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Homogeneous Semiconductors Dopants Use Density of States and Distribution Function to: Find the Number of Holes and Electrons.

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Presentation on theme: "Homogeneous Semiconductors Dopants Use Density of States and Distribution Function to: Find the Number of Holes and Electrons."— Presentation transcript:

1 Homogeneous Semiconductors Dopants Use Density of States and Distribution Function to: Find the Number of Holes and Electrons.

2 Energy Levels in Hydrogen Atom

3 Energy Levels for Electrons in a Doped Semiconductor

4 Assumptions for Calculation

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8 Density of States (Appendix D) Energy Distribution Functions (Section 2.9) Carrier Concentrations (Sections )

9 GOAL: The density of electrons (n o ) can be found precisely if we know 1. the number of allowed energy states in a small energy range, dE: S(E)dE “the density of states” 2. the probability that a given energy state will be occupied by an electron: f(E) “the distribution function” n o =  band S(E)f(E)dE

10 For quasi-free electrons in the conduction band: 1. We must use the effective mass (averaged over all directions) 2. the potential energy E p is the edge of the conduction band (E C ) For holes in the valence band: 1. We still use the effective mass (averaged over all directions) 2. the potential energy E p is the edge of the valence band (E V )

11 Energy Band Diagram conduction band valence band ECEC EVEV x E(x) S(E) E electron E hole note: increasing electron energy is ‘up’, but increasing hole energy is ‘down’.

12 Reminder of our GOAL: The density of electrons (n o ) can be found precisely if we know 1. the number of allowed energy states in a small energy range, dE: S(E)dE “the density of states” 2. the probability that a given energy state will be occupied by an electron: f(E) “the distribution function” n o =  band S(E)f(E)dE

13 Fermi-Dirac Distribution The probability that an electron occupies an energy level, E, is f(E) = 1/{1+exp[(E-E F )/kT]} –where T is the temperature (Kelvin) –k is the Boltzmann constant (k=8.62x10 -5 eV/K) –E F is the Fermi Energy (in eV) –(Can derive this – statistical mechanics.)

14 f(E) = 1/{1+exp[(E-E F )/kT]} All energy levels are filled with e - ’s below the Fermi Energy at 0 o K f(E) 1 0 EFEF E T=0 o K T 1 >0 T 2 >T 1 0.5

15 Fermi-Dirac Distribution for holes Remember, a hole is an energy state that is NOT occupied by an electron. Therefore, the probability that a state is occupied by a hole is the probability that a state is NOT occupied by an electron: f p (E) = 1 – f(E) = 1 - 1/{1+exp[(E-E F )/kT]} ={1+exp[(E-E F )/kT]}/{1+exp[(E-E F )/kT]} - 1/{1+exp[(E-E F )/kT]} = {exp[(E-E F )/kT]}/{1+exp[(E-E F )/kT]} =1/{exp[(E F - E)/kT] + 1}

16 The Boltzmann Approximation If (E-E F )>kT such that exp[(E-E F )/kT] >> 1 then, f(E) = {1+exp[(E-E F )/kT]} -1  {exp[(E-E F )/kT]} -1  exp[-(E-E F )/kT] …the Boltzmann approx. similarly, f p (E) is small when exp[(E F - E)/kT]>>1: f p (E) = {1+exp[(E F - E)/kT]} -1  {exp[(E F - E)/kT]} -1  exp[-(E F - E)/kT] If the Boltz. approx. is valid, we say the semiconductor is non-degenerate.

17 Putting the pieces together: for electrons, n(E) f(E) 1 0 EFEF E T=0 o K T 1 >0 T 2 >T EVEV ECEC S(E) E n(E)=S(E)f(E)

18 Putting the pieces together: for holes, p(E) f p (E) 1 0 EFEF E T=0 o K T 1 >0 T 2 >T EVEV ECEC S(E) p(E)=S(E)f(E) hole energy

19 Finding n o and p o the effective density of states in the conduction band

20 Energy Band Diagram intrinisic semiconductor: n o =p o =n i conduction band valence band ECEC EVEV x E(x) n(E) p(E) E F =E i where E i is the intrinsic Fermi level

21 Energy Band Diagram n-type semiconductor: n o >p o conduction band valence band ECEC EVEV x E(x) n(E) p(E) EFEF

22 Energy Band Diagram p-type semiconductor: p o >n o conduction band valence band ECEC EVEV x E(x) n(E) p(E) EFEF

23 A very useful relationship …which is independent of the Fermi Energy Recall that n i = n o = p o for an intrinsic semiconductor, so n o p o = n i 2 for all non-degenerate semiconductors. (that is as long as E F is not within a few kT of the band edge)

24 The intrinsic carrier density is sensitive to the energy bandgap, temperature, and m *

25 The intrinsic Fermi Energy (E i ) For an intrinsic semiconductor, n o =p o and E F =E i which gives E i = (E C + E V )/2 + ( kT / 2 )ln(N V /N C ) so the intrinsic Fermi level is approximately in the middle of the bandgap.

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37 Higher Temperatures Consider a semiconductor doped with N A ionized acceptors (-q) and N D ionized donors (+q), do not assume that n i is small – high temperature expression. positive charges = negative charges p o + N D = n o + N A using n i 2 = n o p o n i 2 /n o + N D = n o + N A n i 2 + n o (N D -N A ) - n o 2 = 0 n o = 0.5(N D -N A )  0.5[(N D -N A ) 2 + 4n i 2 ] 1/2 we use the ‘+’ solution since n o should be increased by n i n o = N D - N A in the limit that n i << N D -N A

38 Simpler Expression Temperature variation of some important “constants.”

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44 Degenerate Semiconductors 1. The doping concentration is so high that EF moves within a few kT of the band edge (EC or EV). Boltzman approximation not valid. 2. High donor concentrations cause the allowed donor wavefunctions to overlap, creating a band at E dn. First only the high states overlap, but eventually even the lowest state overlaps. This effectively decreases the bandgap by  E g = E g0 – E g (N D ). ECEC EVEV ++++ E D1 E g0 E g (N D ) for N D > cm -3 in Si impurity band

45 Degenerate Semiconductors As the doping conc. increases more, E F rises above E C EVEV E C (intrinsic) available impurity band states EFEF EgEg E C (degenerate) ~ E D filled impurity band states apparent band gap narrowing:  E g * (is optically measured) - E g * is the apparent band gap: an electron must gain energy E g * = E F -E V

46 Electron Concentration in degenerately doped n-type semiconductors The donors are fully ionized: n o = N D The holes still follow the Boltz. approx. since E F -E V >>>kT p o = N V exp[-(E F -E V )/kT] = N V exp[-(E g * )/kT] = N V exp[-(E go -  E g * )/kT] = N V exp[-E go /kT]exp[  E g * )/kT] n o p o = N D N V exp[-E go /kT] exp[  E g * )/kT] = (N D /N C ) N C N V exp[-E go /kT] exp[  E g * )/kT] = (N D /N C )n i 2 exp[  E g * )/kT]

47 Summary non-degenerate: n o p o = n i 2 degenerate n-type: n o p o = n i 2 (N D /N C ) exp[  E g * )/kT] degenerate p-type: n o p o = n i 2 (N A /N V ) exp[  E g * )/kT]


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