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20f7
Problem 1 of 4: Answer each question briefly and clearly. (5 pts each, 40
total)
1.1 Assume a diffused resistor with a fixed thickness. What happens to the Sheet
Resistance (Rs) if the doping level doubles (assuring that there is only 1 single
dopant and that mobility does not change)?
. stays the same
doubles
is reduced by 50% Oo: apn viz Wo
_ none of the above. My doub los oo oubles
+ ers duced hy F2%
1.2 Find the ābuilt-in potentialā $3 of a junction that has N, |= 10*fem? and Np=10"emā,
bo Koa
Pa > Gay x T= Mom"? = Yop Wy 7 ED Ome
: A=
Pps = 6OAV 6 2-3E0mY Ya te
lusivy the 60mV rb)
1.3 Place check marks where appropriate to indicate the comrect-region of upecation,
assuming that Vin=-Vap= 1V.
Transistor [Ves [Vos | Vas on Triode Saturation_|
NMOS 2 2 9 Be x
NMOS 2 05 a x
FM0s 2 2 8 x
PMOS 2 05 [8 x
1.4 Chaose the most appropriste answer.
āA nceatively charged ādepletionā region in doped Si is characterized by.
an abundance of electrons and negative ions
ā100 many electrons, too few holes ZL
Ā«severe imbalance between negative fons and holes // jou, et owrly
the complete depletion of charge density have Sxed, neyarioe
/
lors )
oxce
3of7
15 We have a 2-terminal device that accumul i
n lates charge as a function of volta
according to the equation: Q = SV +3V 42 , where Q is in pico Cb and V is in volte
Find the smell signal capacitance of this device in pF at 2V. :
ae eo Co
Cr: - fas. Pleo
ovā C Ā° Ā„+3) v
fr very IB. 23 roth _
by yo 23 v = Up
1.6 Write the expression of a transfer function that corresponds to the Bode plot drawn.
eo et 1
; -to
ooo (I+ jwhoĀ®) a A - Bike
ee
POLE Ā„
(ir 22) (W38) ull ir
to?
jo? jo* Io? on
20f055 (1000) = B08! ted free)
1.7 The value of a voltage source is 5 + 0.02cos(10Ā° + 45Ā°) Volts, Please complete the
list below:
ve 3 Ce shynol
ne 0.07005 (tt us) sel siyuae
- 5 cots Got eas) total siyacā¬
Vee
1.8 For the n-channel MOS transisior shown below, please consider the two ends of the
channel (near the source and near the drain) and indicate whether or not each musl be
inverted so that this device is in saturation
Ā© This end is inverted
This end is NOT inverted
we This end is inverted
This end is NOT inverted
4of7
Problem 2 of 4: Answer each question briefly and clearly. (20 points)
The C/V plot below was taken by āscanningā the gate-to-buik voltage of an MOS
structure. We know (from other measurements) that dy+ = 0.55V and Cox = 2.36F/pm?.
Also, ā¬ = 8.85 10 Flom, Gox = 3.9 ā¬9 and 83) 7 11.7 Fe āUse the graph to calculate:
accumulation es inversion
Minima
Clq" 03
Vo OMY Ont qn O8Ā„
a | F
āa { Figg on
2.1 The oxide thickness tox.
expression _value & unils
7 BR WRI F885 1 Fern 0.0 Wyn
Cox Cox DRONES pry #
2.2 The maximum depth of the depletion region (What is the Vag voltage that yields the
maximum depth of the depletion region?). i ' \
i. typ to op eto DT
Cam fox (Cee Gept ā min Cox
Ć©si
Cool = Ces b
Xdueay
expression value and units
[= bi (Ze - ZL): (sn &) | 6.105 pm
Van when maximum depletion depth is achieved = O6Y @ Show f figure
Sof?
2.3 The doping concentration of acceptors (N,) in the channel
~O.42
ow,
Voss - (Guts) =? fe* ~Vex - Po ve
GOmif
p= bon Ā£04. May Mae yl?
to" Jem
expression _ value and units
~ at VERT ent)
Ne Wi Ogee | 10 Lan |
2.4 The charge density of the inversion layer (in Cb/emt") when Ve is 2V.
Above Vin, sttucrute acta as a brnear lapacitet vy
value = Cox.
Cont 4g 2 Aa: Log AV =-Cow (2Ā„-Ven )
Ineprvlos charge vobage aboe
tlneshold
expression value and units
2
a S/H
foe
Cox C Voe- Ven) 3.27 $cb/y
f
|
she tavenion
layer (& nade ot elections