ShanghaiTech University
School of Information Science and Technology
EE115ALab Experiments
Experiment :MOS Characterization
Student Name: Titian Yang
Student Number: 2023531030
Contents
1 .Plot ID vs. VDS curves with different VGS. Label the cutoff, triode and saturation
regions on the plot. What is the channel length modulation λ: _0.025 �−1 _.
VGS is set as 2.2V. According to the figure, we find that the linearity of the curve is better when
VDS ≥ 0.86V. The formula is obtained by fitting the curve of this interval as Id = (2e-05)*VDS
2
+ 0.0008. R is close to 1, which means good linearity.
Since
iD
1 'W
kn
(VGS Vt )2 (1 VDS )
2
L
The channel resistance can be calculated by
ro (
iD 1
1 W
1
) [ kn ' (VGS Vt ) 2 ]1
vDS
2
L
I D
We find that the the product of ro and I D is a fixed value (defined as VA ), which is the focal
point of the curve and the horizontal axis.
So |�� | can be calculated as 0.0008 40V
2 10 5
λ=
1/40 = 0.025 V 1
2 .What is the transconductance Gm with a bias of VGS = 2.2 V and VDS = 1.5 V: _0.022_S__.
Vov VGS Vt 2.2 2.078 0.122V VDS
So the MOSFET works in saturation region. Neglect CLM, we get
iD
1 'W
1 W
W
1 W
kn
(VGS vgs Vt )2 kn ' (VGS Vt )2 kn ' (VGS Vt )vgs kn ' vgs 2
2
L
2
L
L
2
L
W
iD I D kn ' (VGS Vt )vgs
L
So
Gm k n VOV k n (VGS Vt ) 0.182 (2.2 2.078) 0.022 S
3 .What is the transconductance Gm with a bias of VGS = 2.2 V and VDS = 0.06 V: _0.0109 S_.
Vov VGS Vt 2.2 2.078 0.122V VDS
So the MOSFET works in triode region.
iD kn '
So
W
1
W
W
W
[(VGS vgs Vt )VDS VDS 2 ] kn ' [(VGS vgs Vt )VDS ] kn ' (VGS Vt )VDS vgs kn ' VDS
L
2
L
L
L
W
iD I D kn ' VDS vgs
L
Gm k n VDS 0.182 0.06 0.0109 S
4 .Plot ID^0.5 vs.VG, extract VTH: _2.078 V_ and kn: __ 0.1824 A/V^2_
According to the id-vGS characteristic in saturation region, we get
iD
1
k n ( vGS VTH ) 2
2
iD
kn
vGS
2
kn
VTH
2
So the slope k = (kn/2)^0.5 = 0.302. So kn can be simply calculated by 0.302^2*2, which equals
to 0.1824 A/V^2.
VTH = 0.6276 / 0.302 = 2.078 V.
0
