❖ \

Extrinsic Semiconductor
2
❖n0 p0 = ni , n0 + Na = p0 + Nd so solving we get
(N − N ) Nd − Na 2
( ) + ni
2 2 d a 2
n
❖ 0 − (Nd − N )n
a 0 − ni = 0 ; n0 = +
2 2
❖*For n-type semiconductor, so Nd > Na , if Na > Nduse same solution for p0

❖Typical use case: Nd − Na ≫ ni ; n0 = Nd − Na

−Eg
❖As temperature rises (remember ni = NcNv e 2kT ), nibegins be the dominant term

❖ Once ni dominates over doping devices don’t work. Wide Bandgap semiconductors
are used for high power electronic devices
 Position of Fermi-Level
❖ Assume all the dopant atoms are ionised, so that n0 = Nd
(EF − Ec) Nc
We know that in equilibrium n0 = Nce kT so EF = Ec − kT ln
❖
Nd
(EF − Ei) Nd
Alternatively n0 = nie kT so EF = Ei + kT ln
❖
ni
❖ Similarly for p-type semiconductors
❖ What about the assumption that all dopants are ionized => need to check
❖ What happens when Nd > Nc
 Statistics of Donors and Acceptors
1
Probability that Donor level is occupied (i.e not ionized)= Ed − EF
1
1+ e
❖
kT
2
❖ Factor is 1/2 is because the donor state may be occupied by a spin up or spin
down electron (alse called degeneracy factor g)
1
Probability that an acceptor level is occupied (no hole)
1 EF − Ea
❖
1+ 4
e kT

❖ factor of 4 comes because 2 valence bands in silicon (lh,hh)

EXAMPLE 1–6 Complete Ionization of the Dopant Atoms
In a silicon sample doped with 1017cm–3 of phosphorus atoms, what fraction of
the donors are not ionized (i.e., what fraction are occupied by the “extra”
electrons)?
SOLUTION:
First assume that all the donors are ionized and each donor donates an
electron to the conduction band.
17 –3
n = N d = 10 cm

From Fig. 1–20, Example 1–3, EF is located at 146 meV below Ec. The
donor level Ed is located at 45 meV below Ec for phosphorus (see Table 1–2
and Figure 1–23).
The probability that a donor is not ionized, i.e., the probability that it is
occupied by the “extra” electron, according to Eq. (1.7.1), is
1
Probability of non-ionization ≈ -------------------------------------------
1 ( Ed –EF ) ⁄ kT
1 + --- e
2
1
------------------------------------------------------------------- = 3.9%
1 ( ( 146 –45) meV) ⁄ 26meV
1 + --- e
2
(The factor 1/2 in the denominators stems from the complication that a donor
atom can hold an electron with upspin or downspin. This increases the
probability that a donor state is occupied by an electron.)
Therefore, it is reasonable to assume complete ionization, i.e., n = Nd .

45 meV 146 meV

E d Ec
EF

Ev
 Dopant Behavior at Low Temperatures

Nd
nd = Ed − EF
. At T=0K, the system is in it’s lowest energy state,
1
1+ e
❖
kT
2
which means that the dopant electrons are bound to the dopant atom
❖ This basically means that the term Ed − EF is negative so the exponential
term in the denominator can ⇒ 0 as T ⇒ 0
❖ This phenomena is called Freeze Out
 Nc Nd 1⁄2
n = ------------- e – ( Ec –Ed ) ⁄ 2kT (1.10.2)
2

Freeze Out
Freeze-out is a concern when semiconductor devices are operated at, for example,
the liquid–nitrogen temperature (77 K) in order to achieve low noise and high
CHAPTER 4 The Semiconductor in Equilibrium speed.
Figure 1–25 summarizes the temperature dependence of majority carrier
concentrations. The slope of the curve in the intrinsic regime is Eg/2k, and the slope
Conduction band in the freeze-out
Conduction band
Ec
portion is (Ec – E d )/2k (according to Eq. (1.10.2)). These facts may
" " " " " " "
Ec be used to determine Eg and Ec – Ed.
Ed
! ! ! ! ! ! !
Electron energy

Electron energy
EFi
EFi
" " " " " " " E Intrinsic regime
a
Ev
Ev ! ! ! ! ! ! !
Valence band Valence band
n ! Nd

ln n
(a) (b) Freeze-out regime
Figure 4.12 | Energy-band diagrams showing complete ionization of (a) donor states
and (b) acceptor states.

1/T
Conduction band Conduction band
Ec Ec High Room Cryogenic
EF temp- temperature temperature
Ed
erature
Electron energy
Electron energy

EFi FIGURE 1–25EFi Variation of carrier concentration in an N-type semiconductor over a wide
range of temperature.
Ea
EF
Ev Ev
Valence band Valence band

(a) (b)
 1.11 ● Chapter Summary 29

● Infrared Detector Based on Freeze-Out ●

Often it is desirable to detect or image the black-body radiation emitted by warm
objects, e.g., to detect tumors (which restrict blood flow and produce cold spots), to
identify inadequately insulated building windows, to detect people and vehicles at night,
etc. This requires a photodetector that responds to photon energies around 0.1 eV. For
this purpose, one can use a semiconductor photoconductor with Eg less than 0.1 eV, such
as HgPbTe operating in the mode shown in Fig. 1–11. Alternatively, one can use a more
common semiconductor such as doped Si operating in the freeze-out mode shown in
Fig. 1–26. In Fig. 1–26, conduction electrons are created when the infrared photons
provide the energy to ionize the donor atoms, which are otherwise frozen-out. The
result is a lowering of the detector’s electrical resistance, i.e., photoconductivity.
At long enough wavelength or low enough photon energy hν, light will no
longer be absorbed by the specimen shown in Fig. 1–26. That critical hν corresponds
to Ec – Ed. This is a method of measuring the dopant ionization energy, Ec – Ed.

Electron
Photon Ec
Ed

Ev

FIGURE 1–26 Infrared photons can ionize the frozen-out donors and produce conduction
electrons.
 Degenerate Semiconductors
❖ For low dopant density the donor/acceptor
level is a discrete level
❖ For very high dopant densities the dopant
atoms start to interact and splitting of the
energy level takes place
❖ At still higher doping, a continuous band from
Donor Level to Conduction band is formed
❖ Band-Gap Narrowing
Aside: Silicon Qubit