3.15 Electrical, Optical, and Magnetic Materials and Devices
Caroline A. Ross
Fall Term, 2003
Problem Set #1
Semiconductor Fundamentals
Solutions
Released Tuesday, September 23, 2003
Reference: Chapters 1-3 of Pierret
Problem 1 Definitions
Conduction Band: The band of allowed electron energy states above the band gap
Valence Band: The band of allowed electron energy states below the band gap
Band Gap: The range of disallowed electron energy levels between the top of the
valence band and the bottom of the conduction band.
Effective Mass: A material property that summarizes quantum-mechanical behavior as a
quasi-classical relation in the ratio of electrical force to charge carrier acceleration
Intrinsic Semiconductor: An undoped, extremely pure semiconductor
Extrinsic Semiconductor: A semiconductor whose properties are controlled by doping
Dopant: Impurity atoms added to a semiconductor to increase electron or hole
concentration
Donor: A dopant that ionizes to increase electron concentration in a semiconductor
Acceptor: A dopant that ionizes to increase hole concentration in a semiconductor
N-type material: Extrinsic semiconductor doped predominantly with donors.
P-type material: Extrinsic semiconductor doped predominantly with acceptors
N+ (or p+) material: A degenerately doped semiconductor.
Majority carrier: the most abundant charge carrier in a semiconductor
Minority carrier: the least abundant charge carrier in a semiconductor
Density of states: the distribution of allowed electron states as a function of energy
Fermi function: under equilibrium, the probability of finding an electron at a given
energy level if an allowed electron state exists there.
Fermi energy: the energy at which the Fermi function is equal to ½
Nondegenerate semiconductor: Semiconductor in which the Fermi Energy is in the
energy gap, at least 3kBT from the edge of both the conduction and valence bands.
Degenerate semiconductor: One in which the Fermi Energy is greater than Ec - 3kBT or
less than Ev + 3kBT, disallowing Maxwell-Boltzmann approx. of carrier concentration
Extrinsic temperature region: Temperature range in an extrinsic semiconductor where
the majority carrier concentration and dopant concentration are approximately equal
Intrinsic temperature region: High temperature range in an extrinsic semiconductor
where thermally generated electron-hole pairs exceed dopant concentration
Freeze-out: Low temperature range in an extrinsic semiconductor where thermal energy
is insufficient to ionize all dopant atoms
Scattering: The action of charge carriers colliding with obstacle, decreasing in mobility
Drift velocity: Velocity of a charge carrier’s movement in response to an electric field
Thermal motion: Random motion of a charge carrier due to thermal energy (entropy)
Drift Current: Rate of flow of charge in response to an applied electric field
Current Density: Rate of flow of charge, per unit of cross-sectional area
Mobility: A measure of the ease of charge carriers to move about a material in response
to an electric field, taking into account the effects of scattering, relating the mean time
between charge carrier collisions to the effective mass of the carrier.
Resistivity: Ratio of applied electric field to induced current (resistance to current flow)
Conductivity: The inverse of resistivity
Diffusion: Movement of particles from regions of high concentration to low
concentration in order to achieve equilibrium, induced by random thermal motion
Diffusion Current: The rate of flow of charge due to diffusion
Diffusion Coefficient: Ratio of diffusion flux to carrier concentration gradient times (-1)
Recombination: The annihilation of electron-hole pairs in a semiconductor
Generation: The creation of electron-hole pairs in a semiconductor
Photogeneration: Generation of electron-hole pairs by absorption of a photon
Direct thermal R-G: Band-to-band Recomb./Gen due to a dec/increase of temperature
R-G centers: Lattice defects/impurities that add electron states to the middle of the
energy gap, through which e-h pairs generate & recombine
R-G via R-G centers: The act of charge carriers passing through an R-G center to
bridge the band gap for R-G
Low level injection: Small perturbation in majority carrier concentration, such that it
remains approximately equal to its equilibrium value
Equilibrium: No perturbing forces act on material & all observables are time-invariant
Steady-state: Carrier concentrations in a semiconductor are constant over time
Minority carrier lifetime: Average time an excess minority carrier exists among
majority carriers before recombining
Minority carrier diffusion length: Average distance minority carriers can travel among
majority carriers before recombining
Quasi-Fermi level: Energy levels which specify carrier concentration in a
semiconductor under non-equilibrium conditions
Problem 2
(a) The material is intrinsic, so we use this equation to determine the Fermi level:
Ei = ¾ kBT ln(mp*/mn*) + [(Ec + Ev) /2],
mn* = .1m0; mp* = m0; Eg = .2 eV
So, Ei = ¾ * (.0259) * ln(10) + [(Ec + Ev) /2] = .0447 eV above mid-gap
Recall that the density of electron states has a 3/2 power dependency on charge carrier
mass. Since mn* = .1mp*, the density of holes across the valence band more than 10 times
larger than the density of electrons into the conduction band.
(b) At 0K, there are no carriers in the conduction or valence bands. Above 0K, we
have a finite probability of finding charge carriers in the conduction and valence bands.
Since this is an intrinsic semiconductor, the total concentration of electrons in the
conduction band will be equal to the concentration of holes in the valence band at both
300 and 1000 Kelvin. Due to increased thermal generation, the overall charge carrier
concentration at 1000 K will be much larger than that at 300 K.
Regarding distribution of charge carriers across energy in each band, we know that since
the distribution of electron energy states in the conduction band is less dense than that in
the valence band, electrons in the conduction band will be spread out across a broader
range of energy levels than holes in the valence band.
(c) Carrier concentration increases as temperature increases, causing an increase in
conductivity [up until the point where scattering due to lattice vibration dominates
mobility and decreases conductivity].
A large band-gap semiconductor might be particularly useful if it were doped to behave
extrinsically. One benefit of utilizing extrinsic semiconductors to create devices is that
their charge carrier concentration stays uniform over a wide range of temperature
variation. Once all of the dopant atoms have ionized, further increases within the
extrinsic temperature region do not affect the carrier concentration.
A material with a larger band gap will tend to have an extrinsic temperature region which
extends to a higher temperature than a material that has a smaller band gap. This is
because the amount of thermal energy it takes to create an electron-hole pair in a large
band gap material is greater than that in a small band gap material. Thus, large band gap
semiconductors are able to retain extrinsic behavior (constant carrier concentration) up to
higher temperatures, before the amount of electrons that are thermally excited from the
valence band to the conduction band exceeds the dopant concentration.
Problem 4
(a) Before being doped with Phosphorus, we know that the Si has a resistivity of .2
ohm-cm. Resistivity as a function of doping is well known for Si, and is listed in a plot
on p. 86 of Pierret. Based on this plot, we immediately know that:
-
If the material were n-type, then ND would approximately equal 2*1016 cm-3
If the material were p-type, then NA would approximately equal 1017 cm-3
We then dope the Si with 1017 cm-3 of Phosphorus, a donor, and observe that resistivity
has increased by several orders of magnitude. If the material were n-type before this
doping, then we would expect the new dopant concentration ND = 2*1016 + 1017 =
1.2*1017 cm-3, causing resistivity to drop down to .05 ohm-cm (based on the p.86 plot).
This is inconsistent with our observed rise in resistivity.
Consider what would happen if the material were p-type before this doping. In this case,
NA = 1017 cm-3, ND = 1017 cm-3, and charge neutrality demands that these donors and
acceptors negate one another. To summarize the discussion on pages 58-60 of Pierret:
p – n + ND – NA = 0;
p = ni2 / n;
Æ n2 – n(ND – NA) – ni2 = 0
Æ p = (NA – ND)/2 + [(NA – ND)2/4 + ni2]1/2
p = 0 + [0 + ni2]1/2 = ni
The resultant material is “compensated”, and behaves as if it were undoped, intrinsic Si.
Extra electrons that are donated from the phosphorus will be much more likely to occupy
empty states provided by the acceptor atoms rather than move about in the conduction
band, and thus they become immobilized. Carrier concentration returns to the intrinsic
level, and resistivity in the material increases by several orders of magnitude.
Based on this, we know that the material was p-type before being doped with
Phosphorus.
(b)
Before being doped with Phosphorus, NA = 1017 cm-3
Ef = Ei - kBT ln(NA/ni)
Ei = ¾ kBT ln(mp*/mn*) + [(Ec + Ev) /2]
kBT = .026 mV assuming room temp. 300K.
mp*/mn* = .81 / 1.18 for Si
Æ Ei = .007 + .56 = .567 eV above the valence band.
Æ Ef = .567 - .419 = .148 eV above the valence band Fermi energy, before
phosphorus doping.
After being doped with phosphorus, because of compensation, the Fermi energy
will be equal to the intrinsic Fermi Energy: Ef = .567 eV
(c) Based on the results and discussion given in parts (b) and (a), we know that this
phosphorus-doped, compensated material will have a Fermi-energy and chargecarrier concentration equal to that of intrinsic silicon. This material will behave very
similarly to intrinsic silicon.
Problem 5
(a) To find the amount of current that results from a given applied voltage, we must
compute the resistance. First, note that ND = 6 1016. At this dopant concentration, the
plot on page 86 of Pierret gives an approximate value for resistivity for Si of .1 ohm-cm.
Resistance of this bar is then given by:
R = resistivity * length / area = .1 * 10-3 / 10-7 = 1000 ohms
Current is then given by Ohm’s Law: I = V/R = .005 amps.
(b) Note: The boundary condition of one side of the bar being “grounded” was meant
to imply that carrier concentrations at that point were forced to their equilibrium value n
= 6 1016 cm-3 [by means of an ohmic contact]. Although this boundary condition was not
explicitly defined, most reasonable choices of boundary had little effect on the solution,
and credit was not deducted for other interpretations of the boundary.
We first note that this concentration gradient across the semiconductor will result in a
diffusion current given by:
Idiff = Jdiff,n * A = .005 amps (from part a), (note, A is area)
Since we want to obtain a current of the same magnitude as that in part (a), we consider
the magnitude of Jdiff,n = q Dn(dn/dx), neglecting a factor of (-1). Substitute this into the
equation for Idiff and solve for (dn/dx):
(dn/dx) = Idiff / (q Dn A)
Since this is a linear concentration gradient, we know that
(dn/dx) = (n0 – 6 1016) / L; where n0 is the initial concentration of electrons that
we need to determine, & L is bar length
Substituting this expression into the one preceding and solving for n0 yields
n0 = + [L Idiff / (q Dn A)] + 6 1016
In order to determine this carrier concentration, we first need to find the diffusion
coefficient Dn. We know that ND = 6 1016, and based on what we assumed in part (a), we
know that resistivity = .1 ohm-cm. We can then find mobility:
Mobility = (q * resistivity * ND)-1
= (1.6 10-19 * .1* 6 1016) -1
= 1000 (cm2 / V-sec) Approximately (reading from the chart on Pierret
p. 80 shows that it is actually closer to 900)
Once we know mobility, then the diffusion coefficient is given by the Einstein relation:
Dn / mobility = kBT / q;
Dn = 26 cm2 / sec
Assume room temperature Æ kBT / q = .026 volts Æ
We can then evaluate n0:
n0 = [10-3 .005 / (1.6 10-19 26 10-7)] + 6 1016
= 1.2 1019 cm-3
(c) Based on what we have seen in Pierret and in class, we know that holes are less
mobile than electrons at comparable doping levels. By the Einstein relation, a lower hole
mobility will result in a lower diffusion coefficient, giving a decreased diffusion current
if this charge gradient were made up of holes instead of electrons.