Iowa state University
EE 330 Lab Report Lab 6: Models for MOS Devices
Chongli Cai 2/20/2012
EE 330 Lab Report
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Introduction The objective of this lab is to find out the relationship between the square law model of a
MOS transistor and the BSIM model. In this lab we compared these two models and
show the reason for the square law model is relative less accurate. In this lab, we also
created a schematic circuit in Cadence and measure some data as well as calculate the
model parameters.
Procedure, Result and Analysis
Part 1: Square-law Parameter Extraction
MOS transistor with W=12µm, L=3µ
I. Derive the process parameter VT0
Keeping the VDS constant and used the different VGS to get the two current
values.
Figure (1)
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Measure date:
VGS=2V, ID1=265.2µA; VGS=2.5V, ID1=479.6µA;
From the equation:
211 1
1
212 2
1
( ) (1 )2
( ) (1 )2
D OX GS T DS
D OX GS T DS
WI C V V V
L
WI C V V V
L
Getting the Result:
00
0
2265.20.5498
479.6 2.5
TT
T
VV V
V
II. Derive the process parameter
Keeping the VGS constant and used the different VDS to get the two current
values.
Measure date:
VDS=4V, ID1=265.2µA; VDS=3V, ID2=263.3µA;
From the equation:
211 1
1
212 2
1
( ) (1 )2
( ) (1 )2
D OX GS T DS
D OX GS T DS
WI C V V V
L
WI C V V V
L
Getting the Result:
265.2 1 40.00737
263.3 1 3
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III. Derive the process parameter OXC
Using the circuit shown in Figure (1) and the measurement value derived
in previous two steps to calculate the process parameter OXC
Measurement and derived parameter value from previous:
VGS=2V, VDS=4V, ID=265.2µA, 0.00737 , 0 0.5498TV V
From the equation:
21
1
( ) (1 )2
D OX GS T DS
WI C V V V
L
Getting the result
21
1
( ) (1 )2
DOX
GS T DS
IC
WV V V
L
261.36 /OXC A V
IV. Derive the process parameter
Considering the Bulk Effect, we using the equation:
0 ( )T T BSV V V
Changing the voltage BSV three times (including the condition BSV =0), we can
getting three groups data (because we have three unknown variables)
Figure (2)
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Measurement data:
When VBS=0, VT=744.3mV;
When VBS=-2V, VT=1.212V;
When VBS=-4V, VT=1.46V;
Getting the result:
Using the calculator the solve the equation system
=0.4347
MOS transistor with W=1.5µm, L=0.6µ
To derive the process parameter for another MOS (W=1.5µm, L=0.6µ) transistor,
using the same procedures and calculation method as the previous one.
So, getting the following result:
VT0=0.8147; =0.1861; 264.8 /OXC A V ; =0.6425
Part 2: Comparison with BSIM Model MOS transistor with W=12µm, L=3µ
For the different VGS value, sweeping the VDS from 0 to 5V and measure the
current and getting the following transfer characteristics (Figure (4))
Figure (3)
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Figure (4)
Figure (5)
This is the operating point for the MOS transistor (VDS=4 V, VGS=2Vand VBS=0). AS shown
above, the operating current (ID) is 269.17µA, which is really near to the value
calculated by using the parameters extracted in the previous part. ( The corresponding
value derived in the previous part is 265.2µA. Also, we can measure the slop of this
curve, which is the value of . The value is 1.869µ which has a big difference with the
value derived in the previous part (The value derived in previous part is 0.00737).
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Next, setting the VDS to the operating point and sweeping the VGS from 0 to 5V, we can
the following plot
Figure (6)
Figure (7)
The threshold voltage of this MOS transistor in BSIM model is 733.2mV. The
corresponding value in the square-law model is 549.8mV. The error percentage
approximately is 20.92%. (a little bit difference).
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MOS transistor with W=50µm, L=1um
For the different VGS value, sweeping the VDS from 0 to 5V and measure the
current and getting the following transfer characteristics (Figure (9))
Figure (8)
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Figure (9)
Figure (10)
This is the operating point for the MOS transistor (VDS=5 V, VDS=5Vand VBS=0). AS
shown above, the operating current (ID) is 12.7831mA.
Using the model parameters extracted in the previous part to calculate the operating point:
21
1
50( ) (1 ) 61.36 * (4 0.8)(1 75.17 *5) 9.8
2 1D OX GS T DS
WI C V V V mA
L
12.7831 9.823.34%
12.7831Error
This error is a little bit large.
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Comparing the error calculated for the two different MOS transistor, we can find out that
the error between BSIM model and square-law model when operating near the operating
point and with device dimension close to those used to extract the parameter is relatively
smaller than that of the MOS which is operating at distant operating point with device
dimensions that are substantially different than those used to extract the model parameter.
In my measurement, the error for two different conditions is 20.94% and 23.34%
respectively.
Part 5: Output Conductance Extraction
Using the small signal model to find the
From the equation : 0 DQg I
The equation is shown as Figure (11)
Figure (11)
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Relationship between L and
L I dq(µ) g ds(s) lamada
2.5u 1550 6.886 0.004442581
3.5u 1217 5.106 0.004195563
4.5u 972.9 3.88 0.003988077
5.5u 810.1 3.101 0.003827922
6.5u 708.4 2.632 0.003715415
7.5u 617.9 2.228 0.003605761
8.5u 547.8 1.924 0.003512231
Table (1)
Figure (12)
0.003
0.0032
0.0034
0.0036
0.0038
0.004
0.0042
0.0044
0.0046
0 2 4 6 8
lamada (um)
L (um)
L VS lamada
lamada
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The relationship between L and VDS
Vds (V) Idq (µ) g ds (s) lamada
2.2 261.7 2.212 0.008452
2.4 262.1 2.092 0.007982
2.6 262.5 2.012 0.007665
2.8 262.9 1.956 0.00744
3 263.3 1.916 0.007277
3.2 263.7 1.886 0.007152
3.4 264.1 1.863 0.007054
Table (2)
Figure (13)
Analysis
From the figure (12) and figure (13), the channel length modulation constant of the
MOS transistor is inversely proportional to both the Vds and length L of it.
0.0068
0.007
0.0072
0.0074
0.0076
0.0078
0.008
0.0082
0.0084
0.0086
2 2.5 3 3.5
Vds (v)
L (um) Vds VS lamada
lamada
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Conclusion This lab is useful to make sense about the difference between BSIM model and square-
law model. That is useful for practicing the formula and parameter. The square-law
model has a relatively lager error when comparing with the BSIM model. BSIM model
has 95 process parameters while the square-law model only has 6 process parameters.