B12: Semiconductor Devices
Example Sheet 2: Solutions
Question 1
To get from eq. (5.70) of the notes to the expression given in the examples sheet, we
simply invoke the relations
n0
p0 ,
n0
n0 .
In this case, we have that D0 ⇡ Dp and µ0 ⇡ µp . Furthermore, the analysis of bandto-band recombination yielded the expression R = p0 /⌧p . The stated result follows.
For the second part of the question, it is stated that E = 0 and @ 2 p0 /@x2 = 0. The
ambipolar transport equation is thus given by
@p0
=G
@t
p0
.
⌧p
The complementary solution to this di↵erential equation can be found by setting G = 0,
which yields
p0c (t) = Ae t/⌧p
for some constant A. Considering the steady-state (t ! 1), the di↵erential equation
becomes
p0
0=G
)
p0 (1) = G⌧p .
⌧p
Since p0c (1) = 0, the steady-state solution is simply the particular solution. Putting it
all together yields
p0 (t) = G⌧p + p0c (t).
Invoking the initial condition p0 (0) = 0 reveals that A =
p0 (t) = G⌧p (1
e
t/⌧p
),
t
G⌧p , so the total solution is
0.
The excess minority carrier hole concentration increases monotonically in a uniform
manner across the substrate according to the product of the rate of carrier generation
and the mean time to recombination, approaching this product in the steady-state.
For the final part of the question, one can simply di↵erentiate the given expression
a few times with respect to the position and time variables to show that the ambipolar
transport equation, with G = 0, is satisfied. This solution suggests that, initially, there
is a large concentration of carriers local to x = 0. As time progresses, the excess holes
drift en masse in the positive x direction and simultaneously di↵use symmetrically about
1
the point x = µp Et in a Gaussian-like manner (see Figure 1). As t grows large, the excess
carrier concentration tends to zero in a spatially uniform manner, provided the substrate
has infinite length. Although not treated in this question, it is interesting to note that
in a substrate of finite length, a rather complicated behaviour of the excess minority
carrier hole concentration would be observed, which would arise from the drift/di↵usion
process described above, but where carriers would collect at the right-hand boundary
(say, x = `) whilst simultaneously recombining at the rate p0 (`)/⌧p .
2.4
2
1.6
1.2
0.8
0.4
-0.5
0
0.5
1
1.5
2
2.5
Figure 1: p0 (x, t) vs x for µp Et = 0.1, 0.5, 1.0.
Question 2
The energy band diagram for a zero bias is shown in Figure 2. In a pn junction
under zero bias, the Fermi energy is constant, with the valence band edge situated close
to the Fermi energy in the p region and the conduction band edge situated close to the
Fermi energy in the n region. The intrinsic Fermi level lies roughly at the mid-gap point
across the device, and thus it is position dependent. Di↵usion causes majority carrier
holes in the p region to move into the n region, and majority carrier electrons in the n
region move into the p region. This process creates a charge separation, which results in
an induced electric field oriented from the n region toward the p region (from positive to
negative). Eventually, this field perfectly balances the di↵usion action, and equilibrium
is reached. The field e↵ectively prevents holes in the p region from di↵using across the
metallurgical junction. Equivalently, it pushes electrons in the n region away from the
junction as well. This yields a space charge or depletion region around the junction,
which is largely devoid of carriers. This region is, however, locally charged since there is
a fixed lattice of charged impurity atoms (positively charged donor ions in the n region
and negatively charged acceptor ions in the p region). The potential di↵erence that is
induced across the depletion region is known as the built-in potential and is denoted by
Vbi in the figure. This potential is equal to the sum of the di↵erences in the Fermi and
2
intrinsic Fermi levels in the n and p regions.
p!
metallurgical!
junction!
n!
Ec!
eVbi!
EFi!
EF!
eVFp!
eVFn!
Ev!
Figure 2: Energy band diagram for a pn junction with no bias.
Figure 3 shows the energy band diagram for a device in forward bias mode. The
applied bias counteracts the induced electric field (equivalently, the built-in potential).
Majority carrier electrons in the n region situated beyond the depletion region require
less energy to overcome the potential barrier between the n and p regions, and thus
carriers move more freely between the regions. Current flows from the p to the n region.
Ec!
eVTotal!
EFi!
EFp!
eVFp!
eVA!
eVFn!
EFn!
Ev!
Figure 3: Energy band diagram for a pn junction with a forward bias.
Carrier transport describes current flow in a pn junction. Transport is related to the
excess minority carrier concentrations in the p and n regions. Carrier concentrations are
derived by integrating the product of the density of states function and the Fermi-Dirac
probability function across all energy values in a given band. This yields an expression
that is proportional to an exponential with an argument that quantifies the potential
di↵erence between a given band edge and the Fermi level. When a bias is applied, the
potential associated with the bias is incorporated into the argument of the exponential
by solving the ambipolar transport equation. To obtain the current through the device,
3
the di↵usion equations are solved, which require one to di↵erentiate the excess carrier
concentration functions. Since these are exponentials, their form is retained through the
di↵erentiation and the total current will have an exponential form.
Question 3
For a one-sided junction with Na
Nd , the regions and di↵usion currents resemble
those shown in Figure 4. Notice the small space charge width in the p-type material
and the very low excess minority carrier electron di↵usion current in the p region. The
ideal diode equation is derived in exactly the same way as in the notes. We simply note
that, due to the relation Na
Nd , we have
✓
✓ ◆
◆
qADp n2i
VA
I ⇡ AJp (xn ) =
exp
1 .
Lp Nd
Vt
Applied electric field!
p!
n!
Jp!
Jn!
Induced electric field!
Figure 4: One-sided junction diagram with di↵usion currents.
Question 4
The correct answer should be centred around the discussion given in sections 7.1
and 7.2 of the notes. Specifically, it is important to note that photodiodes are reverse
biased so that, when a photon generates carriers in the depletion region, the current
resulting from the carriers moving in opposite directions can be easily detected. It is
also important to note that the choice of semiconductor will dictate what wavelength of
light can be detected, with the band gap of the semiconductor adhering to
Eg 
1.24
4
eV
with being measured in microns.
The main things to note in relation to the generation of light are that the recombination process results in photon emission in direct band gap semiconductors, but phonons
are typically emitted in indirect band gap materials since a change in momentum is
required for an electron to jump from the bottom of the conduction band to the top of
the valence band. It is also important to note that Si is an indirect band gap semiconductor, whilst GaAs is a direct band gap semiconductor. For this reason, GaAs is a very
important semiconductor for optical applications.
Question 5
(See appended hand-written work.)
Question 6
We adopt the setup depicted in Figure 5. Firstly, we note that the cuto↵ frequency
is defined as
1
fT =
2⇡⌧ec
where
w2
wBC
⌧ec ⇡ ⌧b + ⌧d = B +
.
2DB
vs
From the figure, we can rewrite this as
⌧ec ⇡
2
wB
2x0
+
2DB
vs
2wB
.
vs
To maximise the cuto↵ frequency, we find the base width that yields the minimum of
the emitter-collector delay. This gives
wB =
and, using the figure, we have
✓
2DB
vs
wBC = 2 x0
2DB
vs
◆
.
These optima yield the minimum value for the emitter-collector delay:
✓
◆
2
DB
⌧ec ⇡
x0
.
vs
vs
We immediately see that, all else being constant, if we increase the base di↵usion coefficient whilst maintaining the condition DB  x0 vs /2 lest wBC < 0, we can reduce the
delay and increase the cuto↵ frequency. This is intuitively satisfying. Alternatively, for
a constant DB , ⌧ec has a maximum at vs = 2DB /x0 . So if the B-C depletion region
saturation velocity is above this value, we can design the device such that vs is increased,
which would reduce ⌧ec . Otherwise, if the saturation velocity is less than this stationary
point, we can design the device such that vs is decreased, which would again reduce ⌧ec .
5
n
p
0
n
wB = !
x0 – wBC/2
x0
x0 + wBC/2
Figure 5: Setup for solving problem 6.
6