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.
BÐ
+
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
BÐ
−
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