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Unformatted text preview: Georgia Institute of Technology
School of Electrical and Computer Engineering ECE 304OD&E: Microelectronics Exam 11 NOTE: Please show all your work to obtain partial credit. The exam is closed book and notes.
You may use four handwritten sheets of formulas. NAME Grade: 3 a i 041‘; (\S Problem 1 (15 points) The energy band diagram for a Si step junction with cross sectional area of 10‘4 cm2 maintained
at room temperature (T = 300 K) with a pside doping of NA = 2 x1014 cm'3 and an nside doping
of ND = 3.5 x 1015 cm'3 is shown in the‘ﬁgure below. The other physical parameters are Dp'=11.9 cmz/s, DN=35.l cmz/s, Lp=3.5x10'3cm, and LN=5.9 xlO'3 cm. For Si at room temperature, ni =
1010 cm'3. V (5) a. Find the builtin voltage (Vbi) under equilibrium condition. \I ‘, El , .V
bi 1‘“ “L . lo g (5) b. What is the applied voltage (if any) to the diode? f Tbé loam} benlma «Crum {3 To It .93 i‘vbi"VpAio‘lci get/Mam a (5) b. What is the diode current? 1 __ I...
is: [IL—f PE? Problem 2 (10 points) Draw qualitatively the energy band diagram in an n—ip—i Si device at T = 300 K. Assume that all
regions are long, and the concentrations of donors and acceptors (in the n and p regions,
respectively) are ND = NA = 5x1015 cm'3. For Si at room temperature, ni = 1010 cm'3. Iv. t
(EF‘EAhsM :(E,~..£F) s 2441'!“ Sﬂobiasqev swLL foto Problem 3 (25 points) L Find and plot the transfer characteristics (vO versus vs) for the following circuit. Assume both
diodes are ideal. 190m check: Dz, oFFﬂVDzK" 0
\IDL = *Vo : £v5<o:7Vs>,°
3 D3 OFF#530,
4’93— zzvo =9 v93:2.v0=z..§:\g<o wvgewé? Cases 30AM, Di 0N , Th 4mm CaJ€S mus+cww' ,.;\,e:mpour<wn:mmm H\L remairtédj YAAJQ 3C \./_3‘ Le.) Problem 4 (20 points) Design a fourresistor bias network for an npn transistor (as shown in the ﬁgure below) to obtain IC = 200 HA and VCE = 3 V if Vcc = 12 V and [SF = 75. Assume that in the active region VBE =
0.7 V. You need to ﬁnd RC, RE, R1, and R2. Assume that the voltage drops across RC is twice the
voltage drop across RE and that the current in R1 is 13 times the base current (11 = 13 IE). in ((322) 6'8) (30?: Vcc=+2v ‘1‘ RC Icﬂcw REYES 3 R3 1c
RCI¢+ 12E IE2; llv3zc‘ I; ‘1' RC 1‘ z 7' RE 15 =7 325 IE ‘1 =2 Pale‘3‘“ R' R; éﬂé IC :1 I6: Alecia/Sf} IE; 1000*?‘3, Loos7.5 LZDZ~7MA i3; 7‘5
aﬁcsizgokﬁ RE; 3 *tqqugkﬁ
J
01' 0.202. 7S '
:9 mi =a7+3 :35? .. .3
i :‘l,5.7_3
kVL In ; till»? Enzvéuﬂz :17 £22.12 KVL‘M ‘Qr®: “RII(+V6£+ REEEZQ 1 Problem 5 (30 points)
For the BJ T ampliﬁer circuits shown below, ﬁnd:
(10) a. DC equivalent circuit and Q point (1c and VCE). (6) b. AC equivalent circuit, ampliﬁer conﬁguration (CE, CB or CC), and small signal model
parameters. (4) c. Voltage gain (VD/Vs). ‘ (3) (1. Input resistance R,» as speciﬁed in the circuit. (3) e. Output resistance R0 as speciﬁed in the circuit (4) f. Current gain (lo/Ii) as speciﬁed in the circuit. For the transistor, assume that VEB = 0.7 in the active region, and BF = [30 = 100, and VA = 00. Other parameters are R]: 100 KQ, R2 = 300 KQ, R3 = 22 KS2, R5: 3 KS), R6: 10 K0, R5 = 2
KO, R7 = 100 K0, and VCC =12 V. ' +vcc= 12v kVLi too
In 3
3 3: 7g 1— 0,1.4.[3 (lacttl13 :: 07+l38313e; Igzlo‘ﬁéﬂz‘n' W 'IC. :0466 MIA , l<VL in Mpg): \Z=22x]'c+\/C5.H3,dot 13 » tzsstgz—wce QlVeezéﬂ , WM f \ Extra work on Problem 5 B : Ac €11 awakesxlf circuM'i R :: “:3 “mi;\~c\§!d‘
Hx 12mm 7“; 121:3 [03*2}: \ﬁa'ﬁYﬁ‘ I {loot‘%‘°3F\Aa.q7q=_gﬂz vaW’:: W ‘3." W g/ ’n°§§f'°.$3 a 75 M r ‘WW
3‘ 344 k“; RC =— Emits?ng {Kay2:60.714} } E” 43 l A > \
)R‘“‘Vn+(Po+\)R§” BONUS Problem (7 points) In the circuit shows below, ﬁnd the collector currents in both transistors (Ia and leg), and the
current in the Zener diode (Iz). Assume both transistors have VBE = 0.7 V and B}: = 100, and the
Zener diode has Vz = 4.5 V and R2 = 0. This combination of two transistors is called a
Darlington pair, and has several applications in ampliﬁer design. For this problem, you can
assume that both transistors are in the active region, and the Zener diode is in the Zener regime. kv L {A Maﬁa; ’43 = o~?+o.7+m 1E2, »4 _ . WAM~¢mqquvav «a. e m. .~ .
. appendix l , PHYSICAL'CONSTANTS
' .AND CONVERSION g
FACTORS‘L Avogadro's number NA = 602 X IO” molecnles/molc
Boltzmann's constant k = 1.38 x 10—23 l/K
. = 8.62 x 10" eV/K
Electronic charge (magnitude) q = [.60 x 10‘” C
Electronic rest mass m" = 9.11 x 10‘“ kg
Pemtittivity of free space 60 = 8.85 X 10’” F/cm
= 8.85 x 10‘” F/m
a! Planck's constant I: = 6.63 X 10'” 15
= 4.14 x 10‘” cVs
Room temperature value of [J [J = 0.0259 eV
Speed of light c = 2.998 X 10I0 cm/s
Preﬁxes:
1A(angsxmm) = 10‘I cm milli~, m. = to"
1 pm (micron) = 10" cm / micro, u» = 10'6
1 mil = IO” in. mm, n = 10"
2.54 em = 1 in. pico—, p— : IO"2
leV = 1.6 x 10“": my. k = 10’ mega, M» = l0ls
giga—, G~ = IO9
A wavelength A of l um corresponds to a photon energy of 1.24 eV.
' \ {Since an is used as theunit of length for many semiconductor quantities, caution
must be exercised to avoidunit errors in calculations. When uSing quantities involving
length in fonnulus which contain quantities measured in MKS units, it is usually best to
use all MKS quantities. Conversion to standnnd semiconductor usage involving cm can be
accomplished as a last step. Similar caution is recommended in using 1 and (N as energy
units. ...
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 Fall '07
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