Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Electrical Energy Conversion and Power Systems Universidad de Oviedo Power Electronic Devices Semester 1 Power Supply Systems Lecturer: Javier Sebastián Research Group Power supply Systems (Sistemas Electrónicos de Alimentación) Javier Sebastián Dr. Electrical Engineer (Ingeniero Industrial) Full professor Room 3.1.21 Edificio nº 3, Campus Universitario de Viesques 33204 Gijón (Asturias). Spain Phone (direct): 985 18 20 85 Phone (secretary): 985 18 20 87 Fax: 985 18 21 38 E-mail: [email protected] Web: http://www.unioviedo.es/sebas/ 2 Outline Review of the physical principles of operation of semiconductor devices. Thermal management in power semiconductor devices. Power diodes. Power MOSFETs. The IGBT. High-power, low-frequency semiconductor devices (thyristors). 3 Outline Review of physical principles of semiconductors Power electronics devices D G S 4 Previous requirements Basic electromagnetic theory. Basic circuit theory. The operation of basic electronics devices in circuits. The student must understand the behaviour of the following electronics devices in simple circuits: Diodes. Bipolar Junction Transistors, both PNP and NPN types. Field Effect Transistor, especially enhancement-mode MetalOxide-Semiconductor Field Effect Transistors (MOSFETs), both in N-channel- and P-channel types. 5 Electrical Energy Conversion and Power Systems Universidad de Oviedo Lesson 1 - Review of the physical principles of operation of semiconductor devices Semester 1 - Power Electronics Devices 6 Outline Review of the physical principles of operation of semiconductor devices: Basic concepts about semiconductor materials: band diagrams, intrinsic and extrinsic semiconductors, mechanisms for electric current conduction and continuity equation and its use in simple steady-state and transient situations. Basic concepts about PN junctions: Equilibrium conditions, forward- and reverse-biased operation and calculation of the current flow when biased. Reverse-biased voltage limits of PN junctions. PIN junctions. Conductivity modulation. Transient effects in PN junctions in switching-mode operation. Metal-semiconductor junctions. 7 Energy of electrons Energy level in a semiconductor as a function of inter-atomic spacing At 0 K, empty - - - - - - - - Inter-atomic spacing Actual spacing At 0 K, filled 8 Concept of band diagram Energy of electrons Empty at 0 K 4 states/atom Eg - - - Conduction band Band gap 4 electrons/atom Material Eg [eV] Ge 0.66 Si 1.1 4H - SiC 3.26 GaN 3.39 Valence band Filled at 0 K 9 Band structure for insulators, semiconductors and metals at 0 K Conduction band Conduction band Band gap Eg Band gap Eg Overlap Valence band Valence band Valence band Semiconductor Eg=0.5-2 eV Insulator Eg= 5-10 eV Metals No Eg Conduction band 10 Band structure for semiconductors at room temperature. Concept of “hole” - Si - - - - - - Semiconductor Eg=0.5-2 eV Si + Si - Valence band - - - - + - Si - - Eg - - Conduction band Visualization using the bonding model • Some electrons jump from the valence band to the conduction band. They are charge carriers because they can move from one atom to another. • The empty state in the valence band is referred to as a “hole”. • The holes have positive charge. They are also charge carriers. 11 Concepts of generation and recombination Recombination Generation Eg - Si - - - - Si + Si - Si - - - - - - - - Si - - - - - + Si - + - Si - - - Si - - - - - + - - - Eg 12 - - Why both holes and electrons are electric charge carriers? Si - - - - - -- - - Si + - - - - Si - - - - - - + Si - Si - - - - - - - - Si Si - Si +++++++ - - • In general, there will be electric current due to both electrons and holes • Example: when there is an electric field in the semiconductor lattice 13 How many electrons and holes are there in 1 cm3? • The number of these electrons and holes strongly depend on both Eg and the room temperature. It is called intrinsic concentration and it is represented as “ni”. • The concentration of electrons in the conduction band (negative charge carriers) is represented as “n”. The concentration of holes in the valence band (positive charge carriers) is represented as “p”. • Obviously n = p = ni in this type of semiconductors (intrinsic semiconductors) • Some examples of the value of ni at room temperature: Material Eg [eV] ni [elect./cm3] Ge 0.66 2.4·1013 Si 1.1 1.5·1010 GaAs 1.4 1.8·106 4H - SiC 3.26 8.2·10-9 GaN 3.39 1.9·10-10 Taking into account the number of bonds of valence band electrons in 1cm3 of silicon, only one bond is broken for each amount of 1013 unbroken bonds (at room temperature) 14 Concept of extrinsic semiconductors: doping semiconductor materials • Can we have different concentration of electrons and holes? • The answer is yes. We need to introduce “special” impurities into the crystal: Donors: atoms from column V of the Periodic Table. We obtain an extra electron for each atom of donor. Acceptors: atoms from column III of the Periodic Table. We obtain an extra hole for each atom of acceptor. Si - 3 - Al - - - - Acceptor 1 - - - 2 Si + - - - - - - - - - 3 4 - Sb Si - + - 2 5 Si - - Donor - 1 Si - Si 15 - N-type and P-type semiconductors - - 3 Si - - Al- - - 1 Si + - - - 2 - Acceptor - - - - - - - - - - - - - - 3 4 Si - Sb+ Si - - 2 - 5 1 Donor Si - Si N-type semiconductor: P-type semiconductor: • Majority carriers are electrons. • Minority carriers are holes. • Positively-charged atoms of donor (positive ions). • Minority carriers are electrons. • Negatively-charged atoms of acceptor (negative ions). • Majority carriers are holes. 16 Charges in N-type and P-type semiconductors + Al Al - - Al Al + - + + Al + + P-type silicon Al - Al Acceptor ions (negative ions) Donor ions (positive ions) - N-type silicon - Sb+ Sb+ - Sb+ Sb+ Sb+ - - Sb+ Sb+ - + Sb+ - - Thermal generation Al + + - + - - electron - - hole Al Al + - + - Thermal generation - + Sb+ Sb+ 17 Charge carries in N-type and P-type semiconductors - Al - Sb+ - - Sb+ - Al Sb+ - Sb+ Sb+ • Concentration of majority carriers: pP • Concentration of minority carriers: nP + P-type Sb+ Al + + Al + + Al + - - - Al + - + • Mass action law: pP·nP = ni2 (only valid at equilibrium) • Concentration of majority carriers: nN • Concentration of minority carriers: pN • Mass action law: nN·pN = ni2 (only valid at equilibrium) N-type Very important equations!!! 18 - Static charges in N-type and P-type semiconductors - Sb+ - Sb+ Al - + P-type Sb+ - + Al - + - + + Al Al + Sb+ Sb+ Sb+ - - - Al Al + - + • Concentration of acceptors: NA (only negative static charges in a P-type semiconductor) • Concentration of donors: ND (only positive static charges in a N-type semiconductor) N-type 19 - Neutrality in N-type and P-type semiconductors • Silicon, aluminium and antimony were neutral before being used The extrinsic semiconductor must be neutral, too. - Sb+ - Sb+ • Neutrality: Sb+ Sb+ N-type Al pP = nP + NA - + P-type - + Al - + - + + Al Al + Sb+ Sb+ - - - Al Al + - + • Positive charges in volume V: pP·V • Negative charges in volume V: nP·V + NA·V • Negative charges in volume V: nN·V • Positive charges in volume V: pN·V + ND·V • Neutrality: nN = pN + ND Very important equations!!! 20 - Calculating the concentration of electrons and holes (I) Al - Al Sb+ - - Sb+ 2 known (NA and ni) and 2 unkown (pP and nP) variables can be solved - + P-type Sb+ - + + Al Al + Sb+ Sb+ Sb+ - - + + Al - Al + - + • Neutrality: pP = nP + NA • Mass action law: pP·nP = ni2 • Neutrality: nN = pN + ND • Mass action law: nN·pN = ni2 2 known (ND and ni) and 2 unkown (nN and pN) variables can be solved N-type 21 - Calculating the concentration of electrons and holes (II) • Frequent case: quite heavy doped semiconductors Al - + Al - - Sb+ Sb+ - Sb+ Al Sb+ N-type • NA >> ni • Neutrality: pP NA • Mass action law: nP ni2/NA - + P-type - + + + Sb+ Sb+ - - + Al - Al Al - + + • ND >> ni • Neutrality: nN ND • Mass action law: pN ni2/ND Very useful equations!!! 22 - Mechanisms to conduct electric current: Drift (I) • Semiconductors can conduct electric current due to the presence of an electric field E - - jp + + + - - - - + - + + + + +++++ - E jn jp_Drift = q·p·p·E is the current density of holes due to drift. jn_Drift = q·n·n·E is the current density of electrons due to drift. 23 Mechanisms to conduct electric current: Drift (II) jp_Drift = q·p·p·E jn_Drift = q·n·n·E q = magnitude of the electronic charge (1.6·10-19 coulombs). p = hole mobility. n = electron mobility. p = hole concentration. n = electron concentration. Ge Si GaAs (cm2/V·s) (cm2/V·s) (cm2/V·s) n 3900 1350 8500 p 1900 480 400 E = electric field. 24 Mechanisms to conduct electric current: Diffusion (I) - - n1 - 1 2 jn_Diff n2 < n1 Electrons have migrated due to “diffusion” (you can see the same phenomenon in gases) 25 Mechanisms to conduct electric current: Diffusion (II) • If we maintain a different concentration of electrons, we also maintain the motion of electrons in the lattice - - - - - - - - 2 - - - n1 jn_Diff n - 1 n2 < n1 26 Mechanisms to conduct electric current: Diffusion (III) - - - - - - 2 - - jn_Diff n - - n1 - - - - 1 n2 < n1 The current density is proportional to the electron concentration gradient: jn_Diff = q·Dn· n Dn = electron diffusion coefficient. 27 Mechanisms to conduct electric current: Diffusion (IV) + + + + + + + + + + + + + + + 2 + + p1 jp_Diff p + + + + + 1 p2 < p1 The current density is proportional to the hole concentration gradient: jp_Diff = -q·Dp· p Dp = hole diffusion coefficient. 28 Mechanisms to conduct electric current: Diffusion (V) jp_Diff = -q·Dp· p jn_Diff = q·Dn· n q = magnitude of the electronic charge (1.6·10-19 coulombs). Dp = hole diffusion coefficient. Dn = electron diffusion coefficient. p = hole concentration gradient. n = electron concentration gradient. Ge Si GaAs (cm2/s) (cm2/s) (cm2/s) Dn 100 35 220 Dp 50 12.5 10 29 Summary of conduction mechanisms jp_Drift = q·p·p·E jn_Drift = q·n·n·E jp_Diff = -q·Dp· p jn_Diff = q·Dn· n • Drift currents depend on the carrier concentration and on the electric field. • Diffusion currents do not depend on the carrier concentration, but on the carrier concentration gradient. 30 Continuity equations (I) There are some relationships between spatial and time variations of carrier concentrations because electrons and holes cannot mysteriously appear and disappear at a given point, but must be transported to or created at the given point via some type of ongoing action. The concentration of holes can be time-changing due to: • Different current density of holes across “A” and “B”. • Excess of carriers over the equilibrium (mass action law). • Generation of electron-hole pairs by radiation (light) . A B 31 Continuity equations (II) - • Different current density of holes across “A” and “B”. + + jp_A jp_B A • Excess of carriers over the equilibrium (mass action law). jn_B B + A • Generation of electron-hole pairs by radiation (light) . Light A B + B 32 Continuity equations (III) Taking into account the three effects, we obtain the continuity equation for holes: Total time variation of holes ·jp/q - p/t = GL- [p(t)-p]/p Variation due to the excess of carriers over the equilibrium Variation due to the generation of electron-hole pairs by light GL: rate of generation of electron-hole pairs by light. p: hole minority-carrier lifetime. p: hole concentration in steady-state. Variation due to the different current density of holes across “A” and “B” Similarly, we can obtain the continuity equation for electrons: ·jn/q + n/t = GL- [n(t)-n]/n 33 Time evolution of an “excess” of minority carries (I) • We generate an “excess” of electron-hole pairs by injecting light into a piece of N-type silicon and we reach the steady-state. 0 0 + N + + + + + + ·jp/q p0= GL·p + p - p/t = GL- [p(t)-p]/p + + + + + p0 + • Now the light injected disappears. We want to compute the time evolution of the hole concentration afterwards. + N + + + + + + + + + + + + p0 p(t) p 34 Time evolution of an “excess” of minority carries (II) • We can also compute the time evolution of the hole concentration from the continuity equation: p/t = GL- [p(t)-p]/p 0 ·jp/q - 0 After integrating p(t) = p + (p- p)·e-tp Tangent line p0 Same area p(t) p p p • Physical interpretation: There is an appreciable increase of holes during 3-5 times p. t 35 Spatial evolution of an “excess” of minority carries (I) • We constantly inject an “excess” of holes into a surface of a piece of N-type silicon and we reach the steady-state. No electric field exists and the hole current is due to diffusion. + + + 0 p0 + + + + + + + + + + + + N + + + p xN x 0 0 p/t = GL- [p(t, x)-p]/p ·jp/q - 0 = - [p(x)-p]/p + Dp·2[p(x)-p]/x2 After integrating p(x) = p + C1·e-x/Lp + C2·ex/Lp where : Lp=(Dp· p)1/2 is the minority hole diffusion length 36 Spatial evolution of an “excess” of minority carries (II) p(x) = p + C1·e-x/Lp + C2·ex/Lp xN: length of the N-type crystal Lp: hole diffusion length • Cases: a) Lp << xN (wide crystal): p(x) = p + (p- p)·e-xLp (decay exponentially). b) Lp >> xN (narrow crystal): p(x) = p + (p- p)·(xN-x)/xN (decay linearly). c) Other cases hyperbolic sine evolution. p0 Lp << xN (wide) p(x) p Lp p0 p(x) Tangent line x xN Lp >> xN (narrow) p x xN 37 Concept of PN junction (I) P-type silicon + + Al + Al - Al - Sb+ + Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ - Al - - - Al - + - - - Al Al - - + - Al + + - + N-type silicon + Sb+ Barrier to avoid carrier diffusion What happens if we remove the barrier? 38 Concept of PN junction (II) P-side + + + Al + Al - Al -+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ - Al - - - Al - + - + - - - Al Al - Al + - + N-side + Sb+ Holes begin to diffuse from the P-side to the N-side. Similarly, electrons diffuse from the N-side to the P-side Are all the carriers to be diffused? 39 Concept of PN junction (III) Are all the carriers to be diffused? P-side -+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Non-neutral P-type region, but Non-neutral N-type region, but negatively charged positively charged Is this situation “the final situation”? The answer is no + - Al + - - - - Al + Al - + - + - - + Al - + - Al + - Al + Al - Al N-side 40 Concept of PN junction (IV) P-side Al Sb+ + + Al + Al - Al - - + -+ Sb+ E Sb+ Sb+ Sb+ Sb+ Sb+ - - - Al - + - + - - - Al Al - Al + - + N-side + Sb+ + An electric field appears just in the boundary between both regions (we call this boundary metallurgical junction) 41 Concept of PN junction (V) • Now, we do zoom over the metallurgical junction P-side Al + Al - Al - Al - Sb+ Sb+ Sb+ - Sb+ Sb+ Sb+ + - Al - Al - Due to diffusion () E - Al Sb+ - + - Al N-type Sb+ + Due to drift (electric field) () The electric field limits the carrier diffusion 42 Concept of PN junction (VI) • Steady-state situation near the metallurgical junction - Al + - Al Al - - Sb+ Sb+ - Sb+ Sb+ Al + Neutral P-type region (holes are balanced by negative ions ) - E Sb+ + - Al - Sb+ Sb+ - + Al + - - - Al - - + - Al + Sb+ Neutral N-type region (electrons are balanced by positive ions ) Depletion region, or space charge region, or transition region Unbalanced charge exists because carriers barely exist 43 Concept of PN junction (VII) • Summary and terminology Metallurgical junction Many holes, but neutral + - P-side (neutral) E N-side (neutral) Many electrons, but neutral V0 Depletion, or transition, or space charge region (non neutral) There is space charge and, therefore, there are electric field E and voltage V0. However, there are almost no charge carriers 44 Computing the built-in voltage V0 (I) Net current passing through any section must be zero. As neither holes nor electrons are being accumulated in any parts of the crystal, net current due to holes is zero and net current due to electrons is zero. Due to drift P-side jp_Diff jn_Drift - Due to diffusion -+ -+ -+ +- + N-side + Due to drift -+ - + - Due to diffusion -+ -+ jp_Drift jn_Diff Currents must cancel each other Currents must cancel each other 45 Computing the built-in voltage V0 (II) pP (hole concentration in P-side ) + + + + + + + Due + to diffusion + + + + + + + + + + + + + + Zona P+ + + + + + + - + - + pN + + V0 + + + + + + (hole concentration in N-side ) jp_Diff N-side + Due to drift jp_Drift jp_Drift = - jp_Diff 46 Computing the built-in voltage V0 (III) Equations: jp_Drift = - jp_Diff jp_Drift = q·p·p·E jp_Diff = -q·Dp·dp/dx E = -dV/dx E Therefore: dV = -(Dp/p)·dp/p After integrating : V0 = VN-side – VP-side = -(Dp/p)·ln(pN/pP) = (Dp/p)·ln(pP/pN) Repeating the same process with electrons, we obtain: V0 = (Dn/n)·ln(nN/nP) It could be demonstrated: k = Boltzmann constant. VT = 26 mV at 300 K. Dp/p = Dn/n = kT/q = VT (Einstein relation) 47 + + + + + + + + - + Zona P+ + + + + + P-side: many holes Almost no electrons - + + + + +- + + V0 + Almost no holes or electrons V0 = VT·ln(pP/pN) and also - + - - - - - - - + + - + - + - + + - + + - + - + + - + + - + + - + - Computing the built-in voltage V0 (IV) Summary (I) nN pN pP nP N-side: many electrons Almost no holes V0 = VT·ln(nN/nP) 48 Computing the built-in voltage V0 (V) Summary (II) V0 + - P-side NA, pP, nP If NA >> ni (current case) p P = NA nP = ni2/NA V0 = VT·ln(pP/pN) N-side ND, nN, pN If ND >> ni (current case) nN = ND pN = ni2/ND and also V0 = VT·ln(nN/nP) V0 = VT·ln(NA·ND/ni2), VT = 26 mV at 300 K 49 Depletion width in P-side and in N-side N-side Al Al + Al + NA Al - Al - Al - WP0 Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ Sb+ - Al - Al - P-side WN0 W0 ND Charge neutrality implies: NA·WP0 = ND·WN0 The heavier doped a side, the narrower the depletion region in that side 50 Calculating the electric field E and the total depletion width W0 (I) We need to know: • The electric field E. • The total depletion width W0. rP-side P-side V0 - + E(x) + - - + WP0 WN0 rN-side N-side W0 We already know: • The charge density in both sides inside the depletion region: rP-side = NA·q and rN-side = ND·q • The relative width of the depletion region in the P-side and in the N-side: NA·WP0 = ND·WN0 • The built-in (contact) voltage: V0 = VT·ln(NA·ND/ni2) We will use Gauss’ law and the relationship between electric field and voltage 51 E(x) + P-side - + - Calculating the electric field E and the total depletion width W0 (II) N-side W0 - V0 + r(x) q·ND Charge density x E(x) V(x) Voltage law: ·E(x) = r(x)/e x -Emax0 V0 • Voltage x Electric field • Gauss’ -q·NA and electric field: E(x) = - V 52 + P-side E(x) - + - Calculating the electric field E and the total depletion width W0 (III) N-side W0 - V0 + q·ND Charge density After applying Gauss’ law and the relationship between electric field and voltage, we obtain: x -q·NA Electric field W0 = x -Emax0 Emax0 = Voltage V0 x 2·e·(NA+ND)·V0 q·NA·ND 2·q·NA·ND·V0 e·(NA+ND) 53 Summary of the study of the PN junction with no external bias Electric field at the metallurgical junction Emax0 - V0 + P-side Doped NA P-side: many holes Almost no electrons - + WP0 WN0 N-side Doped ND N-side: many electrons W0 Almost no holes or electrons, but space charge, electric field and voltage V0 = VT·ln(NA·ND/ni2) W0 = 2·e·(NA+ND)·V0 q·NA·ND W0 = WP0 + WN0 Emax0 = Almost no holes Very important equations!!! NA·WP0 = ND·WN0 2·q·NA·ND·V0 e·(NA+ND) 54 Connecting external terminals to a PN junction metal-semiconductor contacts +- P-side + VmP - V0 N-side +V - + Nm i=0 V=0 Therefore: V = 0, i = 0 Hence: VmP – V0 + VNm = 0 And: VmP + VNm = V0 No energy can be dissipated here Conclusion: The built-in voltages across each metalsemiconductor contact cancel out the effect of V0 in such a way that V0 does not appear externally. 55 Biasing the PN junction: forward bias Low resistivity: VP=0 + VmP i0 +- P-side Low resistivity: VN=0 VmP and VNm do not change and, therefore VmP+VNm= V0 N-side -V + j +V Nm V0 becomes Vj now + - Vext Vext = VmP - Vj + VNm = V0 - Vj Therefore: Vj = V0 - Vext Conclusion: The built-in voltage across the junction has decreased Vext volts 56 Biasing the PN junction: reverse bias Low resistivity: VP=0 +- P-side Low resistivity: VN=0 N-side - + + - Vj VmP i0 +V Nm VmP and VNm do not change and, therefore VmP+VNm= V0 - + Vext Vext = -VmP +Vj - VNm = -V0 + Vj Therefore: Vj = V0 + Vext Conclusion: The built-in voltage across the junction has increased Vext volts 57 Biasing the PN junction: notation for a general case +- P-side N-side - + Vj i + = - Vext Conclusion: • Always: Vj = V0 - Vext, being: 0 < Vext < V0 (forward biased) Vext < 0 (reverse biased) 58 Effects of the bias on the depletion region We must replace V0 with Vj, that is, replace V0 with V0-Vext With bias Without bias Vj0 = V0 W0 = Emax0 = Vj (Vext) = V0 - Vext 2·e·(NA+ND)·V0 q·NA·ND W(Vext) = 2·q·NA·ND·V0 e·(NA+ND) Emax(Vext) = Always: V0 = VT·ln(NA·ND/ni2) 2·e·(NA+ND)·(V0-Vext) q·NA·ND 2·q·NA·ND·(V0-Vext) e·(NA+ND) 59 W0 W P-side -- + Effects of the forward bias on the depletion region N-side VV 0-V1 0 r(x) V1 x E(x) • Less spatial charge x • Lower electric field -Emax -Emax0 Vj(x) V0-V1 V0 • Lower built-in voltage x 60 WW0 P-side - - ++ N-side 0 2 V0V+V V2 Effects of the reverse bias on the depletion region r(x) x E(x) x • More spatial charge • Higher electric field -Emax0 -Emax Vj(x) V0 V0+V2 x • Higher built-in voltage 61 Effects of the bias on the neutral regions (I) nNV nN - - - - - - - - - - - - - - - - + + + + - - - P-side V0V-V0ext Zona P - - + - nP nPV No bias: V0 = VT·ln(nN/nP) Forward bias: V0-Vext =VT·ln(nNV/nPV) For holes with forward bias: V0-Vext =VT·ln(pPV/pNV) 62 Effects of the bias on the neutral regions (II) • The quotients nNV/nPV and pPV/pNV strongly change with bias. • In practice, nNV and pPV do not change appreciably (i.e., nNV nN and pPV pP) for charge neutrality reasons. • Therefore the concentration of minority carriers (i.e., pNV and nPV) strongly changes at the depletion region edges. • The values of nPV and pNV can be easily obtained: V0-Vext =VT·ln(nN/nPV) nPV = nN·e -(V0-Vext)/ VT -(V0-Vext)/ VT p = p ·e V0-Vext =VT·ln(pP/pNV) NV P • As nN ND and pP NA, then: nPV = ND·e-(V0-Vext)/ VT pNV = NA·e-(V0-Vext)/ VT 63 Effects of the bias on the neutral regions (III) Non-biased junction -V0/ VT 2 pN = ni /ND = NA·e Non-biased junction nP = ni2/NA = ND·e-V0/ VT Biased junction -Vj/ VT -Vj/ VT nPV = ND·e +- pNV = NA·e Biased junction Zona P pP = NA - Vj + Vj = V0-Vext + Zona N nN = ND =Vext 64 Effects of the bias on the neutral regions (IV) Concentration of minority carriers : -Vj/ VT pNV = NA·e-Vj/ VT nPV = ND·e • Forward bias: The concentration of minority carriers at the depletion region edges increases, because Vj < V0 • Reverse bias: The concentration of minority carriers at the depletion region edges decreases, because Vj < V0 +- P-side - Vj + N-side Vj = V0-Vext • Forward and reverse bias: The concentration of majority carriers in neutral regions does not change 65 Effects of the bias on the neutral regions (V) What happens with the minority carriers along the neutral regions? +- P-side N-side - Vj + • This is a case of injection of an “excess” of minority carriers (see slide #36). • Cases of interest: a) Lp << xN (wide N-side) decay exponentially b) Lp >> xN (narrow N-side) decay linearly 66 Effects of the bias on the neutral regions (VI) The concentration of minority carriers along the neutral regions under forward biasing. Wide P and N sides Narrow P and N sides Vext P-side Minority carrier concentration nPV Vext N-side N-side Minority carrier concentration 0.1 mm pN Length 0 0.001 mm pNV nPV pNV nP 0 P-side nP pN Length Excess of minority carriers It plays a fundamental role evaluating the switching speed of electronic devices. 67 Effects of the bias on the neutral regions (VII) The concentration of minority carriers along the neutral regions under reverse biasing. Wide P and N sides Narrow P and N sides Vext P-side Minority carrier concentration nP 0 nPV Vext N-side P-side N-side Minority carrier concentration 0.1 mm pNV nPV pN Length 0 0.001 mm nP pN pNV Length Deficit (negative excess) of minority carriers 68 Carriers along the overall device Example of a silicon PN junction Properties of Si at 300 K Dp=12.5 cm2/s Dn=35 cm2/s p=480 cm2/V·s n=1350 cm2/V·s ni=1010 carriers/cm3 er=11.8 P -side N-side NA=1015 atm/cm3 ND=1015 atm/cm3 p=100 ns n=100 ns Lp=0.01 mm Ln=0.02 mm Carriers/cm3 V0=0.596 V 1016 Forward biased with Vext = 0.48 V 1014 pP nN 1012 1010 nPV pNV 108 They decay exponentially (log scale) Log scale 106 104 -0.3 -0.2 -0.1 0 0.1 Length [mm] 0.2 0.3 69 Calculating the current passing through a PN junction (I) • We have addressed a lot of important issues related the PN junction: Charge, electric field and voltage across the depletion region. Concentration of majority and minority carriers along the total device. • However, the most important issue has not been addressed so far: How can we compute the current passing through the device? • Fortunately, we already have the tools to answer this question. 70 Calculating the current passing through a PN junction (II) Case of wide P and N sides Vext jtotal 2 P-side several mm jtotal Vj P 3 -+ 3 1 N N-side 2 0.3m jtotal = jp_total(x) + jn_total(x) = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x) • Two questions arise: What carrier must be evaluated to compute the overall current? Where? • Places: Depletion region 1 . Neutral regions far from the depletion region 2 . Neutral regions, but near the depletion region edges 3 . 71 Calculating the current passing through a PN junction (III) Computing the overall current from the current density due to carriers in the depletion region Vext jtotal P-side jtotal Vj -+ P N N-side 0.3m Carriers/cm3 several mm 1016 p N 1014 nPV pNV nP Log scale 1m jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x) Currents due to drift (jp_Drift and jn_Drift) have opposite direction to currents due to diffusion. Both currents have extremely high values (very high electric field and carrier concentration gradient) and cannot be determined precisely enough to guarantee that the difference (which is the total current) is properly computed. Therefore, this is not the right place. 72 Calculating the current passing through a PN junction (IV) Vext Computing the overall current from the current density due to carriers in the neutral regions far from the depletion region jtotal P-side Vj P -+ N jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x) High concentration Weak field 0 jp_Drift(x) = q·p·p(x)·E(x) jp_Diff (x) = -q·Dp·dp(x)/dx jn_Drift(x) = q·n·n(x)·E(x) jn_Diff (x) = q·Dn·dn(x)/dx 0 0 Constant concentration Few electrons in P-side Current is due to drift of majority carriers. However, it cannot be determined properly because we do not know the value of the “weak” electric field. Therefore, these are not the right places. 73 Calculating the current passing through a PN junction (V) Computing the overall current from the current density due to carriers in the neutral regions but near the depletion region edges (I) Vext jtotal P-side Vj P -+ N jtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x) High concentration Weak field jp_Drift(x) = q·p·p(x)·E(x) 0 jn_Drift(x) = q·n·n(x)·E(x) jp_Diff (x) = -q·Dp·dp(x)/dx jn_Diff (x) = q·Dn·dn(x)/dx Few electrons in P-side We cannot compute the total current yet, but we can compute the current density due to minority carriers: jn_total (x) = jn_Drift(x) + jn_Diff(x) jn_Diff (x) = q·Dn·dn(x)/dx 74 Calculating the current passing through a PN junction (VI) Computing the overall current from the current density due to carriers in the neutral regions but near the depletion region edges (II) We can do the same for the holes just in the opposite side of the depletion region Vext Vj jn_total(x) P-side Minority carrier concentration P nPV N-side N pNV nP pN 0 Length jn_total(x) = q·Dn·dnPV(x)/dx Current density -+ jp_total(x) Taking derivatives jn_total(x) 0 jp_total (x)= -q·Dp·dpNV(x)/dx jp_total(x) Length 75 Calculating the current passing through a PN junction (VII) Computing the overall current from the current density due to carriers in the neutral regions but near the depletion region edges (III) What happens with carriers in the depletion region? Vext Vj jn_total(x) P-side Current density of minority carriers P jn_total(x) 0 -+ jp_total(x) N N-side jp_total(x) Length The carrier density currents passing through the depletion region are constant because the probability of carrier recombination is very low, due to the low carrier concentration in that region. 76 Calculating the current passing through a PN junction (VIII) Computing the overall current from the current density due to carriers in the neutral regions but near the depletion region edges (IV) Vext Now the total current density can be easily computed Vj jn_total(x) P-side P -+ jp_total(x) N N-side Current density jtotal jn_total(x) 0 jp_total(x) Very important conclusion!!! Length • The total current density passing through the device can be computed as the addition of the two minority current densities at the edges of the depletion region 77 Calculating the current passing through a PN junction (IX) Summary of the computing of the overall current density in a PN junction jn_total(x) P-side Vj -+ jp_total(x) N-side • We need to know the variation of the minority carrier concentrations at the depletion region edges. • We have to calculate the gradients of these concentrations (taking derivatives). • We have to calculate the current densities due to these minority carriers, which are diffusion currents. • We must add both current densities to obtain the total current density, which is constant along all the device. This is the total current density passing through the device. 78 Calculating the current passing through a PN junction (X) • Once the total current density and the minority-carrier current densities are known, the majority-carrier current density can be easily calculated by difference. jn_total(x) Vj jp_total(x) -+ P-side N-side Total current Current density jtotal Majority-carrier currents, due to both drift and diffusion jn_total(x) 0 jp_total(x) Length Minority-carrier currents, only due to diffusion 79 Current passing through an asymmetrical junction (P+N-) P-side is heavily doped (P+) and wide N-side is slightly doped (N-) and narrow Vext Vj jn_total(x) Concentration P+-side (wide) -+ N--side pNV pN nPV nP Length Current density 0 0 jtotal jp_total(x) This is a case of special interest, because it is directly related to the operation of Bipolar Junction Transistors (BJTs) jn_total(x) Length 80 Qualitative study of the current in a forward-biased PN junction Vext Vj jtotal P-side Minority carrier concentration P nPV -+ N-side N pNV nP pN 0 Length High slope High current density due to electrons in the depletion region High slope High current density due to holes in the depletion region Current density jtotal jn_total(x) 0 jp_total(x) High and positive total current density Length 81 Qualitative study of the current in a reverse-biased PN junction Vext Vj jtotal P-side Minority carrier concentration P nP N-side N pNV nPV 0 pN Length 0 Low slope Low current density due to electrons in the depletion region Current density -+ Low slope Low current density due to holes in the depletion region jn_total(x) jp_total(x) Low and negative total current density Length jtotal 82 Quantitative study of the current in a PN junction (I) Procedure: 1- Compute the concentration of minority holes (electrons) in the proper edge of the depletion region when a given voltage is externally applied. 2- Compute the excess minority hole (electron) concentration at the above mentioned place. It is also a function of the externally applied voltage. 3- Compute the decay of the excess minority hole (electron) concentration (exponential, if the semiconductor side is wide, or linear, if it is narrow). 4- Compute the gradient of the decay of the excess minority hole (electron) concentration just at the proper edge of the depletion region. 5- Compute the diffusion current density due to the above mentioned gradient. 5- Once the current due to minority holes (electrons) has been calculated, repeat the same process with electrons (holes). 6- Add both current densities. 7- Compute the total current by multiplying the current density by the crosssectional area. 83 Quantitative study of the current in a PN junction (II) The final results is: i = IS·(eV /VT ext - 1), where: IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)] (Is is called reverse-bias saturation current) VT = kT/q (Shockley equation) i + Vext P N - where: A = cross-sectional area. q = magnitude of the electronic charge (1.6·10-19 coulombs). ni = intrinsic carrier concentration. Dp = hole diffusion coefficient. Dn = electron diffusion coefficient. Lp = hole diffusion length in N-side. Ln = electron diffusion length in P-side. ND = donor concentration. NA = acceptor concentration. k = Boltzmann constant. T = absolute temperature (in Kelvin). 84 Quantitative study of the current in a PN junction (III) i = IS·(eV /V - 1) IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)] VT = kT/q ext T 100 i [mA] • Forward bias VO > Vext >> VT i IS Vext ·e VT exponential dependence - 0.25 0 0.25 0.5Vext [V] i [nA] • Reverse bias Vext << -VT Vext [V] 0 -0.5 i -IS constant (reverse-bias saturation current) -10 85 Quantitative study of the current in a PN junction (IV) Wide versus narrow P and N sides Wide sides Vext Narrow sides P-side N-side P-side N-side XP XP >> Ln XN XN >> Lp XP XP << Ln XN XN << Lp Minority carrier concentration nPV Minority carrier concentration pNV pNV nPV pN nP 0 Vext Length IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)] Equation i = IS·(eV ext /VT 0 nP pN Length IS = A·q·ni2·[Dp/(ND·XN)+Dn/(NA·XP)] - 1) is valid in both cases 86 Quantitative study of the current in a PN junction (V) The I-V characteristic in a real scale of use i [A] Actual I-V characteristic Equation i = IS·(eV /V - 1) describes the operation in the range VO > Vext > -. However, three questions arise: ext 3 According to Shockley equation • What happens if Vext > VO? Vext [V] 0 -4 VP 0 - P-side + VmP i 0 1 VN 0 + +- + T - + V j Vext N-side + - VNm • How does the temperature affect this characteristic? • What is the actual maximum voltage that the junction can withstand? When Vext appraches V0 (or it is even higher), the current passing is so high that the voltage drop in the neutral regions is not zero. This voltage drop is proportional to the current (it behaves as a resistor). 87 Temperature dependence of the I-V characteristic (I) Reverse bias: i -IS where: IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]. It should be taken into account that ni strongly depends on the temperature. Therefore: Reverse current strongly increases when the temperature increases. It doubles its value when the temperature increases 10 oC. Forward bias: i IS·eV /VT ext = IS·eq·V ext /kT Decreases with T Increases with T In practice, forward current increases when the temperature increases. For extremely high currents, the dependence can become just the opposite. 88 Temperature dependence of the I-V characteristic (II) Forward bias 30 Reverse bias i + i [A] V 370C i [A] P N Vext [V] -0.25 27 0C 27 0C Vext [V] 0 1 37 0C -10 In both cases, the current increases for a given external voltage. 89 Maximum reverse voltage that a PN junction can withstand • There are three different physical processes which limit the reverse voltage that a given PN junction can withstand: Punch-through It will be explained later Zener breakdown This phenomenon does not take place in power devices (two heavily doped regions are needs). Avalanche breakdown • Actual reverse current is higher than predicted due to the generation of electron-hole pairs by collisions with the lattice. • If the electric field is high enough, this phenomenon becomes degenerative. i + - + Vext - + +- - -- - i + + + P Vext N 0 90 Electric field in the depletion region with reverse bias No bias W0 = Emax0= Reverse bias 2·e·(N V A+ND)·V0 UV p e T q·N ·N A D PN 2·q·NA·ND·V0 e·(NA+ND) W(Vrev) = 2·e·(N V A+ND)·(V0+Vrev) UV p e T q·N ·N A D PN 2·q·NA·ND·(V0+Vrev) Emax(Vrev) = e·(NA+ND) W0 P - + V0 W(Vrev) N P - + N V0+Vrev -Emax0 -Emax(Vrev) • As already known, both the electric field and depletion length increase. • When the maximum electric field is high enough, the avalanche breakdown starts. 91 Limits for the depletion region with reverse bias Punch-through limit WPN Emax(Vrev) W(Vrev) P - N + |Vrev| 2·q·NA·ND·|Vrev| e·(NA+ND) • We must design the semiconductor according to: Emax(Vrev) < EBR. • The breakdown voltage is: VBR = EBR2·e·(NA + ND)/(2q·NA·ND). W(Vrev) -Emax -EBR Avalanche breakdown limit V 2·e·(N A+ND)·|Vrev| UV p e T q·NA·ND PN V UV p e T PN • Moreover, W(Vrev) < WPN to avoid the phenomenon called punch-through. • Usually W(VBR) < WPN, which means that practical voltage limit is not due to punch-through, but to avalanche breakdown. 92 What must we do to withstand high-voltage? VBR = EBR2·e·(NA + ND) 2·q·NA·ND VBR = EBR2·e 2q 1 1 ·( + ) ND N A • A high value of VBR is obtained if one of the two regions has been slightly doped (i.e., either NA or ND is relatively low). • However, it should taken into account that low values of ND (NA) implies: Wide WN (WP), which also implies wide XN (XP) to avoid punch-through. Low nN (pP) and, therefore, low conductivity. • If we have long length and low conductivity, then we have high resistivity. • Hence, a trade-off between resistivity and breakdown voltage must be established. P+ NA p P = NA NA >> ND - + N--side ND n N = ND WN XN 93 Maximum electric field Emax with reverse bias Vrev - P+ NA NA >> ND + N--side ND r(x) q·ND -q·NA x x -Emax • Can we increase VBR for a given EBR value? - Yes, we can. We must modify the electric field profile. - The result is the PIN junctions. -EBR VBR Vrev_N Vrev_P Vrev x The main part of the reverse voltage is dropping in the slightly doped region. 94 PIN junctions (I) W(Vrev) P - Vrev + N • Main idea: the voltage across the device is proportional to the dashed area (E(x) = - dV/dx). • Can we have the same area (same voltage across the device) with a lower value of Emax(Vrev)? -Emax(Vrev) • Yes, we can. We need another E(x) profile. -Emax(Vrev) new profile (ideal) -Emax(Vrev) • To obtain this profile, we need a region without space charge (undoped) inside the PN junction. -Emax(Vrev) new profile (real) -Emax(Vrev) 95 PIN junctions (II) Negative space charge Many holes P-side - • It means P-Intrinsic-N Positive space charge A few holes and electrons Intrinsic r(x) Many electrons + N-side q·ND Characteristics: x -q·NA x - Good forward operation due to conductivity modulation. - Low depletion capacitance. - Slow switching operation. -Emax Vrev All these characteristics will be explained later. x 96 Other structure to withstand high voltage: P+N-N+ Heavily doped P P+ NA - q·ND1 Lightly doped N + r(x) N- Heavily doped N ND1 q·(ND2-ND1) + N + N D2 q·ND2 Partially depleted x -q·NA Low reverse voltage Vrev1 x -Emax(Vrev1) -Emax(Vrev2) Reverse voltage Vrev2 Vrev2 > Vrev1 97 Forward-bias behaviour of structures to withstand high voltage P+ Intrinsic Undoped Lightly doped P+ N- N+ PIN In both cases, there is a high-resistivity layer (called drift region) N+ P+N-N+ • This means that, when forward biased, bad behaviour might be expected. • However, a new phenomenon arises and the result is quite better than expected. • This phenomenon is called conductivity modulation. In this case, high-level injection takes place. 98 Injection levels Carriers/cm3 Low-level injection: nN(0+) >> pNV(0+) High-level injection: nN(0+) pNV(0+) 1016 1014 pP pP nN 1012 1010 108 106 104 -0.3 P+-side nPV -0.2 P+-side N--side Log scale 0.2 N--side nPV Not possible! pNV -0.1 0- 0+ 0.1 Length [mm] nN pNV Log scale 0.3-0.3 -0.2 -0.1 0- 0+ 0.1 Length [mm] 0.2 0.3 • Low-level injection has been assumed so far, for PN and P+N- junctions. • In the case of a P+N- junction, this assumption is only valid if the forward bias is not very intense. Else, high-level injection starts. • If the forward voltage is high enough, pNV(0+) approaches nN(0+). In this case, nN does not remain constant any more, but it notably increases. 99 Conductivity modulation Drift region P+ NA = 1019 Holes are injected from the P+-side nP+ 10 N- N+ ND1 = 1014 ND2 = 1019 nN- pN- 1016 P+N-N+ 1014 Electrons are injected from the N+-side 106 pN+ 10 • There is carrier injection from both highly doped regions to the drift region. This is called double injection. • This phenomenon substantially increases the carrier concentration in the drift region, thus dramatically reducing the device resistivity. 100 Semiconductor junctions designed to withstand high voltage PIN P+N-N+ Summary • A high-resistivity region (drift region) is needed to withstand high voltage when the junction is reverse biased. • Fortunately, this high-resistivity “magically” disappears when the junction is forward biased if the device is properly designed to have conductivity modulation. • Due to this, devices where the current is passing through P-type and N-type regions (bipolar devices) have superior performances in on-state than devices where the current always passes through the same type (either P or N) of extrinsic semiconductor (unipolar devices). • Unfortunately, bipolar devices have inferior switching characteristics than unipolar devices. • Due to this, a trade-off between conduction losses and switching losses has to be established frequently selecting power semiconductor devices. 101 Transient and AC operation of a PN junction If we change the bias conditions instantaneusly, can the current change instantaneusly as well? • The answer is “no, it cannot”. • This is due to the fact that the current conducted by a PN junction depends on the minority carrier concentration just at the edges of the depletion region and the voltage withstood by a PN junction depends on the depletion region width. • In both cases, carriers have to be either generated or recombined or moved, which always takes time. These non-idealities can characterize as: • Parasitic capacitances (useful for linear applications) • Switching times (useful for switching applications) 102 Parasitic capacitances: depletion layer capacitance (I) (also known as junction capacitance) This is the dominant capacitance in reverse bias P-side - + N-side Zona N VO+Vext+Vext Vext + Vext r(x) x Carriers are pulled out from the depletion region when Vext is increased in Vext . Additional space charge has been generated. 103 Parasitic capacitances: depletion layer capacitance (II) PN junction Capacitor Vext - P N + Vext + Vext P - - + + Vext Vext + Vext + + + +++++ - - - ----- N • Capacitor: new charges are located at the same distance constant capacitance. • PN junction: new charges are located farther away from each other non-constant capacitance. 104 Parasitic capacitances: depletion layer capacitance (III) dQ -dQ • As it is a non-constant capacitance, static and dynamic capacitances could be defined. The latter is defined as: Cj=dQ/dV=e·A/W(Vext) W(Vext) As: Cj V 2·e·(N A+ND)·(V0-Vext) UV p e T q·NA·ND PN W(Vext) = Then: Vext 0 V Cj = A· UV e·q·NA·ND p e T PN 2·(N A+ND)·(V0-Vext) In an “abrupt” PN junction (as we have considered so far), this capacitance is a K·(V0-Vext)-1/2 -type function 105 Parasitic capacitances: diffusion capacitance (I) This capacitance is the one dominant in forward bias Cj 0 Reverse bias Vext Forward bias • Cj increases when the PN junction is forward biased. • However, depletion layer capacitance only dominates the reactance of a PN junction under reverse bias. • For forward bias, the diffusion capacitance (due to the charge stored in the neutral regions) becomes dominant. 106 Parasitic capacitances: diffusion capacitance (II) Carriers/cm3 1016 pP 1014 nN V=240mV nPV 1012 Increase of minority carriers due to a increase of 60mV in forward bias pNV V=180mV 1010 -3 -2 -1 0 1 2 3 Longitud [mm] • This increase in electric charge is a function of the forward bias voltage. • This means that a capacitive effect takes place in these conditions. • The dynamic capacitance thus obtained is called diffusion capacitance. 107 Switching times in PN junctions (I) Let us consider a PN diode as PN junction. The results obtained can be generalized to PN junctions in other semiconductor devices. R a V1 b V2 i V1/R v -V2 i + Transition from “a” to “b” (switching off) in a wide time scale (ms or s). v t t The diode behaviour seems to be ideal in this time scale. 108 Switching times in PN junctions (II) Transition from “a” to “b” (switching off) in a narrow time scale (s or ns). R a V1 b V2 i + i V1/R trr v - ts = storage time. tf = fall time. trr = reverse recovery time. Reverse recovery peak t ts tf (i= -0.1·V2/R) -V2/R v t -V2 109 Switching times in PN junctions (III) R a i + b V2 V1 i • This is because the junction cannot withstand voltage until all the excess of minoritary carriers disappears from the neutral regions. v - Carriers/cm3 8·1013 V1/R t3 t0 Why does this evolution occur? t4 t t1 t2 pNV nPV 4·1013 t1 -V2/R 0 v t t2 -0.1 -V2 t0 t3 t4 0 Length [mm] 0.1 110 Switching times in PN junctions (IV) R a i + b V2 V1 Transition from “b” to “a” (switching on) in a narrow time scale (s or ns). v - i 0,9·V1/R t0 td td = delay time. 8·1013 t4 tr t2 t3 pNV nPV 4·1013 t1 0,1·V1/R Carriers/cm3 t3 0 t2 t4 t1 t0 tfr tr = rise time. tfr = td + tr = forward recovery time. -0. 1 0 Length [mm] 0.1 111 Trade-off between static and dynamic behaviour in PN junctions P+ NA = 1019 1016 Excess of electrons in N- nP+ N- N+ ND1 = 1014 ND2 = 1019 nN- pN1014 106 Excess of holes in N- pN+ Log scale 10 Excess of electrons in P+ 10 Excess of holes in N+ • P+N-N+ and PIN structures allow us to combine high reverse voltage (due to a wide drift region) and low forward resistivity (due to conductivity modulation). • However, these structures imply large excess of minority carriers (even majority carries due to conductivity modulation). This excess of carriers must be eliminated when the device switches off to allow the device to withstand voltage. • The time to remove this excess of carriers depends on the width of the drift layer. If the drift layer is shorter than a hole diffusion length, then very little charge is stored and the device switches off fast. In this case, however, the device cannot withstand high reverse voltages. • The switching process can be made still faster by purposely adding “recombination centers”, such as Au atoms in Si, to increase the recombination rate. However, this fact 112 can deteriorate the conductivity modulation. Introduction to the metal-semiconductor junctions (I) • Metals have many more electrons than semiconductors. However, metals and semiconductors are different materials. This is not the case of a PN junction, where the two sides (P and N) are made up of the same material. • In a PN junction made up of a given semiconductor, electrons (holes) move from the N-side (P-side) to the P-side (N-side) due to diffusion, until the builtin voltage establishes an equilibrium between diffusion and drift currents. • In a metal-semiconductor junction, the electron movement when the junction is being built strongly depends on the work function of both materials. The higher the work function, the more difficult for the electrons to eject the material. • 4 possibilities exist when you build a Metal-Semiconductor (MS) junction: N-type semiconductor Metal Case #1: an N-type semiconductor transfers electrons to a metal Electrons (thin sheet) -- + + -- + + -- + + - + + N N-type Donor ions 113 Introduction to the metal-semiconductor junctions (II) Case #2: a metal transfers electrons to a P-type semiconductor Lack of electrons (thin sheet ) Recombination between the transferred electrons and the P-side holes takes place in this edge. Metal P-type semiconductor + - + + + - P + P-type + - + + - - Acceptor ions • In Case #1 and Case #2, a depletion region in the semiconductor side has been generated. • This depletion region has a built-in voltage that stops the electron diffusion. • This built-in voltage can be decreased by external forward bias (thus allowing massive electron diffusion) or increased by external reverse bias (avoiding electron diffusion). • The final result is that it works like a rectifying contact (similar to a PN 114 junction). Introduction to the metal-semiconductor junctions (III) Case #3: a metal transfers electrons to an N-type semiconductor N-type semiconductor Metal Lack of electrons (thin sheet) + + + + + + + + - N-type Electrons (thin sheet) Case #4: a P-type semiconductor transfers electrons to a metal P-type semiconductor Metal Electrons (thin sheet) - - + - + + + + + + + P-type Holes (thin sheet) • We have an ohmic contact (non-rectifying contact) in both cases. 115 Rectifying contacts (I) W0 Case #1: an N-type semiconductor transfers electrons to a metal -- + + -- + + Metal -- + + - + + N N-type ND • The width of the depletion region, the maximum electric field and the depletion layer capacitance can be calculated as in the case of a PN junction with an extremely-doped P side (i.e., NA ). • Therefore: W0 = 2·e·V0 q·ND Emax0 = 2·q·ND·V0 e Cj0 = A· However, the built-in voltage and the I-V characteristic depend on the work function of both the semiconductor and the metal. V e·q·N D U V T p e2·V PN 0 116 Rectifying contacts (II) • Built-in voltage: V0 = (Fm - Fs_N)/q, where: Fm = metal work function. FS_n = N-type semiconductor work function. • Barrier voltage to avoid electron diffusion without bias: VB = (Fm - cS_n)/q, where: cS_n = N-type semiconductor electron affinity. To define these concepts properly, • I-V characteristic: we should introduce others. This task, however, is beyond the scope i = IS·(eVext/VT - 1), as in a PN junction. of this course. However, the value of Is has a very different value: IS = A*·A·T2·e-V /V , where: B T A* = Richardson constant (120 amps/(cm2·K2)). 117 Rectifying contacts (III) • There is a type of diode based on the operation of a rectifying contact. It is the Schottky diode. Schottky diodes are widely used in many applications, including RF (telecom) circuits and low-voltage, medium-power power converters. • Main features: Lower forward voltage drop than a similar-range, PNjunction diode. They are faster than PN diodes because minority carriers hardly play any role in the current conduction process. They are majority carrier devices. However, they always have a higher reverse current (this is not a big problem). When they are made up of silicon, the maximum reverse voltage (compatible with reasonable drop voltage in forward bias) is about 200 V. However, Schottky diodes made up of wide-bandgap materials (such as silicon carbide and gallium nitride) reach breakdown voltages as high as 600-1200 V. Schematic Symbol 118 Ohmic contacts • There are two different possibilities to obtain ohmic contacts: a) According to the previous slides, to have an MS junction corresponding to Case #3 or to Case #4. b) To have MS junctions corresponding to Cases #1 or #2, but with an extremely-doped semiconductor side (1019 atoms/cm3). In this situation, electrons can flow in both directions by a tunneling process. Beyond the scope of this course, as well. Ohmic contacts Ohmic contacts P+ P NA1 = 1019 NA2 = 1016 Rectifying contacts N ND1 = 1016 N ND1 = 1016 N+ PN diode ND2 = 1019 Ohmic contacts N+ ND2 = 1019 Schottky diode 119