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CHAPTER ONE SEMICONDUCTORS Copyright, 2006 © Ahmed S. Bouazzi 2e A G.I. Module Energie Solaire المدرسة الوطنية للمهندسين بتونس.

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Presentation on theme: "CHAPTER ONE SEMICONDUCTORS Copyright, 2006 © Ahmed S. Bouazzi 2e A G.I. Module Energie Solaire المدرسة الوطنية للمهندسين بتونس."— Presentation transcript:

1 CHAPTER ONE SEMICONDUCTORS Copyright, 2006 © Ahmed S. Bouazzi 2e A G.I. Module Energie Solaire المدرسة الوطنية للمهندسين بتونس

2 The Crystal Lattice of Silicon Each silicon atom is situated at the center of a tetrahedron and connected to four other atoms occupying the summit of the tetrahedron.

3 A two dimension representation of the silicon crystal structure Intrinsic silicon Si Each silicon atom is situated at the center of four other atoms.

4 N-type Semiconductors n-doped silicon P P P P Si

5 P-type Semiconductors p-doped silicon Si B B B

6 Energy Interatomic distance Eg Permitted levels (a) (b) The Gap

7 Fermi Level [ F(E) is the probability for an electron to be in the E energy level] In intrinsic silicon, E F is situated in the middle of the gap. In doped silicon, the Fermi level goes up or down depending on the electron concentration.

8 Semiconductor Doping The doping atoms create localized levels inside the band gap.

9 Energy v E E F c E c E v E Semiconductor Metal Electron-hole Pairs

10 Intrinsic Carrier Concentration = 4 = 4 for silicon at 300 K, n i 2 = 2x10 20 cm -6 np = (m e m h ) 3/2 exp

11 In n-type silicon: Minority and Majority Carriers n = N D p no N D = n i 2 p no = n i 2 / N D In p-type silicon: p = N A n po N A = n i 2 n po = n i 2 / N A

12 In n-type silicon: Minority and Majority Carriers in Excess n n0 = N D ≈ 10 16 – 10 18 cm -3 p n0 N D = n i 2 ; p no = n i 2 / N D ≈ 2×10 4 – 2 × 10 2 cm -3 Creating  n =  p ( ≈ 10 11 – 10 14 cm -3 ) electron- hole pairs will give: n n = n n0 +  n ≈ N D and p n = p n0 +  p ≈  p

13 Minority and Majority Carriers in Excess In p-type silicon: p p0 = N A ≈ 10 16 – 10 18 cm -3 n p0 N A = n i 2 ; n po = n i 2 / N A ≈ 2×10 4 – 2 × 10 2 cm -3 Creating  n =  p ( ≈ 10 11 – 10 14 cm -3 ) electron- hole pairs will give: p p = p p0 +  p ≈ N A and n p = n p0 +  n ≈  n

14 Lifetime and Recombination N = N o exp In the bulk: At the surface: J sur = q(n p - n po )S  is the lifetime of the minority carriers. S is the surface recombination velocity of minority carriers.

15 Diffusion Length L = The diffusion length is the free path of the minority carriers before recombination.  is the lifetime of the minority carriers. D is the diffusion constant of minority carriers.

16 Absorption Coefficient The absorbed quantity of photons at the depth x is:  0 is the flux of photons arriving at the surface of the semiconductor and x is the depth. x 00 0

17 Drift Minority Carriers Current in a Semiconductor Electrons: J n = qn p µ n E and  n = qn p µ n

18 Holes: J p = qp n µ p E and  p = qp n µ p

19 J n = qD n  n p (x,y,z) J p = – qD p  p n (x,y,z) Diffusion Minority Carriers Current in a Semiconductor

20 J = J n + J p Electrons: J n = qn p µ n E + qD n  n p (x,y,z) Holes: J p = qp n µ p E – qD p  p n (x,y,z) Total Minority Carriers Current

21 E c E v E F E g p n Junction plane p-n Junction Depletion region

22 PHOTOVOLTAIC EFFFECT (1) The p-n Junction

23 E c E v E f Electrons current Holes current The photocurrent under illumination PHOTOVOLTAIC EFFFECT (2)

24 PHOTOVOLTAIC EFFFECT (3) The photovoltage under illumination

25 Depletion Region Where the built-in voltage is defined by:

26 Polarization of a p-n Junction direct inverse

27 I-V Characteristic

28 Saturation Current np =np =; p n =; L k = J o =

29 Metal/semiconductor junction E c E v Semiconductor E F metal E c E F E v (a) (b) (c) Schottky Diode and Ohmic Contacts


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