Question

Question 2: a) Consider a bar of silicon, by some unspecified process the electrons are injected in the middle of the bar. The concentration profile is shown below. Write the equation of current flow? Electron Concentration b) I made a diode with following impurities in intrinsic silicon - +5) Phosphorous Boron 1. What kind of device will it make? 2. Draw the schematic. 3. Label the terminals. c) If the voltage is kept constant, the drift velocity of electron is higher than of holes in intrinsic silicon a) Explain Why? b) Find the resistance of intrinsic silicon bar that is 10 um long, 3 um wide, and 1 um thick.

Answer 1

- from given election concentration profile, elections are injected at middle of the semiconductor ncorna) Fя (1) x - x -> at X0 - nox) = maximum as x->20 =n(x) = 0 na) election concentration as a funchon of Tn - n'ca)= n) - h. -> (from fig()) dnor) = nC) - n - 0 dt (an-carrier che lifetime) - Solution for equation & in n'a) -n. +no) n'= n'co) e lin -> In (X) = Aq Pon(o) -44% Inx) - Aq Dn [nco) noJe /

(b) phosphovory has 5 elections in valence orbit in Valence of bit Boron - has 3 elections in Phosphorous - n-type Boron -p-type (1) This will make a pon (2) & (3) Junction diode Anode cathode P-side -boron will be present naride -phosphorous will be present P- ride -> Anode N-ride - Cathode 67607 VaRME Ce a un tup) va = (Mn + Up) E voltage u kept constant u va a intep (va - drift velocity of Carriers) - at room temperature for silicon (300k) Mn= 14000m/v. s o Up= 450 cm (v-g @ as mobility of elections are higher than holes, drift of elechong is also higher for elections only.

B) R = 11 [giren la loum As wit W = 3um tolum & Az 3x156x1x106 A = 3x10thm R= 3.4*105 2-cm x 106 3x1012 = 3.4x105x102R-mxlo - 3x1012 3 x - z 1.133 x 10's

resistivity is a standard value at room temperature. which is 3.4*10^5ohm-cm.

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