n-type p-type n-type semiconductors In these, the charge carriers are negative p-type semiconductors. In these, the charge carriers are positive

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2 Extrinsic Semiconductors Extrinsic semiconductors : impurity atoms dictate the properties Almost all commercial semiconductors are extrinsic Impurity concentrations of 1 atom in is enough to make silicon extrinsic at room T! Impurity atoms can create states that are in the band gap. In most cases, the doping of a semiconductor leads either to the creation of donor or acceptor levels n-type semiconductors In these, the charge carriers are negative n-type p-type semiconductors. In these, the charge carriers are positive p-type

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4 Band Diagram: Acceptor Dopant in Semiconductor For, add a group III element to accept an electron and make p-type (more positive holes ). E C Missing electron results in an extra hole, with an acceptor energy level E A just above the valence band E V. Holes easily formed in valence band, greatly increasing the electrical conductivity. E F E V p-type E A Fermi level E F moves down towards E V. Typical acceptor elements are Boron, Aluminum, Gallium, Indium.

5 Band Diagram: Donor Dopant in Semiconductor Increase the conductivity of a semiconductor by adding a small amount of another material called a dopant (instead of heating it!) For group IV, add a group V element to donate an electron and make n-type (more negative electrons!). Extra electron is weakly bound, with donor energy level E D just below conduction band E C. Dopant electrons easily promoted to conduction band, increasing electrical conductivity by increasing carrier density n. Fermi level E F moves up towards E C. Typical donor elements that are added to or Ge are phosphorus, arsenic, antimonium. E C E F E V n-type E gap ~ 1 ev E D

6 licon n-type semiconductors: Bonding model description: Element with 5 bonding electrons. Only 4 electrons participate in bonding the extra e - can easily become a conduction electron p-type semiconductors: Bonding model description: Element with 3 bonding electrons. nce 4 electrons participate in bonding and only 3 are available the left over hole can carry charge P B

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9 The Mass Action Law This relationship is valid for both intrinsic and extrinsic semiconductors. In an extrinsic semiconductor the increase in one type of carrier (n or p) reduces the concentration of the other through recombination so that the product of the two (n and p) is a constant at a any given temperature. The carriers whose concentration in extrinsic semiconductors is the larger are designated the majority carriers, and those whose concentration is the smaller the minority carriers. At equilibrium, with no external influences such as light sources or applied voltages, the concentration of electrons,n 0, and the concentration of holes, p 0, are related by o p n 2 n i denotes the carrier concentration in o i intrinsic silicon n =

10 A material is defined as intrinsic when it consists purely of one element and no outside force (like light energy) affects the number of free carrier other than heat energy. In intrinsic, the heat energy available at room temperature generates approximately 1.5x10 10 carriers per cm 3 of each type (holes and electrons). The number of free carriers doubles for approximately every 11 C increase in temperature. This number represents a very important constant (at room temperature), and we define n i = 1.5x10 10 cm -3 where n i denotes the carrier concentration in intrinsic silicon at room temperature (constant for a given temperature).

11 Based on charge neutrality, for a sample doped with N D donor atoms per cm -3 and N A acceptor atoms per cm -3 we can write n o + N A = p o + N D which shows that the sum of the electron concentration plus the ionized acceptor atoms is equal to the sum of the hole concentration plus the ionized donor atoms. The equation assumes that all donors and acceptors are fully ionized, which is generally true at or above room temperature. Given the impurity concentration, the above equations can be solved simultaneously to determine electron and hole concentrations. In electronic devices, we typically add only one type of impurity within a given area to form either n-type or p-type regions.

12 In n-type regions there are typically only donor impurities and the donor concentration is much greater than the intrinsic carrier concentration, N A =0 and N D >>n i. Under these conditions we can write n o N D where n o is the free electron concentration in the n-type material and N D is the donor concentration (number of added impurity atoms/cm 3 ). nce there are many extra electrons in n-type material due to donor impurities, the number of holes will be much less than in intrinsic silicon and is given by, p o = n i2 / N D where p o is the hole concentration in an n-type material and n i is the intrinsic carrier concentration in silicon.

13 milarly, in p-type regions we can generally assume that N D =0 and N A >>n i. In p-type regions, the concentration of positive carriers (holes), p o, will be approximately equal to the acceptor concentration, N A. p o = N A and the number of negative carriers in the p-type material, n o, is given by n o = n i2 / N A Notice the use of notation, where negative charged carriers are n, positive charged carriers are p, and the subscripts denote the material, either n-type or p-type. This notation will be used throughout our discussion of p-n junctions and bipolar transistors. The above relationships are only valid when N D or N A is >> n i, which will always be the case in the problems related to integrated circuit design.

14 Example Calculate the conductivity and the resistivity of a n-type silicon wafer which contains electrons per cubic centimeter with an electron mobility of 1400 cm 2 /Vs. Solution: The conductivity is obtained by adding the product of the electronic charge, q, the carrier mobility, and the density of carriers of each carrier type, or: As n-type material contains almost no holes, the conductivity equals: σ= q μ n n = 1.6 x x 1400 x = /Ωcm. The resistivity equals the inverse of the conductivity or: ρ = and equals ρ = 1/σ = 1/2.24 = Ωcm. ( μ n p) n μ σ = q + 1 = σ q 1 p ( μ n p) n + μ p

15 Example A sample is doped with 10-4 atomic% of P donors. Assuming complete ionisation of donors at room temperature, calculate the charge carrier concentration and conductivity at room temperature. [For : ρ = 2330 kg m -3, atomic weight = 28, μ e = 0.15 m 2 V -1 s -1, μ h = 0.05 m 2 V -1 s -1, n i = 1.5x10 10 carriers per cm 3 ] Solution: 1) Calculate the fraction of donor atoms (phosphorus atoms per silicon atom) where N number of atoms per unit volume 6 N 2) Calculate the number of silicon atoms per unit volume = ρ N A Avogadro Kg. m = Kg. mol ) Calculate the number of donors atoms (phosphorus) 23 atoms. mol 1 = 5 10 N N 28 D =10 atoms. m N D. 22 = 5 10 P atoms m 3 3

16 4) As N A =0 and N D >>n i, then we can safely assume that n o =N D and p o is very small ~ zero σ = n σ = o q μ ( 22 3 ) ( 19 ) ( P atoms. m C 0. 15m V s ) σ = 1200Ω 1 e. m 1

17 Consider the following equations: Mass Action: Neutrality: n = o n + N = p + 2 po ni o A o D N Solving the equations simultaneously Electron concentration (n-type semiconductor) Hole concentration (p-type semiconductor) n o = 2 2 ( N ) + ( ) D N A N D N A n i p o p o = 2 2 ( N ) + ( ) A N D N A N D n i 2 2 i n n i o = = po no n

18 Solving the problem again. N A =0 n o = 2 2 ( N ) + ( ) D N A N D N A n i n o = ND + ND + 4n 2 i = ( 5 10 ) + 4( ) 2 n o = p ( 16 ) n = m 2 i o = =. 22 no 5 10 σ = q σ = ( n μ + p μ ) o ( ) [( ) ( ) ( ) ( )] C 5 10 P atoms. m 0. 15m V s σ = 1200Ω 1 e. m 1 o h

19 Example An n-type piece of silicon of length L = 10 micron has a cross sectional area A = cm 2. A voltage V = 10 Volt is applied across the sample yielding a current I = 100 ma. What is the resistance, R of the silicon sample, its conductivity, σ, and electron density, n? μ n = 1400 cm 2 /Vs Solution The resistance of the sample equals R = V/I = 10/0.1 = 100 Ω. nce R = L /(σa) the conductivity is obtained from: σ = L/(R A) = 0.001/(100 x 0.001) = /Ωcm. The required electron density is related to the conductivity by: σ = q n μ n so that the density equals: n = σ/(q μ n ) = 0.01/(1.6 x x 1400) = 4.46 x cm -3.

20 Example A sample at room temperature is doped with As atoms/cm 3. What are the equilibrium electron and hole concentrations at 300 K? Solution nce the N A is zero we can write, n o p o = n i 2 And n o2 N D n o n i2 = 0 n o + N A = p o + N D Solving this quadratic equations results in n 0 = 1.02x10 11 [cm -3 ] and thus, p 0 = n i2 / n 0 = 2.25x10 20 / 1.02x10 11 p 0 = 2.2x10 9 [cm -3 ] Notice that, since N D >n i, the results would be very similar if we assumed n o =N D =10 11 cm -3, although there would be a slight error since N D is not much greater than n i.

21 Semiconductors Thermal excitation of electrons Fermi level now lies in the gap VB E F CB pure solid Fermi-Dirac distribution E F n-type E F p-type

22 Conductivity Intrinsic semiconductor (Germanium, licon). For every electron, e, promoted to the conduction band, a hole, h, is left in the valence band (+ charge). The conductivity is determined by the number of electron-hole pairs. Total conductivity σ = σ e + σ h = nqμ e + pqμ h For intrinsic semiconductors: n = p & σ = nq(μ e + μ h ) Extrinsic semiconductor (doping). n-type. The number of electrons in the conduction band far exceeds the number of holes in the valence band (or n>>p). σ = σ e = nqμ e p-type. The number of holes in the valence band far exceeds the number of electrons in the conduction band (or p>>n) σ = σ h = pqμ h

23 Temperature variation of conductivity Intrinsic Semiconductors σ = n q μ e + p q μ h Strong exponential dependence of carrier concentration in intrinsic semiconductors Temperature dependence of carrier mobility is weaker. Eg n = p A exp 2kBT Eg σ C exp 2kBT

24 Temperature variation of conductivity - Intrinsic Semicoductor n = p A exp σ C exp E 2k g B E 2k g B T T ln(n) = ln(p) ln(a) - E g /2 kt The constant A is related to the density of states and the effective masses of electrons and holes. Plotting log of σ, p, or n vs. 1/T produces a straight line. Slope is E g /2k B ; gives band gap energy. Δln p Δ = E ( 1 T ) 2kB g

25 Temperature variation of conductivity Extrinsic Semiconductor Extrinsic semiconductors low T: all carriers due to extrinsic excitations mid T: most dopants ionized (saturation region) high T: intrinsic generation of carriers dominates

26 At very high temperatures, n i increases beyond N D ln(n) Intrinsic slope = -E g /2k provided N D >> n i the number of carriers is dominated by n d dopants are activated as T > K so carrier concentration increases ln(n d ) T i Extrinsic T s Ionization slope = -ΔE/2k n i (T) 1/T Fig. 5.15: The temperature dependence of the electron concentration in an n-type semiconductor. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002)

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28 Electron and hole mobility in silicon The mobility of electrons and holes in silicon at room temperature is shown in the figure below. Fig. Electron and hole mobility versus doping density for silicon The electron and hole mobilities have a similar doping dependence: For low doping concentrations the mobility is almost constant and is primarily limited by phonon scattering. At higher doping concentrations the mobility decreases due to ionized impurity scattering with the ionized doping atoms. The actual mobility also depends on the type of dopant.

29 These are empirical relations obtained by fitting experimental values. The above figure is for phosphorous and boron doped silicon and is calculated using: And

30 Example A silicon wafer contains cm -3 phosphor atoms. Using the data in the table; calculate the resistivity and conductivity of the material. Repeat for arsenic and boron atoms. Solution Plugging the values from table into the following equation one obtains a mobility of 277 cm 2 /V-sec for phosphorus- doped material, 284 cm 2 /V-sec for arsenic-doped material and 153 cm 2 /V-sec for boron-doped material, corresponding to a resistivity of 22.6, 22.0 and 40.9 mωcm and a conductivity of 44.3, 45.4 and /Ωcm.

31 Semiconductor: Dopant Density via Hall Effect Why Useful? Determines carrier type (electron vs. hole) and carrier density n for a semiconductor. How? Place semiconductor into external B field, push current along one axis, and measure induced Hall voltage V H along perpendicular axis. Carrier _ Density _ n = ( Carrier ( Current _ I)( Magnetic _ Field _ B) _ Ch arge _ q)( Thickness _ t)( Hall _ Voltage _ V H ) Derived from Lorentz equation F E (qe) = F B (qvb). F = qv B Hole B + charge charge Electron

32 The Hall Effect and the Lorentz Force The basic physical principle is the Lorentz force. When an electron (e - ) moves along a direction perpendicular to an applied magnetic field (B), it experiences a force acting normal to both directions and moves in response to this force and the force effected by the internal electric field. For an n-type, bar-shaped semiconductor shown in Fig.1, the carriers are predominately electrons of bulk density n. We assume that a constant current I flows along the x-axis in the presence of a z-directed magnetic field (B). Electrons subject to the Lorentz force drift away from the current line toward the negative y-axis, resulting in an excess surface electrical charge on the side of the sample. This charge results in the Hall voltage, a potential drop across the two sides of the sample. This transverse voltage is the Hall voltage V H and its magnitude is equal to IB/qnd, where I is the current, B is the magnetic field, d is the sample thickness, and q (1.602 x C) is the elementary charge. In some cases, it is convenient to use layer or sheet density (n s = nd) instead of bulk density.

33 Semiconductors Devices Semiconductors (thermistors) p-n junctions (diodes) Transistors bipolar junction transistors (BJT s) field effect transistors (FET) Optoelectronic and photonic devices: light emitting diodes (LED s) semiconductor lasers photodetectors solar cells Pentium 4 has 42 MILLION transistors in one processor!

34 Impurities Put Allowed Levels in the Band Gap of licon p Type n Type Conduction Band Many ELECTRONS! Conduction Band Acceptor Level Many HOLES! Valence Band Donor Level Valence Band = where thermal electrons can easily go Boron Doped Phosphorous Doped

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36 Majority Carrier and Current Flow in p-type licon + p-type licon - Hole Flow Current Flow Majority Carrier and Current Flow in n-type licon + n-type licon - Electron Flow Current Flow

37 The p-n Junction p n 0 Volts Hole Diffusion Electron Diffusion Holes and Electrons Recombine at the Junction A Depletion Zone (D) and a Barrier Field Forms at the p-n Junction Barrier Field 0 Volts Acceptor Ions p -- D ++ n Hole (+) Diffusion Donor Ions Electron (-) Diffusion The Barrier Field Opposes Further Diffusion (Equilibrium Condition)

38 Depletion Region When a p-n junction is formed, some of the free electrons in the n-region diffuse across the junction and combine with holes to form negative ions. In so doing they leave behind positive ions at the donor impurity sites.

39 In the p-type region there are holes from the acceptor impurities and in the n-type region there are extra electrons. When a p-n junction is formed, some of the electrons from the n-region which have reached the conduction band are free to diffuse across the junction and combine with holes. Filling a hole makes a negative ion and leaves behind a positive ion on the n-side. A space charge builds up, creating a depletion region which inhibits any further electron transfer unless it is helped by putting a forward bias on the junction.

40 Equilibrium of junction Coulomb force from ions prevents further migration across the p-n junction. The electrons which had migrated across from the N to the P region in the forming of the depletion layer have now reached equilibrium. Other electrons from the N region cannot migrate because they are repelled by the negative ions in the N region and attracted by the positive ions in the N region.

41 Reverse bias An applied voltage with the indicated polarity further impedes the flow of electrons across the junction. For conduction in the device, electrons from the N region must move to the junction and combine with holes in the P region. A reverse voltage drives the electrons away from the junction, preventing conduction.

42 Forward bias An applied voltage in the forward direction as indicated assists electrons in overcoming the coulomb barrier of the space charge in depletion region. Electrons will flow with very small resistance in the forward direction.

43 Forward Bias of a p-n Junction p - + n + Volts - Volts Current Applied voltage reduces the barrier field Holes and electrons are pushed toward the junction and the depletion zone shrinks in size Carriers are swept across the junction and the depletion zone There is a net carrier flow in both the P and N sides = current flow! -Volts Reverse Bias of a p-n Junction p --- D +++ n Current + Volts Applied voltage adds to the barrier field Holes and electrons are pulled toward the terminals, increasing the size of the depletion zone. The depletion zone becomes, in effect, an insulator for majority carriers. Only a very small current can flow, due to a small number of minority carriers randomly crossing D (= reverse saturation current)

44 Due to diffusion, electrons move from n to p-side and holes from p to n-side. Causes depletion zone at junction where immobile charged ion cores remain. Results in a built-in electric field or potential, which opposes further diffusion. p-n Junction: Band Diagram p-n regions touch & free carriers move n-type E electrons C E F E V E V holes p-n regions in equilibrium E C E F Depletion Zone p-type E F

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46 Built-in potential For example:

47 pn Junction: IV Characteristics Current-Voltage Relationship ev / kt I = I [ e 1] o Forward Bias: current exponentially increases. Reverse Bias: low leakage current equal to ~I o. Ability of p-n junction to pass current in only one direction is known as rectifying behavior. Reverse Bias Forward Bias