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UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences
Professor Oldham Spring 1999
EECS105 — Midterm 1
Thursday, 25 February 1999
Your name: ______________ ____________________
Your discussion TA: r Allan Chang r Lily Tam
- This is a closed book exam, but you may use your page of notes. - Please do all your work on the pages of this exam. Ask if you need extra paper. - Full credit will be given only when you indicate the source of your answer, such as a table, graph, or calcula- tion. - Please write your name in the above space - Special notes: 1. SOME GRAPHS AND FORMULAS ARE GIVEN AS APPENDICES TO THIS EXAM. BE SURE TO LOOK THESE OVER. 2. SOME PARTS OF THE EXAM ARE GRADED WITH NO PARTIAL CREDIT. They are noted. You may wish to double check your answers on those parts. 3. ONCE IN A WHILE SOME EXTRA CREDIT IS POSSIBLE FOR CLEVER INSIGHT. Again, these places are noted. But we will not answer questions about these problems. Just be very clear in your work.
SCORE
Problem 1 (20 pts.) Problem 2 (25 pts.) Problem 3 (30 pts.) Problem 4 (25 pts.) TOTAL (100 pts.)
first last
Yes, I have looked these over. (Check box)
Problem 1 (20 pts.)
a. [No partial credit] In a certain process, a thick layer of n-type silicon (doping = is created over a p-type substrate. It is to be used for the purpose of making integrated circuit resistors. What is the sheet resistance of this layer? (Units must be .)
b. Using the layers of (a), above, you need to make a resistor with value of. It is wide. What must its length be (ignoring contact effects)?
c. Someone properly points out to you that the layer in part a), though it is physically thick, is electri- cally somewhat thinner, because there must be a depletion layer at the n-p interface. (You are to ignore this in part a.) Suppose the doping in the p region is also (but acceptors instead of donors). At zero applied voltage between the n and p regions (i.e., in thermal equilibrium), just what is the net electrical thickness of the n-region? (Thickness minus depleted portion.)
2 μm 2 × 10 15 cm 3 ⁄
Ω ⁄square
200 KΩ 5 μm
2 μm
2 × 10 15 ⁄cm 3
( μm)
Problem 2 (25 points)
The structure below is made with 4 masking steps.
gate oxide
/
source (^) n+ polysilicon gate interconnect (^) /
drain
bulk interconnect
oxide
Four-mask layout and cross section of an integrated n-channel MOSFET.
a. Name the masks (use the terminology within the figure in the names, if possible).
b. For each mask name point to two edges in the cross-section which come about as a result of the application
of that mask (photolithography and etching.) The gate edges are identified as an example.
C. If this structure represents the n-mos transistor in my^ 450MHz^ pentium processor (made with a so-called
“quarter-micron process,” can you give me approximate values for the size of the following:
c.1)
c.2)
c.3)
c.4)
c.5)
L
L (^) dl.ff
gate oxide thickness
metal thickness
n+ layer thickness
d. Why is Ldiff so much bigger than L? Is this not wasted space? [Extra credit possible]
e. If this is an n-channel MOS transistor, what is the purpose of the p+region?
d. Now we adjust the flat-band voltage (and thus the threshold voltage) with an ion implant just at the bottom of the gate oxide. We set the implant value to get a final threshold of 0.5 V. In testing the device we short source to body, i.e.,.
d.1) [No partial credit] What is of this device if we set?
d.2) Neatly sketch the I-V characteristics on the linear axes below for three cases: , and. Cover the range. Assume the electron mobility in the channel is. You must put a scale on the current axis. (Note that partial credit will only be possible if you very carefully show your work, including giving any formulas you are using before evaluation.)
V (^) Tn V (^) SB = 0
V (^) DSAT V (^) GS = 2 V
V (^) GS = 0.5 V , V (^) GS = 1 V V (^) GS = 2 V V (^) DS = 0 to 5 V 500 cm 2 ⁄Vsec
Drain Current
Drain-Source Voltage
Drain Current vs VDS
Problem 4 (25 points)
A p-n junction capacitor is made in an integrated circuit with the cross-section as shown below. The p-region is
very heavily doped compared to the n-region (it is a p +n junction) and the n-region doping is. The
junction area is , or.
a) Make a sketch of the charge density (C/cm 3 ), the electric field (V/cm), and the potential versus for this structure at 5V reverse bias. The “graph paper” is provided on the page opposite. As part of the calcula- tion to prepare these graphs, please compute the following:
a.1) The built in voltage
a.2) The depletion width at 5V reverse bias
a.3) The peak electric field at 5V reverse bias
b) What is the small-signal capacitance at 5V reverse bias?
1 × 1014 cm 3 ⁄ 100 × 200 μm 2 10 –^4 cm 2 ×
p
n metal contact to n side
metal contact ( Na ) to p side
( N (^) d )
x = 0
(b)
n type
p type^
p type
x
x
(a)
ρ E φ x
(V)
( μm)
(V/cm)
Formula (in terms of known quantities)
F
Value pF
A.
Appendix A:
Appendix B:
10 13 10 14 1015 1016 1017 1018 1019 1020
1400
1200
1000
800
600
400
200
0
holes
Nd + Na total dopant concentration (cm−^3 )
electrons
mobility (cm
2 /Vs)
Log-linear plot of electron and hole mobilities at room temperature, as functions of the total doping concentration. Note that electron and hole diffusivi- ties are given by:.
N (^) d +N (^) a D =kT ⁄qμ
φ (mV)
φ (mV)
p-type intrinsic
po , equilibrium hole concentration (cm−^3 )
no , equilibrium electron concentration (cm−^3 )
n-type
p-type intrinsic n-type
1019
101 102 104 106 108 1010 1012 1014 1016 1018 1019
1018 1016 1014 1012 1010 108 106 104 102 101
− 550 − 550
− 480 − 480
− 360 − 360
− 240 − 240
− 120 − 120
0 0
120 120
240 240
360 360
480 480
550 550
A.
Appendix C: Formulas
(EQ 2.7)
(EQ 2.34)
(EQ 2.55)
(EQ 2.56)
(EQ 2.67)
(EQ 3.1)
(EQ 3.4)
(EQ 3.56)
(EQ 3.88)
(EQ 3.89)
(EQ 3.95)
(EQ 4.18)
(EQ 4.19)
(EQ 4.59)
(EQ 4.60)
(EQ 4.67)
n (^) o ⋅ p (^) o = n (^) i^2 ( T)
v (^) dn = –μn E
J (^) n J (^) ndr^ + J (^) ndiff qnμn E qD (^) n dn dx
J (^) p J (^) pdr^ + J (^) pdiff qpμp E qD (^) p dp dx
R
qN (^) dμn t
L
W
R L
W
= = ^ ^ = R N
dE dx
ρ ε
E x( ) dφ^ ( )x dx
X (^) do x (^) no + x (^) po
2 εsφB q
N (^) a
N (^) d
V (^) FB = –(φ (^) n+ – φp)
Q (^) G ( V (^) GB =V (^) FB) = 0
V (^) Tn V (^) FB – 2 φp
C (^) ox = + -------- 2 qεs N (^) a (– 2 φp)
V DS
SAT = V (^) GS – V (^) Tn
I D SAT
W
2 L
μ n C^ ox (^ V^ GS – V^ Tn)
2 W
2 L
μ n C^ ox V^ DS (^) SAT
= =^2
I (^) D = 0 A (^ V^ GS ≤V^ Tn)
I (^) D = ( W ⁄L)μn C (^) ox [V (^) GS – V (^) Tn – (V (^) DS ⁄ 2 )] ( 1 +λn V (^) DS)V (^) DS ( V (^) GS ≥V (^) Tn ,V (^) DS ≤V (^) GS – V (^) Tn)
I (^) D (^) SAT = ( W ⁄ 2 L)μn C (^) ox ( V (^) GS – V (^) Tn)^2 ( 1 +λn V (^) DS) ( V (^) GS ≥V (^) Tn ,V (^) DS ≥V (^) GS – V (^) Tn)
V (^) Tn = V (^) TOn +γn ( – V (^) BS– 2 φp– – 2 φp)
g (^) m
W
L
≅ (^) ----- μn C (^) ox ( V (^) GS – V (^) Tn) 2
W
L
= ----- μn C (^) ox I (^) D
