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# MICROELECTRONICS TECHNOLOGY ECSE 2210

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This 217 page Class Notes was uploaded by Immanuel Brakus PhD on Monday October 19, 2015. The Class Notes belongs to ECSE 2210 at Rensselaer Polytechnic Institute taught by E. Schubert in Fall. Since its upload, it has received 223 views. For similar materials see /class/224773/ecse-2210-rensselaer-polytechnic-institute in ELECTRICAL AND COMPUTER ENGINEERING at Rensselaer Polytechnic Institute.

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Date Created: 10/19/15

Chapter 112 Deviations from the ideal The measured characteristics deviates slightly from the ideal characteristics discussed We will discuss some of the non idealities of the BJ T characteristics BaseWidth modulation Punchthrough Avalanche multiplication and breakdown Others base resistance depletion region recombinationgeneral Review BJT in cutoffquot Review BJ T in saturation Base Width modulation W B E C WB Emitter A Base A A Collector 1930 72 quot1930 C0 CO qVEB 1C N 0 qADB pBO e kT WB WB Base Width modulation When the reverse bias applied to the CB junction increases the C B depletion Width increases and W decreases Thus the collector current 1C increases This is also known as Early Effect More prominent in narrowbase transistors IC 3H 3 mA 2mA ZHA lmA IHA Punchthrough Punchthrough can be Viewed as base width modulation carried to the extreme ie punchthrough occurs when W gt O For C B voltage beyond punchthrough the EB barrier lowers and results in large increase in carrier injection from emitter to collector Large increase in collector currents at high VCEO occurs due to two reasons Punchthrough or Avalanche multiplication Example 1 ltWVB ltWVB 1019 1016 1016 1019 1015 1016 Two transistors are identical except that the base doping is different A Which transistor will have higher baseWidthmodulation effect B Which one will have higher punchthrough voltage Approximate value of punchthrough voltage can be obtained by equating the depletion layer Width on the base side to the base width W3 1 NC 5 239 NC 5 28s1 WB x CB 1 Vb CB VCB VCB n q NBNCNB 1 q NBNCNB Other effects Base series resistance Current crowding Recombinationgeneration current Modern BJ T structures Heterojunction bipolar transistor HBT Chapter 34 RG statistics RG statistics is the mathematical characterization of RG processes An important generation process in device operation is photo generation If the photon energy hv is greater than the band gap energy then the light will be absorbed thereby creating electronhole pairs Photogeneration The intensity of monochromatic light that passes through a material is given by I 10 eXp or x where 10 is the light intensity just inside the material at x O and or is the absorption coef cient Note that or is material dependent and is a strong function of 7 Since photogeneration creates equal s of holes and electrons and each photon creates one eh pair we can write an 6p ocx Eilight Eilight GLOW GLOe Where GL0 is the photogeneration rate cm3 s at x 0 Question What happens if the energy of photons is less than the band gap energy Light absorption and transmittance Consider a slab of semiconductor of thickness Z lt Z gt gtlt It 10 eXp ocl Where I0 is light intensity at x 0 and It is light intensity at x Z Absorption coef cient vs wavelength in semiconductors 105 I llllL 104 103 IIIII lllllll a cm1 102 IIIIIII I l GaAs l 10 InP GaP I IIIITIII lllllllll I IIIIIII llllllll 1 I I 04 06 08 10 12 14 16 18 1 pm Figure 320 Bandgaps of common semiconductors l l E 5 a E quotU v r a 3 O a G H quot4 gt4 3 CD gt Infrared 39 Visible Ultraviolet GaAs GaP InSb Ge Si CdSe CdS SiC ZnS 5k 39 e I j f E eV 0 1 2 3 4 A lt 5 m 05 035 M Indirect thermal recombinationgeneration n0 p0 under thermal equilibrium n p under arbitrary conditions functions of 2 An n 720 An and Ap are deviations in carrier concentrations Ap p p om their equilibrium values An and Ap can be 0 both positive or negative Nt is the of RG centerscm3 Lowlevel iniection condition is assumed Change in the majority carrier concentration is negligible For example in ntype material Apltlt710 nno R G statistics Consider ntype silicon under perturbation We look at only minority carriers and in this case holes In general 6p 6p 6p G x at at R at lG L rate of loss gain external hole due to due to such as build up recomb generation light lt11 at should be proportional to p and Nt Why R 0p EiR Clep RG statistics continued Under thermal equilibrium GL O and dpdt O G R Cptho So under arbitrary conditions when GL O 0p A p de ning IP 2 TP CpNt 6p 6 6A Slnce A 6t 62 po 19 6t RG statistics continued We can write For holes in ntype Similarly For electrons in ptype Tp or In is called minority carrier lifetime indicating the average time an excess minority carrier will survive in a sea of majority carriers Minority carrier lifetime is an important material parameter Depends strongly on the concentration of deeplevel of impurities crystalline quality etc Varies strongly from material to material Varies from a few ns to few ms in silicon depending on the quality What happens when the perturbation is removed at t O 8A1 A T For holes 1n an ntype sem1conductor gt P Ap The solutlon for t gt 0 1s Ap0 Ap Ap0 exp t Ip The excess carrier concentration exponentially decays to zero when the external perturbation is removed This fact is used to measure lifetimes using photoconductivity decay technique See Sect 334 Photoeonduetivity decay measurement Light g pdws A Vi R g S VA RL UL m V Figure 322 Scope 11 Photoeonduetivity decay measurement system GPIB t0 Ebe gt Computer 39 E TEK11401 11 5 L C P55004 I FG5010 DM501O II II I II III III 56 asss 8 n 5508 INNIIHIIIIHHHIIIIIIIIIII 3939 1 Stroboscobe Figure 323 12 Photoconductivity transient response 1E 1 150 17 cc 1slope 39 1E 2 50 j 00 02 04 06 08 10 00 02 04 06 08 10 x1E 3 x1E 3 SECONDS SECONDS a b VOLTS VOLTS Figure 324 13 Chapter 162 MOS electrostatic Quantitative analysis In this class we will Derive analytical expressions for the charge density electric eld and the electrostatic potential Expression for the depletion layer Width Describe delta depletion solution Derive gate voltage relationship Gate voltage required to obtain inversion Electrostatic potential bot De ne a new term lx taken to be the potential inside the semiconductor at a given point x The symbol 4 instead of Vusecl in MOS work to avoid confusion with externally applied voltage V WC 2 lEibu1k Ei 36 Potential at any point x q I 1 E1 bulk Ei surface Surface potential S q 1 l IF l related to doping F g E1 b11110 EF concentration F gt 0 means ptype IF lt 0 means ntype Electrostatic parameters S IS is positive if the band bends downward Oxide I Semiconductor 0 60 EC Pf IS 2 at the myE Tq p E depletioninversion F transition point I Ev F Bulk 0 b Example 1 Consider the following bF and bs parameters Indicate whether the semiconductor is ptype or ntype specify the biasing condition and draw the energy band diagram at the biasing condition i F 12 kTq 15 12 kTq F 12 kTq means that Ei EF in the semiconductor is 12 kT a positive value So ptype N A ni exp Ei EF kT s12 kTq means Ei bulk Eisurface 12 kT ie the band bends downward near the surface Example 1 continued ii F 9 kTq S 18 kTq here bF 9 kTq means Eibulk EF 9 kT ie Ei is below EF Thus the semiconductor is ntype s 18 kTq means that Ei bulk Eisurface 18 kT So band bends upwards near the surface The surface is inverted since the surface has the same number of holes as the bulk has electrons Deltadepletion solution MO S Consider pthe silicon S Accumulatlon cond1tlon VGltO JT 1 The accumulation charges Accumulation tholes are mobile holes and appear close to the surface and falloff I rapidly as x increases Assume that the free carrier x concentration at the oxide semiconductor interface is a 8 function Charge on metal QM Charge on semiconductor charge on metal I QAccumulationl I Delta depletion solution cont Consider p pe Si depletion condition Apply VG such that 1S lt 2 1F Charges in Si are immobile M O S ions results in depletion pSi layer similar to that in pn V gt O J 1 junction or Schottky diode G T mmAmpu QM If surface potential is 1S with respect to the bulk then the depletion layer Width E E WWill be S 4 Depletion of holes id MZ ng W qN A l 2 E W and ESi 2 WM 12 At the start of inversion pg 2 bF and W 2 WT 285i 24m qNA 7 Depletion layer Width W and E eld For a pn junction or a MS nSi junction the depletion layer Width is given by band bending Vbi in Volts is numerically equal to the amount of band bending in eV 12 qN 2qN Emax DW 2 DVbi 8s1 8s1 qN D For MOS the same equation applies except that Vbi is replaced by pg 2 N 12 2 N 12 Emaxan Si Elk or Ma 8s1 8s1 ntype ptype Delta depletion solution cont MO S Consider pSi strong inversion pSi Once inversion charges VGgtgtO 1 appear they remain close to the surface since they are mobile Any additional voltage to the gate results gt QM Depletion of in extra QM in gate and get holes compensated by extra Inversion electrons inversion electrons in 5functi0nlike semiconductor So depletion layer does not have to increase to balance the charge on the metal Electrons appear as 5function near the surface Maximum depletion layer Width W WT Gate voltage relationship Applied gate voltage will be equal to the voltage across the oxide plus the voltage across the semiconductor Consider ptype Si VG Abox AlSemi M O S S1 VGgt 0 i p 1 Acbsem w 0 ltbu1k T t S A 0X A semi AIOX xOX Equot0X Since the interface does not have any charges up to inversion we can say that 80X on asi 81 fox 3s1 30x 51 Gate voltage relationship cont 12 ESi qNAW WA s for 0lt SltZ F 881 8Si qNA 12 S 881 VG bs xOXEOX 881 bs xox ESi OX 12 8 2 N Z sxox Slq A sj for Og s32 F 80X 8Si Gatevoltage relationship Alternative method Consider ptype silicon VG Abox i Semi A OX QMC0X QSC0X Where Cox is oxide capacitance and QS is the depletion layer charge in semiconductor Qs q A NAW Cox 2 EOXA xox OX Ad QANAW 28s1x qNA 0X goxAxox 8ox 8Si ox 8Si 8ox 8Si same as before 12 12 8 N 8 2N VGZ S Sl xoxq AWZ S Sl xox q A Sj Chapter 52 Quantitative electrostatic relationships We make the analysis in l dimension even though actual diode as shown may not be a onedimensional system This makes the analysis simple The metallurgical junction is located at x 0 13 Ohmic contacts nsilicon a b Figure 58 Quantitative analysis Electric eld E P Where 8K8 g S 0 NA ND p 2 n gNA xpltxlt0 8 xp 0 x11 q END Oltxltxn O 39 W xgtx11 xlt xp y I v N q Axpx xprSO 8 Emmi q NAxps N q Dxnx Oltxgxn qNDxns 8 O xlt xp xgtx11 Relationship between X and x12 Emmi q NAxp e q NDxn e N A xp ND x11 Net charge on pside Net charge on nside Depletion layer Width W x11 xp xn 2 WA xp WA NAl ND NAlND IfNA gtgt ND then Wz x11 and ifNA ltlt ND then Wz xp Builtin voltage Vm xdx P Vbi area under E versus x curve 1 2 M 1 8 g c NAND NAND 12 WW NDxn8 q28 NDon NAND NA ND W2 since xn W 12 NA NAND Quantitative analysis Electrostatic potential 1V qNA a 8 xpx xprSO quDxnx 033can 8 with the reference potential at x xp set to zero AV qNA 2 Vx x x x SxSO 28 p j p KJVbi Vbi qND Jen x2 OSxan x 28 Step iunction with Vii O VA dropped here Negligible voltage drop lowlevel injection Negligible VOItagc drop lowlevel injeCtion Negligible voltage drop Negligible voltage drop ohmic contact ohmic contact P N I VA Figure 510 The equation for W is similar to the earlier equation except that Vbi is replaced by Vbi VA VA is restricted to VAlt Vbi 12 2 W 8 M VbiVA q NAND Effects of forward and reverse bias b x d Figure 51 1 PN iunction energyband diagrams b Forward bias VA gt 0 c Reverse bias VA lt 0 Figure 512 Example 1 Consider the following diode Calculate the maximum electric field Emax the location of Emax the depletion layer Width W x11 and xp and the builtin voltage Vbi Carefully plot the charge density electric field and the potential as a function of x ND2gtlt1016cm393 NA3gtltlOl7cm393 NA lgtlt1016cm393 ND2gtltlOl7cm393 Also calculate the depletion layer Width and the maximum electric field if a reverse voltage of 10 V is applied across the diode What Will be Wif VA 05 V Example 2 Consider the diode of Example 1 Calculate the depletion layer width and the maximum electric eld if a reverse voltage of 10 V is applied across the diode Calculate the depletion layer Width in the nside and pside under this biased condition What Will be Wif VA 05 V What happens if we apply VA gt Vbi ECSE 2210 Microelectronics Technology Class Activity 28 Solution 1 Two MOS capacitors are made from different silicon substrates A and B All parameters are identical except the substrate doping Substrate A is doped with 1016 cm 3 acceptors and substrate B is doped with 1014 cm 3 acceptors a b Which one will have the higher VT VT 2 F m K Ks Sox Substrate A will have higher VT The higher the doping the higher is by So we have to apply more gate voltage to get the bands in the surface region to bend down below Ei Hence the threshold voltage is larger for the MOSC with higher substrate doping Qualitatively draw the high frequency C G VG characteristics for both m 2 An ideal MOSC is made from ptype Si with a dopant concentration of 2 X 1015 cm 3 The oxide thickness is 01 pm and the junction areaA 1 cmz a Fquot C Determine VT for this deVice 2 F st0x 4qNA F KOX KS 8OX E1 EF 71 ex 1 P kT J 2x1015 1010 exp 00259 2x1015 1010 719 15 VT06323x01x10 41W0316 1235V 10 Determine the oxide capacitance 80x 39 X 885 gtlt10714 Cox xox 01gtlt10 4 Determine the semiconductor capacitance when VG VT VT p F Ei EF 00259111 0316V 345 x10 8 Fcm2 WT 219581 2 0628unn qNA e12 CDX i 4 159x10 8 F2 WT 0628x10 Em d Detetnntne the tota1 htgh frequency steadyrstate gate eapaettanee when VG V1 and VGgt VT P1ot the CG39VG eharaetensttes sews VTCDXC CDXC107 gtlt1039Elcm2 Law equancy Htgn frequency ttt tn e Suppose the gate vo1tage ts zeto at t 0 and mptdzy changes torn VG 0 to Detetnntne the total gate eapaettanee at t 0 tht Fttst ftnd 4 usmg equauon 16 28 Then nd Wusmg equauon1615 Ifwe subsutute x for 65 equauon 16 28 beeonnes 5 28 0 758 x 9x 5 19V and Qs36V W 1 5 m Cs 6 66 x 103 Fenn2 CG CWCS CMr cg 5 5 gtlt10399 Fenn2 f Fot VG gt VT an tnvetston ehanne1 forms The tnvetston ehanne1wt11eonststo e1eettons n ehoose one If the MOSFET ts made usmg the above subsuate the devtee ts ea11eo1 nrchannel hoose one MOSFET g Suppose a gatervoltage of 5V ts apphed to the MOSFET gate Assume VD ts e1ose to zeto Detetnntne the tnvetston 1ayet charge gm tn c tht gm ts a1nnost zeto when V T7 1 24 V When V57 V7 the tnvetston 1ayet tust statts to form VG gt VT thete wtu be nnotetnvetston1ayet ehatges than dopant atonns tn the tntetfaee e C t VG VT tnt Qm5e1235x 34 Nate gm eonesponds 5 x10 e129 gtlt1077 Cenn2 to 7 8 x 10 e1eetxonsenn2 h Suppose a drain voltage of 4 V is applied while the gate voltage is 5 V What will be the inversion layer charge Qinv at the source end What will be inversion layer charge Qim at the drain end At the source end Qim will be the same as above At the drain end the difference between the gate voltage and the drain side is less than the threshold voltage ie VG7 VD 5 V 7 4 V l V lt VT Therefore the inversion layer charge is zero Pinchoff condition Chapter 21 Semiconductor models The subatomic particles responsible for charge transport in metallic wires 7 electrons The subatomic particles responsible for charge transport in semiconductors 7 electrons amp holes In this chapter we will study these topics The quantization concept Semiconductor models Carrier properties State and carrier distributions Equilibrium carrier concentrations g Quantization concept In 1901 Max Planck showed that the energy distribution of the black body radiation can only be explained by assuming that this radiation ie electromagnetic waves is emitted and absorbed as discrete energy quanta photons The energy of each photon is related to the wavelength of the radiation EhVhC where h Planck s constant h 663 X 10 34 Js v frequency Hz s l c speed of light 3 X 108 ms 9 wavelength m Example Our eye is very sensitive to green light The corresponding wavelength is 0555 um or 5550 A or 555 nm What is the energy of each photon 662 x10 34 Js gtlt 3x108 ms Ehv 76 357gtlt10 19J 0555 gtltlO In These energies are very small and hence are usually measured using a new energy unit called electron Volts 1eV16x10 19CV16X10 19J A new unit of energy Since the energies related to atoms and photons are very small EGREEN LIGHT 357 lt 10 19 J we have de ned a new unit of energy called electron Vol or eV One eV is the energy acquired by an electron when accelerated by a 10 V potential difference 1V l Vl6X10 19J e Energy acquired by the electron is qV Since q is 16 x 10 19C the energy is 16 x 1019 De ne this as 1 eV Therefore EGREEN LIGHT 2236V i av 1 gt lt 16x1049 Cfv i 6gtlt101399 I g Quantization concept continued 2 Niels Bohr in 1913 hypothesized that electrons in hydrogen was restricted to certain discrete levels This comes about because the electron waves can have only certain wavelengths i e n 211quot where r is the orbit radius Quantization Based on this one can show that 4 4 EH m0 1 2 03 2 for n1 23 24nsohn 8807 h h where h 2 and h Planck39s constant 11 Bohr s hydrogen atom model Flguro 21 A numerical example 4 E m0 q4 911x10 311ltggtlt16gtlt10 19 C n 2 2 2 2 2 8 8o n 1 8x 885x10 12 Fm gtlt1gtlt662gtlt10 34 Js 217gtlt10 19 J 135 eV for the 111 orbit For the n 2 orbit E2 34 eV and so on The number n is called the principal quantum number which determines the orbit of the electron Since Hydrogen atom is 3D type we have other quantum numbers like 1 and m within each orbit So in atoms each orbit is called a shell See AppendixA in text for the arrangement of electrons in each shell and also for various elements in the periodic table Atomic configuration of Si So an important idea we got from Bohr model is that the energy of electrons in atomic systems is restricted to a limited set of values The energy level scheme in multielectron atom like Si is more complex but intuitively similar Ten of the 14 Siatom electrons occupy very deep lying energy levels and are tightly bound to the nucleus The remaining 4 electrons called valence electrons are not very strongly bound and occupy 4 of the 8 allowed slots Con guration for Ge is identical to that of Si except that the core has 28 electrons six Allowed levers mm cnt rgy Two allowed level ni we nnugy n l 2 Llcumns quot 3 Figure 22 SLIIL IIIHUU Npl w39quotl1llm on nu imlnlud s mom Bond model Consider a semiconductor Ge si or c Ge si and c have four nearest neighbors each has 4 electrons in outer shell Each atom shares is electrons with its nearest neighbor This is called a covalent bonding No electrons are available for conduction in this covalent s cture so the material is and should be an insulator at 2dimensional 2D semiconductor bonding model Line represents a shared valence electron Circle represents the core of a semiconductor e g Si atom Figure 23 No electrons are available for conduction Therefore Si is an insulator at T O K Simplified 2D representation of Si lattice o How many atomneighbors has each Si atom in a Si lattice o How many electrons are in the outer shell of an isolated Si atom How many electrons are in the outer shell of a Si atom with 4 neighbors a Point defect b Electron quot Flgure 24 At higher temperatures eg 300 K some bonds get broken releasing electrons for conduction A broken bond is a de cient electron of a hole At the same time the broken bond can move about the crystal by accepting electrons from other bonds thereby creating a hole Energy band model An isolated atom has its own electronic structure with n l 2 3 shells When atoms come together their shells overlap Consider Silicon Si has 4 electrons in its outermost shell but there are 8 possible states When atoms come together to form a crystal these shells overlap and form bands We do not consider the inner shell electrons since they are too tightly coupled to the inner core atom and do not participate in anything Development of the energyband model No mm an pram mm 1N Hum mu 4Neleclmns om Flgure 25 Energy band model At T0K No conduction can take place since there are no carriers in the conduction band Valence band does not contribute to currents since it is full Actually valence electrons do move about the crystal No empty energy state available For every electron going in one direction there is another one going in the opposite direction Therefore Net current ow in lled band 0 Both bond model and band model shows us that semiconductors behave like insulators at UK Visualization of carriers usin ener bands Empty E Completely lled a N0 carriers 6 gt E Ec Ev T1 h The electron c The hole Figure 27 Insulators semiconductors and metals electrons EC E E0 142 eV GaAs Wide G 8eVSi02 E5 lZeVSi 56quot 5 6V Diamond E EC 066 ev Ge Thermal quot Room temperature E excitation quot moderately easy a Insulator b Semiconductor Very i EC E narrow E quot T quot or Ec c Metal Figure 28 Chapter 10 Bipolar junction transistor fundamentals Invented in 1948 by Bardeen Brattain and Shockley Contains three adjoining alternately doped semiconductor regions Emitter E Base B and Collector C The middle region base is very thin compared to the diffusion length of minority carriers Two kinds npn and pnp Schematic representation of pnp and npn BJ Ts Finure 101 Emitter is heavily doped compared to collector So emitter and collector are not interchangeable The base width is small compared to the minority carrier diffusion length If the base is much larger then this will behave like backtoback diodes BJ T circuit symbols 0 mm Figure 102 IEIBIC and VEBVBCVCE0 VCE VEC As shown the currents are positive quantities when the transistor is operated in forward active mode BJ T circuit confi urations and out ut characteristics A C J E 1 I B B P v B 13 V V P N mm V H N mifx 5quot 144 K33 lt5 g 3 E Q E c c L I Iquot 40 volls Vac min a Cnmmon base on Common cmiucr BJT biasing modes Table 101 Biasing Modes Biasing Biasing Polarity Biasing Polarity Mode E B Junction C73 Junction Active Forward Reverse I Inverted Reverse Forward Cutoff Reverse Reverse v VnLMWI Active Sumanon Vca Imp Vm39 non a Common base Cross sections and sim li ed models ofdiscrete and 1C npn BJ Ts E E E B he 7 LL B A my w my waes ym W 39 P 1 Wide mo pm msnmmm i C C m n 5 mm a L 50 39 I I M m P quotmm s Em V 7 Made J W quotNb nHm mIaya wwgm Am WNW h ngurelns Electrostatic variables for a pnp BJ T at equilibrium p charge density 8 KS 80 Qualitative discussions of operation Consider two diodes one forward biased and one reverse biased quotU amp m ICVCI SC 8 E 5 ll TT quot g Qualitative discussion of transistor action Combine the two diodes VF VR IIIII J l M l Pa V 4 No transistor action Consider ve thin base width Transistor action EP BN i CP 4M Hole concentration is zero here reverse biased inf 001E VF VR The collector current 1c is almost equal to IE and collector current is controlled by the EBjunction bias The loss ie a lt 1 corresponds to the recombination of holes in base PNP under forward active mode P N P IEP E 39 E 2 113R 59 4 71 EN lt 71BE lm 113 E B C IEIEPIEN125 1BIBRIBEIBRIEN 45 1c IEP IBR 2 9 holes 9 electrons 1c I Electron ux Hole ux and current Current components 1 hole current lost due to recombination in base 1BR 2 hole current collected by collector N 1c 1 2 hole part of emitter current IEP 5 electrons injected across forward biased EB junction 7 IBE same as electron part of emitter current 7 IEN 4 electron supplied by base contact for recombination with holes lost 7 1BR 1 3 thermally generated e amp h making up reverse saturation current of reverse biased CB junction generally neglected Performance parameters g Consider pnp 1 Neglect the reverse leakage electron current of CB junction Emitter ef ciency 1 I y i 3 Fraction of emltter current carrled by holes IEPIEN IE Wewantycloseto 1 Base transport factor 17C Fraction of holes collected by the collector a T We want aT close to l IEp Common base dc current gain 1c TIEP 0LTij Oldch ado aT 7 Note that DtlS less than 10 but close to 10 eg 0 099 13 Performance parameters 1 Consider pnp 1 Common emitter dc current gain dc 1c dclg But C adCIE adc 1C 1B C d0 1 l adc Bdc ado QTY Note that is large eg 8 100 10de 10LT For npn transistor similar analysis can be carried out However the emitter current is mainly carried by electrons EP etc EP 1 EN Example 7 Chapter 162 MOS electrostatic Quantitative analysis In this class we will Derive analytical expressions for the charge density electric eld and the electrostatic potential Expression for the depletion layer width Describe delta depletion solution Derive gate voltage relationship 7 Gate voltage required to obtain inversion Electrostatic potential 91 x1 De ne a new term x taken to be the potential inside the semiconductor at a given point x The symbol Q instead of V used in MOS work to avoid confusion with externally applied voltage V 00 iEibu1kEix Potential at any point x q 5 i E bulk Eisurface Surface potential 1 l DP l related to doping F Ei bu1k EF concentration DE gt 0 means ptype F lt 0 means ntype Electrostatic parameters ts is positive if the band bends downward 15 2 at the depletioninversion transition point Example 1 Consider the following DP and ts parameters Indicate whether the semiconductor is ptype or ntype specify the biasing condition and draw the energy band diagram at the biasing condition i F 12 kTq tbs 12 kTq F 12 kTq means that ErEF in the semiconductor is 12 kT a positive value So ptype NA ni exp Ei iEF kT S12 kTq means Ei bulk 7Eisurface 12 kT ie the band bends downward near the surface Example 1 continued ii F 9 kTq pg 18 kTq here DP 9 kTq means Eibulk iEF 9 kT ie E is below EF Thus the semiconductor is ntype ts 18 kTq means that E bulk 7Eisurface 718 kT So band bends upwards near the surface The surface is inverted since the surface has the same number of holes as the bulk has electrons Deltadepletion solution M O S Cons1der ptype s111con Accumulation condition VGlt 0 The accumulation charges are mobile holes and appear close to the surface and falloff rapidly as x increases Assume that the free carrier l x concentration at the oxide semiconductor interface is a 5 function Accumulation of holes Charge on metal QM Charge on semiconductor charge on metal iQAccumulationi Delta depletion solution cont 1 Consider ptype Si depletion condition Apply VG such that 15 lt 2 DF Charges in Si are immobile M O S ions results in depletion Si layer similar to that in pn p V gt 0 junction or Schottky diode G T qNAA m QMl QM I Jr l lt Depletion of oles If surface potential is 15 with respect to the bulk then the depletion layer Width E Wwill be Elsi d dx inASsi 2s N W SI S and 55qu X qNA S 2s At the start of 1nvers1on b5 2 F and W WT SI 2 F qN A 7 De letion la er width Wand Efield For a pn junction or a MS nSi junction the depletion layer width is given by 2 Where Vbi is related to the amount of W Vbi band bending Vbi in Volts is numerically l W D equal to the amount of band bending in eV qN 2qN Emax DW DVbi 8s1 8s1 For MOS the same equation applies except that Vbi is replaced by b 2 N 2 N Emaxon Sl1j l sl or J ml 851 851 n39tYPe ptype Delta depletion solution gcont 1 MO 5 Cons1der Sl stron 1nvers1on I pSi Once inversion charges VGgtgt0 T appear they remain close to the surface since they are gt mobile Any additional QM voltage to the gate results in extra QM in gate and get compensated by extra inversion electrons in semiconductor So depletion layer does not have to increase to balance the charge on the metal Electrons appear as 5function near the surface Maximum depletion layer width W WT Inversion electrons 5functionlike Gate voltage relationship Applied gate voltage will be equal to the voltage across the oxide plus the voltage across the semiconductor Consider ptype Si VG Apox AlSemi M O S I pSi VG gt 0 Mm x 0 ltbu1k T N A 0X AlSemi S xox Eox A ox Since the interface does not have any charges up to inversion we can say that 80x fox ssi Esi Eox SSi Sox Si Gate voltage relationship cont 1 qNA 851 2851 qNA qNA 851 S l 851 VG 5 xoxfox 851 5 xox 8 ESi W 51 5 for 0 lt 5 lt 2 F qgtsxox qk for 0S SS2 F 8ox 8Si Gatevolta e relationshi Alternative method Consider ptype silicon VG A ox AlSemi Aoox QMC0x QsC0x where C UK is oxide capacitance and QS is the depletion layer charge in semiconductor Q5 q A NAW Cox SOXA xox AN W 8 39 N A OX q AixoxqAW SoxAxox 8ox SSi s qN s ZqN VG SixOX AW S ixox A s 8ox SSi 8ox SSi same as before 12 Physical constants aB 05292 A Bohr radius aB 05292 X 10 10 m 80 88542 X 10 12 AsVm absolute dielectric constant e 16022 X 10719 C elementary charge c 29979 X 108 ms velocity of light in vacuum ERyd 13606 eV Rydberg energy g 98067 m s2 acceleration on earth at sea level due to gravity G 66873 X 10 11 m3 kg s2 gravitational constant F GM mr2 h 66261 X 10 34 Js Planck constant h 41356 X 10 15 eVs h 10546 X 10 34 Js h h2n h 65821 X 10 16 eVs k 13807 X 10 23 JK Boltzmann constant k 86175 X 10 5 eVK 110 12566 X 10 6 VsAm absolute magnetic constant me 91094 X 10 31 kg free electron mass N Am 60221 X 1023 mol 1 Avogadro number R kNAVO 83145 J K71 mol 1 ideal gas constant Note The dielectric permittivity of a material is given by s Sr 80 where Sr and so are the relative and absolute dielectric constant respectively The magnetic permeability of a material is given by p pr 110 where ur and 110 are the relative and absolute magnetic constant respectively Useful conversions 1 eV 16022 X 10 cv 16022 X 10 J E hv hclt 12398 eVltnm kT 2586 meV at T 300 K kT 2525 meV at T 20 C 29315 K Chapter 172 MOSFET smallsignal equivalent circuit Last class we discussed the dc characteristics of MOSFETs The dc characteristics for NMOS are reviewed below Z V2 D in CoxVG VT VDs ES 0 lt VDS lt VDSsat VG gt VT Z C IDsat ZLOX VG VT 2 VDS gt VDSsat VG gt VT Linear region Saturation region F VGgt V1 VG lt vT MOSFET ac response MOSFET ac response is routinely expressed in terms of small signal equivalent circuits This circuit can be derived from the twoport network shown below G D input MOSFET output S S The input looks like an open circuit except for the presence of the gate capacitor At output we have a current ID which is controlled by VG and VDS ID fUG VDS MOSFET smallSi nal e uivalent circuit Any ac signal in VG or VDS Will result in corresponding ac variation in ID 81 81 ND 8413 AVG 8413 AVDS VG VDS VDs VG BID BID 1 v v and gd d gm g gd d Where gm E st BVDS VG gm transconductance g01 drain or channel conductance Nate id vg and vd are smallsignal ac currents and voltages They are different from ID VG and VDS which are do currents and voltages Smallsignal eguivalent circuit So the equivalent circuit at lowfrequency looks like neglecting the gate capacitance low frequency a lt a I c v DA Km u N u s For highfrequency we have to include the capacitive effects MOSFET smallsignal parameters When VDS lt VDSYsat ie below pinchoff or linear region Z C gd HquotL 0X VG VT VDs Z C gm VDS When VDS gt VDSYSKL ie above pinchoff or saturation region gd0 Z C gm ltVG VT Note The parameters depend on the dc bias VG and VDS Frequency response of MOSFET The cutoff frequency fT is de ned as the frequency When the 1 current gain is Input current ijGvG VG here is ac signal CGS is approximately equal to the Z L C Output current g mVG gate capacitance ox ngG 1 275 fT XCGsVG So at fT CC VC characteristics MOSC versus MOSFET CG vs VG characteristics of a MOSFET at high frequency looks similar to the lowfrequency response unlike the MOSC This is because even at high frequency the source and drain can supply the minority carriers required for the structure to follow the ac uctuations in the gate potential when the device is inversion biased 3 CG vs VG characteristics of a MOSFET with VDS 0 Enhancement mode MOSFETs The devices we discussed so far are called enhancementmode MOSFETs For NMOS VT is positive and one has to apply a positive gate voltage to turn on the device At zero gate voltage the device will be off For PMOS VT is negative and one has to apply a negative gate voltage to turn on the device At zero gate voltage the device will be off Exercise Draw the ID VDS characteristics for NMOS and PMOS enhancementmode devices Next class we will discuss depletionmode devices 8 Semiconductors A general introduction Classi cation of Materials in terms of electrical resistivity Insulators 1010 1018 2 cm Semiconductors 10 4 108 2 cm Conductors 10 6 10 4 2 cm The uniqueness of semiconductors is that their conductivity can be varied by us over a Wide range e g by adding minute quantities of impurities by applying electric field illumination Introduction Conducting lton Current I Non conducting oft Voltage V We can use this as a switch Example Digital computers Abbreviated periodic table of the elements 4 5 6 7 8 Be B C N O 12 13 14 15 16 Mg Al Si P S 30 31 32 33 34 Zn Ga Ge As Se 48 49 50 51 52 Cd In Sn Sb Te 80 81 82 83 84 Hg Tl Pb Bi Po Elements ESi 11 eV EGe 067eV Compounds EGaAS 143eV EGaN 34eV The most common semiconductor is Silicon IA Periodic table of the elements Periodic System of Elements VIII 1 101 2 400 Dlmttrl Mendeleev I869 Hydrogen Helium 3 IIA IIIA IVA VA VIA VIIA 32 3 6194 4 901 5 108 6 2 0 7 140 8 160 9 190 10 202 c e Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon 2339 232 2p39 2172 2173 2 p5 2p6 ll 330 12 243 13 27014 281 15 310 16 321 17 355 18 400 5 d39 AlAl S l39 Pl 1 S If Chl Ar r L 0 1l1111 Lllnllluln 1 Con IDS IDYOLIS u LU 011115 on 33 M39992 IHB IVB VB VIB VIIB K VIIIB x IB IIB 3 r p 3 3 5 g 316 19 391 20 401 21 450 22 479 23 509 24 520 25 549 26 559 27 589 28 587 29 635 30 654 3 697 32 726 33 749 34 790 35 799 36 838 r Fe 0 1 u 1 Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton 4539 4S2 3aquot4s2 31342 3113432 3d54r39 3115432 3d64s2 3d74s3 3dS4SZ 3d3904s39 3d39 4sZ 41239 4p2 4 3 4p4 4p5 4p6 37 855 38 876 39 889 40 912 41 929 42 959 43 98 44 101 45 103 46 106 47 108 48 112 49 115 50 119 5 122 52 128 33 127 54 131 0 Sn e Rubidiuin Strontium Yttrium Zirconium Niobium Molybdenum Techiietium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon 5339 5x3 443953Z 4612532 4d 5539 Aid75 quot 40555Z 4cl75339 4185339 4d390 4d3905s39 4aquotquot53Z Sp39 apz 5p3 5 55 I33 56 I37 57 139 72 I78 73 181 74 184 5 186 76 190 77 192 78 195 79 197 80 201 8 204 82 207 83 209 84 209 85 539 a H Pb l Cesium Barium Lanilianum Hafnium Tantalum Tungsten Rlienium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Poloiiium Astatine Radon 6339 r 5d3632 5213956S2 51 5652 5d 6s39 5d3906539 5d3906s2 6p39 6p2 6 3 6 6 6126 87 223 88 226 89 227 Coinage emem Halcgens Noble 3 metals semiconductors gases Francium Radium Actinium Explanmmn 7539 6d7S2 3 4 11lt1 23045 Ammlc nghl Metals Nonmetals Alkaline Alkaline anh 410mm quotumber meta netals 1 340151010115 Sodium 331 Outer shell elec tron contiguration 58 140 59 14160 144 61 145 62 150 63 152 64 157 65 159 66 163 67 157 68 167 69 169 70 173 71 175 Lanthanides C PI39 Pm m Eggsenalrslh Cerium iaseodymiui chuu yitiiuiii um rliium Samarium Eurupium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ylterbium Lutetium 4f395d39631 M3632 4632 4P6s2 4f5631 M7632 4 754139 632 4fquot6xz 4f39 0652 4f39 1632 12631 419633 4 14632 55139 652 90 232 91 231 92 238 93 237 94 244 95 243 96 247 97 247 98 251 99 252 100 257 10 258 102 259 103 260 Actinides Th Np Cf Es Thorium Protactinium Uranium Neptu ium Plutonium Amei39iciuin Cui ium Bei kelium f 39 39 39 erinium Mendelevium Nobelium Lawrencium 6d17s2 swarm Wech 5 643973392 567Z 5fth s edlm 5f97s2 5fquotquot7s2 sfllnll sfllnZ 5f7s2 5111751 awn2 Note s eectron shell can be occupied by at most 2 electrons p electron shell by at most 6 electrons delectron shell by at most 10 electrons fe1ectron shell by at most 14 electrons Noble gases have 2 He 10 Ne 18 Ar 36 Kr 54 Xe and 86 Rn electrons Semiconductor materials Table 11 Semiconductor Materials General Semiconductor Classi cation Symbol Name 1 Elemental Si Silicon Ge Germanium 2 Compounds a IVIV SiC Silicon carbide b IIIV AlP Aluminum phosphide AlAs Aluminum arsenide Ale Aluminum antimonide GaN Gallium nitride GaP Gallium phosphide 39 GaAs Gallium arsenide GaSb Gallium antimonide InP Indium phosphide InAs Indium arsenide InSb Indium antimonide c IIVI ZnO Zinc oxide ZnS Zinc sul de ZnSe Zinc selenide ZnTe Zinc telluride CdS Cadmium sul de CdSe Cadmium selenide CdTe Cadmium telluride HgS Mercury sul de d IV VI PbS Lead sul de PbSe Lead selenide PbTe Lead telluride 3 Alloys a Binary Si1xGex b Ternary Alx Gal xAs or GaxAles A1x Inl x As or In1 x A1x As Cdl x GaAs1 x P Gax Inl xAs or In1xGaxAs GaXInxP or InlxGaxP H gl x Cdx Te c Quaternary A1x Gal x Asy Sb GaxIn1xAslyP ly Crystalline solids The fact that one can alter the properties of semiconductors over a Wide range may have something to do With the atomic arrangement of atoms in these materials So let us look at the crystal structure a Amorphous b Polycrystalline 39 c Crystalline No recognizable Completely ordered Entire solid is made up Of longrange order in segments atoms in an orderly array Figure 11 Crystalline solids Lattice Periodic arrangement of atoms The atomic arrangement determines the macroproperties of the crystal Examples Amorphous Si thin lm transistors used as switching devices in LCDs Polycrystalline Si used as gate in MOSFETs Actual active region of MOSFET is fabricated in crystalline Si Unit cell concept The unit cell is a small portion of any given crystal that could be used to reproduce a crystal Figure 12 b la 1 4 atom each corner Two different ways of representing a unit cell Simple 3D unit cells m Sunplt cubic m hcc m m Flguve L x all cube m Bunlywnlcmd mm mm mu uh Face ccmcmd hm unucull Crystal structure of Si and Ge and other common semiconductors 2 FCC lattices displaced by 14 a 14 a 14 a along body diagonal 8 atoms per unit cell Diamond lattice also called zincblende if interpenetrating FCC lattices are made of different elements like in GaAs Each atom is bonded to 4 other atoms tetrahedral bonding structure The lattice constant or cubic edge is a Generally a is expressed in Angstroms l A 10 8 cm 10 10 m Diamond and zincblende lattice unit cells b a c Figure 14 11 Diamond lattice detail a 12 Example What is the number of Si atoms in 1 cm3 of Si Given is the lattice constant a 543 A 8at0ms leozz atoms 3 cm3 What is the density of Si Atomic weight of Si 281 ie 1 mole NA 6023 X 1023 atoms of Si has a mass of281 g 5x1022 Longs x 281 g m0 6 233 g 6 02x1023 atoms 0 39 mole 13 Density 2 Chapter 23 States and carrier distributions So far we have concentrated on carrier properties of qualitative nature We also need to Determine the carrier distribution with respect to energy in different bands Determine the quantitative information of carrier concentrations in different bands Two concepts will be introduced to determine this Density of states FermiDirac distribution and FermiLevel Density of states There are 4 states per atom or 4 x 5x1022 cm3 states in each of conduction and valence bands of Si The distribution of these states in the bands are not uniform but follows a distribution function given by the following equations m2mgE EC W mgxzmggEv E gvE T E 5 EV 266 26b Dependence of DOS near band edges EC I 805 gt I l E I l I I E gvE gt Figure 214 More on density of states DOS gcE dE represents the of conduction band statescm3 lying in the energy range between E and E dE gVE dE represents the of valence band statescm3 lying in the energy range between E and E dE More states are available available away from the band edges similar to a seating arrangement in a football eld Units for gcE and gV E per unit volume per unit energy ie cm3 eV Energy bands are drawn with respect to electron energies FermiDirac distribution and the Fermilevel Density of states tells us how many states exist at a given energy E The Fermi function fE speci es how many of the existing states at the energy E will be lled with electrons The function fE speci es under equilibrium conditions the probability that an available state at an energy E will be occupied by an electron It is a probability distribution function 1 fa 1 eE EFkT EF Fermi energy or Fermi level k Boltzmann constant 138 x 10 23 JK 86 X 105 eVK T absolute temperature in K Distribution function for gas molecules Example Gas molecules follow a different distribution function The MaxwellBoltzmann distribution E kT2 E2 E1 2 2 e 2 e kT 721 E1 6 kT Let us look at the FermiDirac distribution more closely FermiDirac distribution Consider T gt 0 K 1 FOIEgtEF 1 1 ForEltEF fEltEF 1eXpOO FermiDirac distribution Consider T gt 0 K IfEEF then fEF 12 1r EZEF3kT then emf25F gtgt 1 E EF Thus the following approximation is valid fE exp ie most states at energies 3kT above EF are empty E EF If E sEF 3kT then exp kT ltlt 1 E E Thus the following approximation is valid f E 1 eXp kT F So 1 fE Probability that a state is empty decays to zero So most states Will be lled kT at 300 K 0025eV EgSi 11eV so 3kT is very small in comparison Temperature dependence of FermiDirac distribution 1 l Hf zeE EFkr L x l 2 2 fEreU EFkr 111111 0 bk I 0 E39 b E EF 3kT EF3kT an OK bTgt0K Figure 215 0 02 015 01 Exercise 23 015 02 Example Assume that the density of states is the same in the conduction band CB and valence band VB Then the probability that a state is lled at the conduction band edge EC is equal to the probability that a state is empty at the valence band edge Where is the Fermi level located 1 l l fltECgt1 fltEVgt W EMF le kT le kT le H E E EC EFzEF EV EF This corresponds to intrinsic material Where the of electrons at EC of holes empty states at EV Note that the probability Within the band gap is finite but there are no states available so electrons cannot be found there 11 Equilibrium distribution of carriers Distribution of carriers DOS gtlt probability of occupancy gE 113 Where DOS Density of states Total number of electrons in CB conduction band no I gcEfEdE Total number of holes in VB valence band p0 I EV gv E 1fEdE E Bottom Fermi level positioning and carrier distributions Energy band diagram Density of states 8305 8 15 quotquot gcE gVE W gcE EF E T Occupancy factors 3 EF above midgap b EF near midgap I C EF below midgap Figure 216 Carrier distributions p gcEfE gvEl fE 13 Visualization of carrier distribution One way to convey the carrier distribution is to draw the following diagram This diagram represents ntype material since there are more electrons than holes O O 0 EC 00 Ev 0 Figure 217 14 Visualization of carrier distribution continued Another more useful way to convey the carrier distribution is to draw the following band diagrams The position of EF with respect to Ei is used to indicate whether is n type p type or intrinsic EC I EC EC EF El quotquotquot quot39 quot39Ei quot quot quotquotquotquotquotquot39Ei EF EV Ev EV Intrinsic ntype ptype Figure 218 15 ChapterSZ F 39 39 39 39 quot We make the analysis in 1 dimension even though actual diode as shown may not be a onedimensional system This makes the analysis simple The metallurgical junction is located at x 0 lt I V r Dmmrmnmns msilimn Arm cmlr Flam 53 uantitative anal sis Electric field E d E B Where 81580 dx 8 iiNA 7xpltxlt0 a 1ND 0ltxltxn 7 E W 7 0 xgtxn xlt7xp E00 qNA xpx exp 3x50 5 maxqNAxpE qND xn 7x 031can 7qNDxnE a 0 xlt7xp xgtxn Relationship between x3 and xR m qNAxps qNDxns NAxpNDxn Net charge on pside Net charge on nside Depletion layer width W xn xp N N xn W A xp 2 W D NAND NAND IfNA gtgt ND then W x xn and ifNA ltlt ND then W x xp Builtin voltage Vm E C or Vbi If xdx Vbi area under E versus x curve 12WqNDxn8l qZSHNDxn W g W2 since xn W NANND 28 N N W A q NAND Quantitative analysis Electrostatic potential dV qNA x x SxSO dx 8 p qD xn x OSxan with the reference potential at x xp set to zero V 9N A 2 V x x x S x S 0 28 p xp Vbi Vbi qNDxn x2 OSXan x 28 5 Step junction with VA 0 VA chopped here VA Applied Voltage H I nu Ara I 1 l csuglulc a r J Negligible voltage dmp t Negligible voltage drop ohmic contact ohmic Contact Flgure 51a The equation for W is similar to the earlier equation except that Vbi is replaced by Vbi VA VA is restricted to VAlt Vbi W E JVAJrNDVin VA q NAND Effects of felward and reverse bias 1 m Figure 51 PN junction energyband diagrams Le Ravens max VA lt u Fiuure 512 Example 1 Consider the following diode Calculate the maximum electric eld Em the location of Em the depletion layer width W Xu and xp and the builtin voltage Vbi Carefully plot the charge density electric eld and the potential as a function of x ND 2gtlt1015cm 3 NA 3gtlt1017cm 3 NAlgtlt1015cm 3 ND2gtlt1017cm 3 Also calculate the depletion layer width and the maximum electric eld if a reverse voltage of 10 V is applied across the diode What will be Wif VA 05 V Example 2 Consider the diode of Example 1 Calculate the depletion layer width and the maximum electric eld if a reverse voltage of 10 V is applied across the diode Calculate the depletion layer width in the nside and pside under this biased condition What will be Wif VA 05 V What happens if we apply VA gt Vbi Chapter 22 Carrier properties Mass like charge is a very basic property of electrons and holes The mass of electrons in a semiconductor may be different than its mass in vacuum Effective mass concept dv F m F m q 0dr q ndt Effective mass Electrons moving inside a semiconductor crystal will collide with semiconductor atoms there by causing periodic deceleration of the carriers In addition to applied electric eld the electrons also experience complex eld forces inside the crystals The effective mass can have different values along different directions The effective mass will be different depending on the property we are observing So you can have conductivity effective mass density of states effective mass etc Table 21 Density of States Effective Masses at 300 K Materlal m imo my lmo Si 18 0 81 Ge 055 036 GaAs 0066 052 Carrier numbers in intrinsic materials Intrinsic semiconductor or pure semiconductor has equal numbers of electrons and holes at a particular temperature Number of electronscm3 n number of holescm3 p Why is n p This is an intrinsic propertV of the 39 J and is called intrinsic carrier concentration ni At T 300 K ni 2 gtlt105cm3 in GaAs lgtlt1010cm3 in Si 2 gtlt1013cm3 in Ge How large is this compared to the number of Si atomscm3 What happens to ni at higher temperature At 0 K Extrinsic semiconductors Table 22 F mm n W1 n Dnnan39 Arr mu U my I opams Donorx Electronincreasing dopams Acceptorr Holeincreasing dapants 1 u 5 l Column V Ga Column 111 As elements In elements Sb Al Elements in column V of the periodic table have 5 electrons in their outer shell one more than Si This can be easily released thus increasing the net free electrons in the Si crystal Elements of column 111 of the periodic table have only 3 electrons in their outer shell one less than Si To complete the bond the atom can accept an electron by breaking a bond somewhere else thus creating a broken bond or a hole Twodimensional representation of Si lattice Visualization of a donors and b acceptors Phosphorus P atom Boron B atom El ill l a b Figure 210 Pseudohvdrogen atom model for donors Instead of mo we have to use 1715 Instead of 80 we have to use Ks 80 K5 is the relative dielectric constant of Si KS Si 118 Figure211 4 EH 2 2 136 6V see page 24 oftext 247t80h m 4 m 8 2 Ed n q2 136eV n 0 z 005eV 247rKs80h m0 K880 This is an approximate value More accurate values are given next 8 Binding energies for dopants 1 23 5151 E Donors IEB I Acceptors IEB I Sb 0039 CV B 0045 CV P 0045 eV Al 0067 eV WWW In 016 eV Questions How much energy is required to break a Si Si bond How much energy is required to break the 5th electron from As in Si How much energy is required to break a Si Si bond When that bond is adjacent to a B atom Does the freeing of an electron from a donor atom create an extra hole Energy band model for donors A l 39 y a I 0045 CV Ly 39 39 L Ec P T r 1 l ED 1 Si4 112eV I Doncrsite L Ev I I x Figure 212 10 Temperature effects on donors and acceptors o o o o o o o o o o E k 00003900 oo oot o E T gt 0K Increasing T Room temperature a EC s o 0090 EA f E 0 o oo o o o o o 0 V T 0K Increasing39l39 Room temperature 0 Figure213 Some items to remember Ionization energy of electrons in hydrogen Bond and band models of semiconductors Intrinsic semiconductors and intrinsic carrier concentrations Extrinsic semiconductors Dopants Donors and acceptors ntype material ptype material Majority carriers Minority carriers Band models of donors and acceptors Band gap energies of Si Ge and GaAs Chapter 62 Carrier iniection under forward bias Last class we established the excess minority carrier concentration pro le under biased conditions The excess minority carrier concentration at the edge of the depletion layer will increase under forward biased condition The minimum carrier concentration decreases exponentially with distance from the depletion layer edge xi x I I L Anpltx Anplt0e Ln Apno Amos P Carrier iniection under forward bias At equilibrium of holes diffusing t0 nside equals of holes drifting m nside When we apply external forward voltage VA holes diffusing injection to nside from p side increases exponentially This increases the hole concentration at the edge of the depletion layer on nside amp pnxn pnO e kT amp Apnxn pnxnpn0 pnO 6 kT 1 Similarly amp Anp xp npo e H l Minority carrier concentration pro le under bias W02 Ame 71p I lpo AHPQCquot a pm pno Apnocr pnO p0 x x Anpocquot Anpltogt e Ln Apno39 Apn0e Lp Carrier iniection under forward bias continued Change of axes to x39 and xquot see graph xquot axis amp Anp0 npo e H 1 H X n L Anpx Anp0e 1 H IVA x npo e H 1e Ln x39 axis amp Apn0 pnO 6 kT 1 I x L Apnx Apn0e p amp L pn0e kT 16 p I x General current and minority carrier diffusion equations Simpli ed equations 1 Jpx qu p d Jpx qupEqu p dx 3 dn Jnx qnilnqi anE 2 2 6A a A A pDp2ppGL OZDPHamp 6t ax 1p 5x2 1p aAn 82ml An D G at n 8x2 In L Current eguations applied to a diode Snppcomact Anp xp Amosquot Apnncunmct dA Jpoc39 qD p dAn Jnocquot qD p dx n dxquot Find JH and Jp at the edge of the depletion layer and add them to get the total current Assumption No generation or recombination inside the depletion layer Current equations applied to a diode Jp Jpn D d A O x L l Qp pn0pne p dx D qL p Apn0exLP P Therefore D D amp Jpltx39 0 Apn0 pno e W 1 P Diode current equations D I1 Current due to electrons will be along positive x39 direction And total current equals Shockley equation Forward and reverse bias characteristics Large forward bias VA gtgt kTq J g 96 W IVA Large reverse bias VA ltlt 7 kTq J JO 6 1 Example 1 log scale Figure 63 is a dimensioned plot of the steady state carrier concentration inside a pn junction diode maintained at room temperature 21 x 102 6gtlt10 2cm Figure 53 l a Is the diode forward or reverse biased Explain b Do lowlevel injection conditions prevail in the quasineutral region of the diode Explain 0 Determine the applied voltage VA d Determine the hole diffusion length Lp ECSE 2210 Microelectronics Technology Class Activity 21 Solution 1 The gure below shows the minority carrier concentrations in the emitter base and the collector region of an npn transistor Answer the following questions Most questions do not need any extensive calculations N P N 13 3 10 cm AreaA 5 X 10392 cm2 Pco pEO nBo Emitter Base Collector pm 10 cm393 7130 5 X 103 cm393 pco 105 cm393 LE05 um WB08 um Lc3um DE 5 cmzs DB 25 cmzs Dc 10 cmzs TB 10397 s a Is the baseemitter junction forwardbiased 0r reversebiased What is the voltage applied to the EB junction Forward biased since the minority carrier concentration is higher than the equilibrium concentration at the depletion layer edges From the gure it is seen that the minority carrier concentration in the base 71130 1013 cm393 is higher than the equilibrium concentration nBo 5 X 10 cm39 quotB30 quot120 X W q V1334 10 cm 5 X 10 cm X exp VBE00259 V VBE 00259 x In 1013 5 x 103 VBE 0555 V Fquot Is the collectorbase junction forwardbiased 0r reversebiased Can we calculate the voltage applied to the CB junction with the available data Reverse biased as the concentration at the edge of the depletion region is much less than the equilibrium concentration Calculating the voltage applied to the CB junction is not possible since we don t know the exact value of the carrier concentration at the CB junction What is the value of the collector current c qA DB 7130 7 0 W3 0 3 1 D quot1 W Pquot F l6gtlt10 19 gtlt25gtlt1013 0810 4 1c 25 mA I C What is the value of the base current due to recombination in base Call it IBR 1BR QBTB BR 7130 TB 16gtlt10 19 x5 gtlt10392 gtlt1013 gtlt08gtlt10 4 2 x10 BR 1BR What is the value of the base current due to the injection of holes into the emitter Call it 11315 13 DEpEo X exp LE 16gtlt10 19 x5 gtlt10 2 X5gtlt10gtlteXp055500259 BE 4 05 X10 13 16 What is the value of the total base current B IBR 13 32 16 48 MA What is the value of dc for this transistor dc IcIB 25 mN48 uA 520 What is the value of the electron component of the emitter current JEN 1c IBR 25mA32uA 25032 mA What is the value of the base transport factor wT Ic JEN 25 25032 09987 What is the value of the hole component of the emitter current This is the same as IBE What is the value of the emitter injection efficiency 9 IEN JEN 1151 25032 mA 25032 mA 16 uA 09993 Suppose the lifetime of minority carriers in the base is increased to 2 X 10397 s What will be the value of dc now IBRwill be reduced by a factor of 2 IBR 32 uA 2 16 MA New IBE is the same as before and equal to l6uA Hence new 13 16 16uA 32 uA dc 25113 25 mA 1616 uA 780 Chapter 21 Semiconductor models The subatomic particles responsible for charge transport in metallic Wires electrons The subatomic particles responsible for charge transport in semiconductors electrons amp holes In this chapter we will study these topics The quantization concept Semiconductor models Carrier properties State and carrier distributions Equilibrium carrier concentrations Quantization concept In 1901 Max Planck showed that the energy distribution of the black body radiation can only be explained by assuming that this radiation ie electromagnetic waves is emitted and absorbed as discrete energy quanta photons The energy of each photon is related to the wavelength of the radiation Ehvhct where h Planck s constant h 663 x 10 34 Js v frequency Hz s l c speed oflight 3 x 108 ms 7 wavelength m Example Our eye is very sensitive to green light The corresponding wavelength is 0555 um or 5550 A or 555 nm What is the energy of each photon 662 gtlt1034 Js gtlt 3x108 m Ehv 6 357gtlt10 19J 0555 gtltlO m These energies are very small and hence are usually measured using a new energy unit called electron Volts 1 eV 16 x 10 19 CV 16 x 10 19 J A new unit of energy Since the energies related to atoms and photons are very small EGREEN LIGHT 357 x 10 19 J we have de ned a new unit of energy called electron Volt or eV One eV is the energy acquired by an electron when accelerated by a 10 V potential difference leVl6 xlO 19J Energy acquired by the electron is qV Since q is 16 x 10 19 C the energy is 16 x 10 19 J De ne this as 1 eV Therefore EGREEN LIGHT Z 223CV leV 1 16x1019 CV161gtlt1390 19 J Quantization concept continued Niels Bohr in 1913 hypothesized that electrons in hydrogen was restricted to certain discrete levels This comes about because the electron waves can have only certain wavelengths i e n 27w where r is the orbit radius Quantization Based on this one can show that 4 4 EH 2 L2 2 for n123 2 4780 h n 8 80 n h h where h 2 and h Planck s constant 7 Bohr s hydrogen atom model LA o n1 n3 l36 6V EH 151 eV n 2 34 eV Figure 21 A numerical example 4 E mO q4 9llgtltlO31kggtltl6gtltlO19 C n822h2 12 2 34 2 8on 8x885gtlt10 Fm gtltlgtlt662gtltlO Js 217gtlt10 19J 135 eV for the nl orbit For the n 2 orbit E2 34 eV and so on The number n is called the principal quantum number which determines the orbit of the electron Since Hydrogen atom is 3D type we have other quantum numbers like 1 and m Within each orbit So in atoms each orbit is called a shell See Appendix A in text for the arrangement of electrons in each shell and also for various elements in the periodic table Atomic con guration of Si So an important idea we got from Bohr model is that the energy of electrons in atomic systems is restricted to a limited set of values The energy level scheme in multielectron atom like Si is more complex but intuitively similar Ten of the 14 Siatom electrons occupy very deep lying energy levels and are tightly bound to the nucleus The remaining 4 electrons called valence electrons are not very strongly bound and occupy 4 of the 8 allowed slots Con guration for Ge is identical to that of Si except that the core has 28 electrons n E 3 EllL39ElJ39UTIS El allowed levels at same encrgy Tum allowed levels al 53m energy n L quotJ Electrons Figure 22 Schcnmlic rcpl39cwnlzllimn nl39 m isnlzllud Si allnm Bond model Consider a semiconductor Ge Si or C Ge Si and C have four nearest neighbors each has 4 electrons in outer shell Each atom shares its electrons with its nearest neighbor This is called a covalent bonding No electrons are available for conduction in this covalent structure so the material is and should be an insulator at 0 K 2dimensional g2D semiconductor bonding model Line represents a shared valence electron Circle represents the core of a semiconductor e g Si atom Figure 23 No electrons are available for conduction Therefore Si is an insulator at T O K Simpli ed 2D representation of Si lattice How many atomneighbors has each Si atom in a Si lattice How many electrons are in the outer shell of an isolated Si atom How many electrons are in the outer shell of a Si atom with 4 neighbors a Point defect b Electron generation m a b Figure 24 At higher temperatures e g 300 K some bonds get broken releasing electrons for conduction A broken bond is a de cient electron of a hole At the same time the broken bond can move about the crystal by accepting electrons from other bonds thereby creating a hole 13 Energy band model An isolated atom has its own electronic structure with n 1 2 3 shells When atoms come together their shells overlap Consider Silicon Si has 4 electrons in its outermost shell but there are 8 possible states When atoms come together to form a crystal these shells overlap and form hands We do not consider the inner shell electrons since they are too tightly coupled to the inner core atom and do not participate in anything Development of the energyband model E N Isolated tiff Siatoms E gy El0 gt 4N allowed states Mostly p 931 E5 conduction band empty E a o r 5 E n 3 g No states gt b 2 13 8 4N allowed states Lu 6N pstates total E3 valence band 2N sstates total 4N electrons total ngstamne Si Mm l a E p 4N empty g states t E S FL I decreasing i I 7 atom spacing 39 Isolated Sl 1att1ce Si atoms spacing Fi gu re 2 5 15 Energy band model AtTOK No conduction can take place since there are no carriers in the conduction band Valence band does not contribute to currents since it is full Actually valence electrons do move about the crystal No empty energy state available For every electron going in one direction there is another one going in the opposite direction Therefore Net current ow in lled band 0 Both bond model and band model shows us that semiconductors behave like insulators at 0K Visualization of carriers using energy bands Empty EV Completely lled a No carriers 0 l C EC EV b The electron lt O c The hole Figure 27 17 Insulators semiconductors and metals electrons EC E EG 142 eV GaAs Wide EG8eVSiOZ E0 ll2eSi EC 5 EV Diamond E EC 066 ev Ge Thermal quot Room temperature E excitation quot moderately easy a Insulator b Semiconductor very i 0 0 E narrow E V T quot or Ec 4 c Metal Figure 28 Chapter 171 MOSFET MOSFET based ICs have become dominant technology in the semiconductor industry We will study the following in this chapter Qualitative theory of operation Quantitative IDversusVDS characteristics Smallsignal equivalent circuits N channel MOSFET Substrate ptype Si Qualitative discussion NMOS 0 lt VGlt VT VDS small or large no channel no current VGgt VT VDS 0 ID increases with VDS VGIgt VT VDS small gt 0 ID 1ncreases wrth VDS but rate of 1ncrease decreases VG gt VT VDS pinchoff ID reaches a saturation value IDat The VDS value is called VDSsat VGgt VT VDSgt VDSsat ID does not increase further saturation region iS characteristics for NMOS derived from Qualitative discussions I Linear Saturation D VDsat Figure I73 Dim characteristics expected from a long channel 1AL ltlt L MOSFET gn channel for various values of VG In VG increasing ngre um Threshold voltage for NMOS and PMOS When VG VT 15 2 DP using equation 1628 we get expression or VT ESi 2 4 NA Ideal nchannel VT 2 l F xox 8 2 F psilicon device X 3 both terms positive 7 7 ssi 2 4ND Ideal pchannel VT 7 M1 XOX Esi l2 Fl n silicon device both terms negative 83 Box erys eo eryoxeo 119 39 3 55 119 relative dielectric constant of Si an absolute dielectnc constant 885 x 10quot2 A sVm 5 Quantitative IQVDS relationships G VG Let 1 be the potential along the channel For VG lt VT Inversion layer charge is zero For VG gt VT Qny QG Cox VG l VT In general J q uh n E When the diffusion current is neglected Here current ID is the same everywhere but Jn current density can vary from position to position Device structure dimension and coordinate orientations assumed in the quantitative analysis Figure 176 7 Quantitative 121m relationships 1 Shockley model d d Jn Jny qunnf qunnd s1nce Ey d j To nd current we have to multiply the above with area but Jn 71 etc are functions of x and 2 Hence d ID HJnydxdz ZjJnydx Zund jqndx do Zund Qny Qny chargeunlt area y Integrating the above equation and noting that ID is constant we get Z VDs Since we know expression for Qny in ID ZMII J0 Q11y dq terms of I we can integrate this to getD 8 Quantitative DiDS Relationships cont 1 2 Zu V ID H C0X VG VTVDS 0 lt VDS lt V1335 VG gt VT L ID will increase as VDS is increased but when VGi VDS VT pinch off of channel occurs and current saturates when VDS is increased further This value of VDS is called VDSvsm ie VDSYSm VGi VT and the current when VDS mm is called IDSvsm u Z C Dsat TOX VG VT 2 VD gt VDSsat VG gtVT Here Cox is the oxide capacitance per unit area C0x sox x0x Example 1 Plot the ID vs VDS characteristics for an NMOS with the following parameters Substrate doping 1016 cm 3 Oxide thickness 100 nm Gate width 15 um Gate length l um Assume u 500 cmzVs Find COX C0x saxxox 333 nFcm2 VT 2 XOXE 2 F 215v OX 1 Z C Dsat ZLOX VG VT 2 VDS gtVDSsat VG gtVT VDSsat VG VT Find IDsalt for different values of VG and plot the graph Chapter 51 PNiunction electrostatics In this chapter you will learn about pn junction electrostatics Charge density electric eld and electrostatic potential existing inside the diode under equilibrium and steady state conditions You will also learn about Poisson s Equation BuiltIn Potential Depletion Approximation StepJunction Solution PN junction fabrication NA NA diffused 0139 ND ND original 7x I Metallurgical junction ND NA 0 a NDNA k l Figure 51 b PN junctions are created by several processes including 1 Diffusion 2 Ionimplantation 3 Epitaxial deposition Each process results in different doping pro les Ideal stepjunction doping pro le V Actual pro le Step junction idealization a Figure 52 Equilibrium energV band diagram for the p11 iunetion ND NA ND E1 Ei x n 2 mi eXp NA kT Ec r E1 EF 39 1 p ni eXp Ev kT pSlde n 51de a quot EF same everywhere Ei Ev under equlhbrlum b x E Join the two sides of the Ei Ev band by a smooth curve c Figure 53 Electrostatic variables for the equilibrium pn junction d Home 54 Potential V lq EC Eref So potential difference between the two sides also called builtin voltage Vbi is equal to lqAEC l V ECEref l dEC l dE1 E 61 dx 61 dx 193 p p charge density a 2 Conceptual pniunction formation P N 9 9 9 ea em em em Bo El 1369 I369 9E1 em ea ea p and n I39CglOHS B E B E GE GE GE GE E E E E GE GE GE GE 0 before Junct1on format1on E6 B E E GE GE GE GE E E9 39 B GE GE GE GE Ew Ew Ew Ew UE UE am am He B 56 E6 em GE GE GE 69 0 Holes and electrons Will diffuse 9 towards opposite directions a SS SE35 E g 85 S 3 uncovering ionized dopant atoms He me Be a 3 AW Be Be 9 9 E 393 9B 9E 9E Th1s w1ll bu11d up an electr1c eld which will prevent further p movement of carriers 7 K x j g d ngre55 The builtin potential Vm When the junction is formed electrons from the nside and holes from the pside will diffuse leaving behind charged dopant atoms Remember that the dopant atoms cannot move Electrons will leave behind positively charged donor atoms and holes will leave behind negatively charged acceptor atoms The net result is the build up of an electric eld from the positively charged atoms to the negatively charged atoms ie from the n side to pside When steady state condition is reached after the formation of junction how long this takes the net electric field or the built in potential will prevent further diffusion of electrons and holes In other words there will be drift and diffusion currents such that net electron and hole currents will be zero Equilibrium conditions Under equilibrium conditions the net electron current and hole current will be zero E eld NA1017cm 3 ND1016cm 3 hole diffusion current net 0 hole drift current lt electron diffusion current gt oppos1te to electron ux net 2 0 electron drift current lt opposite to electron ux The builtin potential Vm EC 0 E V E1 E1 kT1n EC quot1 q Vbi Ei EFpside EF Einside Ei E V EF Ei 14115 1 The builtin potential Vm The builtin potential Vbi measured in Volts is numerically equal to the shift in the bands expressed in eV Vbi 1 Q Ei EFpside Er Einside k T1n k Tln 6 n1 61 i n kT 1n p P211 Q Hi Where pp 2 hole concentration on p side and nn 2 electron concentration on n side 39 p n g V An interesting fact p n exp bl p11 np Mai oritV and minority carrier concentrations p side N A ND nside Pp nl l Pn quotp gt x x x Builtin potential as a function of doping concentration for an abrupt pn or np junction 4 095 A 09 a 5 gt x 085 1 08 Si 300K pln anc np diodes 075 3 i i 1014 1015 1016 1017 NA or ND cm 3 Figure E51 12 Depletion approximation P 4WD NA E d p P01sson equatlon dX K580 i l in x quot 0 1 ND NA for xpran a K880 p 0 everywhere else l We assume that the free 39xE 0 x gt x carrier concentration inside i Exact the depletion region is zero Depl Approx b Figure 56 13 Example 1 A pn junction is formed in Si with the following parameters Calculate the builtin voltage Vbi ND 1016 cm 3 NA1017cm 3 Calculate majority carrier concentration in nside and p side Assume n11 ND 1016 cm 3 andppNA 1017 cm 3 I l pp n z 2 ij q q 721 Plug in the numerical values to calculate Vbi 14 Example 2 A pn junction is formed in Si with the following parameters Calculate the builtin voltage Vbi ND 2 gtlt1016 cm393 NA 3 gtlt1017 cm393 NA1016cm393 ND 2 gtlt1017 cm393 Calculate majority carrier concentration in nside and p side nn effective ND 1016 cm393 pp effective N A 1017 cm 3 kT p n kT N N Here N A and ND Vbi 1n In A are effective or q ni q ni net values Plug in the numerical values to calculate Vbi Chapter 16 l MOS fundamentals Metaloxidesemiconductor FET is the most important device in modern microelectronics In this chapter we Will study Ideal MOS structure electrostatics MOS band diagram under applied bias Gate voltage relationship capacitancevoltage relationship under low frequency and under high frequency MOSFET ID N channel MOSFET pSi NMOS B uses ptype substrate MOSFET operation Pinchoff VG3 VGZ VGl gt VD VG3 gt VGZ gt VGl When a positive voltage VG is applied to the gate relative to the substrate mobile negative charges electrons gets attracted to Si oxide interface These induced electrons form the channel For a given value of VG the current ID increases with VD and finally saturates Ideal MOS capacitor Let us consider a simple MOS capacitor and call it ideal Oxide has zero charge and no current can pass through it No charge centers are present in the oxide or at the oxide sem1conductor mterface Semiconductor is uniformly doped X EC EFFB Insulator Metal Semiconductor 4 Eauilibrium energy band diagram for an ideal MOS structure Figure 54 Effect of an applied bias Let us ground the semiconductor and start applying different voltages VG to the gate VG can be positive negative or zero with respect to the semiconductor EF metal iEF semiconductor 7 q VG Since electron energy q V when Vlt 0 electron energy increases Slnce ox1de has no charge d foxide eld inside the oxide is constant dx ps 0 ie the E Consider pthe Si appr Vpilt0 Negative voltage attracts holes to the Sioxide interface This is called accumulation condition Ei EF should increases near the surface of Si a E 1 85 3lde O gt foxide const 2 g ax1 The oxide energy band has constant slope as shown No current ows in Si EF in Si is constant Accumulation condition VGlt O pthe Si m Ml o l E charge density small Accumulation of holes near silicon surface and electrons Sheet of near the metal surface electrons Similar to a parallel plate capacitor structure Consider pthe Si appr Vpigt O Depletion condition p 2 positive Finite depletion layer Width p0 p 2 negative Consider pSi apply VG gtgt O glnversion condition Immobile acceptors Mobile electrons Inversion condition If we continue to increase the positive gate voltage the bands at the semiconductor bends more strongly At sufficiently high voltage Ei can be below EF indicating large concentration of electrons in the conduction band We say the material near the surface is inverted The 1nverted layer is not gotten by doping but by applying Efield Where did we get the electrons from When Eisurface 7Eibulk 2 EF 7Eibulk the condition is start of inversion and the voltage VG applied to gate is called VT threshold voltage For VGgt VT the Si surface is inverted Energy band diagrams and charge density diagrams describing MOS capacitor in ntype s1 Flgure 155 Ener band dia rams and char e densi dia rams describin MOS ca acitor in t e Si Energy I Dan W J Arrl39 a dc V n V n V V n I I u I u u I values I l I I Exposed Q Q i 9E acceptors Charge I i diagram Q I 7Q i l I Electrons Name Flathand Accumulation Depletion Inversion Flgure 166 Example 1 Construct line plots that visually identify the voltage ranges corresponding to accumulation depletion and inversion in ideal n type Si ie pchannel and ptype Si ie nchannel MOS devices Answer INV InFPI A I I E Vr ntVDe H V ACC iDEPLj le 1 I I VG l Hm5 I 0 V1 Fl ureEI J g 14 Chapter 7 Smallsignal admittance We will study the small signal response of the pn junction diode A small ac signal Va is superimposed on the DC bias This results in ac current 139 Then admittance Y is given by YivaGjxC Speci cally the following parameters will be studied Reverse bias junction or depletion layer capacitance Forward bias diffusion or charge storage capacitance Forward and reverse bias conductance Ca acitance measurements li gt YijG Model for a diode under ac Y admittance L G ij Vac i and Val depend on the applied DC bias Reverse bias 39unction ca acitance A pn junction under reverse bias behaves like a capacitor Such capacitors are used in le as voltagecontrolled capacitors Depletion layer Width under small ac superimposed on DC bias voltage Looks similar to a parallel plate capacitor Where W is the depletionlayer Width under DC bias Reverse bias junction capacitance 2 N N 12 Esi A D W 7 VeV q NAND In A Foranunction 1 2 28si Vb 7 V For ptn or pn r junction Where NB is qNB 39 A the doping on the lightly doped side 12 8siA 8siIINB C A J W 2 Vbi iVA For asymmetrically doped Junction CJ increases With NB Z CJ decreases With applied reverse bias Parameter extractionprofiling C V data from a pn junction is routinely used to determine the doping pro le on the lightly doped side of the junction 12 851 A 851 qNB Slope 2 C A N A J W 2Vbi VA q 213831 1 2 Vbi VA 0 AZqNBsSi 1 1Cj2 F Z If the doping on the lightly I I I doped side is uniform a plot 710 75 0 0 of lCJ2 versus VA should be a VA Volts l stra1ght 11ne w1th a slope Intercept Vbi inversely proportional to NB and an extrapolated lCJ2 0 intercept equal to Vbi Forward bias diffusion capacitance C 2 CD is also called the charge storage capacitance The variation of the injected minoritycarrier charge which is a function of the applied bias results in the diffusion capacitance Both C J and CD are always present but for the forwardbias case CD becomes dominant W U Origin of diffusion 6 9 pnO 1 p0 x n For a pn junction I Qpip where Qp is total excess charge in nside Dpr qVA Qp Tp qA L pnO exp CT 1 z qALp pnOe p dQ V C piAL eX qA il t D dV qu Ppno pl kT H P 6 Forward bias conductance dV L W E Assumes mp ltlt1 Complicated at higher frequencies VApplied VA Equivalent circuit for a diode Example Problem Consider a pn junction forward biased such that the forward current is 1 mA Assume the lifetime of holes is 10 7 s Calculate the diffusion capacitance and the diffusion resistance Solution CD 386 nF rd lGD 259 9 The current through the depletion layer will mostly be carried by holes electrons choose one Plot the current carried by the holes and electrons through the n type region assuming that the diffusion length of holes is l um Chapter 111 Detailed Quantitative Analysis The goal is to relate transistor performance parameters 7 ar dc etc to doping lifetimes baseWidths etc Assumptions pnp transistor steady state lowlevel injection Only drift and diffusion no external generations One dimensional etc General approach is to solve minority carrier diffusion equations for each of the three regions 2 2 aApszaAp GL and 6An 6A An 81 8x2 Tp ax Tn General Quantitative Analysis Under steady state and when GL O 2 82A A 6 An An pp0 and DH 0 8x2 Tp 8x2 Tn Dp For the base in pnp we are interested only in holes 0 D 6x2 Tp P The rigorous analysis is carried out in chapter 11 but we are going to take a more simpli ed approach Review Operational Parameters IE IEP gt IC a 1 BR k 4 31 EN lt BE IBR lIB Injection Ef ciency 7 IEP IEP IEN Base transport factor ar C IEP Collector to emitter current gain aDC ar 7 Collector to base current gain DC aDC 1 aDC These parameters can be related to device parameters such as doping lifetimes diffusion lengths etc Review of PN Junction Under Forward Bias Ame Area Q11 5 Area Qp pBO Review of PN Junction Under Forward Bias cont I11 q A DE dAnde q A DELE AnEO 1p q A DB dApde q A DBLB Ame Total current I IP IN because XE and xB point in opposite directions 2 q A DBLB ApB0 q A DELE AnE 0 q A DBLBI9B0CXP q VEB H 1 q A DELE nE0eXp q VEB kT 1 s q A DBLB pBO exp q VEBkT q A DELE E0 exp q VEBkT Note I Ip and III can also be calculated based on the fact that QP has to be replaced every TB seconds Ip QPTB and III QnTE and IE IP IN Simpli ed Analysis Consider the carrier distribution in a forward active pnp transistor Emitter E0 E0 Base pBO Collector Simpli ed Analysis cont nEO pBO and mm equilibrium concentration of minority carr1ers 1n em1tter base and collector nEO 19130 and nCO minority carrier concentration under forward active conditions at the edge of the respective depletion layers AnE O ApBO and Anc0 Excess carrier concentration at the edge of the depletion layers Simpli ed Analysis cont A71E 0 E 0 E0 2 quotE0 exp q VEB k7 1 A19B 0 pB 0 pB0 pBO exp q VEB k7 1 By taking the slopes of these minority carrier distribution at the depletion layer edges and multiplying it by qAD we can get hole and electron currents Note that I11 q A D11 dndx and p q A Dp dp dx Calculation of Currents Collector current IC 10 q A DB de slope must be taken at end of base ZQADB pBOOWB 2 QA DB pBO WB C q A DBWE pBO exp q VEB k7 quotquot A only hole current if we neglect the small reverse saturation current of reverse biased CB junction Calculation of Currents cont Emitter Current IE IE is made up of two components namely IEP and EN IEP 13 current lost in base due to recombination 13 excess charge stored in base TB C q A WB Ame 2 239B W q A D BWB pBO exp qVEB k7 q A WE2 z39B pBO exp q VEB kn quot39 B Assuming exp q VEB k7 1 z exp q VEB k7 when VEB is positive ie forward biased Calculation of Currents cont Emitter Current c0nt EN corresponds to electron current injection from base to emitter since EB junction is forward biased EN 2 QA DELE E0 exp q VEBk 1 z qA D E LE E0 exp q VEB 7 quotquot39 C Calculation of Currents cont Base Current IB supplies electrons for recombination in base supplies electrons for injection to emitter 1B quB0 WB 27B exp WEB 73 qADELE E0 exp qVEB k7 recombination electron injection to emitter Now we can nd transistor parameter easily Calculation of Currents cont Base transport factor on T aT C IEP qADB qVEB WB PM P H 1 A V AW V 2 qW DB pBO 4 q2 B pBO equ 1W B B TB kT 2L123 same as eq 1142 in text Emitter injection ef ciency y 7 IEPIEPIEN 211IENIEP 11CB 1 DE E0LE DBPBO WB Calculation of Currents cont 1 1 7 Hm HM Eq1141intextb00k DBLEpBO DBLEN E nEO 2 nizNE doping in emitter pB0 2 nizNB doping in base DC Z aDC 1 aDC Chapter 33 Relating diffusion coefficients and mobilities In this class we will learn about Constancy of Fermilevel Indicates equilibrium conditions Current ow under thermal equilibrium Net hole current diffusion drift should be zero Net electron current diffusion drift should be zero Einstein relationship relates diffusion coefficient to mobility Introduction to generationrecombination Constancv of Fermi level Under equilibrium conditions dEF dx O the Fermi level inside a material or a group of materials in intimate contact is invariant as a function of position EF appears as a horizontal line on equilibrium energy band diagram If the Fermi level is not constant With position charge transfer will take place resulting in a net current ow in contrast to the assumption of equilibrium conditions Constancy of Fermi level b Figure 314 Doping concentration varies With position This results in a gradient in carrier concentration EC EF represents the change in carrier concentration With position The graph as drawn indicates that there Will be an electric eld This eld causes drift current The concentration gradient gives diffusion current These two currents exactly cancel each other so that the net current is zero A nonhorizontal Fermi level means there Will be a continuous movement of carriers from one side to the other indicating current ow against the assumption Current ow under equilibrium conditions The total current under equilibrium conditions is equal to zero Total electron current JH and total hole current Jp must also be zero Why J diff Jnldrift and Jpldiff Jpldrift 11 Under equilibrium conditions both drift and diffusion components will vanish only if E O and dn dx dp dx 0 Even under thermal equilibrium conditions nonuniform doping will give rise to carrier concentration gradient a builtin E fz ela and nonzero current components Einstein relationship Consider a nonuniformly doped semiconductor under equilibrium I l E eld J Id Net current O Jnldiff gtx dn JndriftJndiff Wunf ana 0 and n qu nieEF EikT Under equilibrium EF const and lJnld lJndiff Einstein relationship Manipulation of the above equations leads to Einstein relationship for electrons Einstein relationship for holes If un 1350 cmzVs then Then D11 kTqu11 00256 V X 1350 cmzVs 35 cmZs Always be careful about units Example 31 Exercise 32 Plot electrostatic potential V and E eld versus x for the case shown below Sq P gt E electrons E holes lt l Example 32 Consider the diagram shown in example 31 How can you say that the semiconductor is in equilibrium What is the electron current density Jn and hole current density JP at x i L2 Roughly sketch n and p versus x inside the sample Is there an electron diffusion current at x i L2 If so What is its direction Is there an electron drift current at x i L2 If so What is its direction Example 32 Log scale x L2 xL2 Jnldiff gt lt 11 J drift Recombination and generation processes Recombination a process whereby electrons and holes are destroyed or annihilated Generation a process whereby electrons and holes are created Under equilibrium conditions the generation rate and the recombination rate exactly cancel In the steady state the generation rate and recombination rate can be different Can you give an example In both equilibrium and steady state there will be a steadystate carrier concentration Recombinationgeneration processes RG processes play a major role in shaping the characteristics exhibited by a device Various RG processes Bandtoband recombination generally results in light emission Bandtoband generation direct thermal generation generation by light absorption photogeneration RG center recombination generation Involves an RG center indirect generation recombination Auger recombination Generation Via impact ionization Various R G processes 1 EC n n n Photon Light Xlt O Ev a Band to band recombination Ec g ETThennal f 1 energy Ev O 18 b R G center recombination 8 Ev c Auger recombination 39I39hermal EC energy or Light W 0 EV d Band to band generation EC T39herrnal E energy T o Ev e R G center generation Ev f Carrier generation Via impact ionization Flgure 31 5 Midgap energy levels due to atomic impurities Ec Figure 316 Recombination in Si is mainly Via RG centers introduced by various unwanted impurities l3 Momentum considerations t Egt T EA Ef a Direct semiconductor b Indirect semiconductor Figure 317 E vs k diagrams for direct and indirect semiconductors 14 E k Visualizations of recombination in direct and indirect semiconductors W Photon a Direct semiconductor Figure 318 E N Phonons W gt A gtk b Indirect semiconductor 15 Chapter 24 Equilibrium carrier concentrations Equilibrium electron concentration is given by Etop no jgcltEfltE dE EC Equilibrium hole concentration is given by Ev p0 we 1fEdE EB ottom This integral cannot be solved in closed form but we can make some approximations to get a closedform solution Approximate solutions When EC EF gt 3kT or EF lt EC 3kT the solution for the total free electron concentration can be expressed as ECEF J 2 kT 32 n Noe kT Where Nc Eq216a h When EF EV gt 3kT or EF gt EV 3kT the solution for the total free hole concentration can be expressed as 32 anp H Eq 216b h2 2 EFEV p NVe kT Where NV 2 2 Effective density of states Nc is called effective density of states in the conduction band NV is called effective density of states in the valence band In Si NC 251 x 1019 m11quot m032 cm 3 28 x 1019 cm 3 NV 251 x 1019 mpm032 cm3 10 x 1019 cm3 What do you get when you multiply n with p The result gives an intrinsic property of the semiconductor Degenerate and nondegenerate semiconductors 3kT I EF here Degenerate Ec semiconductor t 1 EF here Nondegenerate j semiconductor u 4 EV EF here Degenerate 3 kT semiconductor Figure 219 Alternative expression for n and p Manipulation of the previous expressions Will give us more useful expressions as given below n nieEFEikT p niewiEpvkr 2196 219b Note from previous class that At T 300K ni 2 x 106 cm 3 in GaAs lx10100m 3 in Si 2 x 1013 cm 3 in Ge npn 222 t An intrinsic property n nieEFEikT Show that p niewiEpvkr 2196 219b Hint Start With Eq 216 in text Intrinsic carrier concentrations in Ge Si and GaAs VS T 10 6 222 2 22 2 22322 2 22 22 22 2 g 4 1015 22332252222 2222 252222 22 e 222223222Z22 222222222225222 Si 223 E 3232 E E 2 2 iiigffffyf 22 22223E23222232 32 Z 2 E E 332233 T C ni cm 3 1014 0 886 X 108 222 2 2222 22222 2222 22222 2 2 222 1 Si 2 23222322 323223322223 2 5 144 X 109 1 2 332535 10 230 X 109 39 quot quot 39 quot quot 7739 15 362 x 109 10 3 2 3 3 3 2 2322 2 223 2 23 2 2 2 3 2 32 2 3 2 252 3 20 562 x 109 I I I I Z 2 I I I I Zquot Z Z Z I I Z 2 Z 2 Z 55 1322EEIE EEEEEEEEEEEEE 3123233353 22 2 2322 25 860 X 109 p A 1GaAS 30 130 X 100 39E 1012 2 222 222222222 2 32 35 193 X 1010 8 E 33 3 E 353 Eff3 EEEEEEE ZEE39EZEE 3573 3353335 33 f 53 3 E 53 3 33 40 285 x 1010 5 2 Z gL ZZZ quot 45 415 X 1010 E 7 a U u nu u n In n quotn v H u 50 597 X 1010 E l 3 I 2 Z Z Z Z I 2 2 Z Z 2 I Z I 2 Z 21 239 I I I I 8 E 33333 333333533352532533333 EEjEEEEEEEEEEEEEEEEE 300K 100 X 1010 g 2 quotquotquot quotquotquot quot 7quot 39quotquot quotquot quotquotquot quotquot E 1010 2232332 23 22 222222222 22 2 222223 23 22 23233333332 233 3 12222333132 5353231 33323233 222322333312332 GaAs 2 Z T C nicm 3 g 09 22222222222 0 102 x 105 EEEEEEE39E EEEEEE ff32235333523222EE EEEEEEEEZEEE 5 189gtlt15 108 225222253322 33325 15 615 X 105 2222 232222222E39222222 222212322223 2222 20 108 X 106 39 Zh 25 185 X 106 7 r 30 313 x 106 1 35 520 X 106 E39 E f 40 851 X 106 39 quot quot39 quotquot39quotquot quotquotquotquotquot quotquotquotquotquotquotquot 45 137 x 107 106 222232222 22222252222223 232332i2322332222222232222 50 218 X 107 3232 22 EEEEEEEEZEEEEE EEEEEEE EEEEEEEEEEEEEEEE 300K 225 X 106 105 N O O W O O b O 0 VI 0 O O O O J 0 O TK Figure 220 Charge neutrality relationship So far we haven t discussed any relationship between the dopant concentration and the free carrier concentrations Charge neutrality condition can be used to derive this relationship The net charge in a small portion of a uniformly doped semiconductor should be zero Otherwise there will be a net ow of charge from one point to another resulting in current ow that is against out assumption of thermal equilibrium Chargecm3qp qnqND qNA 0 or p nND NA 0 where ND of ionized donorscm3 and N A of ionized acceptors per cm3 Assuming total ionization of dopants we can write p nND NAO 225 Carrier concentration calculations Assume a nondegenerater doped semiconductor and assume total ionization of dopants Then n p r212 electron concentration x hole concentration r212 p n ND N A 0 net charge in a given volume is zero Solve for n and p in terms of ND and N A Weget nizn nND NA O 722 72ND NA ni2 O Solve this quadratic equation for the free electron concentration n From n p r212 equation calculate free hole concentration p Special cases Intrinsic semiconductor NDOandNAO gt pnni Doped semiconductors where ND N A gtgt ni nND NA pnizn if NDgtNA pNA ND nnizp if NAgtND Compensated semiconductor npniwhen nigtgtND NA When ND N Al is comparable to n13 we need to use the charge neutrality equation to determine n and p Fermi level in Si at 300 K VS doping concentration i EC i I notdoped i 5F Do 3kT E 15 I EF Acce ptOfd Oped 3kT E 1 1 1 1 1 1 v 1013 1014 1015 IOIG 1017 Q 018 1019 1020 3 NA orND cm Figure 221 11 Mai oritV carrier temperature dependence 20 1 5 Freeze out I Extrinsic T region I l I I l l l I l I I I I n I 0 ND Intrinsic I I T region I 05 I I ni IND I l I l I I l O l 00 200 300 400 500 600 T K a n 0 n n c ED Dominant Negligible E v O K Moderate T High T b Figure 222 12 Equations to remember n nieEF EikT 2 n n p neEi EpkT p 1 P n ND NA 0 l 222 219a 219b 225 EC EC EC EF quotEi quotquot39quot quot39 quot39Ei quot quotquot39quotquot39quot39Ei EF Ev Ev Ev Intrinsic ntype p type Figure 218 Note Our interest was in determining n and p Free carriers strongly in uence the properties of semiconductors 13 Example 1 a Consider Si doped with 1014 cm 3 boron atoms Calculate the carrier concentration n and p at 300 K b Determine the position of the Fermi level and plot the band diagram c Calculate the the carrier concentration n and p in this Si material at 470 K Assume that intrinsic carrier concentration at 470 K in Si is 1014 cm 3 Refer to gure 220 1 Determine the position of the Fermi level with respect to Ei at 470 K Example 2 Consider a Si sample doped with 3 x 1016 cm 3 of phosphorous P atoms and 1016 cm 3 of boron B atoms a Is the semiconductor ntype or ptype b Determine the free carrier concentration hole and electron concentrations or p and n at 300K c Determine the position of the Fermi level and draw the band diagram

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