14.7 A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

<p><strong>14.7</strong> Energy band gap of the given photodiode,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>g</mi></mrow></mrow></msub></math></span>&nbsp;= 2.8 eV</p><p>Wavelength,&nbsp;<span title="Click to copy mathml"><math><mi>λ</mi></math></span>&nbsp;= 6000 nm = 6000&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>9</mn></mrow></mrow></msup></math></span>&nbsp;m</p><p>The energy of a signal is given by the relation,</p><p><span title="Click to copy mathml"><math><mi>E</mi></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>h</mi><mi>c</mi></mrow></mrow><mrow><mrow><mi>λ</mi></mrow></mrow></mfrac></math></span></p><p>Where</p><p>h = Planck’s constant = 6.626&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>34</mn></mrow></mrow></msup></math></span>&nbsp;Js</p><p>c = sped of light = 3&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>8</mn></mrow></mrow></msup></math></span>&nbsp;m/s</p><p><span title="Click to copy mathml"><math><mi>E</mi></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>6.626</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>34</mn></mrow></mrow></msup><mo>×</mo><mn>3</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>8</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>6000</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>9</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;= 3.313&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>20</mn></mrow></mrow></msup></math></span>&nbsp;J =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>3.313</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>20</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>1.6</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>19</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;eV = 0.207 eV</p><p>Therefore the energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV – the energy band gap of the photodiode. Hence, the photodiode cannot detect the signal.</p>

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