2 Numerical Example: Carrier Concentrations 15 -3 Donor concentration: Nd = 10 cm Thermal equilibrium electron concentration: ≈ 15 –3 no Nd = 10 cm Thermal equilibrium hole concentration: 2 2 ⁄ ≈2 ⁄ ()10 –3 ⁄ 15 –3 5 –3 po =ni no ni Nd ==10 cm 10 cm 10 cm Silicon doped with donors is called n-type and electrons are the majority carriers. Holes are the (nearly negligible) minority carriers. EECS 105 Fall 1998 Lecture 2 Doping with Acceptors Acceptors (group III) accept an electron from the lattice to fill the incomplete fourth covalent bond and thereby create a mobile hole and become fixed negative charges. Example: Boron. B– + mobile hole and later trajectory immobile negatively ionized acceptor -3 Acceptor concentration is Na (cm ), we have Na >> ni typically and so: one hole is added per acceptor: po = Na equilibrium electron concentration is:: 2 no = ni / Na EECS 105 Fall 1998 Lecture 2 Compensation Example shows Nd > Na positively ionized donors + + As As B– − negatively ionized acceptor mobile electron and trajectory ■ Applying charge neutrality with four types of charged species: ρ () ===– qno ++qpo qNd – qNa qpo–no+Nd–Na 0 2 we can substitute from the mass-action law no po = ni for either the electron concentration or for the hole concentration: which one is the majority carrier? answer (not surprising): Nd > Na --> electrons Na > Nd --> holes EECS 105 Fall 1998 Lecture 2 Carrier Concentrations in Compensated Silicon ■ For the case where Nd > Na, the electron and hole concentrations are: 2 n ≅ ≅ i no Nd – Na and po ------------------- Nd – Na ■ For the case where Na > Nd, the hole and electron concentrations are: 2 n ≅ ≅ i p o N a – N d and n o ------------------- N a – N d Note that these approximations assume that | Nd - Na|>> ni, which is nearly always true. EECS 105 Fall 1998 Lecture 2 Carrier Transport: Drift ■ If an electric field is applied to silicon, the holes and the electrons “feel” an electrostatic force Fe = (+q or - q)E. ■ Picture of effect of electric field on representative electrons: moving at the thermal velocity = 107 cm/s ... very fast, but colliding every 0.1 ps = 10-13 s. Distance between collsions = 107 cm/s x 10-13 cm = 0.01 µm (a) Thermal Equilibrium, E = 0 (b) Electric Field E > 0 Electron # 1 Electron # 1 E x x xi f,1 xi xf,1 x Electron # 2 Electron # 2 E xf,2 xi x xf,2 xi x Electron # 3 Electron # 3 E x xf,3 xi x xf,3 xi * xi = initial position * xf, n = final position of electron n after 7 collisions ■ The average of the position changes for the case with E > 0 is ∆x < 0 EECS 105 Fall 1998 Lecture 2 Drift Velocity and Mobility ■ The drift velocity vdn of electrons is defined as: ∆x v = ------ dn ∆t ■ Experiment shows that the drift velocity is proportional to the electric field for electrons µ v dn = – nE, µ with the constant n defined as the electron mobility. ■ µ Holes drift in the direction of the applied electric field, with the constant p defined as the hole mobility. µ vdp = pE How do we know what’s positive and what’s negative? positive: negative: E E x x EECS 105 Fall 1998 Lecture 2 Electron and Hole Mobilities ■ mobilities vary with doping level -- plot is for 300 K = room temp. 1400 1200 electrons 1000 Vs) / 2 800 600 mobility (cm holes 400 200 0 1013 1014 1015 1016 1017 1018 1019 1020 + −3 Nd Na total dopant concentration (cm ) ■ “typical values” for bulk silicon - assuming around 5 x 1016 cm-3 doping µ 2 n = 1000 cm /(Vs) µ 2 p = 400 cm /(Vs) ■ at electric fields greater than around 104 V/cm, the drift velocities saturate --> max. out at around 107 cm/s. Velocity saturation is very common in VLSI devices, due to sub-micron dimensions EECS 105 Fall 1998 Lecture 2 Carrier Transport: Drift Current Density Electrons drifting opposite to the electric field are carrying negative charge; therefore, the drift current density is: dr -2 -1 -2 Jn = (-q) n vdn units: Ccm s = Acm dr µ µ Jn = (-q) n (- n E) = q n n E dr Note that Jn is in the same direction as the electric field. µ For holes, the mobility is p and the drift velocity is in the same direction as the µ electric field: vdp = p E The hole drift current density is: dr Jp = (+q) p vdp dr µ Jp = q p p E EECS 105 Fall 1998 Lecture 2 Drift Current Directions and Signs ■ For electrons, an electric field in the +x direction will lead to a drift velocity in dr the -x direction (vdn < 0) and a drift current density in the +x direction (Jn > 0). electron drift current density dr Jn vdn E x ■ For holes, an electric field in the +x direction will lead to a drift velocity in the dr +x direction (vdp >0) and a drift current density in the +x direction (Jn > 0). hole drift current density dr Jp vdp E x EECS 105 Fall 1998 Lecture 2.