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Calculation of External and Internal Conversion Efficiencies in LEDs, Assignments of Physical Education and Motor Learning

Solutions to homework problems related to the external and internal conversion efficiencies of light emitting diodes (leds). The problems involve calculating the external conversion efficiency of an algaas red led and the overall efficiency of coupling light from an led into multimode and single mode fibers. Additionally, the internal quantum efficiency of an algaas led is determined using the mean lifetimes of radiative and nonradiative recombinations.

Typology: Assignments

2009/2010

Uploaded on 03/28/2010

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EEE 165 CSUS Instructor: Russ Tatro
Chapter 3 Semiconductor Science and Light Emitting Diodes
S.O. Kasap, Optoelectronics and Photonics, Principles and Practices, 2001
Solutions to Chapter 3 Homework.
Problems 7, 10, 11
7 ηext = 0.03125 = 3.125%
10 a) ηoverall = 0.18%
b) part (i) ηoverall = 0.0307% and part (ii) ηoverall = 0.0045%
11 ηint = 66.7%
Poptical(internal) = 97 mW
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Download Calculation of External and Internal Conversion Efficiencies in LEDs and more Assignments Physical Education and Motor Learning in PDF only on Docsity!

EEE 165 CSUS Instructor: Russ Tatro

Chapter 3 Semiconductor Science and Light Emitting Diodes S.O. Kasap, Optoelectronics and Photonics, Principles and Practices , 2001

Solutions to Chapter 3 Homework.

Problems 7, 10, 11

7 ηext = 0.03125 = 3.125%

(^10) a) ηoverall = 0.18% b) part (i) ηoverall = 0.0307% and part (ii) ηoverall = 0.0045%

(^11) ηint = 66.7% Poptical(internal) = 97 mW

3.7 External conversion efficiency The external power or conversion efficiency ηext

is defined as

ηext =

Optical power output Electrical power input

Po IV

One of the major factors reducing the external power efficiency is the loss of photons in extracting the emitted photons which suffer reabsorption in the pn junction materials, absorption outside the semiconductors and various reflections at interfaces.

The total light output power from a particular AlGaAs red LED is 2.5 mW when the current is 50 mA and the voltage is 1.6 V. Calculate its external conversion efficiency.

Solution

ηext =

Po IV

2.5 × 10 −^3 W

(50 × 10 −^3 A)(1.6 V)

3.11 Internal quantum efficiency The internal efficiency ηint gauges what fraction

of electron hole recombinations in the forward biased pn junction are radiative and therefore lead to photon emission. Nonradiative transitions are those in which an electron and a hole recombine through a recombination center such as a crystal defect or an impurity and emit phonons (lattice vibrations). By definition,

ηint =

Rate of radiative recombination Total rate of recombination (radiative and nonradiative)

or ηint =

τ r

τ r

τ nr

where τ r is the mean lifetime of a minority carrier before it recombines radiatively and τ nr

is the mean lifetime before it recombines via a recombination center without emitting a photon. The total current I is determined by the total rate of recombinations whereas the number of photons emitted per second (Φ ph ) is determined by the rate of radiative recombinations.

ηint =

Photons emitted per second Total carriers lost per second

Φ (^) ph I / e

Pop (int) / h υ

I / e

where Pop (int) is the optical power generated internally (not yet extracted).

For a particular AlGaAs LED emitting at 850 nm it is found that τ r = 50 ns and τ nr

= 100 ns. What is the internal optical power generated at a current of 100 mA?

Solution

Consider ηint =

τ r

τ r

τ nr

50 ns 1 50 ns

100 ns

From ηint =

Pop (int) / h υ

I / e

Pop (int) =ηint (

I

e

) h υ =ηint

Ihc

e λ

(100 × 10 −^3 )(6.626 × 10 −^34 )(3 × 10 8 )

(1.6 × 10 −^19 )(850 × 10 −^9 )

= 0.097 W or 97 mW