Recitation 2 Equilibrium and Doping 6.012 Spring 2009 Recitation 2: Equilibrium Electron and Hole Concentration from Doping Here is a list of new things we learned yesterday:1. Electrons and Holes2. Generation and Recombination3. Thermal Equilibrium4. Law of Mass Action5. Doping - (donors and acceptors) and charge neutrality6. Intrinsic Semiconductor vs. Extrinsic Semiconductor7. Majority and Minority carriers1 Electrons and HolesThis refers to the “free”electrons and holes. They carry charges (electron -ve and hole +ve),and are responsible for electrical current in the semiconductor. Concentration of electron(= n) and hole (= p) is measured in the unit of /cm+3or cm−3(per cubic centimeter).Remember in Si the atomic density is 5× 1022cm−3, very useful number2 Generation and RecombinationGeneration is one way to obtain “free” e & h in semiconductors. 1 electron-hole pair (1e +1h) is generated by breaking a bond. Recombination is the reverse process.3 Thermal EquilibriumA concept which will be used very often. Thermal equilibrium is defined as steady state +no extra energy source. Note that we have generation or recombination under thermal equi-librium. It is just that the two rates are equal and cancel each other, so the concentrationsof e & h do not change. n o and p o refer to concentrations in thermal equilibrium.4 Law of Mass ActionAt each temperature T, under thermal equilibrium:no· po= constant = f(T ) (only depends on temperature)no· po= ni2(T )(ni≡ intrinsic carrier concentration)1Recitation 2 Equilibrium and Doping 6.012 Spring 2009 This is like a chemical reaction:H20 H++OH−[H+][OH−]=10−14(mol/L)2at Room T2bond e−+h+no· po= ni(T )=1020(cm−3)2at Room TNote nihas a temperature dependence:EG(T )3/2−2kBTni= A e·A is a constant, T is in Kelvin, T (K) = 273 + T(◦C), and kB=8.62 × 10−5eV/K. EGis the“Bandgap” energy of the semiconductor - it also corresponds to the ease of bond breakage.For Si, EG=1.12 eV.Example 1At room temperature, T = 300 K, ni(300 K) = 1 × 1010cm−3. What is ni(500◦C)?ni(500◦C) = ni(773 K) EG ni(773 K)773 3/2e−2kB(773)e−8.4ni(300 K)=300×EG=4.14 ×e−21.65=3.5 × 106−2kB(300)eTherefore, ni(773 K) = 3.5 × 106× ni(300 K) = 3.5 × 106× 1010cm−3=3.5 × 1016cm−3Something to observe:At room temperature, no= po=1010cm−3for Si. Atomic density is 5 × 1022cm−3. There-10101fore, only a tiny fraction of atoms (5 × 1022=5 × 1012=2× 10−11%) lose an electron inone of their 4 b onds. By heating up to 500◦C, the concentration of free carriers goes up∼ 106(1 million) times, but the percentage is still quite low.25Recitation 2 Equilibrium and Doping 6.012 Spring 2009 Doping and Charge NeutralityDopingfree electrons generated by Asn type dopants: As, P, Sbgive out an electron easily leave behind a positively charged ionDonorconcentration Nd(cm-3)a hole generated due to bond breaking and e-given to B-p type dopants: Bhave one less electron, will grab oneeasily from another place, become negatively charged and thus generate a holeAcceptor concentration Na(cm-3)Figure 1: Types of DopingCharge neutralityAlthough foreign atoms are introduced in Si, the overall semiconductor is charge neutral.Therefore, concentration of positive charges = concentration of negative charges. The pos-itive charges include holes (p) and donors (Nd). The negative charges include electrons (n)and acceptors (Na).po+ Nd− no− Na=036Recitation 2 Equilibrium and Doping 6.012 Spring 2009 Example 2Boron doping, dopant concentration 1017cm−3. At R.T. under thermal equilibrium Nd=? Na=? no=? po=? ni=? p or n type? Boron is an acceptor meaning Na=1017cm−3, Nd=0.ni=1010cm−3at R.T. under thermal equilibrium (material property, doping does not matter)po Na=1017cm−3(because Na ni)1020cm−6∵ no· po=1020cm−6, no=1017cm−3=103cm−3Intrinsic Semiconductor vs. Extrinsic SemiconductorIn the above example, the semiconductor is extrinsic because the carrier concentrationsare determined by the dopant concentrations.Example 3Si at 500◦C, with As doping 1018cm−3, extrinsic or intrinsic?At 500◦C, ni(773 K) = 3.5 × 1016cm−3> Nd It is intrinsic semiconductor even though there is doping. Example 4A semiconductor can have both dopings. If Na=1015cm−3, Nd=1019cm−3= n-type⇒Si, no poeven though we have Na=1015cm−3. When things get complicated, the following relations always work: po+ Nd− no− Na=0no· po= ni2(T )47Recitation 2 Equilibrium and Doping 6.012 Spring 2009 Consider example 2, Nd=0Na=1017cm−3. We said po Na=1017cm−3. How accurateis this approximation?no· po= ni2(T )ni2(T )= no=⇒poplug into charge neutralitypo−ni2(T )+ Nd− Na=0popo2− Na· po− ni2=0=NaNa1+4ni2(T ) po 2±2 N2adiscard, otherwise po< 0= po=Na+Na1+4ni2(T )⇒22 N2a2iis a good approximation since4nN(T )∴ po Na 1 2aMajority and Minority CarriersIn example 2, po Na=1017cm−3 no=103cm−3. Hole is majority carrier and electronis minority carrier.5MIT OpenCourseWarehttp://ocw.mit.edu 6.012 Microelectronic Devices and CircuitsSpring 2009 For information about citing these materials or our Terms of Use, visit:
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