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November 4, 2013 1 ECSE-6230 Semiconductor Devices and Models I Lecture 4 Prof. Shayla Sawyer Bldg. CII,

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Presentation on theme: "November 4, 2013 1 ECSE-6230 Semiconductor Devices and Models I Lecture 4 Prof. Shayla Sawyer Bldg. CII,"— Presentation transcript:

1 November 4, ECSE-6230 Semiconductor Devices and Models I Lecture 4 Prof. Shayla Sawyer Bldg. CII, Rooms 8225 Rensselaer Polytechnic Institute Troy, NY Tel. (518) Fax. (518) November 4, 2013

2 2 Outline Carrier Concentration at Thermal Equilibrium – Introduction – Fermi Dirac Statistics Donors and Acceptors Determination of Fermi Level Dopant Compensation 2 November 4, 2013

3 3 Carrier Concentration Introduction One of most important properties of a semiconductor is that it can be doped with different types and concentrations of impurities Intrinsic material-no impurities or lattice defects Extrinsic-doping, purposely adding impurities – N-type mostly electrons – P-type mostly holes 3 November 4, 2013

4 4 Carrier Concentration Introduction To calculate semiconductor electrical properties, you must know the number of charge carriers per cm 3 of the material Must investigate distribution of carriers over the available energy states Statistics are needed to do so Fermi-Dirac statistics Distribution of electrons over a range of allowed energy levels at thermal equilibrium 4 November 4, 2013

5 5 Fermi-Dirac Distribution Probability that an available energy state at E will be occupied by an electron at absolute temperature T Mathematically, E F (Fermi Energy) is the energy at which f(E) = 1/2 The transition region in (E - E F ) from f(E) =1 to f(E) = 0 is within 3 k T. When T 0, E is discontinous at E = E F.

6 November 4, Fermi-Dirac Distribution To apply the Fermi-Dirac distribution, we must recall that f(E) is the probability of occupancy of an available state at E. Where can we find available states?

7 November 4, Carrier Concentration At Thermal Equilbrium Number of electrons (occupied conduction band levels) given by: Density of states g(E) can be approximated by the density near the bottom of the conduction band where M C is the number of equiv. minima

8 November 4, Carrier Concentration At Thermal Equlibrium The integral can be evaluated as Where N C is the effective density of states in the conduction band given by: For the valence band, consider light and heavy holes for the density of states effective mass for holes (m dh ) and use similar equation

9 November 4, Carrier Concentration At Thermal Equlibrium: Intrinsic For intrinsic material lies at some intrinsic level E i near the middle of the band gap, electron and hole concentrations are Law of mass action: product of maj. and min. carriers is fixed

10 Donors and Acceptors Doping by substituting Si atoms with Column III or V of the Periodic Table. Very dilute doping level, typical to cm -3, results in discrete energy levels. Donor level is neutral if filled with e -, positively charged if empty. e.g., P, As, and Sb in Si. Acceptor level is neutral if empty, negatively charged if filled with e -. e.g., B and Al in Si.

11 Donors and Acceptors Hydrogen-like Model to describe dopant atom ionization. Hydrogen Atom Ground state (n=1) ionization energy of hydrogen is 13.6 eV. To estimate ionization energy of donors, replace m 0 with m* and 0 and S (e.g., for Si). E D = ( 0 / S ) 2 ( m*/ m 0 ) E H ~ eV for Ge, eV for Si, eV for GaAs E A ~ eV for Ge, 0.05 eV for Si, 0.05 eV for GaAs The_Chemistry_of_Gemstones kT~0.026eV Comparable to thermal energies so ionization is complete at room temperature

12 Donors and Acceptor Levels

13 Determination of Fermi Level

14 Determination of Fermi Level Intrinsic Semiconductor -E F ~ E g / 2 Extrinsic Semiconductor -E F adjusted to preserve space charge neutrality Space Charge Neutrality n 0 + N A - = N D + + p 0 Total Neg. Charges = Total Positive Charges electrons and ionized acceptors=holes and ionized donors 100% ionization assumed. Ionized Concentration of Donors When impurities are introduced: whereg D is the ground state degeneracy of donor impurity g D = 2 (i) electrons with either spin (ii) no electrons at all

15 Determination of Fermi Level Ionized Acceptors whereg A is the ground state degeneracy of acceptor impurity g A = 4 for Ge, Si, and GaAs because (i) Acceptor levels can receive electrons with either spin and (ii) Valence band double degeneracy. Space Charge Neutrality N-type Semiconductor is assumed. n=N D + +p ~ N D + therefore

16 Charge Neutrality Since the material must balance electrostatically, the Fermi level must adjust such that charge neutrality remains. The Fermi level therefore can be calculated for a set given N D, E D, N C, and T November 4,

17 Graphical Determination of Fermi Level Graphical solution of the space charge neutrality equation. No need to assume 100% ionization. Need to know the donor (or acceptor) level. Degenerate Doping Impurity levels are broadened into impurity bands, thus reducing the ionization energy of the dopants. Ex. Phosphorus in Silicon. E C - E D (N D ) = x10 -8 (N D ) 1/3 [eV] for N D > cm -3

18 Dopant Compensation When both n- and p-type (donor and acceptor) impurities are present, the space charge neutrality condition n 0 + N A - = N D + + p 0 holds, even when the impurities are deep levels. In an n-type semiconductor where N D >>>N A Fermi level can be obtained from

19 Dopant Compensation In an p-type semiconductor where N A >>>N D Fermi level can be obtained from

20 Example Problem A hypothetical semiconductor has an intrinsic carrier concentration of 1.0 x cm -3 at 300 K, it has a conduction and valence band effective density of states N C and N V both equal to cm -3. a)What is the band gap Eg? b)If the semiconductor is doped with N d = 1x10 16 donors/cm 3, what are the equilibrium electron and hole concentrations at 300K?

21 Example Problem A hypothetical semiconductor has an intrinsic carrier concentration of 1.0 x cm -3 at 300 K, it has a conduction and valence band effective density of states N C and N V both equal to cm -3. c) If the same piece of semiconductor, already having N d = 1x10 16 donors/cm 3, is also doped with N a = 2x10 16 acceptors/cm 3, what are the new equiliblrium electron and hole concentrations at 300 K? a)Consistent with your answer to part (c), what is the Fermi level position with respect to the intrinsic Fermi level, E F – E i ?


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