Solution

HW #1 Solution :


Problem #1, Given the table of NMOS threshold voltages for measured 250nm devices below find the VT mean, standard
deviation, variance, and model the data as a Gaussian function. Find the probability that the threshold voltages will be between
710mV and 690 mV assuming a random Gaussian distribution. The top row shows VT in mV and the bottom row show the
frequency of measured occurrence for 1014 samples.

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Vt (mV) 686 688 690 692 694 696 698 700 702 704 706 708 710 712 714
Frequency (ea) 0 5 10 22 54 81 152 163 169 148 94 67 32 11 6
Total1 (Vt  F) 0 3440 6900 15224 37476 56376 106096 114100 118638 104192 66364 47436 22720 7832 4284
deviation (Vt - m) -15 -13 -11 -9 -7 -5 -3 -1 1 3 5 7 9 11 13
Total2 (deviation  F) 0 -66 -113 -204 -392 -426 -496 -205 125 406 446 452 280 118 76


(1) Vt mean?

 =

(2) Standard Deviation?

 =

(3) Variance?



(4) Model the data as Gaussian function
Gaussian Distribution is g(x) = .
So, by inputting value above, g(x) = =

(5) Probability between 690 to 710 ?
Normalizing by using  Z
690
= -2.42, Z
710
= +1.88
Then, by using below

Where =1/, =2

P(Z
690
Z
710
) = (1.88) - (-2.42) = 0.97 – 0.01 = 0.96



Problem #2
a) Consider a 22 nm technology. In that 22 nm technology, an ion implanter sweeps threshold adjustment dopant atoms into the
top of a substrate to create transistors at a total dose of 1.2 x 1014 atoms/cm2. These dopant atoms for threshold adjustment make
the transistor have the proper threshold in the VLSI circuit. If a minimal size transistor is 22 nm gate length and 66 nm gate width
and assuming all the dopant in the implant goes into the transistor volume how many dopant atoms do you expect under the
transistor gate controlling the transistor threshold?
First, channel area below transistor = 22nm 66nm = 22
Then, channel area


b) For the accepted standard deviation estimation method for this type of dopant distribution is the standard deviation of an area
with N total dopant atoms is N
1/2
. What how many dopant atoms represent one standard deviation in the doping under this
transistor?

If the number of dopant atoms below transistor is 1742atoms, the standard deviation of atoms is atoms.
Thus, =41.74


c) Assume for rough calculation purposes that are threshold is 200 mV for the 22 nm transistor. Assume that the change of the
threshold voltage is i) linearly proportional and ii) proportional to the square-root of the doping variation. What is the standard
deviation of the threshold voltage that you would expect given the doping variations in this transistor for the linear and square-root
cases?


1) Linearly proportional to the doping variation
The ratio of mean of atom to mean of V
t
is the same with the ration of standard deviation of atom to standard deviation
of V
t
.
Thus, (V
t
)

= 41.74 = 4.8mV

2) Proportional to the square-root of the doping variation
The ratio of mean of atom to mean of V
t
is the same with the ration of standard deviation of atom to standard deviation
of V
t
.
Thus, (V
t
)

=  = 0.74mV


d) Now we are going to make 2000 million of these transistors on our IC. What is the highest threshold voltage you expect and the
lowest threshold voltage you expect for transistors on our IC for the linear and square-root cases?

1) Linearly proportional to the doping variation
The highest V
t
= 200+4.8=204.8mV, The lowest V
t
= 200-4.8=195.2mV

2) proportional to the square-root of the doping variation
The highest V
t
= 200+0.74=200.74mV, The lowest V
t
= 200-0.74=199.26Mv


e) Which answer seems more reasonable?
In designing transistor, it is very important to reduce variation of V
t
in order to control it reasonably. Smaller standard deviation of
V
t
means that the distribution of V
t
of transistors more concentrates into mean, which shows that variation of V
t
is small. Thus, I
think that the way to be proportional to the square-root of the doping variation is more reasonable than the way to be linearly
proportional to the doping variation.