Prerequisites
Before studying this topic, make sure you understand:
- p-n Junction Formation - Forward and reverse bias concepts
- Diode Characteristics - Basic diode operation
- Semiconductors - Band theory and energy gaps
The Hook: How Does Your Phone Screen Glow?
Your smartphone screen contains millions of tiny LEDs (Light Emitting Diodes)! Each pixel has red, green, and blue LEDs that can mix to create 16 million colors.
Mind-blowing facts:
- Modern OLED screens have >800 pixels per inch - that’s over 2 million LEDs in a 6-inch screen!
- When you dim your screen to save battery, you’re literally reducing current through billions of p-n junctions
- The solar panel on a calculator is essentially running LEDs in reverse - absorbing light instead of emitting it!
How does a simple forward-biased p-n junction emit light? And how does the reverse process generate electricity from sunlight? Let’s explore these miracles of semiconductor physics!
Interactive Demo
See how LEDs emit light and photodiodes detect it:
Light Emitting Diode (LED)
What is an LED?
LED = Forward-biased p-n junction that emits light when current flows
Basic principle:
- Electrons from n-side cross junction
- Recombine with holes in p-side
- Energy released as photons (light)!
Energy and Wavelength
When electron-hole recombination occurs:
Energy of emitted photon:
$$\boxed{E = h\nu = \frac{hc}{\lambda} = E_g}$$where:
- $E_g$ = band gap energy
- $h$ = Planck’s constant ($6.63 \times 10^{-34}$ J·s)
- $c$ = speed of light ($3 \times 10^8$ m/s)
- $\lambda$ = wavelength of emitted light
Wavelength formula:
$$\boxed{\lambda = \frac{hc}{E_g} = \frac{1240}{E_g} \text{ nm}}$$(when $E_g$ is in eV)
Different semiconductor materials → Different band gaps → Different colors!
| Material | Band Gap ($E_g$) | Wavelength | Color |
|---|---|---|---|
| GaAs (Gallium Arsenide) | 1.4 eV | 886 nm | Infrared (IR) |
| GaAsP | 1.9 eV | 650 nm | Red |
| GaP | 2.2 eV | 560 nm | Green |
| GaN | 2.9 eV | 430 nm | Blue |
| InGaN | 2.7-3.4 eV | 365-460 nm | Violet to Blue |
White LED = Blue LED + Yellow phosphor coating!
Silicon ($E_g = 1.1$ eV) and Germanium ($E_g = 0.7$ eV) don’t work as LEDs - they’re indirect band gap semiconductors (energy goes to vibrations, not light).
LED Characteristics
Advantages:
- High efficiency - 90% energy → light (vs 5% for incandescent bulbs!)
- Long life - 50,000+ hours (vs 1,000 for bulbs)
- Fast switching - nanosecond response time
- Small size - can be microscopic
- Low voltage - typically 1.5-3.5V
- Monochromatic - pure color without filters
Forward voltage ($V_f$):
- Red LED: ~1.8V
- Green LED: ~2.0V
- Blue LED: ~3.0V
- White LED: ~3.2V
Higher energy (shorter wavelength) → Higher voltage needed!
Typical current: 10-20 mA
Power: $P = V_f \times I$ ≈ 20-60 mW
LED Circuit
Never connect LED directly to battery! Always use current-limiting resistor.
+V ——R——┬——▷|——⏚
LED
Resistor calculation:
$$\boxed{R = \frac{V_s - V_f}{I_f}}$$where:
- $V_s$ = supply voltage
- $V_f$ = LED forward voltage
- $I_f$ = desired LED current
Example: Power red LED (1.8V, 20mA) from 5V battery:
$$R = \frac{5 - 1.8}{0.02} = \frac{3.2}{0.02} = 160 \text{ Ω}$$Use standard 180Ω resistor.
Incandescent bulb:
- Heats filament to 3000K → mostly infrared heat!
- Only 5% energy becomes visible light
- 95% wasted as heat
LED:
- Direct band gap recombination → photons!
- 80-90% energy becomes light
- Cool to touch, saves energy
This is why:
- LED bulbs use 1/10th the power
- Last 50× longer
- Global push to replace all incandescent bulbs with LEDs!
Your smartphone screen would be impossible with bulbs - it would melt!
Photodiode
What is a Photodiode?
Photodiode = Reverse-biased p-n junction that detects light
Exactly opposite of LED:
- LED: Current → Light
- Photodiode: Light → Current
Working Principle
In dark (no light):
- Reverse-biased junction
- Only small reverse saturation current $I_0$ flows
- Called dark current (~nA)
When illuminated:
- Photons create electron-hole pairs in depletion region
- Electric field separates them immediately
- Electrons → n-side, Holes → p-side
- Photo-current flows (proportional to light intensity)
Total current:
$$\boxed{I = I_0 + I_L}$$where:
- $I_0$ = dark current (constant)
- $I_L$ = photo-current (proportional to light intensity)
Photo-current vs Illuminance
$$\boxed{I_L \propto \Phi}$$where $\Phi$ = illuminance (light intensity)
Linearity: Photo-current is directly proportional to light intensity over wide range!
Responsivity ($R$):
$$\boxed{R = \frac{I_L}{P_{opt}}}$$Typical values: 0.4-0.6 A/W (amperes per watt of light)
Applications of Photodiode
- Light detection - cameras, light sensors
- Optical communication - fiber optic receivers
- Barcode scanners - detecting reflected light patterns
- Remote controls - IR photodiodes detect TV remote signals
- Safety systems - burglar alarms, automatic doors
- Medical instruments - pulse oximeters (measuring blood oxygen)
Ever noticed your phone screen automatically dims in dark and brightens in sunlight?
That’s a photodiode at work!
- Measures ambient light continuously
- Adjusts screen brightness accordingly
- Saves battery in dark conditions
- Prevents eye strain in bright conditions
Same principle used in:
- Auto-brightness in laptops
- Street lights that turn on at dusk
- Camera exposure control
Photodiode vs LED Comparison
| Property | LED | Photodiode |
|---|---|---|
| Bias | Forward | Reverse |
| Function | Emits light | Detects light |
| Input | Electric current | Light photons |
| Output | Light photons | Electric current |
| Material | GaAs, GaP, GaN | Si, Ge (also GaAs) |
| Current | ~10-20 mA | ~µA (depends on light) |
Beautiful symmetry: LEDs and photodiodes are inverse processes!
Solar Cell (Photovoltaic Cell)
What is a Solar Cell?
Solar cell = p-n junction that generates voltage from light
Key difference from photodiode:
- Photodiode: Requires external battery (reverse bias)
- Solar cell: NO external source - generates its own voltage!
Working Principle
When light falls on p-n junction:
Photon absorption:
- Photons with $E > E_g$ create electron-hole pairs
- Generated throughout p-region, n-region, and depletion region
Carrier separation:
- In depletion region: Built-in field separates them
- Electrons → n-side
- Holes → p-side
- Outside depletion: Minority carriers diffuse to junction, then separate
- In depletion region: Built-in field separates them
Voltage generation:
- Accumulation of electrons on n-side
- Accumulation of holes on p-side
- Creates photo-voltage across junction
Current flow:
- Connect load across terminals
- Electrons flow n → load → p (conventional current p → n)
- Continuous power generation!
Solar Cell Parameters
1. Open Circuit Voltage ($V_{oc}$):
Maximum voltage when no load connected (infinite resistance).
Typically:
- Si solar cell: $V_{oc} \approx 0.5-0.7$ V
- Single cell gives only ~0.6V!
- Panels connect many cells in series
2. Short Circuit Current ($I_{sc}$):
Maximum current when terminals shorted (zero resistance).
$$I_{sc} \propto \text{Area} \times \text{Intensity}$$Typical: 30-40 mA/cm² in full sunlight
3. Fill Factor (FF):
Measure of “squareness” of I-V curve.
$$\boxed{FF = \frac{V_m \times I_m}{V_{oc} \times I_{sc}}}$$where $V_m$, $I_m$ = voltage and current at maximum power point
Typical FF: 0.7-0.85 (70-85%)
4. Efficiency ($\eta$):
$$\boxed{\eta = \frac{P_{out}}{P_{in}} = \frac{V_m I_m}{P_{incident}}}$$Or using fill factor:
$$\boxed{\eta = \frac{FF \times V_{oc} \times I_{sc}}{P_{incident}}}$$Typical efficiencies:
- Commercial Si panels: 15-20%
- Lab record (multi-junction): >45%
- Theoretical limit (Si): ~33% (Shockley-Queisser limit)
Major losses:
- Photons with $E < E_g$: Not absorbed (pass through)
- Photons with $E > E_g$: Excess energy lost as heat
- Reflection: ~4% of light reflects off surface
- Recombination: Some carriers recombine before reaching contacts
- Resistance losses: Wire resistance, contact resistance
Maximum theoretical efficiency for single-junction Si cell: ~33%
To beat this: Multi-junction cells
- Stack materials with different $E_g$
- Each layer absorbs different wavelength range
- Space solar panels achieve >30% efficiency this way!
I-V Characteristics
In dark: Regular diode curve
Under illumination: Curve shifts down by $I_{sc}$
In 4th quadrant (negative current, positive voltage):
- Cell delivers power!
- $P = V \times I$ (both positive in magnitude)
Maximum power point:
- Neither at $V_{oc}$ (I = 0, so P = 0)
- Nor at $I_{sc}$ (V = 0, so P = 0)
- At intermediate point where $P = VI$ is maximum
Solar Panel Construction
Single cell: ~0.6V
Practical panel:
- 60 cells in series: $V = 60 \times 0.6 = 36$ V
- Can charge 12V battery (needs >14V for charging)
- 72 cells: 43V (for 24V systems)
For high power:
- Multiple strings in parallel
- Typical home panel: 300-400W
- Voltage: 30-40V
- Current: 8-10A
Applications of Solar Cells
- Residential power - rooftop solar panels
- Satellites - primary power source in space
- Calculators - small solar cells
- Street lights - solar-powered LED lights
- Water pumps - irrigation in remote areas
- Spacecraft - ISS has 2500 m² of solar panels!
International Space Station:
- Solar array area: 2500 m² (half a football field!)
- Power generated: 120 kW (powers ~50 homes)
- No fuel needed - works for decades!
Your calculator:
- Tiny solar cell (~1 cm²)
- Generates ~1 mW
- Enough to power LCD display!
- Never needs batteries in bright light
Global impact:
- Solar panel costs dropped 90% in 10 years
- Now cheapest source of electricity in many countries
- Can power homes, cities, even spacecraft - all from p-n junctions!
Comparison Table
| Device | Bias | Input | Output | Application |
|---|---|---|---|---|
| LED | Forward | Current (20mA) | Light | Displays, lighting |
| Photodiode | Reverse | Light | Current (µA) | Sensors, detectors |
| Solar Cell | None (self) | Light | Voltage + Current | Power generation |
Common physics: All based on photon ↔ electron-hole pair conversion!
Important Formulas Summary
LED
$$\boxed{\lambda = \frac{hc}{E_g} = \frac{1240}{E_g} \text{ nm}} \quad \text{($E_g$ in eV)}$$ $$\boxed{R = \frac{V_s - V_f}{I_f}} \quad \text{(Series resistor)}$$Photodiode
$$\boxed{I = I_0 + I_L}$$ $$\boxed{I_L \propto \Phi} \quad \text{(Light intensity)}$$Solar Cell
$$\boxed{\eta = \frac{FF \times V_{oc} \times I_{sc}}{P_{in}}}$$ $$\boxed{FF = \frac{V_m I_m}{V_{oc} I_{sc}}}$$Memory Tricks & Patterns
Mnemonic for LED vs Photodiode
“LED Lights Forward, Photodiode Pulls Reverse”
- LED: Forward bias → Light output
- Photodiode: Reverse bias → Receives light
Wavelength Calculation Shortcut
“1-2-4-0 divided by eV”
$$\lambda(\text{nm}) = \frac{1240}{E_g(\text{eV})}$$Example: $E_g = 2$ eV → $\lambda = 1240/2 = 620$ nm (red)
Solar Cell Orientation
“Sun Shines on p-type Surface”
Most solar cells have:
- Thin p-layer on top (light enters here)
- Thick n-layer below
- Light must penetrate to junction!
LED Color Order
“RGB Going Blue needs More Energy”
- Red: 1.8-2.0 eV (lowest)
- Green: 2.0-2.4 eV (medium)
- Blue: 2.6-3.0 eV (highest)
Higher energy → Higher voltage needed!
Pattern Recognition
All three devices involve photons:
- LED: Electrons → Photons (emission)
- Photodiode: Photons → Current (detection)
- Solar cell: Photons → Power (generation)
Voltage values:
- LED forward voltage: 1.5-3.5V (depends on color)
- Solar cell voltage: ~0.6V (fixed for Si)
- Photodiode: Externally biased (typically 5-10V reverse)
Efficiency hierarchy:
- LED: 80-90% (electrical → light)
- Solar cell: 15-20% (light → electrical)
- Round-trip efficiency: ~15% (terrible for energy storage!)
When to Use This
Use LED formulas when:
- Asked about wavelength from band gap
- Calculating series resistor
- Comparing different color LEDs
- Energy efficiency questions
Use photodiode concepts when:
- Light detection problems
- Photo-current proportionality
- Sensor applications
- Reverse bias with light input
Use solar cell formulas when:
- Efficiency calculations
- Fill factor problems
- Power output from illumination
- Panel voltage/current in series/parallel
Quick identification:
- Forward bias + light output → LED
- Reverse bias + light input → Photodiode
- No bias + light input + power output → Solar cell
Common Mistakes to Avoid
Wrong: “Connect LED directly to 5V battery”
Correct: ALWAYS use current-limiting resistor!
Why?
- LED has very low dynamic resistance above threshold
- Without resistor, current skyrockets
- LED burns out in seconds!
JEE scenario: “LED rated 2V, 20mA connected to 5V. What happens?”
- Without resistor: LED destroyed
- With resistor: $R = (5-2)/0.02 = 150$ Ω needed
Wrong: “Photodiode and solar cell are same thing”
Correct:
- Photodiode: Requires external battery (reverse bias), measures light intensity
- Solar cell: Generates its own voltage, provides power
Key difference:
- Photodiode: Consumes power (needs battery)
- Solar cell: Generates power (no battery needed)
JEE trap: “Which device needs external bias?” → Photodiode (not solar cell!)
Wrong: “More light → higher voltage from solar cell”
Correct:
- More light → higher current ($I_{sc}$ increases)
- Voltage ($V_{oc}$) stays nearly constant (~0.6V for Si)
- Intensity mainly affects current, not voltage!
Why?
- $V_{oc}$ depends on band gap (material property)
- $I_{sc}$ depends on number of photons (intensity)
To get higher voltage: Connect cells in series, not increase light!
Wrong: Using $E = hc/\lambda$ with wrong units
Correct: Must be consistent!
Method 1 (SI units):
$$E(\text{J}) = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda(\text{m})}$$Method 2 (eV and nm):
$$\lambda(\text{nm}) = \frac{1240}{E_g(\text{eV})}$$JEE shortcut: Memorize 1240 formula - saves calculation time!
Example: Green LED, $E_g = 2.2$ eV
$$\lambda = \frac{1240}{2.2} = 564 \text{ nm}$$✓ (Green region)
Practice Problems
Level 1: Foundation (NCERT/Basic)
An LED has band gap 1.9 eV. Find the wavelength and color of emitted light.
Solution:
$$\lambda = \frac{1240}{E_g} = \frac{1240}{1.9} = 653 \text{ nm}$$Color identification:
- 620-750 nm = Red region
- 653 nm = Red
Answer: λ = 653 nm, Red color
Material: Likely GaAsP (Gallium Arsenide Phosphide)
A red LED (forward voltage 1.8V) needs 15 mA current. Calculate series resistor for 5V supply.
Solution:
$$R = \frac{V_s - V_f}{I_f}$$ $$R = \frac{5 - 1.8}{0.015} = \frac{3.2}{0.015} = 213.3 \text{ Ω}$$Nearest standard resistor: 220Ω
Answer: Use 220Ω resistor
Power dissipated in resistor:
$$P_R = I^2 R = (0.015)^2 \times 220 = 0.05 \text{ W} = 50 \text{ mW}$$Use 1/4W (250 mW) or higher rated resistor.
Level 2: JEE Main
A solar cell has $V_{oc} = 0.6$V, $I_{sc} = 40$ mA, and fill factor FF = 0.8. If incident power is 100 mW, find: (a) Maximum power output (b) Efficiency
Solution:
(a) Maximum power:
$$P_{max} = FF \times V_{oc} \times I_{sc}$$ $$P_{max} = 0.8 \times 0.6 \times 0.04$$ $$P_{max} = 0.0192 \text{ W} = 19.2 \text{ mW}$$(b) Efficiency:
$$\eta = \frac{P_{max}}{P_{in}} \times 100\%$$ $$\eta = \frac{19.2}{100} \times 100\% = 19.2\%$$Answer: (a) 19.2 mW, (b) 19.2% efficiency
Insight: Good commercial Si cell efficiency!
Compare the wavelengths of red LED ($E_g = 1.8$ eV) and blue LED ($E_g = 2.7$ eV).
Solution:
Red LED:
$$\lambda_R = \frac{1240}{1.8} = 689 \text{ nm}$$Blue LED:
$$\lambda_B = \frac{1240}{2.7} = 459 \text{ nm}$$Ratio:
$$\frac{\lambda_R}{\lambda_B} = \frac{689}{459} = 1.5$$Answer: Red wavelength is 1.5 times longer than blue
Energy relationship:
$$\frac{E_B}{E_R} = \frac{2.7}{1.8} = 1.5$$ $$\frac{\lambda_R}{\lambda_B} = \frac{E_B}{E_R}$$(inverse relationship!)
Key insight: Higher energy → shorter wavelength!
A photodiode has dark current 10 nA. When illuminated, total current is 50 µA. Find photo-current.
Solution:
$$I_{total} = I_0 + I_L$$ $$I_L = I_{total} - I_0$$ $$I_L = 50 \times 10^{-6} - 10 \times 10^{-9}$$ $$I_L = 50 \times 10^{-6} - 0.01 \times 10^{-6}$$ $$I_L = 49.99 \times 10^{-6} \approx 50 \text{ µA}$$Answer: Photo-current ≈ 50 µA
Insight: Dark current (10 nA) is negligible compared to photo-current (50 µA) - ratio of 1:5000!
Level 3: JEE Advanced
A solar panel has 36 cells in series. Each cell has $V_{oc} = 0.6$V and $I_{sc} = 2$A. If FF = 0.75 and incident solar power is 500 W, find: (a) Open circuit voltage of panel (b) Maximum power output (c) Overall efficiency
Solution:
(a) Panel voltage:
Cells in series → voltages add:
$$V_{oc,panel} = 36 \times 0.6 = 21.6 \text{ V}$$(b) Panel current:
Cells in series → same current:
$$I_{sc,panel} = 2 \text{ A}$$Maximum power:
$$P_{max} = FF \times V_{oc} \times I_{sc}$$ $$P_{max} = 0.75 \times 21.6 \times 2 = 32.4 \text{ W}$$(c) Efficiency:
$$\eta = \frac{P_{max}}{P_{in}} \times 100\% = \frac{32.4}{500} \times 100\% = 6.48\%$$Answer:
- (a) 21.6 V
- (b) 32.4 W
- (c) 6.48%
Note: Low efficiency suggests this is an older or small panel. Modern panels achieve 15-20%.
An LED emits at 600 nm. If forward current is 20 mA and voltage is 2.0V, find: (a) Band gap energy (b) Power input (c) If quantum efficiency is 50%, find optical power output
Solution:
(a) Band gap from wavelength:
$$E_g = \frac{1240}{\lambda(\text{nm})} = \frac{1240}{600} = 2.07 \text{ eV}$$(b) Electrical power input:
$$P_{in} = V \times I = 2.0 \times 0.02 = 0.04 \text{ W} = 40 \text{ mW}$$(c) Optical power:
Quantum efficiency = 50% means half the electrons produce photons
External quantum efficiency:
$$\eta_Q = \frac{\text{Number of photons emitted}}{\text{Number of electrons injected}}$$Current of 20 mA means:
$$\text{Electrons per second} = \frac{I}{e} = \frac{0.02}{1.6 \times 10^{-19}} = 1.25 \times 10^{17}$$Photons emitted: $0.5 \times 1.25 \times 10^{17} = 6.25 \times 10^{16}$ per second
Energy per photon: $E = 2.07$ eV $= 2.07 \times 1.6 \times 10^{-19}$ J
$$P_{optical} = 6.25 \times 10^{16} \times 2.07 \times 1.6 \times 10^{-19}$$ $$P_{optical} = 0.02 \text{ W} = 20 \text{ mW}$$Or simply: $P_{optical} = 0.5 \times P_{in} = 0.5 \times 40 = 20$ mW
Answer:
- (a) $E_g = 2.07$ eV
- (b) $P_{in} = 40$ mW
- (c) $P_{optical} = 20$ mW
Power efficiency: 20/40 = 50% (same as quantum efficiency here)
Quick Revision Box
| Device | Bias | $V_f / V_{oc}$ | Current | Main Use |
|---|---|---|---|---|
| LED | Forward | 1.8-3.5V | 10-20 mA | Light emission |
| Photodiode | Reverse | -5 to -10V (ext) | µA (depends on light) | Light detection |
| Solar Cell | Self (none) | ~0.6V | Depends on area/light | Power generation |
Key formula: $\lambda(\text{nm}) = 1240 / E_g(\text{eV})$
JEE Strategy: High-Yield Points
LED wavelength calculation - Given $E_g$, find λ
- Use shortcut: $\lambda = 1240/E_g$
- Remember color ranges: Red 620-750, Green 495-570, Blue 450-495 nm
Series resistor for LED - Always appears!
- $R = (V_s - V_f)/I_f$
- Common trap: Forgetting to subtract $V_f$
Solar cell efficiency - Standard problem
- $\eta = (FF \times V_{oc} \times I_{sc}) / P_{in}$
- Typical Si efficiency: 15-20%
- Maximum theoretical: ~33%
Comparison questions:
- “LED vs photodiode bias?” → Forward vs Reverse
- “Which generates power?” → Solar cell (not photodiode!)
- “Effect of increasing light on solar cell?” → Current increases (voltage ~constant)
Conceptual understanding:
- Why Si doesn’t make good LED? → Indirect band gap
- Why photodiode in reverse bias? → Wider depletion, faster response
- Why solar cell efficiency < 100%? → Photon energy loss, recombination, reflection
Numerical values to remember:
- Si solar cell: $V_{oc} \approx 0.6$ V
- LED forward voltage: 1.8V (red) to 3.2V (white)
- $\lambda = 1240/E_g$ (master formula!)
Time-saving trick: For LED color identification:
- < 500 nm → Blue/Violet
- 500-600 nm → Green/Yellow
600 nm → Orange/Red
Related Topics
Within Electronic Devices
- p-n Junction - Foundation of all these devices
- Diode Characteristics - Basic diode behavior
- Zener Diode - Another special diode
- Transistor - Building on junction concepts
Connected Chapters
- Dual Nature of Matter - Photon energy $E = h\nu$
- Atomic Structure - Energy levels and transitions
- Electromagnetic Waves - Light as wave
Real-world Applications
- Displays: Phone/TV screens (OLED - billions of LEDs)
- Lighting: LED bulbs replacing incandescent (90% energy savings)
- Renewable energy: Solar panels (terawatts of installed capacity)
- Communication: Fiber optics (LED/photodiode pairs)
- Sensors: Cameras, proximity sensors, ambient light detection
Teacher’s Summary
LED (Light Emitting Diode) - Forward bias emits light
- Electron-hole recombination releases energy as photons
- $\lambda = 1240/E_g$ determines color
- Different materials (GaAs, GaN) → different colors
- Always use series resistor: $R = (V_s - V_f)/I_f$
Photodiode - Reverse bias detects light
- Photons create electron-hole pairs → current flow
- $I = I_0 + I_L$ where $I_L \propto$ light intensity
- Used in cameras, sensors, optical communication
- Opposite of LED!
Solar Cell - Self-powered by light
- No external bias needed - generates its own voltage
- $V_{oc} \approx 0.6$V for Si (fixed), $I_{sc}$ depends on light
- Efficiency: $\eta = (FF \times V_{oc} \times I_{sc})/P_{in}$
- Powers satellites, homes, calculators!
All three convert between photons and electricity:
- LED: Electricity → Light (emission)
- Photodiode: Light → Current (detection)
- Solar cell: Light → Power (generation)
Your smartphone uses all three:
- OLED screen: Millions of LEDs
- Camera: Photodiode arrays (image sensor)
- Ambient sensor: Photodiode
- All from p-n junction physics!
“Three simple variations of the p-n junction - one for lighting the world with LEDs, one for detecting light in cameras, and one for powering the future with solar energy!”