Homework 2 Solutions
Problem 1
1) Circuit schematic
Perform .op analysis, the small-signal parameters of M1 and M2 are
shown below.
Small-signal parameters of M1
gds = 9.723u
gm = 234.5u
region = 2
vds = 1.255
vgs = 1.256
vth = 488m
Small-signal parameters of M2
gds = 9.723u
gm = 234.5u
region = 2
vds = 1.255
vgs = 1.256
vth = 488m
M1 and M2 both are in the saturation region.
2) For M1,
Small-signal gain is,
AV ,M1  Gm Rout Gm  234.5u Rout  1/ gds1 R1  9.11k 
So, small-signal gain of M1 is 2.136.
For M2,
Small-signal gain is,
AV ,M1  Gm Rout Gm  234.5u Rout  1/ gds1 R1  9.11k 
So, small-signal gain of M2 is 2.136.
Total voltage gain is,
AV  AV ,M 1 AV ,M 2  4.562
3) Plot of voltage gain vs. output voltage
In order to plot the voltage gain versus output voltage, we need to
create the feedback loop and sweep DC output voltage, the
schematic is shown below.
With the above schematic, we could sweep DC output voltage.
Assume that the differential amplifier is the ideal VCVS with very
large gain, and its maximum and minimum output voltage is 2.5V and
0V respectively. Therefore, DC output voltage is equal to the input of
amplifier and sweeping it is considered as sweeping output voltage.
The input voltage of M1, Vi, should change with sweeping output
voltage, so that we could calculate the voltage gain. (In calculator,
select derive from the special function, for example, use
deriv(vo)/deriv(vi) to calculate the small signal gain). The simulation
results are shown below.
4) Plot of small-signal gain vs. frequency
Since the AC magnitude of input voltage source is 1, output voltage is
directly equal to the voltage gain. Based on the simulation results, the
voltage gain is also 4.565, which is almost the same as that of
calculation results in problem 2.
5. Output impedance is,
Ro  1/ gds 2 R2  9.11k 
6. Schematic to extract output impedance Ro,
Ro = Vx/Ix, if we set AC magnitude Vx=1, then output impedance Ro
= 1/Ix. Based on the schematic shown above, the simulation result of
Ix is shown below.
Ix = 109.72uA, therefore, Ro = 1/109.72u = 9.11kΩ, which is the
same as the calculated output impedance in problem 5.
7) Large-signal swing plot,
The upper limit of large-signal swing is the voltage across the resistor
R2, the maximum value for output voltage swing is VDD, when M2 is
cut-off. The low limit of large-signal swing is the overdrive voltage of
M1. In order to make sure that M1 is in the saturation region, output
voltage should be larger than Vov,M1.
Problem 2
1)
Perform .op analysis, the small-signal parameters of M1 and M2 are
shown below.
Small-signal parameters of M1
gds = 31.98u
gm = 1.561m
gmbs = 235.8u
region = 2
vds = 1.301
vgs = 801.1m
vth = 721.6m
Small-signal parameters of M2
gds = 5.255u
gm = 272.3u
gmbs = 55.1u
region = 2
vds = 2.097
vgs = 796m
vth = 579.8m
M1 and M2 are both in the saturation region.
2) For M1,
Small-signal gain is,
AV ,M 1  Gm Rout Gm  1.561m
Rout  1/ gm1 1/ gmb1 1/ gds1 R1  0.518k 
So, small-signal gain of M1 is 0.809.
For M2,
Small-signal gain is,
AV ,M 2  Gm Rout Gm  272.3u
Rout  1/ gm2 1/ gmb2 1/ gds 2 R2  2.31k 
So, small-signal gain of M2 is 0.629.
Total voltage gain is,
AV  AV ,M 1 AV ,M 2  0.509
3) Plot of voltage gain vs. output voltage
4) Plot of small-signal gain vs. frequency
Since the AC magnitude of input voltage source is 1, output voltage is
directly equal to the voltage gain. Based on the simulation results, the
voltage gain is 0.509, which is the same as that of calculation results
in problem 2.
5) Output impedance is,
Ro  1/ gm2 1/ gmb2 1/ gds 2 R2  2.31k 
6)
Ro=Vx/Ix, if we set AC magnitude as Vx=1, then the output
impedance Ro=1/Ix. Based on the schematic shown above, the
simulation result of Ix is shown below.
Ix=432.53u, therefore, Ro=1/432.53u=2.31kΩ, which is the same as
the calculated output impedance in problem 5.
7) Large-signal swing plot
The upper limit of large-signal swing is the overdrive voltage of M1
and the Gate-to-Source voltage of M2, the maximum value for
output voltage is VDD-Vov,M1-VGS,M2.
(Note that, if we assume that the maximum input voltage is VDD,
the maximum output voltage should be VDD-VGS,M1-VGS,M2.)
The low limit of large-signal swing is very close to the ground,
when M2 is cut-off.
Connect the bulk to the source for both two transistors, and repeat
above 1)
(Note, the size of M1 and M2 are changed, however, if the size is
not changed in the homework, the answer is still acceptable as
long as the simulation is rational.)
1) Perform .op analysis, the results are shown as the following
part.
Small-signal parameters of M1
gds = 13.48u
gm = 597.4u
region = 2
vds = 1.299
vgs = 799.2m
vth = 489.6m
Small-signal parameters of M2
gds = 3.986u
gm = 198.1u
region = 2
vds = 2.094
vgs = 794.7m
vth = 488.1m
2)
For M1,
Small-signal gain is,
AV ,M 1  Gm Rout Gm  597.4u
Rout  1/ gm1 1/ gds1 R1  1.41k 
So, small-signal gain of M1 is 0.84.
For M2,
Small-signal gain is,
AV ,M 2  Gm Rout Gm  198.1u
Rout  1/ gm2 1/ gds 2 R2  3.31k 
So, small-signal gain of M2 is 0.656.
Total voltage gain is,
AV  AV ,M 1 AV ,M 2  0.551
3) Plot of voltage gain vs. output voltage
4) Plot of small-signal gain vs. frequency
Since the AC magnitude of input voltage source is 1, output voltage is
directly equal to the voltage gain. Based on the simulation results, the
voltage gain is 0.551, which is almost the same as that of calculation
results in problem 2.
5)
Output impedance is
Ro  1/ gm2 1/ gds 2 R2  3.31k
6)
The analysis part is the same as the last question. Ix=302.02u,
therefore, Ro=1/302.02u=3.31kΩ, which is the same as the
calculated output impedance in problem 5.
7) Large-signal swing plot
The analysis part is the same as the previous part.