Download Midterm Exam with Answers | Semiconductor Devices | EEL 4351C and more Exams Physics of semiconductor devices in PDF only on Docsity!
Midterm exam
EEE4351C.001U
Semiconductor Devices
June 15, 2015
Full Name:
USFID:
Time: 75 minutes
ฮต (^) r=11.9, ฮต 0 =8.85u 10 -12^ F/m q=1.6u 10 -19^ C
๐ = ๐ (^) ๐ exp[๐ธ๐^ ๐๐โ ๐ธ ๐น] ; ๐ = ๐ (^) ๐ exp[๐ธ๐น๐๐^ โ ๐ธ ๐]
Assume room temperature in all questions.
2 1 /^2
A D
s bi D A N N
(N N )
q
W ฮต^ V ยป ยผ
1 / 2
( )
A A^ ^ D
p s bi^ D N N N
N
q
x^ H^^ V^1 /^2
D A^ ^ D
n s bi^ A N N N
N
q
x^ H^^ V
๐ ๐๐^ (
๐๐^2 )
- In equilibrium @300k, the depletion width in the n region of a silicon pn junction is 10 times larger than the depletion width in the p region. If the doping density in the n region is 10^17 cm-3^ , find: a. The doping density in the p region (10 points) b. Using the graph below estimate the conductivity of electrons in the bulk n part and conductivity of the holes in the bulk p part. (15 points)
(a) Na.xp=Nd.xn ร xn=10xp ร Na=10Nd = 101E17 = 1E18 1/cm (b) ฮผn=700 cm2/Vs and ฮผ (^) p=70 cm2/Vs ฯn=qn ฮผn = 1.6E-191E17700 = 11.2 S/cm ฯp=qp ฮผp = 1.6E-191E18*70 = 11.2 S/cm
- The doping density across a silicon PN junction is N (^) A =10 16 cm-3^ and N (^) D=10 18 cm -3^. The junction is biased at 0.2 V in the Forward mode. a. Draw the band diagram of the junction (10 points) b. Draw the minority carrier concentration versus x. (10 points) c. Calculate p (^) n0 , n (^) p0 , p (^) n(xn), and n (^) p(-xp). (10 points)
p-type n-type (a)
(b)
(c) p (^) n0=ni 2 /Nd=100 cm -3^ n (^) p0 =ni^2 /Na=10000=1E4 cm -
p (^) n= p (^) n0exp(qV/kT) = 100 exp( 0.2/0.025)= 298095= 2.98 E5 cm-
n (^) p= n (^) p0exp(qV/kT) = 10000 exp( 0.2/0.025)= 29809500= 2.98 E7 cm -
Ec
Ev
EF
Ei EF
q(Vbi-V)
qV
Pn
np 0
np 0 *exp(qV/kT)
pn 0 *exp(qV/kT)
- The minority carrier distribution in the bulk n-side of two pn junctions (junction 1 and junction
- are shown in the plots below. N (^) D1, V 1 , L 1 , and JP1 are the doping level, applied voltage, diffusion length and diffusion current density from holes in junction 1, respectively. Similarly, ND2, V 2 , L 2 , and JP2 are the doping level, applied voltage, diffusion length and diffusion current density from holes in junction 2.
From the plots choose the correct answers for part A, B, C, and D. For each case write a line to justify your answer. Without the justification your choice will not be accepted. (20 points) A. a. ND1 > ND2 b. ND1 < ND2 c. ND1 = ND B. a. V 1 > V 2 b. V 1 < V 2 c. V 1 = V 2 C. a. L 1 > L 2 b. L 1 < L 2 c. L 1 = L (^2) D. a. Jp1 > Jp2 b. Jp1 < Jp2 c. Jp1 = Jp
(Junction 1) (Junction 2)
A. Since Pn0 in Junction 2 is lower than that in Junction 1, so Nd2>Nd1. Nd=ni^2/p (^) n B. Point T on the graph is pn0exp(qV/kT). Although pn0 in Junction 2 is lower than that in Junction 1, point T is higher in Junction 2. So V2 should be Larger than V C. The exponent is diffused to larger x values in Junction 1 than 2. So L1 must be larger than L2. Also if Nd2>Nd1mobility and diffusion coefficient in 2 is lower than 1 (see the graph from question 1) when D2<D1, L2is smaller than L1, too. (L=SQR(Dtau)) D. The diffusion current is:
The slope of the curve is much sharper in Junction 2 than 1.
x Point T
x Point T
