Suppose a n-type wafer is‘created by doping Si crystal having 5 x 1028atoms/m3 with 1 ppm concentration of As. On the surface 200 ppm boron is added to create ‘p’ region in this wafer. Considering ni = 1.5 x 1016 m-3, (i) Calculate the densities of the charge carriers in the n and p regions, (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Physics Class 12th Semiconductor Electronics

Suppose a n-type wafer is‘created by doping Si crystal having 5 x 10²⁸atoms/m³ with 1 ppm concentration of As. On the surface 200 ppm boron is added to create ‘p’ region in this wafer. Considering n_i = 1.5 x 10¹⁶ m^-3,
(i) Calculate the densities of the charge carriers in the n and p regions,
(ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

whem As is used then it will create n type semiconductor

N_e=N_D= $\frac{1}{10^{6}} * 5 * 10^{28} = 5 * 10^{22} / m^{3}$

Number of minority carriers n_h= $\frac{{n i}^{2}}{n e} = \frac{{(1.5 * 10^{16})}^{2}}{5 * 10^{22}}$ = 0.45 $* 10^{10} / m^{3}$

But when B is implanted in Si crystal then p type semiconductor is formed

N_h=N_A= $\frac{200}{10^{6}} * (5 * 10^{28}) = 1 * 10^{25} / m^{3}$

Minority carriers created in p type is

N_e= $\frac{{n i}^{2}}{n e} = \frac{({1.5 * 10^{16})}^{2}}{1 * 10^{25}} = 2.25 * 10^{27} / m^{3}$

Minority charge carriers holes in p type would contribute more in reverse saturatiom current.

This is a Long Answer Type Questions as classified in NCERT Exemplarwhem As is used then it will create n type semiconductorNe=ND= <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>×</mo> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> <mo>=</mo> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>22</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> Number of minority carriers nh= <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>n</mi> <mi>i</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> <mi>e</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mn>1.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>22</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 0.45 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> But when B is implanted in Si crystal then p type semiconductor is formedNh=NA= <math> <mfrac> <mrow> <mrow> <mn>200</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>×</mo> <mfenced separators="|"> <mrow> <mrow> <mn>5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>28</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> <mo>=</mo> <mn>1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>25</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> Minority carriers created in p type isNe= <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>n</mi> <mi>i</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> <mi>e</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mo>(</mo> <msup> <mrow> <mrow> <mn>1.5</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>25</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>2.25</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </msup> <mo>/</mo> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> Minority charge carriers holes in p type would contribute more in reverse saturatiom current.

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