Download presentation

Presentation is loading. Please wait.

Published byAlejandro Whitmore Modified over 3 years ago

1
Frequency Response of Amplifier Jack Ou Sonoma State University

2
RC Low Pass (Review) A pole: a root of the denomintor 1+sRC=0S=-RC

3
Laplace Transform/Fourier Transform p=1/(RC) (Fourier Transform) (Laplace Transform) -p Location of the zero in the left complex plane Complex s plane

4
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

5
RC High Pass Filter (Review) A zero at DC. A pole from the denominator. 1+sRC=0S=-RC

6
Laplace Transform/Fourier Transform p=1/(RC) Zero at DC. (Fourier Transform) (Laplace Transform) -p Location of the zero in the left complex plane Complex s plane

7
Zero at the origin. Thus phase(f=0)=90 degrees. The high pass filter has a cut-off frequency of 100.

8
RC High Pass Filter (Review) R 12 =(R 1 R 2 )/(R 1 +R 2 ) A pole and a zero in the left complex plane.

9
Laplace Transform/Fourier Transform (Low Frequency) z=1/(RC) p=1/(R 12 C) (Fourier Transform) (Laplace Transform) -p Location of the zero in the left complex plane Complex s plane -z

10
Laplace Transform/Fourier Transform (High Frequency) z=1/(RC) p=1/(R 12 C) (Fourier Transform) (Laplace Transform) -p Location of the zero in the left complex plane Complex s plane -z

11
Design ω z =1/R 1 C ω p =1/(R 12 )C Note that R 12

12
High Frequency

13
Examples

14
Source Follower

15
Device Setup Gmoverid: Gm=17.24 mS RS=1000 Ohms GMBS=2.8 mS CGS=62.79 fF

16
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

17
Simulation Results

18
Current Mirror Example

19
Gm1=201.3uS GM3=201uS CGS3=CGS4=306.9fF GDS4=3.348uS GDS2=5.119uS Rload=118 Kohms Cload=1 pF Fp1=1.347 MHz Fp2=52.11 MHz Fz=104.2 MHz

20
Magnitude Av DC,matlab =27.52 Av DC,sim =27.45 Fp1 matlab =1.34MHz Fp1 sim =1.23 MHz

21
Phase

22
Transit Frequency

23
Transit Frequency Calculation

24
Understanding Transit Frequency Since f T depends on V GS -V T, f T depndes on g m /I D. f T depends on L.

25
Overdrive Voltage as a function of g m /I D g m /I D =2/(V OV )

26
Transit Frequency as function of g m /I D

27
gm/gds as a function of g m /I D

28
Trade-off of g m /g ds and f T 15-20 fTfT g m /g ds g m /I D

29
Numerical Example L=120ng m /g ds f T (Hz) g m /I D =512.0584.32G g m /I D =1015.7164.05G g m /I D =1517.1943.94G g m /I D =2017.5422.76G g m /I D =2517.050.42 G VDS=0.6 V

30
Numerical Example g m /I D =20g m /g ds f T (Hz) L=0.12um17.5422.7 G L=0.18 um29.8812.6 G L=0.25 um37.357.96 G L=1 um46.00714.4 M L=2 um47.26190.3 M VDS=0.6 V

31
g m /I D Principle

32
Use to gm/ID principle to find capacitance g m /I D (f T,I/W,g m /g ds ) f T =g m /c gg, c gg =c gs +c gb +c gd c gs /c gg is also g m /I D dependent.

33
Example Assume gm/ID=20, L=120 nm, VDS=0.6V, I=100uA. fT=22.76 GHz c gg =g m /f T =13.98 fF c gd /c gg =0.29c gd =4.1 fF c gs /c gg =0.75 C gs =10.5 fF

34
Noise

35
Noise is not deterministic The value of noise cannot be predicted at any time even if the past values are known.

36
Average Power of a Random Signal Observe the noise for a long time. Periodic voltage to a load resistance. Unit: V 2 rather than W. It is customary to eliminate RL from P AV.

37
Power Spectral Density PSD shows how much Power the signal carries at each frequency. S x (f 1 ) has unit of V 2 /Hz.

38
PSD of the Output Noise

40
Output Noise

41
PSD of the Input Noise

42
Input Noise

43
Noise Shaping

44
Correlated and Uncorrelated Sources (How similar two signals are.) P av =P av1 +P av2 Superposition holds for only for uncorrelated sources.

45
Uncorrelated/Correlated Sources (Multiple conversations in progress) (clapping)

46
Resistor Thermal Noise

47
Example Vnr1sqr=2.3288 x 10 -19 Vnr3sqr=7.7625 x 10- 20 Vnoutsqr=3.1050 x10 -19

48
Analytical Versus Simulation

50
Corner Frequency (f co )

51
f co as a function of length

Similar presentations

Presentation is loading. Please wait....

OK

Signals and Systems EE235 Leo Lam © 2010-2013.

Signals and Systems EE235 Leo Lam © 2010-2013.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on film industry bollywood actor Ppt on child labour act in india Free convert pdf to ppt online Ppt on pin diode radiation Ppt on nutrition in hindi Free ppt on teachers day Ppt on christian leadership Ppt on email etiquettes presentation topics Download ppt on android Ppt on immigration