6.012 Recitation 2: Equilibrium Electron and Hole Concentration from Doping

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 Holes 2. Generation and Recombination 3. Thermal Equilibrium 4. Law of Mass Action 5. Doping - (donors and acceptors) and charge neutrality 6. Intrinsic Semiconductor vs. Extrinsic Semiconductor 7. Majority and Minority carriers 1 Electrons and Holes This 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+3 or cm−3 (per cubic centimeter). Remember in Si the atomic density is 5 × 1022 cm−3 , very useful number 2 Generation and Recombination Generation 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 Equilibrium A 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 equilibrium. It is just that the two rates are equal and cancel each other, so the concentrations of e & h do not change. n o and p o refer to concentrations in thermal equilibrium. 4 Law of Mass Action At each temperature T, under thermal equilibrium: n o · po = constant = f (T ) (only depends on temperature) no · po = n2i (T ) (ni ≡ intrinsic carrier concentration) 1 Recitation 2 Equilibrium and Doping 6.012 Spring 2009 This is like a chemical reaction: H2 0  H+ + OH− [H + ][OH − ] = 10−14 (mol/L)2 at Room T 2 bond  e− + h+ no · po = n2i (T ) = 1020 (cm−3 ) at Room T Note ni has a temperature dependence: EG ni = A · (T )3/2 e 2kB T − A is a constant, T is in Kelvin, T (K) = 273 + T(◦ C), and kB = 8.62 × 10−5 eV/K. EG is the “Bandgap” energy of the semiconductor - it also corresponds to the ease of bond breakage. For Si, EG = 1.12 eV. Example 1 At room temperature, T = 300 K, ni (300 K) = 1 × 1010 cm−3 . What is ni (500 ◦ C)? ni (500 ◦ C) = ni (773 K) ni (773 K) ni (300 K)  = 773 300 3/2 EG e 2kB (773) e−8.4 × = 4.14 × −21.65 = 3.5 × 106 EG e − e 2kB (300) − Therefore, ni (773 K) = 3.5 × 106 × ni (300 K) = 3.5 × 106 × 1010 cm−3 = 3.5 × 1016 cm−3 Something to observe: At room temperature, no = po = 1010 cm−3 for Si. Atomic density is 5 × 1022 cm−3 . There1 1010 fore, only a tiny fraction of atoms ( = = 2 × 10−11 %) lose an electron in 5 × 1022 5 × 1012 one of their 4 bonds. By heating up to 500 ◦ C, the concentration of free carriers goes up ∼ 106 (1 million) times, but the percentage is still quite low. 2 Recitation 2 5 Equilibrium and Doping 6.012 Spring 2009 Doping and Charge Neutrality Doping free electrons generated by As n type dopants: As, P, Sb give out an electron easily leave behind a positively charged ion Donor concentration Nd (cm-3) a hole generated due to bond breaking and e- given to Bp type dopants: B have one less electron, will grab one easily from another place, become negatively charged and thus generate a hole Acceptor concentration Na (cm-3) Figure 1: Types of Doping Charge neutrality Although foreign atoms are introduced in Si, the overall semiconductor is charge neutral. Therefore, concentration of positive charges = concentration of negative charges. The positive charges include holes (p) and donors (Nd ). The negative charges include electrons (n) and acceptors (Na ). po + N d − n o − N a = 0 3 Recitation 2 Equilibrium and Doping 6.012 Spring 2009 Example 2 Boron doping, dopant concentration 1017 cm−3 . At R.T. under thermal equilibrium Nd =? Na =? no =? po =? ni =? p or n type? Boron is an acceptor meaning Na = 1017 cm−3 , Nd = 0. ni = 1010 cm−3 at R.T. under thermal equilibrium (material property, doping does not matter) po  Na = 1017 cm−3 (because Na  ni ) 1020 cm−6 ∵ no · po = 1020 cm−6 , no = 17 −3 = 103 cm−3 10 cm 6 Intrinsic Semiconductor vs. Extrinsic Semiconductor In the above example, the semiconductor is extrinsic because the carrier concentrations are determined by the dopant concentrations. Example 3 Si at 500 ◦ C, with As doping 1018 cm−3 , extrinsic or intrinsic? At 500 ◦ C, ni (773 K) = 3.5 × 1016 cm−3 > Nd It is intrinsic semiconductor even though there is doping. Example 4 A semiconductor can have both dopings. If Na = 1015 cm−3 , Nd = 1019 cm−3 =⇒ n-type Si, no  po even though we have Na = 1015 cm−3 . When things get complicated, the following relations always work: po + N d − no − N a = 0 no · po = n2i (T ) 4 Recitation 2 Equilibrium and Doping 6.012 Spring 2009 Consider example 2, Nd = 0 Na = 1017 cm−3 . We said po  Na = 1017 cm−3 . How accurate is this approximation? no · po = n2i (T ) n2i (T ) =⇒ no = po plug into charge neutrality n2i (T ) + Nd − Na = 0 po p2o − Na · po − n2i = 0  4n2 (T ) Na Na 1+ i 2 ± po = 2 2 Na discard, otherwise po < 0  4n2 (T ) Na Na =⇒ po = + 1+ i 2 2 2 Na po − ∴ po  Na is a good approximation since 7 4n2i (T ) Na2 1 Majority and Minority Carriers In example 2, po  Na = 1017 cm−3  no = 103 cm−3 . Hole is majority carrier and electron is minority carrier. 5