# Frequency Response of Amplifier Jack Ou Sonoma State University.

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Frequency Response of Amplifier Jack Ou Sonoma State University

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

Laplace Transform/Fourier Transform p=1/(RC) (Fourier Transform) (Laplace Transform) -p Location of the zero in the left complex plane Complex s 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) A zero at DC. A pole from the denominator. 1+sRC=0S=-RC

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

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) R 12 =(R 1 R 2 )/(R 1 +R 2 ) A pole and a zero in the left complex plane.

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

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

Design ω z =1/R 1 C ω p =1/(R 12 )C Note that R 12 <R 1 If R 2 <<R 1, ω p / ω p =R 1 /R 2 Design for ω p / ω p =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 Rload=118 Kohms Cload=1 pF Fp1=1.347 MHz Fp2=52.11 MHz Fz=104.2 MHz

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

Phase

Transit Frequency

Transit Frequency Calculation

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.

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

Transit Frequency as function of g m /I D

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

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

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

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

g m /I D Principle

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.

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

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 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.

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

PSD of the Output Noise

Output Noise

PSD of the Input Noise

Input Noise

Noise Shaping

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

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

Corner Frequency (f co )

f co as a function of length

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