PSLE-SCI-REALITY-0349
Wait, What? If a Solar Panel Is 22% Efficient, the Other 78% Is Not One Single Thing
A solar-panel specification says module efficiency: 22%. A student looks at the number and says, “Easy. Twenty-two percent of the sunlight becomes electricity, so the other seventy-eight percent becomes waste heat.”
The first half is close to the intended meaning under the stated test conditions. The second half is too simple. Some incoming light is reflected. Some photons carry too little energy to be converted by the semiconductor. Some excess photon energy is lost through thermal processes. Electrical and optical losses occur inside the module. The final energy accounting is more complicated than “electricity or heat”.
This article is not a solar-cell engineering manual. Its job is to train a Primary 5 or 6 learner to read an efficiency percentage as a ratio with a defined numerator, denominator and test condition, then resist the temptation to turn the remainder into an unsupported physical story.
Quick Answer
- Solar-module efficiency compares useful electrical power output with incident solar power falling on the module under specified conditions.
- 22% does not mean the panel produces 22 watts unless the incoming solar power is known.
- 22% also does not mean the remaining 78% follows one single pathway.
- Some incoming light may be reflected or not absorbed; absorbed energy can be lost through several physical processes, many of which ultimately warm the module or surroundings.
- Laboratory efficiency, rated power and real daily energy yield are related but different quantities.
The Exact Learner Job This Page Owns
This page owns one real-world evidence-transfer job: evaluating a photovoltaic efficiency specification without confusing an input-output ratio with rated power, daily energy generation, or a complete breakdown of where every unconverted joule goes.
It does not re-teach the photovoltaic effect, semiconductor band gaps, solar irradiance or electrical power as standalone concepts. Those remain with their existing owners. This page applies existing PSLE Science habits to a product datasheet, comparison chart or advertisement.
- Reality Lab Vol No.180: “Solar Panel Rated 400 W” — Does It Produce 400 W All Day?
- Reality Lab Vol No.210: “Solar Irradiance = 1000 W/m²” — Is Every 1 m² Solar Panel Producing 1000 W?
- How to Design an Indirect Measurement in PSLE Science
Original Reality Lab Case: Two Panels, One Sunny Roof
This is an original composite case. No manufacturer advertisement or examination question has been copied.
A fictional school compares two one-square-metre demonstration panels under the same test irradiance of 1000 W/m².
| Panel | Efficiency | Electrical output under the stated idealised test |
|---|---|---|
| A | 20% | about 200 W |
| B | 22% | about 220 W |
Leonie says, “Panel B is 2% better.” Maren says, “Panel B makes 22% more electricity.” Iona says, “Panel B converts a larger fraction of the same incident solar power into electricity, but we need careful mathematics before describing the size of the advantage.”
Iona is closest. The efficiency rises by two percentage points, from 20% to 22%. Relative to 20%, that is a 10% increase in conversion efficiency under the same conditions. The language chosen changes the numerical claim.
Observed, Specified, Calculated and Inferred
| Layer | What belongs here |
|---|---|
| Observed or tested | Incident irradiance, module area, temperature and electrical output under defined test conditions. |
| Specified | The manufacturer or laboratory reports a module efficiency derived from those quantities. |
| Calculated | Efficiency is useful electrical power output divided by incident solar power, expressed as a fraction or percentage. |
| Inferred | A higher efficiency can allow more electrical output from the same illuminated area under comparable conditions, but it does not by itself determine annual energy yield or every loss pathway. |
The Ratio Check: What Made the 100%?
An efficiency percentage has a numerator and a denominator. For a photovoltaic module, the useful output is electrical power. The input is the solar power incident on the module’s area under the test definition.
efficiency = electrical power out ÷ solar power in
If 1000 W of solar power falls on the active module area and the module supplies 220 W of electrical power under those conditions, the conversion efficiency is 22%. The percentage is therefore a relationship between two energy-flow rates. It is not a label attached to the module independent of conditions.
The Area Check: 22% Does Not Tell You the Panel’s Wattage by Itself
Imagine two panels with the same 22% efficiency. One has twice the area of the other. Under the same irradiance, the larger panel intercepts roughly twice as much solar power and can therefore have roughly twice the electrical output, before other differences are considered.
That is why datasheets contain both efficiency and rated power. Efficiency describes conversion per unit of incident input. Rated power describes an output under defined conditions for that particular module.
The Irradiance Check: The Sun Does Not Deliver a Fixed Input All Day
Standard module ratings are commonly measured under standard test conditions that include a defined irradiance, spectrum and cell temperature. Real roofs move away from those conditions. Clouds reduce irradiance. The Sun’s angle changes. Modules heat up. Shading can affect part of the array. Dust and electrical system losses matter.
A 22% laboratory efficiency therefore does not mean “22% of some fixed daily amount every minute”. The input itself changes through the day.
The Temperature Check: A Hotter Panel Can Produce Less Electrical Power
Photovoltaic modules often become less efficient as cell temperature rises. That creates an important real-world twist: strong sunlight supplies more incident energy, but the same sunlight can also heat the module, changing its electrical behaviour.
This does not mean heat is “bad energy” in a moral sense. It means the photovoltaic device is designed to deliver electrical energy, and temperature affects how effectively it performs that job.
The Remainder Check: Where Can the Unconverted Energy Go?
Now return to the tempting sentence: “The other 78% becomes waste heat.” It captures part of the story but collapses several pathways.
- Reflection: some light never enters the cell because it is reflected from surfaces.
- Transmission or incomplete absorption: depending on the cell and module structure, some wavelengths may not be absorbed usefully.
- Below-band-gap photons: some photons do not carry enough energy to create the charge carriers needed for the intended electrical conversion.
- Excess photon energy: energy above what the semiconductor can use for one electron-hole excitation is rapidly lost through thermalisation.
- Electrical losses: resistance and imperfect charge collection reduce useful electrical output.
Much of the non-electrical energy eventually appears as heat in the module or surroundings, but the scientifically careful learner should not replace a full energy accounting with the claim that the missing percentage is one directly measured bucket called “waste heat”.
The Comparison Check: When Is 22% Fairly Better Than 20%?
A fair efficiency comparison requires comparable definitions and conditions. If one value is module efficiency and another is cell efficiency, they may describe different objects. If one value was measured under a standard laboratory condition and another is a field estimate at a different temperature, the comparison can become misleading.
- same type of efficiency;
- same relevant test standard;
- same or properly normalised irradiance;
- same temperature basis;
- same area definition;
- comparable measurement uncertainty.
Efficiency Is Not the Same as Annual Energy Yield
Suppose Panel A is slightly more efficient but is badly shaded each afternoon. Panel B is slightly less efficient but faces the Sun well and stays unshaded. Which produces more energy over the year?
The efficiency number alone cannot answer. Annual yield depends on irradiance over time, orientation, shading, temperature, downtime, inverter performance and other system conditions. A product specification becomes a performance prediction only after those conditions enter the model.
Alternative Explanations for a Lower Field Output
If a 22% panel produces less power than expected at noon, do not immediately conclude that the efficiency claim is false. Plausible explanations include:
- irradiance below the rating condition;
- higher cell temperature;
- partial shading;
- soiling;
- orientation away from the Sun;
- wiring or inverter losses;
- measurement error;
- normal manufacturing tolerance.
The correct next step is to discriminate among explanations with evidence rather than choosing the most dramatic story.
What Evidence Would Strengthen a Product Comparison?
- The efficiency values use the same recognised test basis.
- Rated power, area and efficiency are internally consistent.
- Temperature coefficients and tolerances are stated.
- Independent certification or laboratory data support the specification.
- Field-energy comparisons account for location, orientation, shading and system losses.
What Would Weaken an Advertisement?
- It compares cell efficiency with another product’s module efficiency.
- It says “22% efficient, therefore 22% more energy” without giving the comparison baseline.
- It treats the remaining 78% as a directly measured heat percentage.
- It uses rated power and efficiency as though they are synonyms.
- It promises a fixed daily output from the efficiency number alone.
Worked Case 1: Same Efficiency, Different Area
Panel A and Panel B are both 22% efficient. Panel B has twice the area. Under the same uniform irradiance, should they have the same electrical power output?
No. The larger panel intercepts more solar power. Equal efficiency means equal conversion fraction, not equal total output.
Worked Case 2: 20% to 22%
An advertisement says, “Our new panel is 2% more efficient because efficiency rose from 20% to 22%.” Is that wording unambiguous?
No. The increase is two percentage points. Relative to the original 20%, the conversion efficiency increased by 10%. A careful communication should state which comparison is meant.
Worked Case 3: Cool Laboratory, Hot Roof
A module meets its datasheet rating in a controlled test but produces less power on a very hot roof even with strong sunlight. Does that prove the datasheet is fraudulent?
No. First compare cell temperature, irradiance and system conditions with the rating conditions. The claim applies under a defined measurement basis.
Worked Case 4: The 78% Claim
A pupil writes, “If efficiency is 22%, exactly 78% of the incoming sunlight is measured as heat in the panel.” What is wrong?
The 78% is the fraction not delivered as useful electrical output under the efficiency definition. It does not identify one directly measured loss pathway. Reflection, spectral non-absorption, thermalisation, resistance and other processes contribute.
Tempting Reasoning That Fails
- “22% efficiency means 22 W.” Output also depends on incident power and area.
- “The remaining 78% is all one kind of loss.” Multiple pathways exist.
- “Higher efficiency always means more yearly energy.” System conditions matter.
- “Rated 400 W means efficiency is 40% at 1000 W/m².” You also need the module area.
- “Efficiency is fixed in all conditions.” Device performance changes with temperature, irradiance and other conditions.
Model and Measurement Limits
Efficiency is a compressed summary of a complex device. Laboratory standards make comparisons possible by fixing important test conditions, but no single number captures spectral response, temperature behaviour, degradation, shading sensitivity, bifacial gain, system losses or lifetime energy production.
The number remains useful precisely because its job is narrower. Scientific literacy means respecting that job rather than demanding that one metric answer every question.
How Far Can the Conclusion Travel?
A verified 22% module efficiency can support the statement that, under the defined test conditions, about 22% of the incident solar power is converted to electrical power at the module output.
It cannot by itself tell us annual energy yield, financial return, durability, total environmental impact, exact roof performance or a complete energy-loss budget.
PSLE-Style Transfer Case
A fictional one-square-metre panel receives 800 W/m² of solar irradiance and produces 160 W electrical power at that moment. A pupil says, “The panel is 20% efficient, so the other 80% must have been measured as heat.”
Explain why only the first part is supported.
Reasoned answer: 160 W divided by 800 W gives 20% electrical conversion efficiency for that moment and area. The remaining input was not converted to useful electrical output, but the information does not show that all of it was directly measured as heat. Other optical and electrical loss pathways can contribute.
Explained Practice
Practice A: Two equal-area panels receive the same irradiance. One is 21% efficient and one 23%. Which should deliver more electrical power under those conditions? The 23% panel, assuming the definitions and conditions are comparable.
Practice B: A 22% panel and a 20% panel have different sizes. Can efficiency alone identify which has the higher rated power? No. Area and test input matter.
Practice C: A high-efficiency panel is shaded for half the day. Does its efficiency number guarantee higher daily energy than a lower-efficiency unshaded panel? No.
Delayed Independent Return: R-A-T-I-O
- R — Ratio: What useful output is divided by what input?
- A — Area: How much solar power reaches the module?
- T — Test conditions: What irradiance, spectrum and temperature define the rating?
- I — Incomplete conversion: Do not invent one loss pathway for the remainder.
- O — Outcome boundary: Efficiency is not the same as daily or yearly energy yield.
Parent and Tutor Teaching Guide
Use 100 counters to represent incoming energy. Put 22 into an “electrical output” cup. Ask the learner where the remaining 78 must go. If the child immediately says “heat”, introduce separate cups labelled reflected light, not-usefully-absorbed light, internal thermal losses and electrical losses. The exact sizes do not matter for this teaching model; the point is that a remainder in a ratio does not name its own mechanism.
Then change the panel area while keeping efficiency fixed. Ask why total output changes. Finally change irradiance while keeping area and nominal efficiency fixed. These three manipulations—loss pathway, area and input—separate concepts that advertisements often compress into one attractive percentage.
Authoritative Sources
- Singapore Examinations and Assessment Board — 2026 PSLE Science Syllabus
- Ministry of Education Singapore — 2023 Primary Science Teaching and Learning Syllabus
- U.S. Department of Energy — Solar Photovoltaic Performance and Efficiency Basics
- U.S. Department of Energy — Solar Photovoltaic Cell Basics
The Quiet Return
Twenty-two percent is meaningful because the ratio is defined.
The remaining seventy-eight percent does not explain itself.
Read the ratio first. Explain the remainder only when evidence identifies the pathways.
