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<h2 class="hd hd-2 unit-title">Light-Emitting Diodes</h2>
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<h2>Light-Emitting Diodes</h2>
<ul>
<li>LEDs are forward-biased diodes. Minority carriers are injected into the depletion region where they recombine, giving off photons.</li>
<li>In LEDs, photons are emitted through spontaneous emission. </li>
<li>The band gap of the active area determines the wavelength of emitted light. </li>
<ul>
<ul>
<ul>
<li>\(\lambda \;(\mu m) = \frac{1.24}{E \; (eV)}\)</li>
</ul>
</ul>
</ul>
<li>A typical LED emissions spectrum is shown below: </li>
</ul>
<p><img src="/assets/courseware/v1/6810784e0bf54e796748b96732201f2b/asset-v1:MITx+3.15.2x+2T2017+type@asset+block/315_LED_Intensity2.jpg" alt="I vs E for an LED. Peak emission intensity is at E_G + kT/2" type="saveimage" target="[object Object]" width="450" height="306" /></p>
<ul>
<li>Heterostructures are employed to prevent light reabsorption.</li>
</ul>
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Exercise 2.1A
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<p>At what energy is the intensity maximum located for a GaAs LED operating at room temperature? The band gap of GaAs is \(1.39 \;eV\).</p>
<p>Energy (in \(eV\)):</p>
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Exercise 2.1B
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<p>What wavelength does this correspond to?</p>
<p>\(\lambda\) (in \(nm\)):</p>
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<h2 class="hd hd-2 unit-title">Semiconductor Lasers</h2>
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<h2>Semiconductor Lasers</h2>
<ul>
<li>Light amplification can occur when stimulated emission > spontaneous emission</li>
<li>We must achieve population inversion for stimulated emission to exceed spontaneous emission.</li>
<li>In a semiconductor laser, population inversion is achieved when there are more electrons in the conduction band than in the valence band. </li>
<li>This is most easily achieved by using a heterostructure to trap carriers in a small region.</li>
<li>We also require a high photon density in the active region to obtain stimulated emission.</li>
<li>A high photon density is achieved by creating a resonant cavity to confine the light.</li>
<li>Interference within the laser cavity limits the wavelengths that can exist within the cavity to those for which twice the cavity length is an integer number of wavelengths (so that constructive interference occurs):</li>
<ul>
<ul>
<ul>
<li>\(N(\frac{\lambda}{n}) = 2d\)</li>
<li>where \(N\) is an integer, \(\lambda\) is the free-space wavelength of light, \(n\) is the refractive index of the laser cavity, \(d\) is the length of the laser cavity, and \(\lambda/n\) is the wavelength of the light within the laser cavity.</li>
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Example 2.2A
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<p>A heterojunction laswer has an output wavelength of \(600 \; nm\). The index of refraction of the laser cavity is \(3.3\), and the cavity length is \(300 \; \mu m\). </p>
<p>What was the wavelength of the emitted light while it was in the laser cavity?</p>
<p>Wavelength (in \(nm\)):</p>
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Exercise 2.2B
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<p>What is the mode number \(N\) of the emitted wavelength?</p>
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Exercise 2.2C
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<p>What is the difference in wavelength between the emitted mode and the next-higher energy mode?</p>
<p>Distance (in \(nm\)):</p>
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<h2 class="hd hd-2 unit-title">Materials Selection for Heterojunction LEDs</h2>
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<h2>Materials Selection for Heterojunction LEDs</h2>
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<li>The band gap of the active layer determines the emission wavelength of the LED:</li>
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<ul>
<li>\(\lambda (\mu m) = \frac{1.24}{E_g (eV)}\)</li>
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<li>The cladding has a larger band gap than the active layer; therefore, the light produced in the active layer is not absorbed by the cladding</li>
<li>The cladding must be lattice matched to the active layer.</li>
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Exercise 2.3A
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><p>Assume that you want to create a red LED with an output wavelength of \(670 \; nm\).</p>
<p>Which of the following materials would make an appropriate active layer?</p>
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Exercise 2.3B
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><p>Assume that we instead choose to make our active layer using AlGaAs. Which of the following would make an appropriate cladding layer?</p>
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<h2 class="hd hd-2 unit-title">Drawing Heterojunction Band Diagrams</h2>
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<h2>Drawing Heterojunction Band Diagrams</h2>
<p>For complete instructions, as well as a chart for determining \(\Delta E_v\) for a variety of materials combinations, please see this week's handout.</p>
<p>Steps for drawing heterostructure band diagrams:</p>
<ol>
<li>Draw the basic structure with \(\Delta E_v\) and \(\Delta E_c\) going in the correct direction. \(\Delta E_v\) and \(\Delta E_c\) are dependent on the particular combination of materials, and can be looked up in tables and charts, such as the one given in this week's handout.</li>
<li>Redraw the band structure with the correct doping levels. Make sure that the Fermi level is correctly positioned and is flat (since the device is currently unbiased). Include any band bending that occurs at the interfaces to accommodate doping.</li>
<li>To forward bias the structure, raise the energy level of the more negative side so that the Fermi level is no longer flat. This changes the bending at the interfaces, leading to current flow. </li>
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Exercise 2.4A
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<p>Consider a heterojuncation made from Material A and Material B. Material A has a bandgap of \(1.8 \; E_v\). Material B has a band gap of \(1.3 \;eV\). The conduction band offset \(\Delta E_c = 0.3 \;eV\). (The conduction band steps down as we move from A to B.)</p>
<p>What is the valence band offset?</p>
<p>\(\Delta E_v\) (in \(eV\)):</p>
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Exercise 2.4B
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<p>Begin by drawing the basic band structure of the heterojunction. When you're finished, click on the "Show Answer" button to check your work. </p>
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Exercise 2.4C
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Exercise 2.4D
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<p>Finally, redraw the junction under a forward-bias condition. When you've completed your drawing, click on the "Show Answer" button to check your work.</p>
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