1

Frequency Response

(Part 1)

Dr. José Ernesto Rayas Sánchez

Most of the figures of this presentation were taken from the web site of the authors of the book:

A.S. Sedra and K.C. Smith, Microelectronic Circuits. New York, NY: Oxford University Press, 1998.

2Dr. J.E. Rayas Sánchez

Frequency Response

! Introduction

! s-Domain Analysis

! The Amplifier Transfer Function

! Low Frequency Responses

! The FET and BJT Hybrid-πmodels

! High Frequency Responses

! Frequency Response by Computer Simulation

! Cascode Configuration

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Introduction

! Frequency response analysis is crucial to amplifier´sperformance

! Exact frequency response analysis can be very complicated(CAD)

! The Time Constant Method will be used

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s-Domain Analysis

Voltage gain as a transfer function of the complex frequency s = jω

)()()(

svsvsT

i

o≡

011

1

011

1)(bsbsbs

asasasasT nn

n

mm

mm

++++++++= −

−

−−

L

L

n is the order of the circuit (m < n)

For a stable circuit, all the roots of the denominator polynomial must have negative real parts

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s-Domain Analysis (cont.)

Poles and Zeros

)())(()())(()(

21

21

n

mm PsPsPs

ZsZsZsasT−−−−−−=

L

L

Z1, ..., Zm are the transfer function zeros, or transmission zeros

P1, ..., Pm are the transfer function poles,transmission poles or natural modes

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s-Domain Analysis (cont.)

First-Order Functions

0

01)(ω++=

sasasT

low-pass first-order network0

0)(ω+

=s

asT

0

1)(ω+

=s

sasT high-pass first-order network

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Assignment

Read appendix F in the textbook, on Single Time Constant (STC) circuits

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Bode Plots

(a pole or zero term)ωjaas +=+

22|| ω+=+ aas2)/(1|| aaas ω+=+

dB)/(1log20|| 2

dBa

aas ω+=+

)/(tan)( 1 aas ω−=+∠

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Bode Plots (continue)

(a zero)

dB)/(1log20|| 2

dBa

aas ω+=+

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Bode Plots (continue)

)/(tan)( 1 aas ω−=+∠

(a pole)

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Bode Plots, an Example

)10/1)(10/1(10)( 52 ss

ssT++

=

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Bode Plots, an Example (continue)

)10/1)(10/1(10)( 52 ss

ssT++

=

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Assignment

Solve problems 7.1, 7.5 and 7.10 from the textbook

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The Amplifier Transfer Function

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The Amplifier Transfer Function (continue)

AM midband gain

ωL cutoff low frequency, 3-dB low frequency

ωH cutoff high frequency, 3-dB high frequency

Bandwidth (BW)

)Hz(LH ffBW −= )rad/sec(LHBW ωω −=

usually ωH >> ωL, BW ≈ ωH

Gain-Bandwith Product (GB)

GB = AM ωH

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The Gain Function

)()()()()( sFsFA

svsvsA HLM

i

o ==

AM midband gainFL(s) low frequency response

FH(s) high frequency response

When ωH >> ω >> ωL, A(s) ≈ AM

When ω >> ωL, FL(s) ≈ 1

When ω << ωH, FH(s) ≈ 1

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Low Frequency Response

)())(()())((

)(21

21

L

L

PnPP

ZnZZL sss

ssssF

ωωωωωω

++++++

=L

L

nL number of poles (or zeros) of FL(s)

then,,,,,if 121 LL ZnZPnPP ωωωωω KK>>

11

and)(

)( PLP

L sssF ωωω

≈+

≈ (ωP1 is a dominant pole)

If there is no dominant pole

)(2 222

21

222

21 LL ZnZZPnPPL ωωωωωωω +++−+++≈ KK

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Low Frequency Response, Example

Calculate ωL for )25)(100()10()(++

+=ss

sssFL

1) ωP1 = 100 is a dominant pole, then ωL ≈ 100 rad/sec

rad/sec102)10(225100)2 222 =−+≈Lω

3) Using Bode plots... ωL ≈ 105 rad/sec

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High Frequency Response

)/1()/1)(/1()/1()/1)(/1(

)(21

21

H

H

PnPP

ZnZZH sss

ssssF

ωωωωωω

++++++

=L

L

nH number of poles (or zeros) of FH(s)

then,,,,,if 121 LL ZnZPnPP ωωωωω KK<<

11

and)/1(

1)( PHP

H ssF ωω

ω≈

+≈ (ωP1 is a dominant pole)

If there is no dominant pole

)/1/1/1(2/1/1/11

222

21

222

21 HH ZnZZPnPP

H ωωωωωωω

+++−+++≈

KK

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High Frequency Response, Example

Calculate ωH for F)104/1)(10/1(

10/1)( 44

5

×++−=

ssssH

1) ωP1 = 104 is a dominant pole, then ωH ≈ 104 rad/sec

rad/sec800,910/2)1016/(110/1/1)2 1088 =−×+≈Hω

3) Using Bode plots... ωH ≈ 9,537 rad/sec

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Assignment

Solve problems 7.14, 7.21 and 7.27 from the textbook