Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15  Problem Set 5 Solutions
Problem 3
a.
R = rnp
= 10 10 (1.8 106 )2 cm3 s
= 324cm 3 s 1
1
cm
3
b.
Each photon is 1.42 eV approximately. Power = 460 eV/cm3 s
1.6 10 19 Joule ) Power = 7 10 17 W/cm3 .
1
. One eV =
c.
2 m piece of GaAs will emit 1.4 10
2
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15  Problem Set 2 Solutions
Problem 1
a. Light produces excess carriers at surface.
intrinsic so n = p.
Area under curve decreases due to recombination. Curve spreads due to diusion.
@n
@n
=
+ diusion + R&G
@t
@t drift
@n
@t drift
= 0: No elds
2
Diusio
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15  Problem Set 4 Solutions
Problem 1
a.
b.
The suggested way of looking at this problem was to realize that photocurrent
is proportional to the light intensity. The intensity ration will then just be the
ratio of short circuit currents.
ISC,illum
3.1A
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15  Problem Set 6 Solutions
Problem 1
a.
1.55 m ! band gap of active region = 0.8 eV.
Choose active: (AlGa)Sb with 10% Al
cladding: (AlGa)Sb with more Al, eg 20  40 % to avoid too much mismatch
while having higher Eg Mention refractive index too.
Subt
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15  Problem Set 7 Solutions
Problem 1
a.
Amperes Law says:
Z
Hdl = ni
5Hcore + Hgap = 104
Force we need to exert is 2 104 N.
Max. magnetization of car = Bs = 1T. B = 0 (H + M ).
So:
1
1
Ms =
Bs =
A/m = 0.08 107 A/m or800kA/m
0
410 7
Force on car is: 0
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15  Problem Set 3 Solutions
Problem 1
a.
Forward Bias: e
injected into p side, leads to high e concentration near
depletion region, decays away with characteristic length Ln (10s of m) slight
reduction of electron concentrtion near depletion region of
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15  Problem Set 1 Solutions
2.
a. (i)
(ii)
b.
p
Nc Nv exp
Eg
2kT
For Si: ni = 1010 cm3, Eg = 1.12eV
ni =
Therefore:
For Ge: ni
p
Nc Nv
= 1010 exp(1.12/2 0.0258)
= 2.167 1019 cm 3
= 2.167 1019 exp (0.67/2 0.0258)
= 6.1 1013 cm 3
This is 6000 times larg
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
Sample Exam 1  Solutions
Problem 1
a.
i.
ii.
b.
p
Nc Nv exp
Eg
2kT
For Si: ni = 1010 cm3 , Eg = 1.12eV
p
Therefore: Nc Nv = 1010 exp(1.12/2 0.0258)
= 2.167 1019 cm 3
ni =
For Ge: ni
= 2.167 1019 exp (0.67/2 0.0258)
= 6.1 1013 cm 3
1
This is 6000 times la
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
2006 3.15 Exam 1
Problem 1
= 105 s, D = 40cm2 s
1
a.
Concentration at x = 0: 1010 /cm2 s
ated carriers.
1
104 /cm
105 s = 109 /cm3 of photogener
Therefore:
p = 1018 + 1049 1018 cm
n = 10 + 10
2
L=
b.
p
Dt =
p
40 10
49
5
10 cm
90
p
= 2 10
1
3
at surfa
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
Sample Final  Solutions
Problem 1
a.
Material is multidomain.
Domains in Co line up with caxis. Hoop shape will depend on angle between
coaxis and eld.
b.
Add a linear term mu0 H.
c.
Assume thermal instability when kT energy needed to reverse particle.
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
3.15 Exam 2  2006
Problem 1
a.
Gate at zero volts: as VD becomes positive it puts the junctions into fwd bias
! large current ows to D.
For negative VD , junctions are in reverse bias, can get pincho.
b.
A positive VG puts the jn in to reverse bias init
Electrical, Optical & Magnetic Materials and Devices
MSE 3.15

Spring 2013
Sample Exam 2 Solutions
Problem 1
a.
Material masked in black is oxide (SiO2 )
VG negative: initially electrons depleted from S/C, eventually goes into inversion (becomes p type). Metal gate has depletion of electrons close to interface.
VG positive: elec