Prof. Jasprit Singh
Fall 2001
EECS 320
Homework 5
This homework is due on October 18.
In this homework, I have included a sample midterm exam. Please look at
the trial exam after the October 18 lecture and try to solve the problems before
looking at the solutions.
Problem 1: An abrupt silicon p-n diode at 300 K has a doping of Na =
1018 cm;3 ; Nd = 1015 cm;3 . Calculate the built-in potential and the depletion
widths in the n and p regions.
Note that the depletion width falls primarily on the lightly doped side.
Problem 2: Consider a p+n Si diode with Na = 1018 cm;3 and Nd =
16
10 cm;3 . The hole diusion coecient in the n-side is 10 cm2 /s and p =
10;7 s. The device area is 10;4 cm2 . Calculate the reverse saturation current
density and the forward current density at a forward bias of 0.7 V at 300 K.
At what forward bias does the current density reach 104 cm;2 ?
Problem 3: A GaAs LED has a doping prole of Na = 1017 cm;3; Nd =
18
10 cm;3 at 300 K. The minority carrier time is n = 10;8 s; p = 5 10;9 s. The electron diusion coecient is 100 cm2 s;1 while that of the holes
is 20 cm2 s;1 . Calculate the ratio of the electron-injected current (across the
junction) to the total current.
Problem 4: The diode of Problem 3 has an area of 1 mm2 and is operated
at a forward bias of 1.2 V. Assume that 50% of the minority carriers injected
recombine with the majority charge to produce photons. Calculate the rate of
the photon generation in the n- and p-side of the diode.
Problem 5: Compare the dark currents (i.e., reverse saturation current)
in p-n diodes fabricated from GaAs and Si. Assume that both diodes are doped
at Nd = Na = 1018 cm;3 . The material parameters are (300 K):
GaAs : n = p = 10;8 s; Dn = 100 cm2 =s; Dp = 20 cm2 =s
Si : n = p = 10;7 s; Dn = 30 cm2 =s; Dp = 15 cm2 =s
When p-n diodes are used as light detectors, the reverse dark current is a noise
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source.
SOME IMPORTANT ISSUES DISCUSSED THIS WEEK
This week we have started our discussions on the p ; n diodes. We have
focused on the current ow mechanisms in the diode.
What is a diode supposed to do?: The diode is a device which has a very
non-linear current-voltage (I-V) characteristics. In particular it has a conducting state when it is biased one way (forward-biased), and a non- conducting
state when the biasing is reversed (reverse-biased). Due to this special I-V
characteristics, the diode can be used for a number of useful applications in information processing systems. These applications include rectication, mixing
of two signals, logic applications, use as a tuning element etc.
Why is the diode called a bipolar device?: In a diode, the current is carried
by both electrons and holes. This occurs because we have a p-type material and
an n-type material making up the device. Devices where the current is carried
by only one type of carrier are called unipolar devices.
Diode under equilibrium: When the diode is under equilibrium, there is no
net current owing in the device. However, this does not mean that there is no
movement of electrons and holes. To understand the various current paths we
note that the p-n region can be divided into the following regions:
i)The neutral p- and n- regions: Here we have a density np ; pp of electrons
and holes on the p-side and nn ; pn on the n-side.
ii)The depletion region across which the built-in potential Vbi drops. In this
region there is a high electric eld present even in equilibrium.
As we know from our previous discussions, electrons and holes are moving
with random velocities under equilibrium. Some of the electrons in the p ; side
wander into the depletion region. These electrons are swept into the n-side by
the strong eld present in the depletion region. The same happens to the few
holes that wander into the depletion region from the n ; side. These current
components are called drift components in the depletion region since the current
is owing due to the high eld present in the depletion region.
Also some of the electrons in the n ; side are able to overcome the builtin potential barrier and diuse across the depletion region into the p ; side.
Similarly, holes in the p ; side are able to overcome the built-in potential barrier
and diuse across the depletion region into the n ; side. This ow is due to the
diusion process.
At equilibrium, the electron drift and diusion currents balance out exactly.
So do the hole drift and diusion currents. Note that the drift current involves
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moving down the hill (for electrons) while the diusion current involves only
those carriers that can overcome the hill. The operation of the p-n diode depends
upon changing the height of the \hill" by applying a bias across the diode.
Depletion widths and electric elds in the device: By solving the Poison
equation under the depletion approximation we are able to calculate the electric
eld and depletion region width. These are given by Eqns. 5.17 and 5.21 for
the elds and Eqns. 5.27 | 5.29 for the depletion widths. Several points are
worth pointing out:
i) Even at equilibrium the average eld in most diodes is quite large (10
kV/cm). As a result in the drift current the electrons and holes move through
the depletion region with saturation velocity.
ii) The depletion region is primarily in the lightly doped side of an assymetrically doped diode.
Current ow in a p-n diode under non-equilibrium: In non-equilibrium
conditions, the drift current that we have talked about above, is essentially
unaected by the external bias. This is an important point to keep in mind for
the diode performance.
The diusion current essentially goes to zero when a reverse bias is applied
(negative bias on the p-side). This is because the reverse bias increases the
barrier that the carriers have to overcome to ow across the depletion region.
As a result, in reverse bias the current is very small and is due to the electrons
wandering from the p-side into the depletion region and being swept into the
n-side. Also the reverse current is due to the holes wandering into the depletion
region from the n-side and being swept out.
In the forward bias, the drift part of the current is unchanged, but the
diusion part changes dramatically. The electrons on the n-side now see a
reduced barrier in their path and diuse across into the p-side. Also the holes
on the p-side see a reduced barrier and diuse across to the n-side.
Since the fraction of electrons and holes that have energies greater than a
certain barrier depend exponentially on the barrier height (in the Boltzmann
approximation), the diusion current in the forward bias increases exponentially
with the applied bias.
Note that (see Eqns. 5.42, 5.43 and 5.45) the diusion current due to carrier
injection over the barrier is made up of electron and hole injection currents.
The relative ratios of these two depends upon the relative doping of the n- and
p-sides. Go over Example 5.7 of the text to see how this fact is exploited in
LEDs for high eciency light emission.
Majority and minority currents: The electrons that are injected into the
p-side by diusion in forward bias recombine with holes on the p-side. Thus
the electron current in the p-side (the minority current) gradually decreases as
we go deeper in the p-side. Holes are injected from the contact on the p-side to
compensate for the holes that recombine with the electrons. The hole current
on the p-side is the majority current.
The overall current in the device is ofcourse constant everywhere in the
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device, but the relative share of the minority and majority currents changes
with position. Please go over section 5.3.2 to see how this occurs quantitatively.
Trade-os in the diode design: In an ideal diode, the reverse current should
be zero and the forward current should be innity. Ofcourse, this is not possible.
We can get a low value of I0 , the reverse current by using very heavily n- and psides. This makes np and pn very small. However, the problem with this is that
the depletion width can be very small and as a result the eld in the depletion
region can reach a very high value. This can cause the diode breakdown at very
small applied bias. Thus one has to optimize the doping for a given operating
voltage specications.
Another important issue in the diode operation is the response speed of the
diode to external bias changes. In the forward bias case, minority charge is
injected into a majority region. This charge has a certain lifetime. The lifetime
determines the time it takes to modify the minority charge in a region. In pure
Si, the minority carrier lifetime is of the order of a microsecond. In pure GaAs,
it is of the order of a nanosecond. To speed up the recombination process, one
often puts impurities such as gold atoms in Si. Another way to ensure a faster
response is to use a narrow diode.
In a narrow diode the distance from the depletion region edge to the contact
is smaller than the diusion length. As a result, minority charge recombines
very fast at the contact.
The disadvantage of the short diode is that the reverse current becomes large.
TOPICS TO BE COVERED NEXT WEEK
Next week we will discuss the following topics:
We will discuss the non-ideal diode and the speed issues in p ; n diodes.
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A TRIAL MIDTERM EXAM WITH SOLUTIONS
You may use any relevant piece of information from the textbook or any
other source.
Problem 1:(6 points)
Consider an n-type Si sample with a conductivity of 0.1 (
cm);1 at 300 K.
Calculate the Fermi level in the material. Also calculate the Fermi level in a
p-type material with the same conductivity.
n = 1000 cm2 =V ; s; p = 300 cm2 =V ; s
We have
= ne = 0:1 (
cm);1
This gives for the n-type sample
cm);1
14 ;3
n = (1:6 10;019:1(
C )(1000cm2=V ; s) = 6:25 10 cm
The Fermi level position is
25 1014 EF = Ec + 0:026ln 26::78
1019 = Ec ; 0:27 eV
For the p-type material we have
cm);1
15 ;3
p = 1:6 10;019:1(
C )(300cm2=V ; s) = 2:08 10 cm
This gives for the Fermi level
2:08 1015 EF = Ev ; 0:026ln 9:84 1018 = Ev + 0:22 eV
Problem 2:(6 points)
Consider an undoped Si sample with area 10;4 cm2 and length 10:0 m.
Calculate the resistance of the sample.
It is found that when light with a certain intensity shines on the samplem
(creating an equal number of excess electrons and holes), the resistance decreases
by a factor of 105. Calculate the electron and hole quasi-Fermi levels for this
case.
For the undoped sample we have
n = p = ni = 1:5 1010 cm;3
5
The resistance is
L =
1:0 10;3cm
6
R = A
10;4cm2 [1:5 1010cm;3 1:6 10;19C (1000 + 300)cm2=V ; s] = 3:210 When light falls on the sample, the resistance decreases by a factor of 105
which means the electron and hole densities increase by 105. We have
n = p = 1:5 1015 cm;3
To nd the quasi-Fermi levels we use the Boltzmann approximation
5 1015 = E ; 0:256 eV
EFn = Ec + 0:026ln 21::78
1019 c
Similarly
5 1015 = E + 0:229 eV
EFp = Ev ; 0:026ln 91::84
1018 v
Problem 3: (8 points)
Consider a Si long p-n diode with the following parameters:
n ; side doping = 1017 cm;3
p ; side doping = 1017 cm;3
minority carrier lifetime n = p = 10;7 s
electron diffusion constant = 30cm2=s
hole diffusion constant = 10cm2=s
diode area = 10;4 cm2
carrier lifetime in the depletion region : 10;8 s
Calculate the diode current at a forward bias of 0.5 V and 0.6 V at 300 K.
What is the ideality factor of the diode in this range.
For this structure we nd the following values For this diode structure we
have the following:
np = 2:25 103 cm;3
pn = 2:25 103 cm;3
Ln = 17:32 m
Lp = 10:0 m
Vbi = 0:817V
The prefactor in the ideal diode equation is
D p D n I0 = e Lp n + Ln p
p
n
= 9:83 10;16 Amperes
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The prefactor to the recombination generation current is
0 = eAWni
IGR
2
where is the lifetime in the depletion region.
The depletion width at a forward bias of 0.5 V is found to be
W (0:5 V ) = 9:2 10;6 cm
The depletion width at 0.6 V forward bias is
W (0:6 V ) = 7:6 10;6 cm
The prefactors to the recombination generation current is
The current is now
0 (0:5 V ) = 1:1 10;10 A
IGR
0 (0:6 V ) = 9:1 10;11 A
IGR
I (0:5 V ) = 9:83 10;16e 000265 + 1:1 10;10e 000525
= 2:2 10;7 + 1:66 10;6 A
= 1:88 10;6 A
:
:
:
:
and
I (0:6 V ) = 9:83 10;16e 000266 + 9:1 10;11e 000526
= 1:034 10;5 + 1:1 10;5 A
= 2:13 10;5 A
:
We can write the diode current as
:
I = I00 exp mkeV T
B
Thus
:
:
I (V2 ) = exp e(V2 ; V1 )
I (V2 )
mkB T
Using this relation we nd that
m = 1:6
Problem 4: (5 points)
Consider a narrow diode with the same parameters as are given above. Calculate the total electron and hole injected charge in the n- and p- sides at a
forward bias of 0.4 V.
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The widths of the n- and p-sides are both 1.0 m.
Using the parameters given in Problem 3 we nd that at a forward bias of
0.4 V the depletion region on the n-side is
Wn = 5:3 10;6 cm
The neutral region width is then
1:0 ; 0:026 = 0:947 m
The excess hole density at the depletion region edge is
eV p(Wn ) = np0 exp k T
B
= 1:08 1010 cm;3
The hole charge injected into the n-side is (using the linear decay model)
Qp = eAp(Wn )(2Wln ; Wn )
= 8:4 10;18 C
Since the device is symmetrically doped the same amount of electron charge
is injected into the p-side.
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