Download Old Test III with Answer Key - Microelectronic Circuits | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040B Microelectronic Circuits
Exam 3
April 12, 2001
Dr. W. Alan Doolittle
°
Solutions
Print your name clearly and largely:
Test Average 53.52174
Test Standard Deviation 16.32413
Broken down by question
Question] 12,3 4 = 5 6 7 8 9 1 1 14
% Score per Question 100 71.7 73.9 56.5 68.1 71 60.9 72.6 65.1 38.7 46.7 28.5
1st 25% 62.81159
2nd 25% 65.94203
3rd 25% 68.84058
4th 25% 28.52174
apes
06 08 OL os os Ov oe 0z
ooL
Number of Test Scores
N o > a
ab
7.) (B-points)
g.) Voltage, Current, Transconduc’
h.) Input Impedance should be Hi
i.) Output Impedance should be Hi;
Rk
i,
4 ‘out
e —
i, SR SR, Rain RS toe
Second 25% Short Answer:
8.) (10-points) Draw the cross-sectional view (view from the side) of a NMOS transistor
biased in linear and in saturation modes (two separate drawings). Label the source, gate,
drain, channel, substrate and indicate the doping type of the source, drain and substrate.
Linear:
Channel Inversion layer (n-type)
P-type Substrate
Saturation:
Vost Yn
S '———_- t+ D
—T os NF
\
9.) (15-points) Draw and label the energy band diagram of a NMOS Enhancement mode
Capacitor in equilibrium, depletion and deep inversion (3 drawings) labeling the fermi,
intrinsic, conduction and valence energies.
All regions are shown below (including accumulation, which was not asked for). In each
picture, the left most material is the metal, then the oxide, with the rightmost material
being the semiconductor.
Ec
Energy Ef a:
band ke Ef
diagram Ef be wie aE cot ey
ef ---—
Pm Ey Ef
Ev
Applied
dc Wo=0 Vg<0 Vz>Vg>0
voltage
Flat band Accumulation Depletion Inversion
Third 25% Problems (3° 25%)
10.) (10-points) Determine the voltage gain, input resistance and output resistance of the
following circuit. You may assume that the Op-Amps are ideal.
Rl Va
ini
Vout
= 0.5Vin
Va=Vin Rl
Rl+R2
Vb= (: Va =2Va
Rl
Ve=Vin
Ri Ri
Vout = -—Vb-—Ve
Ri Ri
Vout =-2Va—-Vin
Vout =—Vin—Vin
Vout
——=-2 VIV
Vin
RK, = ak |
Kour =O [Vino = V,2U20 |
Extra work can be done here, but clearly indicate with problem you are solving.
Vos, = 10 666.8 (10k)
2 333V DO Ves-Vp = 3-1 RT
r t '
een ot Saturanon 18 va lial,
Asbu my
Do current Sonece
2s Since rhe
vem * carreny Me
Cans tans
‘
GULeCS A ‘
- r be in $& Fura e107 «
meee wy Los,= 2mFt
TY a VK= amACik) = av’
Ves2 B33V- 2V
= L33Vv
G5, ~ VT a
2
Los3 Amit
mae -~ AOI2A Afv
? Ves, —Vr ~ (33-1 ° (
» a
| | =
l= SETI = GBBT pe OL
EG > IEG
fo, = 720, 000 $2.
Extra work can be done here, but clearly indicate with problem you are solving.
| Steph: Wen = 5s v
/ Vesa + Im VGsa (lla) 2 9m, V5, (« IIe)
Vour = Jra Vo52 (Glas)
Vous Vout ys 2
Th? eae VI 51
Vain Vo5% VW95, Vn
= (ams Ge! (Il) [ aE | @
Nout . 2 -94Gll fe) Jra(k Meo)
UU tam, ile)
Standard ee Rl
CF gain Standarl CY 5a
You could hove wiitten tAIS doi Th
1 PL
ate? A. “pg to
