### High Frequency Analysis and Noise

```Frequency Response of
Amplifier
Jack Ou
Sonoma State University
RC Low Pass (Review)

= 1/(1 + )

A pole: a root of the denomintor
1+sRC=0→S=-RC
Laplace Transform/Fourier
Transform
ω
(Laplace Transform)

() = 1/(1 + )

Complex s plane
| ω+p|
= ω
(Fourier Transform)

(ω) = 1/(1 + ω)

ω

-p
p=1/(RC)
|(ω)|=1/| ω+p|
Phase((ω))=-tan-1(ω/p)
Location of the zero in the left complex
plane
Rules of thumb: (applicable to a pole)
Magnitude:
1. 20 dB drop after the cut-off frequency
2. 3dB drop at the cut-off frequency
Phase:
1. -45 deg at the cut-off frequency
2. 0 degree at one decade prior to the cut-frequency
3. 90 degrees one decade after the cut-off frequency
RC High Pass Filter (Review)

= /(1 + )

A zero at DC.
A pole from the denominator.
1+sRC=0→S=-RC
Laplace Transform/Fourier
Transform
ω
(Laplace Transform)

() = /(1 + )

Complex s plane
| ω+p|
= ω
(Fourier Transform)

(ω) = ω/(1 + ω)

p=1/(RC)
Zero at DC.
|(ω)|=| ω|/| ω+p|
Phase((ω))=90-tan-1(ω/p)
ω

-p
Location of the zero in the left complex
plane
Zero at the origin.
Thus phase(f=0)=90 degrees.
The high pass filter has a cut-off frequency of 100.
RC High Pass Filter (Review)

1 1 + 1
=

1 + 2 1 + 12
R12=(R1R2)/(R1+R2)
A pole and a zero in the left complex plane.
Laplace Transform/Fourier
Transform (Low Frequency)
ω
(Laplace Transform)

1 1 + 1
() =

1 + 2 1 + 12
Complex s plane
| ω+p|
ω
| ω+z|
= ω
(Fourier Transform)

1 1 + ω1
(ω) =

1 + 2 1 + ω12
z=1/(RC)
p=1/(R12C)

-p
-z
Location of the zero in the left complex
plane
At low frequencies, | ω+p|>| ω+p|.
Laplace Transform/Fourier
Transform (High Frequency)
ω
(Laplace Transform)

1 1 + 1
() =

1 + 2 1 + 12
Complex s plane
| ω+z|
ω
| ω+p|
= ω
(Fourier Transform)

1 1 + ω1
(ω) =

1 + 2 1 + ω12
z=1/(RC)
p=1/(R12C)

-p
-z
Location of the zero in the left complex
plane
At high frequency, | ω+p|is almost equal to | ω+p|.
Design
•
•
•
•
•
ωz=1/R1C
ωp=1/(R12)C
Note that R12<R1
If R2<<R1, ωp/ ωp=R1/R2
Design for ωp/ ωp=1000
High Frequency
Examples
Source Follower
Device Setup
Gmoverid:
Gm=17.24 mS
RS=1000 Ohms
GMBS=2.8 mS
CGS=62.79 fF
Small Signal Parameters
Design Constraints:
1. 1/(gm+gmbs)=50 Ohms
2. Large R1 to minimize Q
R2=58 Ohms
R1=1102 Ohms
L=4.013 nH
Simulation Results
Current Mirror Example
Gm1=201.3uS
GM3=201uS
CGS3=CGS4=306.9fF
GDS4=3.348uS
GDS2=5.119uS
Fp1=1.347 MHz
Fp2=52.11 MHz
Fz=104.2 MHz
Magnitude
AvDC,matlab=27.52
AvDC,sim=27.45
Fp1matlab=1.34MHz
Fp1sim=1.23 MHz
Phase
Transit Frequency
Transit Frequency Calculation
Understanding Transit Frequency
Since fT depends on VGS-VT, fT depndes on gm/ID.
fT depends on L.
Overdrive Voltage as a function
of gm/ID
gm/ID=2/(VOV)
Transit Frequency as function of
gm/ID
gm/gds as a function of gm/ID
fT
gm/gds
gm/ID
15-20
Numerical Example
L=120n
gm/gds
fT(Hz)
gm/ID=5
12.05
84.32G
gm/ID=10
15.71
64.05G
gm/ID=15
17.19
43.94G
gm/ID=20
17.54
22.76G
gm/ID=25
17.05
0.42 G
VDS=0.6 V
Numerical Example
gm/ID=20
gm/gds
fT(Hz)
L=0.12um
17.54
22.7 G
L=0.18 um 29.88
12.6 G
L=0.25 um 37.35
7.96 G
L=1 um
46.00
714.4 M
L=2 um
47.26
190.3 M
VDS=0.6 V
gm/ID Principle
Use to gm/ID principle to find
capacitance
• gm/ID→(fT,I/W,gm/gds)
• fT=gm/cgg, cgg=cgs+cgb+cgd
• cgs/cgg is also gm/ID dependent.
Example
• Assume gm/ID=20, L=120 nm,
VDS=0.6V, I=100uA.
• fT=22.76 GHz
• cgg=gm/fT=13.98 fF
• cgd/cgg=0.29→cgd=4.1 fF
• cgs/cgg=0.75 →Cgs=10.5 fF
Noise
Noise is not deterministic
The value of noise cannot be predicted at any time even
if the past values are known.
Average Power of a Random
Signal
resistance.
Observe the noise
for a long time.
It is customary to
eliminate RL from
PAV.
Unit: V2 rather than W.
Power Spectral Density
Sx(f1) has unit of V2/Hz.
PSD shows how much
Power the signal carries
at each frequency.
PSD of the Output Noise
PSD of the Output Noise
Output Noise
PSD of the Input Noise
Input Noise
Noise Shaping
Correlated and Uncorrelated
Sources
Pav=Pav1+Pav2
Superposition holds for only for
uncorrelated sources.
(How similar two signals are.)
Uncorrelated/Correlated
Sources
(Multiple conversations in progress)
(clapping)
Resistor Thermal Noise
Example
Vnr1sqr=2.3288 x 10-19
Vnr3sqr=7.7625 x 10-20
Vnoutsqr=3.1050 x10-19
Analytical Versus Simulation
as a function of length
Corner Frequency (fco)
fco as a function of length
```