Two-Port Models, Frequency Response - CppSim

M.H. Perrott

Analysis and Design of Analog Integrated CircuitsLecture 2

Two-Port Models, Frequency Response

Michael H. PerrottJanuary 25, 2011

Copyright © 2012 by Michael H. PerrottAll rights reserved.

M.H. Perrott

Review: Basics of One-Port Modeling

Vth computed as open circuit voltage at port nodes Ith computed as short circuit current across port

nodes Zth computed as Vth/Ith- All independent voltage and current sources are set to

zero value

Zth

Vth Ith Zth

Thevenin Equivalent Norton EquivalentLinear Network

2

M.H. Perrott 3

Basics of Two-Port Modeling (Unilateral)

We now include a dependent current or voltage source

Zin- Solve using 1-Port analysis at input

Zout- Solve using 1-Port analysis at output with V1 = 0

GM- Short circuit output current as a function of V1

Av- Open circuit output voltage as a function of V1

No IndependentSources

ZL

Zs

Vin

ZoutGmV1

Linear Network

ZinV1 ZL

Zs

Vin

Zout

AvV1 ZLZinV1

Zs

Vin

OR

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Analysis of Cascaded Blocks


Block 2


Block 3

ZLVb


Block 1

Vin Va Vc

Linear NetworkLinear NetworkLinear Network

ZLZoutGmVbZinVbZoutGmVaZinVaZoutGmVinZinVin Vc

Zout,effective

Vth,effective Zin,effectiveVb

Analysis carried out without solving simultaneous equations!

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Problem: Most Circuits are Very Nonlinear!


Block 2


Block 3

ZLVb


Block 1

Vin Va VcNonLinearNetwork

NonLinearNetwork

NonLinearNetwork

Can we still leverage two-port modeling?

Thevenin/Norton modeling only applies to linear networks

Direct analysis of nonlinear networks is challenging

M.H. Perrott 6

Small Signal Modeling Allows Us to Linearize


Block 2


Block 3

ZLVb


Block 1

Vin Va VcNonLinearNetwork


Block 2


Block 3

ZLVb


Block 1

Vin Va Vc

Linear NetworkLinear NetworkLinear Network

NonLinearNetwork

NonLinearNetwork

Linearization

Small signal model is only valid about a specific operating point

M.H. Perrott

Small Versus Large Signal Modeling

Sketch Vout versus Vin as the amplitude of Vin is increased

7

ZLVin Vout

NonLinear Gain Block

Vin

Vout

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Impact of Operating Point on Small Signal Modeling

8

ZLVin Vout


Vin

Vout

Vin_dc

Vout_dc

Sketch Vout versus Vin as the DC operating point is changed

M.H. Perrott

Achieving a Small Signal Model

9

ZLVin Vout


Vin

Vout

Vin_dc

Vout_dc

Create a two port model of the above block

M.H. Perrott

Including Impedances in Two-Port Models

Compute Vout as a function of Vin

10

Zout

AvV1 ZLZinV1

Zs

Vin Vout

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Example of Two-Port Derivation

Compute Zin, Zout, and Gm- Assume rin = infinity, gmb = 011

RD

RS

RG

-gmbvsvgs

vs

rogmvgs voutvin

Device Small Signal Model

rin

g d

s

ZoutGmV1ZinV1 RD

RG

Vin

g d

vout

M.H. Perrott

Frequency Domain Modeling of Impedances

Determine Laplace Transform of Impedances Below:

12

L

C R

CR

ZoutCRZin

M.H. Perrott

Example: Transfer Function of Two-Port Circuit

Derive the transfer function Vout(s)/Vin(s) Label the poles and zeros of the transfer function

13

roGmV1V1Vin

g d

voutCin Cload

Rsrc

M.H. Perrott

Frequency Response

Frequency response is readily derived from a transfer function:- For w (rad/s), you substitute s = jw- For f (Hz), you substitute s = j2f- Note that j = sqrt(-1)

Example, for the transfer function on the previous page, the frequency response (in f (Hz)) is:

14

M.H. Perrott

Bode Plot Basics

The magnitude and phase of the frequency response is often depicted in the form of a Bode plot

Example:

- Log of magnitude (dB):

Taking the log allows the poles and zeros to be plotted separately and then added together

- Phase:

Phase of poles and zeros can also be plotted separately and then added together

15

M.H. Perrott

Plotting the Magnitude of Poles

Plot the magnitude response of pole wp1

- For w << wp1:

- For w >> wp1:

16

ω

20log|H(ω)|

0 dB

-20 dB/decadeωp1

M.H. Perrott

Plotting the Magnitude of Zeros

Plot the magnitude response of pole wz

- For w << wz:

- For w >> wz:

17

ω

20log|H(ω)|

0 dB

20 dB/decade

ωz

M.H. Perrott

Putting It All Together

Example Frequency Response:

- Assume wz << wp1 << wp2

18

ω

20log|H(ω)|

0 dB

20 dB/dec

ωz

0 dB/dec

-20 dB/dec

ωp1 ωp2

What happens if wp1 << wz << wp2 ?

H(w) =Vout(w)

Vin(w)=

1 + jw/wz(1 + jw/wp1)(1 + jw/wp2)

M.H. Perrott

Changing the Order of Poles and Zeros


- Assume wp1 << wz << wp2

19

H(w) =Vout(w)

Vin(w)=

1 + jw/wz(1 + jw/wp1)(1 + jw/wp2)

ω

20log|H(ω)|

0 dB

-20 dB/dec

ωp1

0 dB/dec

-20 dB/dec

ωz ωp2

M.H. Perrott

Changing the DC Gain from 1 to K


- Assume wp1 << wz << wp2

20

H(w) =Vout (w)

Vin(w)= K

1 + jw/wz(1 + jw/wp1)(1 + jw/wp2)

ω

20log|H(ω)|

20log(K) dB

-20 dB/dec

ωp1

0 dB/dec

-20 dB/dec

ωz ωp2

Two-Port Models, Frequency Response - CppSim

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