Presentation on theme: "Power Electronic Devices"— Presentation transcript:
1 Power Electronic Devices Electrical Energy Conversion and Power SystemsUniversidadde OviedoPower Electronic DevicesSemester 1Power Supply SystemsLecturer: Javier Sebastián
2 (Sistemas Electrónicos de Alimentación) Research GroupPower supply Systems(Sistemas Electrónicos de Alimentación)Javier Sebastián Dr. Electrical Engineer (Ingeniero Industrial) Full professorRoom Edificio nº 3, Campus Universitario de Viesques Gijón (Asturias). Spain Phone (direct): Phone (secretary): Fax:Web:
3 OutlineReview 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).
4 Power electronics devices OutlineReview of physical principles of semiconductorsPower electronics devicesGDS
5 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 Metal- Oxide-Semiconductor Field Effect Transistors (MOSFETs), both in N-channel- and P-channel types.
6 Semester 1 - Power Electronics Devices Electrical Energy Conversion and Power SystemsUniversidadde OviedoLesson 1 - Review of the physical principles of operation of semiconductor devicesSemester 1 - Power Electronics Devices
7 OutlineReview 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.
8 Energy level in a semiconductor as a function of inter-atomic spacing At 0 K, emptyInter-atomic spacingEnergy of electronsActual spacing-At 0 K, filled
9 Concept of band diagram Empty at 0 KEg4 electrons/atom-4 states/atomEnergy of electronsBand gapValence bandConduction bandFilled at 0 KMaterialEg [eV]Ge0.66 Si1.14H - SiC3.26GaN3.39
10 Band structure for insulators, semiconductors and metals at 0 K Band gap EgConduction bandValence bandSemiconductorEg=0.5-2 eVBand gap EgConduction bandValence bandInsulatorEg= 5-10 eVMetalsNo EgValence bandConduction bandOverlap
11 Visualization using the bonding model Band structure for semiconductors at room temperature. Concept of “hole”Si-EgConduction bandValence band-+-+-SemiconductorEg=0.5-2 eVVisualization using the bonding modelSome 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.
12 Concepts of generation and recombination Eg-+Eg-+--Si-Si-----++12
13 Why both holes and electrons are electric charge carriers? In general, there will be electric current due to both electrons and holesExample: when there is an electric field in the semiconductor lattice+-Si----++
14 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:MaterialEg [eV]ni [elect./cm3]Ge0.66 2.4·1013Si1.11.5·1010GaAs1.41.8·1064H - SiC3.268.2·10-9GaN3.391.9·10-10Taking 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)
15 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.-Si123AlAcceptor-Si1234SbDonor+-5--+
16 N-type and P-type semiconductors Si123Al-+Acceptor-Si1234Sb+5DonorN-type semiconductor:Majority carriers are electrons.Minority carriers are holes.Positively-charged atoms of donor (positive ions).P-type semiconductor:Majority carriers are holes.Minority carriers are electrons.Negatively-charged atoms of acceptor (negative ions).
17 Charges in N-type and P-type semiconductors +Acceptor ions (negative ions)Al--+Thermal generationP-type siliconholeelectron+-Donor ions (positive ions)Sb+-+Thermal generationN-type silicon
18 Charge carries in N-type and P-type semiconductors Al--+Concentration of majority carriers: pPConcentration of minority carriers: nPMass action law: pP·nP = ni2(only valid at equilibrium)Sb+-+N-typeConcentration of majority carriers: nNConcentration of minority carriers: pNMass action law: nN·pN = ni2(only valid at equilibrium)Very important equations!!!
19 Static charges in N-type and P-type semiconductors Al--+Concentration of acceptors: NA(only negative static charges in a P-type semiconductor)Sb+-+N-typeConcentration of donors: ND(only positive static charges in a N-type semiconductor)
20 Neutrality in N-type and P-type semiconductors Silicon, aluminium and antimony were neutral before being used Þ The extrinsic semiconductor must be neutral, too.P-typeAl--+Positive charges in volume V: pP·VNegative charges in volume V: nP·V + NA·VNeutrality: pP = nP + NASb+-+N-typeNegative charges in volume V: nN·VPositive charges in volume V: pN·V + ND·VNeutrality: nN = pN + NDVery important equations!!!
21 Calculating the concentration of electrons and holes (I) P-typeAl--+Neutrality: pP = nP + NAMass action law: pP·nP = ni22 known (NA and ni) and 2 unkown (pP and nP) variables Þ can be solvedSb+-+N-typeNeutrality: nN = pN + NDMass action law: nN·pN = ni22 known (ND and ni) and 2 unkown (nN and pN) variables Þ can be solved
22 Calculating the concentration of electrons and holes (II) Frequent case: quite heavy doped semiconductorsP-typeAl--+NA >> niNeutrality: pP » NAMass action law: nP » ni2/NAND >> niNeutrality: nN » NDMass action law: pN » ni2/NDSb+-+N-typeVery useful equations!!!
23 Mechanisms to conduct electric current: Drift (I) Semiconductors can conduct electric current due to the presence of an electric field EE----++++----++++jpjnjp_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.
24 Mechanisms to conduct electric current: Drift (II) jp_Drift = q·p·p·Ejn_Drift = q·n·n·Eq = magnitude of the electronic charge (1.6·10-19 coulombs).p = hole mobility.n = electron mobility.p = hole concentration.n = electron concentration.E = electric field.
25 Mechanisms to conduct electric current: Diffusion (I) ---------------12n1n2 < n1---jn_DiffElectrons have migrated due to “diffusion” (you can see the same phenomenon in gases)
26 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-----------------------------12n1n2 < n1---njn_Diff
27 Mechanisms to conduct electric current: Diffusion (III) --------------------12n1n2 < n1--njn_DiffThe current density is proportional to the electron concentration gradient:jn_Diff = q·Dn· nDn = electron diffusion coefficient.
28 Mechanisms to conduct electric current: Diffusion (IV) ++++++++++++++++++++12p1p2 < p1++pjp_DiffThe current density is proportional to the hole concentration gradient:jp_Diff = -q·Dp· pDp = hole diffusion coefficient.
29 Mechanisms to conduct electric current: Diffusion (V) jp_Diff = -q·Dp· pjn_Diff = q·Dn· nq = magnitude of the electronic charge (1.6·10-19 coulombs).Dp = hole diffusion coefficient.Dn = electron diffusion coefficient.Ñp = hole concentration gradient.Ñn = electron concentration gradient.
30 Summary of conduction mechanisms jp_Drift = q·p·p·Ejn_Drift = q·n·n·EDrift currents depend on the carrier concentration and on the electric field.jp_Diff = -q·Dp· pjn_Diff = q·Dn· nDiffusion currents do not depend on the carrier concentration, but on the carrier concentration gradient.
31 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) .AB
32 Continuity equations (II) jp_Ajp_Bjn_BABDifferent current density of holes across “A” and “B”.--++++ABExcess of carriers over the equilibrium (mass action law).+-LightABGeneration of electron-hole pairs by radiation (light) .-+
33 - + ·jp/q p/t = GL- [p(t)-p]/p ·jn/q Continuity equations (III)Taking into account the three effects, we obtain the continuity equation for holes:·jp/q -p/t = GL- [p(t)-p]/pVariation due to the different current density of holes across “A” and “B”Total time variation of holesVariation due to the generation of electron-hole pairs by lightVariation due to the excess of carriers over the equilibriumGL: rate of generation of electron-hole pairs by light.tp: hole minority-carrier lifetime.p: hole concentration in steady-state.Similarly, we can obtain the continuity equation for electrons:·jn/q +n/t = GL- [n(t)-n]/n
34 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.·jp/q -p/t = GL- [p(t)-p]/pÞ p0= GL·p + p+N+pp0NNow the light injected disappears. We want to compute the time evolution of the hole concentration afterwards.p0N+++++p(t)++++p
35 Time evolution of an “excess” of minority carries (II) We can also compute the time evolution of the hole concentration from the continuity equation:·jp/q -p/t = GL- [p(t)-p]/pAfter integrating Þ p(t) = p+(p-p)·e-tppTangent linepp0p(t)tSame areaPhysical interpretation: There is an appreciable increase of holes during 3-5 times tp.
36 where : Lp=(Dp· p)1/2 is the minority hole diffusion length 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.xxN+Np0+p·jp/q -p/t = GL- [p(t, x)-p]/pÞ 0 = - [p(x)-p]/p + Dp·2[p(x)-p]/x2After integrating Þ p(x) = p + C1·e-x/Lp + C2·ex/Lpwhere : Lp=(Dp· p)1/2 is the minority hole diffusion length
37 p(x) = p + C1·e-x/Lp + C2·ex/Lp Spatial evolution of an “excess” of minority carries (II)p(x) = p + C1·e-x/Lp + C2·ex/LpxN: length of the N-type crystalLp: hole diffusion lengthCases:a) Lp << xN (wide crystal): Þ p(x) = p+(p-p)·e-xLp (decay exponentially).b) Lp >> xN (narrow crystal): Þ p(x) = p+(p-p)·(xN-x)/xN (decay linearly).c) Other cases Þ hyperbolic sine evolution.Lp >> xN (narrow)Lp << xN (wide)p(x)pp0xxNp(x)pp0xxNTangent lineLp
38 What happens if we remove the barrier? Concept of PN junction (I)P-type silicon-+Al-N-type siliconSb+-+Barrier to avoid carrier diffusionWhat happens if we remove the barrier?
39 Are all the carriers to be diffused? Concept of PN junction (II)Al-P-side-+N-sideSb+-+Holes begin to diffuse from the P-side to the N-side. Similarly, electrons diffuse from the N-side to the P-sideAre all the carriers to be diffused?
40 - - - - - - P-side N-side Is this situation “the final situation”? Concept of PN junction (III)Are all the carriers to be diffused?P-sideN-sideAl-Al-+Al-Al--Sb+Sb+Sb+--Sb++++-+-+Al-Al-Al-Al-Sb+Sb+Sb+Sb+-Non-neutral P-type region, but negatively chargedNon-neutral N-type region, but positively chargedIs this situation “the final situation”?The answer is no
41 Concept of PN junction (IV) Al-P-side-+N-sideSb+-++-EAn electric field appears just in the boundary between both regions (we call this boundary metallurgical junction)
42 Concept of PN junction (V) Now, we do zoom over the metallurgical junctionAl-P-sideN-typeSb++-+-+-EDue to diffusion (® ¬)Due to drift (electric field) (¬®)The electric field limits the carrier diffusion
43 Steady-state situation near the metallurgical junction Concept of PN junction (VI)Steady-state situation near the metallurgical junctionNeutral P-type region (holes are balanced by negative ions )Al-+Al-Sb++-ENeutral N-type region (electrons are balanced by positive ions )Sb+-Depletion region, or space charge region, or transition regionUnbalanced charge exists because carriers barely exist
44 Summary and terminology Concept of PN junction (VII)Summary and terminologyMetallurgical junctionP-side(neutral)N-side+ -EV0Many holes, but neutralMany electrons, but neutralDepletion, 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
45 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.- +P-sideN-side+Due to diffusionjp_Diff+Due to driftjp_Drift-Due to diffusionjn_Diff-Due to driftjn_DriftCurrents must cancel each otherCurrents must cancel each other
46 Computing the built-in voltage V0 (II) pP (hole concentration in P-side )+(hole concentration in N-side ) pN+N-sideZona PV0++Due to diffusionjp_Diff+Due to driftjp_Driftjp_Drift = - jp_Diff
47 Computing the built-in voltage V0 (III) Equations:jp_Drift = - jp_Diffjp_Drift = q·p·p·Ejp_Diff = -q·Dp·dp/dxE = -dV/dxETherefore: dV = -(Dp/mp)·dp/pAfter 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)k = Boltzmann constant.VT = 26 mV at 300 K.It could be demonstrated:Dp/p = Dn/n = kT/q = VT (Einstein relation)
48 V0 = VT·ln(pP/pN) and also V0 = VT·ln(nN/nP) Computing the built-in voltage V0 (IV)pP+-nN-nPSummary (I)pN++-V0Zona PP-side: many holesN-side: many electronsAlmost no holesor electronsAlmost no electronsAlmost no holesV0 = VT·ln(pP/pN) and also V0 = VT·ln(nN/nP)
49 If NA >> ni (current case) If ND >> ni (current case) Computing the built-in voltage V0 (V)Summary (II)P-sideN-sideV0If NA >> ni (current case)pP = NA nP = ni2/NANA, pP, nPIf ND >> ni (current case)nN = ND pN = ni2/NDND, nN, pNV0 = VT·ln(pP/pN) and also V0 = VT·ln(nN/nP)V0 = VT·ln(NA·ND/ni2), VT = 26 mV at 300 K
50 Charge neutrality implies: NA·WP0 = ND·WN0 Depletion width in P-side and in N-sideP-sideN-sideW0WN0Sb+-NDWP0Al-+NACharge neutrality implies: NA·WP0 = ND·WN0The heavier doped a side, the narrower the depletion region in that side
51 Calculating the electric field E and the total depletion width W0 (I) V0-+rN-sideWe need to know:The electric field E.The total depletion width W0.rP-sideE(x)+-P-sideN-side-+W0WN0WP0We already know:The charge density in both sides inside the depletion region:rP-side = NA·q and rN-side = ND·qThe relative width of the depletion region in the P-side and in the N-side:NA·WP0 = ND·WN0The built-in (contact) voltage: V0 = VT·ln(NA·ND/ni2)We will use Gauss’ law and the relationship between electric field and voltage
52 Gauss’ law: ·E(x) = (x)/e Calculating the electric field E and the total depletion width W0 (II)E(x)+-P-sideN-side-+W0- V0 +(x)Charge densityxq·ND-q·NAGauss’ law: ·E(x) = (x)/eE(x)-Emax0Electric fieldxV(x)V0VoltagexVoltage and electric field: E(x) = - V
53 Calculating the electric field E and the total depletion width W0 (III) E(x)+-P-sideN-side- V0 +W0After applying Gauss’ law and the relationship between electric field and voltage, we obtain:-Emax0Electric fieldxCharge densityq·ND-q·NAV0Voltage2··(NA+ND)·V0W0 =q·NA·ND·(NA+ND)Emax0 =2·q·NA·ND·V0
54 Summary of the study of the PN junction with no external bias Electric field at the metallurgical junction Emax0- V0 +P-sideDoped NAN-sideDoped ND-+P-side: many holesAlmost no electronsW0WN0WP0N-side: many electronsAlmost no holesAlmost no holes or electrons, but space charge, electric field and voltageVery important equations!!!V0 = VT·ln(NA·ND/ni2)W0 = WP0 + WN0NA·WP0 = ND·WN02··(NA+ND)·V0W0 =q·NA·ND·(NA+ND)Emax0 =2·q·NA·ND·V0
55 Connecting external terminals to a PN junction metal-semiconductor contactsV0-+P-sideN-side+ -VmP-+VNm-+V = 0i = 0No energy can be dissipated hereTherefore:V = 0, i = 0Hence:VmP – V0 + VNm = 0And:VmP + VNm = V0Conclusion:The built-in voltages across each metal-semiconductor contact cancel out the effect of V0 in such a way that V0 does not appear externally.
56 Biasing the PN junction: forward bias Low resistivity:VN=0VP=0P-sideN-side+ -VmP and VNm do not change and, therefore VmP+VNm= V0VmP-+VNmVj-+i 0V0 becomes Vj nowVext-+Vext = VmP - Vj + VNm = V0 - VjTherefore: Vj = V0 - VextConclusion:The built-in voltage across the junction has decreased Vext volts
57 Biasing the PN junction: reverse bias Low resistivity:VN=0VP=0P-sideN-side+ -VmP-+VNmVj-+VmP and VNm do not change and, therefore VmP+VNm= V0i 0Vext-+Vext = -VmP +Vj - VNm = -V0 + VjTherefore: Vj = V0 + VextConclusion:The built-in voltage across the junction has increased Vext volts
58 Biasing the PN junction: notation for a general case Vext-+=VjP-sideN-side+ -iConclusion:Always: Vj = V0 - Vext, being:0 < Vext < V0 (forward biased)Vext < 0 (reverse biased)
59 Effects of the bias on the depletion region We must replace V0 with Vj, that is, replace V0 with V0-VextWithout biasWith biasVj0 = V0Vj (Vext) = V0 - Vext2··(NA+ND)·V0W0 =q·NA·ND2··(NA+ND)·(V0-Vext)W(Vext) =q·NA·ND·(NA+ND)Emax0 =2·q·NA·ND·V0·(NA+ND)Emax(Vext) =2·q·NA·ND·(V0-Vext)Always: V0 = VT·ln(NA·ND/ni2)
60 Effects of the forward bias on the depletion region P-side-+N-sideV0W0V0-V1P-side-+N-sideV1WEffects of the forward bias on the depletion region(x)xE(x)-Emax0Vj(x)V0-EmaxV0-V1Less spatial chargeLower electric fieldLower built-in voltage
61 Effects of the reverse bias on the depletion region P-side-+N-sideV0W0V0+V2V2P-side-+N-sideWEffects of the reverse bias on the depletion region(x)xE(x)-Emax0Vj(x)V0V0+V2-EmaxMore spatial chargeHigher electric fieldHigher built-in voltage
62 Effects of the bias on the neutral regions (I) nPVnNVnNnP- +Zona PP-side---V0V0-VextNo 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)
63 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) ÞV0-Vext =VT·ln(pP/pNV) ÞnPV = nN·e-(V0-Vext)/ VTpNV = pP·e-(V0-Vext)/ VTAs nN » ND and pP » NA, then:nPV = ND·e-(V0-Vext)/ VTpNV = NA·e-(V0-Vext)/ VT
64 Effects of the bias on the neutral regions (III) pN = ni2/ND = NA·e-V0/ VTNon-biased junctionnP = ni2/NA = ND·e-V0/ VTNon-biased junctionpNV = NA·e-Vj/ VTBiased junctionnPV = ND·e-Vj/ VTBiased junctionVext-+=VjVj = V0-VextZona PZona N+ -pP = NAnN = ND
65 Effects of the bias on the neutral regions (IV) pNV = NA·e-Vj/ VTnPV = ND·e-Vj/ VTConcentration of minority carriers :Forward bias: The concentration of minority carriers at the depletion region edges increases, because Vj < V0Reverse bias: The concentration of minority carriers at the depletion region edges decreases, because Vj < V0Vj-+P-sideN-side+ -Vj = V0-VextForward and reverse bias:The concentration of majority carriers in neutral regions does not change
66 What happens with the minority carriers along the neutral regions? Effects of the bias on the neutral regions (V)What happens with the minority carriers along the neutral regions?Vj-+P-sideN-side+ -This is a case of injection of an “excess” of minority carriers (see slide #36).Cases of interest:Lp << xN (wide N-side)Þ decay exponentiallyb) Lp >> xN (narrow N-side)Þ decay linearly
67 Excess of minority carriers Effects of the bias on the neutral regions (VI)The concentration of minority carriers along the neutral regions under forward biasing.VextP-sideN-sideWide P and N sidesVextP-sideN-sideNarrow P and N sides0.1 mm0.001 mmLengthpNVnPVMinority carrier concentrationpNnPLengthpNVnPVMinority carrier concentrationnPpNExcess of minority carriersIt plays a fundamental role evaluating the switching speed of electronic devices.
68 Deficit (negative excess) of minority carriers Effects of the bias on the neutral regions (VII)The concentration of minority carriers along the neutral regions under reverse biasing.VextP-sideN-sideWide P and N sidesVextP-sideN-sideNarrow P and N sides0.1 mm0.001 mmLengthpNVnPVMinority carrier concentrationpNnPLengthpNVnPVMinority carrier concentrationnPpNDeficit (negative excess) of minority carriers
69 pP nN nPV pNV Carriers along the overall device P -side N-side Properties of Si at 300 KDp=12.5 cm2/sDn=35 cm2/sp=480 cm2/V·sn=1350 cm2/V·sni=1010 carriers/cm3r=11.8Example of a silicon PN junctionNA=1015 atm/cm3p=100 nsLp=0.01 mmND=1015 atm/cm3n=100 nsLn=0.02 mmP -sideN-sideV0=0.596 VCarriers/cm3104101210141016-0.3-0.2-0.10.10.20.3Length [mm]1010108106pNVpPnPVnNForward biased with Vext = 0.48 VThey decay exponentially(log scale)Log scale
70 How can we compute the current passing through the device? 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.
71 Calculating the current passing through a PN junction (II) Case of wide P and N sidesseveral mmVextVj0.3mPN- +P-sideN-sidejtotal23321jtotal = 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 regionNeutral regions, but near the depletion region edges
72 Calculating the current passing through a PN junction (III) Computing the overall current from the current density due to carriers in the depletion regionseveral mmVextVj0.3mPN- +P-sideN-sidejtotalCarriers/cm3nPpN10141016pNVnPVLog scale1mmjtotal = 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.
73 Constant concentration Calculating the current passing through a PN junction (IV)Computing the overall current from the current density due to carriers in the neutral regions far from the depletion regionVextjtotalVjP-side- +PNjtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)High concentrationWeak field» 0Constant concentrationjp_Drift(x) = q·p·p(x)·E(x)jn_Drift(x) = q·n·n(x)·E(x)jp_Diff (x) = -q·Dp·dp(x)/dxjn_Diff (x) = q·Dn·dn(x)/dx» 0» 0Few electrons in P-sideCurrent 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.
74 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)VextVjPN- +P-sidejtotaljtotal = jp_Drift(x) + jp_Diff(x) + jn_Drift(x) + jn_Diff(x)High concentrationWeak fieldjp_Drift(x) = q·p·p(x)·E(x)jn_Drift(x) = q·n·n(x)·E(x)jp_Diff (x) = -q·Dp·dp(x)/dxjn_Diff (x) = q·Dn·dn(x)/dx» 0Few electrons in P-sideWe 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
75 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 regionVextVjPN- +P-sideN-sidejn_total(x)jp_total(x)LengthMinority carrier concentrationpNnPpNVnPVjn_total(x) = q·Dn·dnPV(x)/dxTaking derivativesjp_total (x)= -q·Dp·dpNV(x)/dxLengthCurrent densityjp_total(x)jn_total(x)
76 What happens with carriers in the depletion region? 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?VextVjPN- +P-sideN-sidejn_total(x)jp_total(x)jp_total(x)LengthCurrent density of minority carriersjn_total(x)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.
77 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)VextVjPN- +P-sideN-sidejn_total(x)jp_total(x)Now the total current density can be easily computedjtotaljp_total(x)LengthCurrent densityjn_total(x)Very important conclusion!!!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
78 Calculating the current passing through a PN junction (IX) Summary of the computing of the overall current density in a PN junctionVjP-sideN-side- +jn_total(x)jp_total(x)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.
79 Majority-carrier currents, due to both drift and diffusion 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.VjP-sideN-side- +jn_total(x)jp_total(x)Total currentjtotaljp_total(x)LengthCurrent densityjn_total(x)Majority-carrier currents, due to both drift and diffusionMinority-carrier currents, only due to diffusion
80 Current passing through an asymmetrical junction (P+N-) P-side is heavily doped (P+) and wideN-side is slightly doped (N-) and narrowVextVj- +P+-side (wide)N--sidejn_total(x)LengthpNConcentrationpNVnPVnPThis is a case of special interest, because it is directly related to the operation of Bipolar Junction Transistors (BJTs)LengthCurrent densityjtotaljp_total(x)jn_total(x)
81 High and positive total current density Qualitative study of the current in a forward-biased PN junctionVextVjPN- +P-sideN-sidejtotalLengthMinority carrier concentrationpNnPpNVnPVHigh slope Þ High current density due to electrons in the depletion regionHigh slope Þ High current density due to holes in the depletion regionjp_total(x)LengthCurrent densityjn_total(x)jtotalHigh and positive total current density
82 Low and negative total current density Qualitative study of the current in a reverse-biased PN junctionVextVjPN- +P-sideN-sidejtotalLengthMinority carrier concentrationpNnPpNVnPVLow slope Þ Low current density due to holes in the depletion regionLow slope Þ Low current density due to electrons in the depletion regionLow and negative total current densityLengthCurrent densityjn_total(x)jtotaljp_total(x)
83 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 cross- sectional area.
84 + - i Vext Quantitative study of the current in a PN junction (II) The final results is:i = IS·(eVext/VT - 1), where:IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)](Is is called reverse-bias saturation current)VT = kT/qwhere: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).PN+-iVext(Shockley equation)
85 Quantitative study of the current in a PN junction (III) i = IS·(eVext/VT - 1)IS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]VT = kT/qVext [V]1000.25- 0.25i [mA]0.5Forward bias VO > Vext >> VTi » IS·eVextVTÞ exponential dependence-10-0.5i [nA]Vext [V]Reverse bias Vext << -VTi » -ISÞ constant(reverse-bias saturation current)
86 Wide versus narrow P and N sides Quantitative study of the current in a PN junction (IV)Wide versus narrow P and N sidesVextP-sideN-sideWide sidesVextP-sideN-sideNarrow sidesXPXP >> LnXNXN >> LpXPXP << LnXNXN << LpLengthpNVnPVMinority carrier concentrationpNnPLengthpNVnPVMinority carrier concentrationnPpNIS = A·q·ni2·[Dp/(ND·Lp)+Dn/(NA·Ln)]IS = A·q·ni2·[Dp/(ND·XN)+Dn/(NA·XP)]Equation i = IS·(eVext/VT - 1) is valid in both cases
87 Quantitative study of the current in a PN junction (V) The I-V characteristic in a real scale of use1-43i [A]Vext [V]Equation i = IS·(eVext/VT - 1) describes the operation in the range VO > Vext > -. However, three questions arise:What happens if Vext > VO?How does the temperature affect this characteristic?What is the actual maximum voltage that the junction can withstand?Actual I-V characteristicAccording to Shockley equationVj-+VmPVNmi 0P-sideN-side+ -VN ¹ 0VP ¹ 0VextWhen 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).
88 Temperature dependence of the I-V characteristic (I) Reverse bias: i » -ISwhere: 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·eVext/VT = IS·eq·Vext/kTDecreases with TIncreases with TIn practice, forward current increases when the temperature increases. For extremely high currents, the dependence can become just the opposite.
89 + - Temperature dependence of the I-V characteristic (II) Forward bias 301i [A]Vext [V]Forward bias-0.25-10Vext [V]i [mA]Reverse biasPN+-iV370C37 0C27 0C27 0CIn both cases, the current increases for a given external voltage.
90 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-throughZener breakdownAvalanche breakdownIt will be explained laterThis phenomenon does not take place in power devices (two heavily doped regions are needs).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.iVexti+ Vext -PN+-+-+-
91 Electric field in the depletion region with reverse bias No biasReverse biasW(Vrev) =2··(NA+ND)·(V0+Vrev)q·NA·NDEmax(Vrev) =·(NA+ND)2·q·NA·ND·(V0+Vrev)W0 =2··(NA+ND)·V0q·NA·NDEmax0=·(NA+ND)2·q·NA·ND·V0V0PN-+W0-Emax0W(Vrev)NP-+V0+Vrev-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.
92 Avalanche breakdown limit Limits for the depletion region with reverse biasPunch-through limitEmax(Vrev) »·(NA+ND)2·q·NA·ND·|Vrev|W(Vrev)NP-+» |Vrev|-EmaxWPNWe must design the semiconductor according to: Emax(Vrev) < EBR.The breakdown voltage is:VBR = EBR2·e·(NA + ND)/(2q·NA·ND).W(Vrev) »2··(NA+ND)·|Vrev|q·NA·NDAvalanche breakdown limit-EBRMoreover, 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.
93 What must we do to withstand high-voltage? 2·q·NA·NDVBR =EBR2·e·(NA + ND)NA2qVBR = ·( )EBR2·eNDÞA 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+-N--side+NDNANA >> NDpP = NAnN = NDWNXN
94 Maximum electric field Emax with reverse bias Vrev P+-N--side+NDNA(x)xq·ND-q·NANA >> NDCan we increase VBR for a given EBR value?- Yes, we can. We must modify the electric field profile.- The result is the PIN junctions.-Emaxx-EBRVBRVrevxVrev_PVrev_NThe main part of the reverse voltage is dropping in the slightly doped region.94
95 PIN junctions (I)W(Vrev)NP-+Vrev-Emax(Vrev)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)?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
96 It means P-Intrinsic-N PIN junctions (II)It means P-Intrinsic-NNegative space chargePositive space chargeA few holes and electronsMany holesMany electronsP-side-N-side+Intrinsicq·ND(x)x-q·NACharacteristics:- Good forward operation due to conductivity modulation.- Low depletion capacitance.- Slow switching operation.All these characteristics will be explained later.-EmaxxVrevx96
97 - + P+ N- N+ x x Vrev2 > Vrev1 Other structure to withstand high voltage: P+N-N+Heavily doped PLightly doped NHeavily doped NP+-+N-N+NAND1ND2Partially depleted-q·NAq·ND1(x)xq·(ND2-ND1)q·ND2Reverse voltage Vrev2Low reverse voltage Vrev1x-Emax(Vrev1)-Emax(Vrev2)Vrev2 > Vrev197
98 Forward-bias behaviour of structures to withstand high voltage P+N+IntrinsicUndopedPINIn both cases, there is a high-resistivity layer(called drift region)P+N+N-Lightly dopedP+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.
99 High-level injection: Injection levelsLow-level injection:nN(0+) >> pNV(0+)High-level injection:nN(0+) » pNV(0+)Log scaleCarriers/cm3104101210141016-0.3-0.2-0.10- 0+0.10.20.3Length [mm]1010108106P+-sideN--sideLog scale-0.3-0.2-0.10- 0+0.10.20.3Length [mm]P+-sideN--sidepPpNVnPVnNpPpNVnNNot possible!nPVLow-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.
100 P+ N+ N- P+N-N+ nN- » pN- nP+ pN+ Conductivity modulation NA = 1019 Drift regionP+N-N+NA = 1019ND2 = 1019ND1 = 1014Holes are injected from the P+-sidenN- » pN-Electrons are injected from the N+-side101610141010610There 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.
101 Semiconductor junctions designed to withstand high voltage PINP+N-N+SummaryA 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.
102 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)
103 This is the dominant capacitance in reverse bias Parasitic capacitances: depletion layer capacitance (I)(also known as junction capacitance)This is the dominant capacitance in reverse biasVextP-sideVO+Vext-+Zona NVO+Vext+Vext-+N-sideVext + Vextx(x)Carriers are pulled out from the depletion region when Vext is increased in Vext . Additional space charge has been generated.
104 Parasitic capacitances: depletion layer capacitance (II) PN junctionCapacitor-+PNVextVextVext + VextVext + Vext-+PNCapacitor: new charges are located at the same distance Þ constant capacitance.PN junction: new charges are located farther away from each other Þ non-constant capacitance.
105 Parasitic capacitances: depletion layer capacitance (III) W(Vext)-dQdQAs it is a non-constant capacitance, static and dynamic capacitances could be defined. The latter is defined as:Cj=dQ/dV=·A/W(Vext)As:W(Vext) =2··(NA+ND)·(V0-Vext)q·NA·NDVextCjThen:Cj = A·2·(NA+ND)·(V0-Vext)·q·NA·NDIn an “abrupt” PN junction (as we have considered so far), this capacitance is a K·(V0-Vext)-1/2 -type function
106 This capacitance is the one dominant in forward bias Parasitic capacitances: diffusion capacitance (I)This capacitance is the one dominant in forward biasVextCjReverse biasForward biasCj 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.
107 Parasitic capacitances: diffusion capacitance (II) 1010101210141016Carriers/cm3-3-2-1123Longitud [mm]pPpNVnNnPVV=240mVV=180mVIncrease of minority carriers due to a increase of 60mV in forward biasThis 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.
108 The diode behaviour seems to be ideal in this time scale. 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.abV1V2Riv+-Transition from “a” to “b” (switching off) in a wide time scale (ms or s).ivtV1/RThe diode behaviour seems to be ideal in this time scale.-V2
109 Switching times in PN junctions (II) Transition from “a” to “b” (switching off) in a narrow time scale (ms or ns).abV1V2Riv+-ivtV1/R-V2/Rts-V2trrtf (i= -0.1·V2/R)ts = storage time.Reverse recovery peaktf = fall time.trr = reverse recovery time.
110 Switching times in PN junctions (III) abV1V2Riv+-Why does this evolution occur?This is because the junction cannot withstand voltage until all the excess of minoritary carriers disappears from the neutral regions.vtipNVnPVCarriers/cm38·10134·1013-0.10.1Length [mm]V1/R-V2/Rt0t0t3t4t1t2-V2
111 Switching times in PN junctions (IV) abV1V2Riv+-Transition from “b” to “a” (switching on) in a narrow time scale (ms or ns).pNVnPVCarriers/cm38·10134·1013-0. 10.1Length [mm]it4tr0,9·V1/Rtd0,1·V1/Rtfrt1t3t0t2td = delay time.tr = rise time.tfr = td + tr = forward recovery time.
112 Trade-off between static and dynamic behaviour in PN junctions 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.1016106101014P+N+N-NA = 1019ND2 = 1019ND1 = 1014nP+pN+nN- » pN-Excess of electrons in N-Excess of holes in N-Log scaleExcess of holes in N+Excess of electrons in P+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 can deteriorate the conductivity modulation.
113 Case #1: an N-type semiconductor transfers electrons to a metal 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 built- in 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 semiconductorMetalNCase #1: an N-type semiconductor transfers electrons to a metal+-N-typeDonor ionsElectrons (thin sheet)
114 Lack of electrons (thin sheet ) Introduction to the metal-semiconductor junctions (II)Case #2: a metal transfers electrons to a P-type semiconductorP-type semiconductorMetalPAcceptor ionsLack of electrons (thin sheet )-+P-typeRecombination between the transferred electrons and the P-side holes takes place in this edge.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 junction).
115 Case #4: a P-type semiconductor transfers electrons to a metal Introduction to the metal-semiconductor junctions (III)Case #3: a metal transfers electrons to an N-type semiconductorN-type semiconductorMetalN-type+-Electrons(thin sheet)Lack of electronsCase #4: a P-type semiconductor transfers electrons to a metalP-type semiconductorMetalP-type+-Electrons(thin sheet)HolesWe have an ohmic contact (non-rectifying contact) in both cases.
116 Case #1: an N-type semiconductor transfers electrons to a metal Rectifying contacts (I)W0MetalN+-N-typeCase #1: an N-type semiconductor transfers electrons to a metalNDThe 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:2··V0W0 =q·NDEmax0 =2·q·ND·V0Cj0 = A·2·V0·q·NDHowever, the built-in voltage and the I-V characteristic depend on the work function of both the semiconductor and the metal.
117 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, we should introduce others. This task, however, is beyond the scope of this course.I-V characteristic:i = IS·(eVext/VT - 1), as in a PN junction.However, the value of Is has a very different value:IS = A*·A·T2·e-VB/VT , where:A* = Richardson constant (120 amps/(cm2·K2)).
118 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, PN- junction 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 V.Schematic Symbol
119 Ohmic contacts P+ N+ N P N+ N 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 contactsOhmic contactsP+N+NNA1 = 1019ND2 = 1019ND1 = 1016NA2 = 1016PPN diodeRectifying contactsOhmic contactsN+NND2 = 1019ND1 = 1016Schottky diode
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