Page 128 - Photonics Essentials an introduction with experiments

Light-Emitting Diodes

122 Photonic Devices

Responsivity of a red LED is 2 W per ampere times the quan-
tum efficiency, or 0.08
Current needed = optical output power/responsivity = 12.5 mA
electrical power needed = 2 volts × 12.5 mA = 25 mW
Each LED has a diameter of 5 mm, so 20 LEDs = radius of the
red stoplight
Total number of LEDs to fill space = ·20 1250. Actually,
2
it is about 10% less because the LEDs are circular, and do
not entirely fill the space. To be safe, we will use only 1000
LEDs.
Total emitted red light = 1000 × 1 mW = 1 W
This is the end of part one. We conclude that there can be a
reasonable balance between the light emitted from a light
bulb and the light emitted from an array of LEDs. However,
the light bulb consumes 100 Watts of electrical power and
the LED light consumes only 25 Watts.
2. Cost of electrical power

Assume electricity costs 5 cents/kW-h
Stoplights run 24 hours a day, 365 days a year: 24 · 365 9000
hours
Number of kW-hours for lightbulb = 900 kW-h = $45.00 per year
Number of kW-hours for LED = 225 kW-h = $12.00 per year
Manhattan is an island approximately 10 avenues wide and 200
streets long. There are 2000 intersections. There are on the
average 10 stoplights at each intersection, so we estimate
20,000 stoplights.
The cost of running an incandescent stoplight per year is a com-
bination of electricity ($45.00) plus maintenance ($40.00)
The cost of running an LED light is only electricity = $12.00.
Cost of operating 20,000 red lights per year:
incandescent = $1,700,000
LED = $240,000
Savings by using LEDs = $1,460,000 per year.
Savings per light = $73.00
3. How much should you pay for the changeover?
Assume that an LED light lasts 10 years. (It should last forever.
Furthermore, 10% of the LEDS can fail, but the lamp will still be
usable.) You would like for the savings to pay for the cost of the

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