.819 = 655 \text{ W/m}^2 ] -
Calculate the area of the panels.
Power = Irradiance × Area × Efficiency.
Rearrange to find Area:
[ \text{Area} = \frac{\text{Power}}{I_{\text{effective}} \times \text{Efficiency}} ] [ \text{Area} = \frac{5000 \text{ W}}{655 \times 0.18} \approx \frac{5000}{117.9} \approx 42.4 \text{ m}^2 ] -
Compute the actual DC power output.
[ P_{\text{DC}} = I_{\text{effective}} \times \text{Area} \times \text{Efficiency} ] [ P_{\text{DC}} = 655 \times 42.4 \times 0.18 \approx 5000 \text{ W} ]
Result – The array will produce roughly 5 kW DC when irradiance is 800 W/m² and the panels are at 35° tilt.
Note: In practice, you’d also subtract losses (temperature, wiring, inverter efficiency) to estimate AC output.
4. Quick Revision Summary
| Category | Key Term | Quick Mnemonic | |