Download ENEE 313, Spr `09 Midterm II Solution

ENEE 313, Spr ’09 Midterm II Solution
PART I—DRIFT AND DIFFUSION, 30 pts
1. We have a silicon sample with non-uniform doping. The sample is 200 µm long: In the figure,
L = 200 µm= 0.02 cm. At the x = 0 edge, the sample is doped at an acceptor density of
NA (0) = 1017 1/cm3 . At the x = L edge, the sample is doped at NA = 5 × 1016 1/cm3 . In
between, the doping density varies linearly, as shown in the figure (y-axis not to scale).
(a) (2 pts.) What type is the material?
The material is p-type.
(b) (10 pts.) Sketch the equilibrium hole concentration p0 (x). Approximately calculate
and roughly sketch the equilibrium electron concentration n0 (x). (Hint: The doping
concentration can be written in terms of x as NA (x) = 1017 − 2.5 × 1018 x, where x is
given in units of cm.)
p0 (x) ≈ NA (x)
n2i
2.1 × 1020
n0 (x) =
= 17
p0
10 − 2.5 × 1018 x
⇒ n0 (0) = 2.1 × 103 ; n0 (L) = 4200
(See next page for the sketch.)
(c) (8 pts.) Assume this sample is at equilibrium. Mark the directions of diffusion for the
majority and for the minority carriers. Mark the direction of the electric field Ebi (x)
which must be built-in to oppose this diffusion current at equilibrium.
The majority carriers are the holes, and they will diffuse towards +x̂, as the h+
concentration decreases in that direction.
The minority carriers are the electrons, and they will diffuse towards −x̂, as the
e− concentration decreases in that direction.
This will create a hole diffusion current and an electron diffusion current both in
the +x̂ direction, and an opposing drift current must balance this for the sample
to be at equilibrium. Therefore the built-in electric field will point in the −x̂
direction.
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(d) (10 pts.) Calculate the magnitude of this electric field Ebi (0) at x = 0.
At equilibrium:
Jp,dif f + Jp,drif t = 0
⇒ |Jp,dif f | = |Jp,drif t | ; at x = 0
dp0 (x)
|x=0 ]| = qµp p0 (0)E(0)
⇒ |[qDp
dx
⇒
kT
µp × 2.5 × 1018 = 101 7µp E(0)
q
⇒ E(0) = 0.026 × 2.5 × 1018 /1017 = 0.65 V/cm
(e) (BONUS: 5 pts.) Calculate the built-in electric field Ebi (x) throughout the sample
and sketch it.
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PART II—EXCESS CARRIER CONCENTRATIONS, 20 pts.
2. A uniformly-doped n-type silicon sample has ND = 1018 1/cm3 . At room temperature,
ni = 1.45 × 1010 1/cm3 . Use µn =1200 cm2 /V.s, µp =400 cm2 /V.s. The carrier lifetimes are
τn = τp = 10−7 sec.
Let a fixed excess hole concentration ∆pn = δpn (x = 0) be injected and maintained at one
edge of the sample, at x = 0. Then the excess hole concentration profile throughout the
sample will be given by
δp(x) = ∆pn exp (
−x
)
Lp
(a) (10 pts.) Find the ∆pn level required to maintain a hole diffusion current density
Jp,dif f = 10−5 A/cm2 =10 µA/cm2 at the edge x = 0 at room temperature (T = 300
K, kT /q ≈ 0.026 V).
dδp(x)
dp(x)
= −qDp
(since p0 is uniform.)
dx
dx
−1 (−x/Lp )
−qDp ∆pn
e
Lp
Dp
q ∆pn = 10−5 A/cm2
Lp
0.026 × 400 = 10.4 cm2 /sec
√
10.4 × 10−6 = 1.02 × 10−3 cm
10.4
∆pn = 10−5
1.6 × 10−19
0.00102
10−5 × 0.00102
= 6.13 × 109 1/cm3
1.6 × 10−19 × 10.4
Jp,dif f = −qDp
=
=
Dp = Vthermal µp =
Lp =
q
Dp τ p =
⇒ Jp,dif f =
⇒ ∆pn =
(b) (10 pts.) While maintaining this excess injection level, a constant electric field E is
applied across the sample. Starting from the hole current continuity equation, derive
the differential equation that will govern δp(x) throughout the sample. (Hint: You do
not need to insert numbers in this derivation, so simply use δp(x) to show the excess
electron concentration. Remember that p(x) = p0 + δp(x), where p0 is the equilibrium
hole concentration. Account for all hole current components.)
Current continuity equation for holes:
1 dJp (x, t) δp(x, t)
dp(x, t)
= −
−
dt
q dx
τp
We have to define the hole current Jp to use in this equation. Since there is both a
non-uniform hole distribution and an electric field in question here, both diffusion
and drift current components must be present:
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Jp (x, t) = Jp,drif t (x, t) + Jp,dif f (x, t)
dp(x, t)
= qµp p(x, t)E(x, t) − qDp
dx
dδp(x, t)
= qµp p(x, t)E(x, t) − qDp
,
dx
since p(x, t) = p0 + δp(x, t), and p0 is constant in x and in t for a uniformly-doped
sample. This also implies dp(x, t)/dt = dδp(x, t)/dt in the left-hand side of the
current continuity equation; therefore that equation becomes:
dδp(x, t)
1 d
dδp(x, t)
δp(x, t)
= −
[qµp (p0 + δp(x, t))E(x, t) − qDp
]−
dt
q dx
dx
τp
2
d δp(x)
d
δp(x, t)
−
[µ
δp(x)E(x)]
−
= Dp
p
dx2
dx
τp
The electric field is specified to be constant throughout the sample; therefore the
full differential equation which governs δp(x, t) in this system is
dδp(x, t)
d2 δp(x)
dδp(x, t) δp(x, t)
= Dp
− µp E
−
2
dt
dx
dx
τp
PART III—PN-JUNCTION DIODES, 35 pts.
3. Consider pn-junction structures.
(a) (7 pts.) Sketch an example junction, built-in electric field (indicating its direction) and
band energy diagram at equilibrium. Indicate conduction and valence bands, intrinsic
Fermi level and Fermi level in your sketch.
(b) (7 pts.) Qualitatively explain the formation of the depletion region and the built-in
potential/electric field in the pn-junction. Use terms such as ”diffusion,” ”dopants” and
”recombination” in your explanation.
At the junction of the two materials, holes diffuse from the p-type material towards
the n-type material, and recombine with the electrons diffusing from the n-type material towards the p-type material, leaving behind uncompensated ionized dopants
in a depletion region (“depleted” of mobile charges): Positively-charged donors on
the n-side and negatively-charged acceptors on the p-side. These revealed charges
set up an electric field between them, pointing from the n-side of the depletion
region to the p-side. This electric field direction sets up a drift current for carriers
of either type which opposes their diffusion current, and its magnitude, at equilibrium, is the right magnitude to completely balance the diffusion current. Thus, at
equilibrium there is no net current flowing through the junction.
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For the rest of this question, consider a silicon pn-junction diode with the p-type bulk
doped uniformly at NA = 5×1018 1/cm3 and n-type bulk doped uniformly at ND = 5×
1015 1/cm3 . ni = 1.45×1010 1/cm3 at room temperature. Dn = 18 cm2 /s in the p-side,
and Dp = 12 cm2 /s in the n-side. τn = τp = 1 µs on either side. Assume the bulk regions
are much longer compared to the diffusion lengths.
(c) (7 pts.) Calculate the built-in potential V0 and the equilibrium depletion region width
WD at room temperature. Which side of the junction is the wider part of the depletion
region? Why? (You do not need to calculate the depletion region sizes on the two sides
of the junction, just comment on the relative size.)
V0 =
s
kT
NA ND
5 × 1018 × 5 × 1015
ln (
)
=
0.026
ln
(
) = 0.843 V
q
n2i
2.1 × 1020
s
2Si V0 NA + ND
1 × 2.07 × 10−12 × 0.843
WD =
=
× 2.002 × 10−16
q
NA ND
1.6 × 10−19
= 4.67 × 10−5 cm = 46.7 µm
(1)
The n-side is doped three orders of magnitude lighter than the p-side; therefore
the much greater part of this depletion region width is on the n-side.
(d) (4 pts.) Remember that we calculated the ideal diode current by capturing the minority
diffusion currents at the depletion region edges. Show which is more significant in size
for this particular diode—electron diffusion current at the p-side depletion region edge,
or hole diffusion current at the n-side depletion region edge? (Hint: Identify which term
in the J0 expression comes from which current. Also, Dn /Ln will have the same order
of magnitude as Dp /Lp .)
J0 = q(
5
Dn
Dp
pn0 +
np 0)
Lp
Ln
n2i
= 0.42 × 105 1/cm3
ND
n2i
= 0.42 × 102 1/cm3
=
NA
> np0
Here, pn0 =
np0
⇒ pn0
⇒ The hole injection into the n-side will yield a higher diffusion current than the
electron injection into the p-side.
(e) (10 pts.) What is the current density at 0.5 V applied bias? At -0.2 V applied bias?
(Hint: You may ignore the component that you found to be less significant in part (c).)
From the discussion above, we
can ignore the (Dn /Ln )np0 component. Then, with
q
2
Dp = 18 cm /sec and Lp = Dp τp = 3.46 × 10− 3 cm, Dp /Lp = 3468.21 cm/sec
and
J0 = 1.6 × 10−19 × 3468.21 × 0.42 × 105 = 2.33 × 10−11 A/cm2
⇒ At VA = 0.5 V, JD = J0 (exp (0.5/0.026) − 1) = 0.0052 A/cm2
At VA = −0.2 V, JD ≈ −J0 = −2.33 × 10−11 A/cm2
.
(2)
(f) (BONUS: 5 pts.) A diode to be used in a high-frequency circuit should respond quickly
to any change in bias. This implies the excess minority carrier concentrations injected to
either side of the junction should be built up or dispelled quickly. This can be achieved
by having the injected minority concentrations, δpn and δnp , as small as possible for
a given bias. Remembering that these injected concentrations are proportional to the
equilibrium minority concentrations on either side, if you could alter the doping for the
p-side, what would you choose for a faster-switching diode? A higher doping would
mean smaller ∆np , which would help the diode switch faster.
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PART IV—15 pts.
4. Fill in the blanks, select the correct answer or indicate “True” or “False” for the statements
below. Each blank or true/false answer is worth one point. Blanks might contain more than
one word. You may choose 15 out of 20 to answer; if you get more than 15 answers
right in this section your grade will be 15.
(a) A non-uniform carrier distribution across a semiconductor sample creates a diffusion
current.
(b) (TRUE/FALSE) For a semiconductor sample with a uniform constant acceptor doping
level NA , if a uniform donor concentration ND > NA is added to the sample, the hole
diffusion current increases.: FALSE
(c) Under reverse bias, the net current of the pn-junction depends on the generation,
OR generation rate of minority carriers approximately within a diffusion length of
the depletion region, OR junction. Therefore, this “saturation current” does not
change with bias.
(d) Zener breakdown in a pn-junction is based on the phenomenon of tunneling. . It occurs
in reverse bias.
(e) The depletion region width narrows with higher forward bias. This is why the diode
responds to very high bias voltages as if it was a resistor.
(f) Under reverse bias, the electric field across the depletion region of a pn-junction will
be higher than the same field in a no-bias case.
(g) It is possible to turn a rectifying metal-semiconductor junction to a tunneling OR
conducting junction by doping the semiconductor much higher.
(h) TRUE/FALSE: With the electron concentration increasing in the +x direction, the
electron diffusion current is in the −x direction: FALSE
(i) If the donor doping of the n-side of a pn-junction is half of the acceptor doping of the
p-side, the depletion region width in the n-side is ........................ that in the p-side.
(a) a quarter of (b) half of (c) equal to (d) twice
(j) Under reverse bias, there is more charge stored in the “capacitor” formed by the depletion OR junction capacitance of a pn-junction.
(k) The charge stored by the “capacitor” for the depletion (junction) capacitance of a pnjunction is the ionized dopants left behind in the depletion region after all the
mobile carriers have been depleted from there.
(l) Avalanche breakdown happens when electrons drifting across the depletion region can
gain enough kinetic energy to knock out another electron-hole pair, which then also begin
to accelerate. For this to be more likely, the diode should be designed with a longer
depletion region, which is attained by using lower doping.
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