ELEN30011 Electrical Device Modelling Page 1 of 6
THE UNIVERSITY OF MELBOURNE
Department of Electrical & Electronic Engineering
Semester 2 Assessment — 2022
ELEN30011 ELECTRICAL DEVICE MODELLING
Reading time: 30 minutes
Writing time: 180 minutes
Submission time: 30 minutes
This examination paper has 6 pages
Authorized materials:
1) A device or computer running Zoom, with working video camera and microphone.
2) Any hardcopy material and/or electronic media stored on an offline PDF reader.
3) Any paper, pens, pencils, rulers, and any dedicated handheld calculator (e.g. FX-100).
4) A mobile phone or tablet, for use as a scanner during submission time.
Instructions to students:
1) This is an open book and Zoom supervised exam. It contains four (4) ques-
tions, each with multiple parts. A maximum of 100 marks is available, with 25 marks
allocated per question. Question parts may be worth unequal marks.
2) Your answers must be handwritten on paper and scanned. Writing with an
electronic device is not permitted.
3) Your answers must be entirely your own work. Assistance, collaboration, col-
lusion, or plagiarism is not permitted, by any means.
4) During reading time: Writing is not permitted, in any form. You may print the
exam paper. Alternatively, you may download the exam paper to your PDF reader,
which must then be disconnected from the internet.
5) During writing time: Attempt all questions. Write your answers using pen / pencil
on paper. Commence each question on a new page. Ensure that all your handwritten
pages, hardcopy material, and your PDF reader remain visible to supervisors at all
times. Interaction with your device or computer running Zoom is not permitted, unless
directed by supervisors.
6) During submission time: Cease writing immediately. Assemble your pages in ques-
tion number order. Scan and combine them into a single PDF file, using your mobile
phone or tablet. Check that all pages are included and clearly readable. Submit your
PDF file using the Canvas Assignment window corresponding to this exam. Confirm
your submission with supervisors.
(OVER)
ELEN30011 Electrical Device Modelling Page 2 of 6
Question 1 (25 marks)
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charge density
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Solid cylinder
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⇢ = ⇢(r)
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⇢(r)
Figure 1: Infinitely long solid cylinder and
supported charge density (Question 1).
An infinitely long solid cylinder of radiusR (units m) is
orientated along the z axis, with circular cross-section
as shown in Figure 1. The surrounding medium is a
perfect insulator of permittivity (units F m−1) which
carries zero net charge, i.e. is electrically neutral. The
charge density ρ(r, φ, z) = ρ(r) in the interior of the
solid cylinder and in the surrounding insulator is given
in cylindrical coordinates (r, φ, z) by
ρ(r) =

λ
pi R2
[
exp
( r
R
)
− 1
]
, r ∈ [0, R],
0, r > R,
(1)
with units C m−3, in which λ > 0 is a constant param-
eter. This charge density is shown in Figure 1, noting
in particular that it is independent of φ and z.
(a) By evaluating a suitable surface or volume integral involving the charge density (1), show that
the total charge per unit length stored in the solid cylinder is λ (with units C m−1).
[HINT:
∫ R
0
r exp( r
R
) dr = R2.] (6 marks)
(b) Using Maxwell’s equations, show that the electric field E at a point P = (x, y, z) outside the
cylinder, at a distance r > R from its axis (see Figure 1), is given in rectangular coordinates by
E = E(x, y, z) =
λ
2pi
[
x
x2 + y2
xˆ +
y
x2 + y2
yˆ + 0 zˆ
]
V m−1. (2)
Include all working in your answer, along with a clear statement of which of Maxwell’s equations
are applied and how. (6 marks)
(c) By evaluating appropriate partial derivatives, compute the divergence of the electric field E of
(2) at P. What is the significance of your answer, if any? (6 marks)
(d) Is the electric field E of (2) conservative? Justify your answer via a suitable computation and
explanation. (7 marks)
(OVER)
ELEN30011 Electrical Device Modelling Page 3 of 6
Question 2 (25 marks)
A very long solenoid, of length ` (units m) and cross-sectional area A (units m2), is wound with two
separate, insulated windings as illustrated in Figure 2, using n1 and n2 turns per metre respectively.
Currents i1 and i2 flow separately in the two windings, corresponding to the voltages v1 and v2 as
shown. The material used for the core of the solenoid has permeability µ (not labelled in the figure).
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i1
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i2
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--
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`length
area
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A
magnetic field
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H
Solenoid
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(turn
s)m
1
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zˆ
Figure 2: A very long solenoid (Question 2).
With the aid of suitable arguments, calculations, and diagrams where appropriate, determine whether
each of the following statements is true or false:
( Unjustified answers of true or false will receive zero marks. )
(a) In the infinite length case ` =∞,
(i) Ampere’s law can be used to show that the magnetic field H in the core of the solenoid is
given by
H = (n1 i1 + n2 i2) zˆ A m
−1 ; (7 marks)
(ii) The total magnetic flux out through the left-hand end of the solenoid core is given by
Φ = µ (n1 i1 − n2 i2)A V s . (3 marks)
(b) In the finite length case ` <∞,
(i) Faraday’s law can be used to show the voltages v1 and v2 across each of the two windings
are approximately related to the currents i1 and i2 flowing in them by
v1 = (µn1 `A)
di1
dt
+ (µ
√
n1 n2 `A)
di2
dt
V,
v2 = (µ
√
n1 n2 `A)
di1
dt
+ (µn2 `A)
di2
dt
V ;
(5 marks)
(ii) Faraday’s law can be used to show that the A.C. voltage gain v2/v1 provided by the solenoid
is positive and depends only on the turns ratio n2/n1; (5 marks)
(iii) [HARDER] The energy stored in the solenoid is proportional to (n1 i1 + n2 i2)
2 (units J).
(5 marks)
(OVER)
ELEN30011 Electrical Device Modelling Page 4 of 6
Question 3 (25 marks)
TX
RX
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`
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`
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parallel

wires
M O
A
B
Figure 3: Comms sys-
tem (Question 3).
A digital communication system shown in Figure 3 consists of a transmitter
TX connected to a receiver RX via two cascaded identical spans of parallel
wires, each of length `
.
= 20 m, and connected at M. A third span of length
x of the same type of parallel wires is also connected at M, but with no load
(i.e. an open circuit) connected to its other end at O.
Each pair of wires consists of two identical solid cylindrical copper cores of
radius a
.
= 0.25 mm, separated by a high density polyethylene insulator of
thickness w
.
= 0.25 mm. The cores may be assumed to be lossless, while the
insulator has relative permittivity r
.
= 2.6 and negligible conductivity. The
permittivity of free space is nominally 8.8542×10−12 F m−1. The capacitance
per unit length for each pair is given by
Ĉ
.
=
pi 0 r
log
(
2 + w
a
) F m−1. (3)
A rising edge transmitted by TX has a duration of tSW = 1 ns, and requires
tD = 107 ns to propagate from A to M. The output impedance of TX is
negligible.
(a) Compute numerical values for the capacitance per unit length, induc-
tance per unit length, and characteristic impedance of the parallel wires
used in the configuration shown. (5 marks)
(b) By computing numerical values for the knee and ringing frequencies involved, i.e. fKNEE and
fRING, show that the signal transmitted by TX is likely to be distorted without an appropriate
termination strategy. Verify your answer by computing the length of the rising edge, i.e. `SW.
(5 marks)
(c) For sufficiently large x > 0, it is impossible to eliminate distortion by simply connecting a resistor
at O and designing an appropriate input impedance for RX at B. Explain why. Support your
answer with numerically evaluated reflection coefficients at TX, RX, and M, and a discussion of
the reflections generated and their expected amplitudes. (8 marks)
(d) Suppose O is left unterminated, i.e. as an open circuit as shown in Figure 3, and that the span
M-B is correctly terminated. It may be shown that the reflection coefficient ρM at M, as seen by
the signal propagating from the transmitter TX to M, varies as a function of frequency f (units
Hz) and length x `SW (units m) according to
|ρM(f, x)| = 1√√√√√1 + 1( f
fKNEE
)2(
x
lSW
)2 . (4)
(i) Interpret and discuss the physical ramifications of (4). (3 marks)
(ii) By assuming that the frequency content (i.e. spectrum) of the signal transmitted by TX
is strictly limited to the interval [0, 2 fKNEE] (units Hz), find the maximum length of the
unterminated span M-O that would ensure that the magnitude of the reflection coefficient
ρM is at most 0.05 for all frequency components of the signal. (4 marks)
(OVER)
ELEN30011 Electrical Device Modelling Page 5 of 6
Question 4 (25 marks)
An NPN BJT is operating in the forward active mode, due to the application of a constant base-
emitter junction voltage VBE > 0 and a constant base-collector junction voltage VBC < 0. This gives
rise to a hole distribution in the p-type base that is approximated by
pp(x) = p¯p − n¯p + (n¯p + ∆nVBEp )
(
1− x
Wb
)
(cm)−3, (5)
∆np
VBE = n¯p
[
exp
(
VBE
VT
)
− 1
]
(cm)−3, VT
.
= k T
q
V, (6)
for all x ∈ [0,Wb], in which Wb Lnp is the base width, Lnp is the diffusion length of electrons in the
p-type base, ∆np
VBE is the excess electron concentration injected at the emitter end of the base, and
p¯p and n¯p denote (respectively) the equilibrium concentrations of holes and electrons in the p-type
base material with no bias applied. [Note that p¯p n¯p = n¯
2
i , where n¯i is the equilibrium concentration
of electron-hole pairs (EHPs) in the corresponding intrinsic semiconductor material.]
Base
Collector
Emitter
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p¯p +n
VBE
p
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p¯p n¯p
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pp(x)
Hole
concentration
AAAIi3icfZXNbhMxE Me35aMhUGjhyGVFVAkhVCUFBEIc2gJSD6WUijRps1Hl9XoTq157ZTtNUstvwRXei7fB+5UmdqilSOPZn2fG/3HsMCVYyGbz78rqnbv37q/VHtQfPlp//GRj8+mpYCMOURsywng3BA IRTFFbYklQN+UIJCFBnfDyc/a9c4W4wIz+lNMU9RMwoDjGEEjjOg+EiFVHq1BfbDSa2818+K7RKo2GV47ji821yyBicJQgKiEBQvRazVT2FeASQ4J0PRgJlAJ4CQaoZ0wKEiT6Ki9Z +1vGE/kx4+ZHpZ9751cokAgxTUJDJkAOhf0tcy79lnm4iMVCfhUm2o4Qiqmub/n2WhBOrNJl/KGvME1HElFYVB6PiC+ZnwnqR5gjKMnUGABybDbvwyHgAEoje70eUDSGLEkAjVQQA u2bYQw0wNRIxcFU60UGVQwys4pYRMSAGkapQKKJ5InK5jn0BZlOcPTN7ON7ijiQjL9SAeCDBFNtgg2C15l1GwgmFWis20IKCWSJ5ub/0IIrEGsfUuheqz/biZBTglSjpW1JIpTgUh QpVBAbcVVLqx0HPMmpPGDWyzBUJ9pRL40ko0j3dorMebQgzY4tIH6efn6aJbHLWRIgKldGsyVWVpSKm+LMBBNG7cAsLfSYbQACskwPlg6NmpV0YxwhM8/cBWzRsFMmnoUVkJs/vcPt aYfac6l9l9p3qTOXOnOptku1XerI6eqRCx26oQ5d6mCRyvQ9cKmutiJ1XebcZs4dJmTaynXonsbxsOz5XCeXtHwsFzGJSbT0v8LIDcjMzZ89DCW3CF7F1WGbCRsvCziieC53mMzYw JTqlyduyRpxsyi/k5OqikV0kh3lWVOyhWrixJu60NSBrl3o2t12hI0og/nGmMqM0+3wiBMLy102N+AgqjgKQgKc64IldjrjcZUwT3F1pxilLsqTFRf3iXmeW/Zj7BqnO9utd9vNH2 8bu/vlQ13znnsvvJdey3vv7XoH3rHX9qBHvV/eb+9Pbb32pvax9qlAV1fKNc+8hVH7+g8gNwaU
Wb
AAAIgnicfZXNThsxEMcXSkuafkF77GXVCKlCFcrSVu2hB6A9c KAUUEMC2Qh5vd6Nhdde2Q5JsPYJem0frm9T71dI7BRLkcazv5mx/+PYQUqwkO3235XV B2sPH603HjefPH32/MXG5stzwUYcog5khPFeAAQimKKOxJKgXsoRSAKCusH11/x79w ZxgRn9KacpGiQgpjjCEEjtOm1fbbTaO+1iuLbhVUbLqcbJ1eb6tR8yOEoQlZAAIfpeO 5UDBbjEkKCs6Y8ESgG8BjHqa5OCBImBKlaauVvaE7oR4/pHpVt45yMUSISYJoEmEyCH wvyWO5d+yz1cRGKhvgqSzMwQiGnW3HLNWBBMjKXL6PNAYZqOJKKwXHk0Iq5kbq6jG2K OoCRTbQDIsd68C4eAAyi12s2mT9EYsiQBNFR+ADJXD22gGFMtFQfTLFtkUM0gPauJRU TEVDNK+RJNJE9UPi+gb0h3gqPveh8/UsSBZHxb+YDHCaaZThb773LrPhBMalBb96UUE sgKLcz/oSVXIsY+pMj63mC2EyGnBKmWl5mShCjBlShSKD/S4iovU7sWeFZQRcK8l0G gzjJLvTSUjKKsv1tWLrL5aX5sAXGL8vPTvIi5nCUJwioynIUYVVEq7hanJ5gwaiZmaa nHbAMQkGV6sHSo1aylG+MQ6XnuLmGDht2q8CytgFx1bW4/s6h9mzqwqQOburCpC5vq2 FTHpo6trh7b0JGd6simDhepXN9Dm+plRqaezVyazKXFBCwzah3Zp3E8rHo+18klLR/L RUxiEi79rzByBzJ94efvQcUtgjdRfdhmwkbLEo4onqsdJDPW10t1qxO3JEbcBRV3clK vYhGd5Ed51pQ8UE2sfFMbmlrQrQ3d2tsOsRYlnm+MXpl22h0ecWJghcvkYg7CmqMgIM C6LlhiltMeWwkhovpO0UpdVScrKu8T/Tx75mNsG+e7O97Hnfbph9beQfVQN5zXzhvnr eM5n5w959A5cToOdJDzy/nt/GmsNbYbXuN9ia6uVDGvnIXR+PIP/5gCGQ==
0
Base current
AAAIi3i cfZXdbtMwFMez8bFSGGxwyU1ENQkhNLUDBEJcbAWkIY0xJrp2W6rJcZzUmmNHtru2s/wW3MJ78TY4adK1dsFSpeOTn885/h/XDjOChWw2/6ys3rp95+5a 7V79/oP1h482Nh+fCDbkEHUgI4z3QiAQwRR1JJYE9TKOQBoS1A0vP+bfu1eIC8zoDznJUD8FCcUxhkAa11kgRKy+aNXWFxuN5nazGL5rtEqj4ZXj6GJz7 TKIGBymiEpIgBDnrWYm+wpwiSFBuh4MBcoAvAQJOjcmBSkSfVWUrP0t44n8mHHzo9IvvPMrFEiFmKShIVMgB8L+ljuXfss9XMRiIb8KU21HCMVE17d8ey 0Ix1bpMn7XV5hmQ4konFYeD4kvmZ8L6keYIyjJxBgAcmw278MB4ABKI3u9HlA0gixNAY1UEALtm2EMlGBqpOJgovUigyoGmVlFLCIioYZRKpBoLHmq8nk BfUKmExx9Nfv4liEOJOMvVAB4kmKqTbAkeJlb/wPBuAKN9b+QQgJZooX5L3TKTRFrH1Lo81Z/thMhJwSpRkvbkkQoxaUoUqggNuKqllY7DnhcUEXAvJdh qI61o14WSUaRPt+ZZi6iBVl+bAHxi/Tz0zyJXc6SAFG5MpotsbKiTNwUZyaYMGoHZtlUj9kGICDL9GDZwKhZSTfCETLz3D2FLRp2y8SzsAJy1XW5Pe1Qe y7Vdqm2S5261KlLdVyq41KHTlcPXejADXXgUvuLVK7vvkv1tBWp5zJnNnPmMCHTVq4D9zSOBmXP5zq5pOUjuYhJTKKl/xVGbkBmbv78YSi5RfAqrg7bTN h4WcAhxXO5w3TGBqZUvzxxS9aIm0XFnZxWVSyi4/woz5qSL1RjJ97EhSYOdO1C1+62I2xESeYbYyozTrfDQ04srHDZXMJBVHEUhAQ41wVL7XTG4yphnu LqTjFKXZQnK57eJ+Z5btmPsWuc7Gy33mw3v79u7LbLh7rmPfWeec+9lvfW2/X2vSOv40GPej+9X97v2nrtVe197cMUXV0p1zzxFkbt81+qXQZm
IB

+
+
recombination current
AAA Ii3icfZXbbhMxEIa35dAQKLRwyc2KqBJCqEoKCIS46AGkIpVSKtKk7UaV1+tNrHrtle00SS2/BbfwXrwN3kPSrB1qKdJ49vPM+B/HDlOChWw 2/y4t37l77/5K7UH94aPVx0/W1p+eCDbkELUhI4x3QyAQwRS1JZYEdVOOQBIS1Akv97LvnSvEBWb0p5ykqJeAPsUxhkAa11kgRKy+arWnL9Ya zc1mPnzXaJVGwyvH0cX6ymUQMThMEJWQACHOW81U9hTgEkOCdD0YCpQCeAn66NyYFCRI9FResvY3jCfyY8bNj0o/986vUCARYpKEhkyAHAj7 W+Zc+C3zcBGLSn4VJtqOEIqJrm/49loQjq3SZfyhpzBNhxJRWFQeD4kvmZ8J6keYIyjJxBgAcmw278MB4ABKI3u9HlA0gixJAI1UEALtm2EM1 MfUSMXBROsqg6YMMrMpUUVEnxpGqUCiseSJyuY59BmZTnD0zezje4o4kIy/UgHg/QRTbYL1g9eZdRsIxlPQWLeFFBLIEs3N/6EFVyDWPqTQ5 63ebCdCTghSjZa2JYlQgktRpFBBbMRVLa22HPA4p/KAWS/DUB1rR700kowifb5VZM6jBWl2bAHx8/Tz0yyJXc6CAFG5MpotsbKiVNwUZyaYMG oHZmmhx2wDEJBFerB0YNScSjfCETLzzF3AFg07ZeJZWAG56rjcjnaoHZfadaldlzp1qVOXartU26UOna4eutCBG+rApfarVKbvvkt1tRWp6z JnNnPmMCHTVq4D9zSOBmXP5zq5oOUjWcUkJtHC/wojNyAzN3/2MJRcFbyKp4dtJmy8KOCQ4rncYTJjA1OqX564BWvEzaL8Tk6mVVTRcXaUZ03 JFqqxE2/iQhMHunaha3fbETai9OcbYyozTrfDQ04sLHfZXJ+DaMpREBLgXBcssdMZj6uEeYqnd4pR6qI8WXFxn5jnuWU/xq5xsrXZerfZ/PG 2sb1bPtQ177n3wnvptbz33ra37x15bQ961Pvl/fb+1FZrb2ofa58KdHmpXPPMq4zal3+yfAZn
IC
current
AAAIi3icfZXdbtMwFMez8bFSGGxwyU1 ENQkhNDUDBEJcbINJQxpjTHTttlST4zitNceObHdtZ/ktuIX34m1w0r Rr7DJLlY5Pfj7n+H9cO8oIFrLZ/Lu0fOfuvfsrtQf1h49WHz9ZW396 ItiAQ9SCjDDeiYBABFPUklgS1Mk4AmlEUDu6/Jx/b18hLjCjP+U4Q90 U9ChOMATSuM5CIRL1Vas9fbHWaG42i+G7RlAaDa8cRxfrK5dhzOAgRV RCAoQ4D5qZ7CrAJYYE6Xo4ECgD8BL00LkxKUiR6KqiZO1vGE/sJ4ybH 5V+4Z1foUAqxDiNDJkC2Rf2t9y58Fvu4SIRlfwqSrUdIRJjXd/w7bUg Glmly+RDV2GaDSSicFJ5MiC+ZH4uqB9jjqAkY2MAyLHZvA/7gAMoje z1ekjRELI0BTRWYQS0b4YxUA9TIxUHY62rDJoyyMymRBURPWoYpUKJR pKnKp8X0BdkOsHRN7OP7xniQDL+SoWA91JMtQnWC1/n1m0gGE1BY90W UkggS7Qw/4dOuAli7UMKfR50ZzsRckyQagTaliRGKS5FkUKFiRFXBVp tOeBxQRUB815GkTrWjnpZLBlF+nxrkrmIFmb5sQXEL9LPT/MkdjkLAs Tlyni2xMqKMnFTnJlgwqgdmGUTPWYbgIAs0oNlfaPmVLohjpGZ5+4J bNGwXSaehRWQq7bL7WiH2nGpXZfadalTlzp1qZZLtVzq0OnqoQsduKE OXGq/SuX67rtUR1uROi5zZjNnDhMxbeU6cE/jsF/2fK6TC1o+lFVMYh Iv/K8wcgMyc/PnD0PJVcGrZHrYZsImiwIOKJ7LHaUzNjSl+uWJW7BG3 Cwq7uR0WkUVHeVHedaUfKEaOfHGLjR2oGsXuna3HWMjSm++MaYy43Q7 PODEwgqXzfU4iKccBREBznXBUjud8bhKmKd4eqcYpS7Kk5VM7hPzPAf 2Y+waJ1ubwbvN5o+3je3d8qGuec+9F95LL/Dee9vevnfktTzoUe+X9 9v7U1utval9rH2aoMtL5ZpnXmXU9v4BwroGaQ==
IE
NPN BJT
AAAIgnicfZXN ThsxEMcXSkuafkF77GXVCKlCFcrSVu2hB6A9cKAUUEMC2Qh5vd6Nhdde2Q5JsPYJem0frm9T71dI7BRLkcazv5mx/+PYQUqwkO3235XVB2sPH603HjefPH32/MXG5s tzwUYcog5khPFeAAQimKKOxJKgXsoRSAKCusH11/x79wZxgRn9KacpGiQgpjjCEEjtOh1ebbTaO+1iuLbhVUbLqcbJ1eb6tR8yOEoQlZAAIfpeO5UDBbjEkKCs6Y8ES gG8BjHqa5OCBImBKlaauVvaE7oR4/pHpVt45yMUSISYJoEmEyCHwvyWO5d+yz1cRGKhvgqSzMwQiGnW3HLNWBBMjKXL6PNAYZqOJKKwXHk0Iq5kbq6jG2KOoCRTbQD Isd68C4eAAyi12s2mT9EYsiQBNFR+ADJXD22gGFMtFQfTLFtkUM0gPauJRUTEVDNK+RJNJE9UPi+gb0h3gqPveh8/UsSBZHxb+YDHCaaZThb773LrPhBMalBb96UUEs gKLcz/oSVXIsY+pMj63mC2EyGnBKmWl5mShCjBlShSKD/S4iovU7sWeFZQRcK8l0GgzjJLvTSUjKKsv1tWLrL5aX5sAXGL8vPTvIi5nCUJwioynIUYVVEq7hanJ5gw aiZmaanHbAMQkGV6sHSo1aylG+MQ6XnuLmGDht2q8CytgFx1bW4/s6h9mzqwqQOburCpC5vq2FTHpo6trh7b0JGd6simDhepXN9Dm+plRqaezVyazKXFBCwzah3Zp3E 8rHo+18klLR/LRUxiEi79rzByBzJ94efvQcUtgjdRfdhmwkbLEo4onqsdJDPW10t1qxO3JEbcBRV3clKvYhGd5Ed51pQ8UE2sfFMbmlrQrQ3d2tsOsRYlnm+MXpl22 h0ecWJghcvkYg7CmqMgIMC6LlhiltMeWwkhovpO0UpdVScrKu8T/Tx75mNsG+e7O97Hnfbph9beQfVQN5zXzhvnreM5n5w959A5cToOdJDzy/nt/GmsNbYbXuN9ia6u VDGvnIXR+PIPxkYCUQ==
h
AAAIgnicfZXNThsxEMcXSkuafkF77GXVCKlCFcrSVu2hB6 A9cKAUUEMC2Qh5vd6Nhdde2Q5JsPYJem0frm9T71dI7BRLkcazv5mx/+PYQUqwkO3 235XVB2sPH603HjefPH32/MXG5stzwUYcog5khPFeAAQimKKOxJKgXsoRSAKCusH1 1/x79wZxgRn9KacpGiQgpjjCEEjtOh1fbbTaO+1iuLbhVUbLqcbJ1eb6tR8yOEoQl ZAAIfpeO5UDBbjEkKCs6Y8ESgG8BjHqa5OCBImBKlaauVvaE7oR4/pHpVt45yMUSI SYJoEmEyCHwvyWO5d+yz1cRGKhvgqSzMwQiGnW3HLNWBBMjKXL6PNAYZqOJKKwXHk 0Iq5kbq6jG2KOoCRTbQDIsd68C4eAAyi12s2mT9EYsiQBNFR+ADJXD22gGFMtFQfT LFtkUM0gPauJRUTEVDNK+RJNJE9UPi+gb0h3gqPveh8/UsSBZHxb+YDHCaaZThb77 3LrPhBMalBb96UUEsgKLcz/oSVXIsY+pMj63mC2EyGnBKmWl5mShCjBlShSKD/S4i ovU7sWeFZQRcK8l0GgzjJLvTSUjKKsv1tWLrL5aX5sAXGL8vPTvIi5nCUJwioynIUY VVEq7hanJ5gwaiZmaanHbAMQkGV6sHSo1aylG+MQ6XnuLmGDht2q8CytgFx1bW4/s 6h9mzqwqQOburCpC5vq2FTHpo6trh7b0JGd6simDhepXN9Dm+plRqaezVyazKXFBC wzah3Zp3E8rHo+18klLR/LRUxiEi79rzByBzJ94efvQcUtgjdRfdhmwkbLEo4onqs dJDPW10t1qxO3JEbcBRV3clKvYhGd5Ed51pQ8UE2sfFMbmlrQrQ3d2tsOsRYlnm+M Xpl22h0ecWJghcvkYg7CmqMgIMC6LlhiltMeWwkhovpO0UpdVScrKu8T/Tx75mNsG +e7O97Hnfbph9beQfVQN5zXzhvnreM5n5w959A5cToOdJDzy/nt/GmsNbYbXuN9ia 6uVDGvnIXR+PIPQBcCYA==w
Figure 4: Hole concentration in the p-type base of an NPN BJT (Question 4).
(a) The first task is concerned with deriving (5).
(i) Under the stated bias conditions, the excess electron concentration in the p-type base is
known to satisfy the boundary conditions δnp(0) = np(0) − n¯p = ∆npVBE and δnp(Wb) =
np(Wb)− n¯p = −n¯p, in which np(x) is the electron concentration at position x in the base.
Explain the physical origin of these boundary conditions, and why the excess electron
concentration δnp(x) can be approximated for x ∈ [0,Wb] by the linear function
δnp(x) = −n¯p + (n¯p + ∆npVBE)
(
1− x
Wb
)
(cm)−3, (7)
provided that Wb Lnp . (5 marks)
ELEN30011 Electrical Device Modelling Page 6 of 6
(ii) Given δnp(x) in (7), argue in terms of charge redistribution times why the excess hole
concentration δpp(x) in the p-type base should satisfy δpp(x) ≈ δnp(x). (5 marks)
(iii) Using (i) and (ii), verify that (5) holds, and subsequently that
pp(x) ≥ p¯p − n¯p ≈ NA
[
1−
(
n¯i
NA
)2]
0, x ∈ [0,Wb], (8)
where NA is the acceptor ion concentration in the p-type material. (5 marks)
(b) The second task is concerned with estimating the resistance of the base, as seen by the base
current IB supplying holes to the device.
(i) Using (5), the conductivity σp(x) due to holes in the p-type base is given by
σp(x) = q µp
[
p¯p − n¯p + (n¯p + ∆nVBEp )
(
1− x
Wb
)]
(Ω m)−1. (9)
What is the corresponding conductivity σn due to electrons in the p-type base? Comparing
with (9), do you expect the resistance of the base as seen by holes to be significantly greater
than or significantly less than the resistance of the base as seen by electrons? Explain.
[HINT: Use (8).] (5 marks)
(ii) Using the conductivity (9), explain how the resistance of the base, as seen by holes entering
the base at the top of Figure 4, could in-principle be estimated via an integral over the
base region. [HINT: Parallel slices.] (5 marks)
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(END)

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