Homework_Chapter_03

Homework_Chapter_03 - EEE 165 CSUS Instructor: Russ Tatro...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 ext = 0.03125 = 3.125% 10 a) overall = 0.18% b) part (i) overall = 0.0307% and part (ii) overall = 0.0045% 11 int = 66.7% Poptical(internal) = 97 mW 3.7 External conversion efficiency The external power or conversion efficiency ext is defined as ext = Optical power output Electrical power input = P o 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....
View Full Document

Page1 / 4

Homework_Chapter_03 - EEE 165 CSUS Instructor: Russ Tatro...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online