Question

(iv) Explain what process(es) will be different at the higher temperature producing the different features from (ii) to (iii). (2) (b) A structure that we sometimes encounter in electronic devices is a so-called p-i-n structure, consisting of a p-doped layer, an intrinsic layer and then an n-doped layer as shown in the drawing below. Assume Si as the semiconductor: P US- [Intrinsic 5 XION n kx 10 cm Assume a temperature of 300 K and with the doping as indicated in the sketch calculate for each layer: (i) the majority carrier concentration, (ii) the minority carrier concentration (iii) and the electrical conductivity. (9) p layer Intrinsic layer n layer Majority carrier concentration Minority carrier concentration Electrical Conductivity

Answer 1

P Instrinsic 15 5x10 cm 2x10 cm Alsume a temperature of 300k and with doping as indicated Calculate (0) Majority Charge carries ("? P-type: In p-type semiconductor cassies = holes Majority charge = No of acceptor atoms added Majority charge cassies in P-type = Na = 5x105 cm (ii) Intóinsic In instrinsic Semiconductor no. of é and holes aje equal there is majority (minority charge carier Intrinsic carsies - 1.5X10'emet at 300k mo 10 le ho tybe In n-type semiconductor majority charge cassies = e majority chaye = e = Nd = No- Majority charge carriers in n-type = Nd - 2X7016 cm) of dones atom

2) Minority charge carriers. (6 P type nis instrinsic carriers Na In p-type minosity chaye carries are è and is given by coufe = (ne)? = (1.5X1010) 4:125x105 4.5x10 cil 2 no of e 581015 i) Instrinsic There is no majority minority carriess iú Instrinsic I ni= me=up 1.5x10 cm) Semiconductor TO (ii) n-type Minosity charge carriess in a type are holes no. of holes (16) 2 (1.5x100) 2 Nd 1,125 X 10 20-16 ax106 =1.125x10 4 no. of holes = 1.125x104

Electrical Conductivity (op-type for p type nh : p=neun & » ne where len= mobility of holes Mn = Na 480 cm2/v.s. op= Nae Mn = 5x10' x 1.681019x480 en = 3840X16-4 = 0.3840 2-cmy -19 (i) In Intrinsic Me= Mobility of e- = 1350 cmPlus of nie (Unt Me) = 1.5X1010 y 1-6X10 * ( 480+ 1350) = 1.581010x1 -6X1619x1830 z 4399 X10-9 a = 4.392x10-6 2cmt (ii) for n-type on = Nae Me =Qxiolóx 1•6X1519X1350 z 4320X183 = 4.320 stemt Tn

i n Majosity carriers concentration 15 5x10 cm lo 1.58 10 cm) 16 2010 chil Minority carrier Loucents ation 4.58104 chet 10 1.5x10 cm 4 1•125 x 10 cmy Electsical Conductivity 0.3840 2 cm 4.392x1026 ntcmy 4.320.02tant