lecture 2: basic op amp design - iowa state universityclass.ece.iastate.edu/ee435/previous course...

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EE 435

Lecture 2:

Basic Op Amp Design- Single Stage Low Gain Op Amps

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How does an amplifier differ from an operational amplifier?

Review from last lecture:

?

Operational Amplifier used in feedback applications

Amplifier used in open-loop applications

Op Amp Amplifier

3

Conventional Wisdom Does Not Always Provide Correct Perspective –


even in some of the most basic or fundamental areas !!

• Just because its published doesn’t mean its correct

• Just because famous people convey information as fact doesn’t mean they are right

• Keep an open mind about everything that is done and always ask whether approach others are following is leading you in the right direction

4

What is an Operational Amplifier?

Textbook Definition:

• Voltage Amplifier with Very Large Gain−Very High Input Impedance−Very Low Output Impedance

• Differential Input and Single-Ended Output


5


• Amplifier with Very Large Gain


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Textbook Definition:

• Voltage Amplifier with Very Large Gain−Very High Input Impedance−Very Low Output Impedance

• Differential Input and Single-Ended Output


7


•Amplifier with Very Large Gain

• Differential Input and Single-Ended Output ?

8

Are differential input and single-ended outputs needed?

Consider Basic Amplifiers

Inverting Amplifier

Noninverting AmplifierOnly single-ended input is needed for Inverting Amplifier !Many applications only need single-ended inputs !

9

Basic Inverting Amplifier Using Single-Ended Op Amp

Inverting Amplifier with Single-Ended Op Amp

10

Fully Differential Amplifier

• Widely (almost exclusively) used in integrated amplif

• Seldom available in catalog parts

2

2

1

1

IN OUT

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Basic Op Amp Design Outline

• Fundamental Amplifier Design Issues

• Single-Stage Low Gain Op Amps

• Single-Stage High Gain Op Amps

• Two-Stage Op Amp

• Other Basic Gain Enhancement Approaches

12

Single-Stage Low-Gain Op Amps

• Single-ended input

• Differential Input

(Symbol not intended to distinguish between different amplifier types)

13

Single-ended Op Amp Inverting Amplifier

VOUT

VIN

VYQ

VXQ

Slope = -A

( )INOUT VfV =

Consider:

Assume Q-point at {VXQ,VYQ}

( )( )OUT IN xQ YQV A V V V= − − +

When operating near the Q-point, the linear and nonlinear model of the amplifier are nearly the same

If the gain of the amplifier is large, VXQ is a characteristic of the amplifier

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Single-ended Op Amp Inverting Amplifier( )( )O 1 XQ YQ

1 21 O IN

1 2 1 2

V = -A V -V +VR RV = V + V

R +R R +R

⎫⎪⎬⎪⎭

( ) 1 20 0 IN XQ YQ1 2 1 2

R RV = -A V + V -V +VR +R R +R⎛ ⎞⎜ ⎟⎝ ⎠

( ) 11 1

21 2

0 iSS INQ XQ YQ1 1 1

1 2 1 2 1 2

R-AR R AV = V +V V V

R R RA 1+A AR R R +R R R

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟+⎝ ⎠⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

If we define ViSS by VIN=VINQ+ViSS

VOUT

VIN

VYQ

VXQ

Slope = -A

Eliminating V1 we obtain:

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( )2 2O inss XQ XQ inQ1 1

R RV = - V + V + V -VR R

( ) 11 1

21 2

0 iSS INQ XQ YQ1 1 1

1 2 1 2 1 2

R-AR R AV = V + V V V

R R RA 1+A AR R R + R R R

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟+⎝ ⎠⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

But if A is large, this reduces to

Note that as long as A is large, if VinQ is close to VXQ

2O inss XQ

1

RV - V + V R

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( )( )O 1 XQ YQ1 2

1 O IN1 2 1 2

V = -A V -V +VR RV = V + V

R +R R +R

⎫⎪⎬⎪⎭

( )2 2O inss XQ XQ inQ1 1

R RV = - V + V + V -VR R

What type of circuits have the transfer characteristic shown?

VOUT

VIN

VYQ

VXQ

Slope = -A

Summary:

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Single-stage single-input low-gain op amp

Basic Structure Practical Implementation

Have added the load capacitance to include frequency dependence of the amplifier gain

18

Solution of Differential Equations

Set of Differential Equations

Circuit Analysis KVL, KCL

Time Domain Circuit

i

OUT

Review of ss steady-state analysis

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20


21


22

Dc and small-signal equivalent elements

Element ss equivalent dc equivalnet

VDC VDC

IDC IDC

R RR

dc Voltage Source

dc Current Source

Resistor

VAC VACac Voltage Source

ac Current Source IAC IAC


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Dc and small-signal equivalent elementsReview of ss steady-state analysis

C

Large

CSmall

LLarge

LSmall

C

L

Simplified

Simplified

Capacitors

Inductors

MOStransistors

Diodes

Simplified

Element ss equivalent dc equivalnet

24

Dc and small-signal equivalent elementsReview of ss steady-state analysis

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Summary of Sinusoidal Steady-State Analysis Methods for Linear Networks

Linear s-domainEquations

Time-Domain Circuit

Differential EquationsCircuit

AnalysisLaplace

Transform

( ) ( ) ( )OUT INs =T s sV V

( )OUT sV

Transfer Function of Time-Domain Circuit:

( ) ( )( )OUT

IN

sT s =

sVV

Key Theorem:

If a sinusoidal input VIN=VMsin(ωt+θ) is applied to a linear system that has transfer function T(s), then the steady-state output is given by the expression

( ) ( ) ( )( )MV T jω sin ωt+θ+ T jωoutv t = ∠


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oL

mv gsC

gA+

−=

Small Signal Models

21

1

ooL

mv ggsC

gA++

−=

dc Voltage gain is ratio of overall transconductance gain to output conductance

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Observe in either case the small signal equivalent circuit is a two-port of the form:

VOUTVin

All properties of the circuit are determined by GM and G

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V1 GMV1 G V2

I2

V1AVV1

G

V2

I2

VOUTVin

All properties of the circuit are determined by AV and G

Small Signal Model of the op amp

Alternate equivalent small signal model obtained by Norton to Thevenintransformation

MV

GA = G

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Mv

L

-GA = sC +G

GB and AVO are two of the most important parameters in an op amp

Mv0

-GA = G

L

GBW = C

M ML L

G GGGB = =G C C

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

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mv

L 0

-gA = sC +g

The parameters gm and go give little insight into design

mv0

0

-gA = g

0L

gBW = C

0m m0 L L

gg gGB = =g C C

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

for notational convenience

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How do we design an amplifier with a given architecture in general or this architecture in particular?

⎭⎬⎫

⎩⎨⎧

DQI,LW1

1

What is the design space?

Generally VSS, VDD,CL (and possibly VOUTQ )will be fixed

Must determine { W1, L1 and IDQ }

Thus there are 3 degrees of freedom

But W1 and L1 appear is a ratio in almost all performance characteristics of interest

Thus the design space generally has only two independent variables or two degrees of freedom

Thus design or “synthesis” with this architecture involves exploring the two-dimensional design space

⎭⎬⎫

⎩⎨⎧

DQI,LW1

1

32

How do we design an amplifier with a given architecturein general or this architecture in particular?

⎭⎬⎫

⎩⎨⎧

DQI,LW1

1

What is the design space?

Generally VSS, VDD,CL (and possibly VOUTQ )will be fixed

Must determine { W1, L1 and IDQ }

Thus there are 3 degrees of freedom

But W1 and L1 appear is a ratio in almost all performance characteristics of interest

Thus the design space generally has only two independent variables or two degrees of freedom

Thus design or “synthesis” with this architecture involves exploring the two-dimensional design space

⎭⎬⎫

⎩⎨⎧

DQI,LW1

1

1. Determine the design space

2. Identify the constraints

3. Determine the entire set of unknown variables and the Degrees of Freedom

4. Explore the resultant design space with the identified number of Degrees of FreedomParameter domains for characterizing the design space are not unique!

5. Determine an appropriate parameter domain

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How do we design an amplifier with a given architecture ?




5. Explore the resultant design space with the identified number of Degrees of Freedom


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Parameter Domains for Characterizing Amplifier Performance

• Should give insight into design• Variables should be independent• Should be of minimal size• Should result in simple design expressions• Most authors give little consideration to

either the parameter domain or the degrees of freedom that constrain the designer

35


Small signal parameter domain :

Degrees of Freedom: 2

Consider basic op amp structure

Small signal parameter domain obscures implementation issues

mv

L 0

-gA = sC +g

mv0

0

-gA = g

mL

gGB = C

{ }m 0g , g

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What parameters does the designer really have to work with?


Consider basic op amp structure

Call this the natural parameter domain

mv

L 0

-gA = sC +g

mv0

0

-gA = g

mL

gGB = C

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

37


DQOXEBOX

EB

DQm IL

WCVLWµC

V2I

g µ===

mL

gGB = C

Consider basic op amp structure Natural parameter domain

How do performance metrics AVO and GB relate to the natural domain parameters?

DQo λIg =

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

mv0

0

-gA = g

38



Natural design parameter domain:


• Expressions very complicated• Both AV0 and GB depend upon both design paramaters• Natural parameter domain gives little insight into design and has complicated expressions

mv

L 0

-gA = sC +g

{gm,g0}

mv0

0

-gA = g

mL

gGB = C

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

OXV0

DQ

W2µCLA =

λI

OX DQ

L

W2µC ILGB =

C

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End of Lecture 2

40





Process Dependent

Architecture Dependent

Process Dependent


mv0

0

-gA = g

mL

gGB = C

{gm,g0}

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

OXV0

DQ

W2µC LA =λ I

⎡ ⎤⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

OXDQ

L

2µC WGB = IC L

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

41





Alternate parameter domain:P=power=VDDIDQ VEB=excess bias =VGSQ-VT

mv0

0

-gA = g

mL

gGB = C

{gm,g0}

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

OXV0

DQ

W2µC LA =λ I

⎡ ⎤⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

OXDQ

L

2µC WGB = IC L

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

{ }EBV,P

DQMV0

0 EB DQ EB

2Ig 1 2A = = = g V λI λV

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠DQM

L EB L DD L EB

2Ig 1 2 PGB = = = C V C V C V

⎛ ⎞ ⎡ ⎤⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

42





Alternate parameter domain: { }, EBP V

Process Dependent

mv0

0

-gA = g

mL

gGB = C

{gm,g0}

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

OXDQ

L

2µC WGB = IC L

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

OXV0

DQ

W2µC LA =λ I

⎡ ⎤⎢ ⎥⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

43





DQ

OXVO

ILW

CA

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

λµ2

DQL

OX ILW

CC

GB⎥⎥⎦

⎤

⎢⎢⎣

⎡=

µ2

Alternate parameter domain:


{ }EBV,P

mv0

0

-gA = g

mL

gGB = C

{gm,g0}

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

44





DQ

OXVO

ILW

CA

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

λµ2

DQL

OX ILW

CC

GB⎥⎥⎦

⎤

⎢⎢⎣

⎡=

µ2


• Alternate parameter domain gives considerable insight into design• Alternate parameter domain provides modest parameter decoupling• Term in box figure of merit for comparing architectures

mv0

0

-gA = g

mL

gGB = C

{gm,g0}

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

{ }EBP,VV0

EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

45



• Design often easier if approached in the alternate parameter domain

• How does one really get the design done, though? That is, how does one get back from the alternate parameter domain to the natural parameter domain?

W = ?L = ?IDQ = ?

{ }EBP,V

46


DDDQ V

PI =



• Design often easier if approached in the alternate parameter domain

• How does one really get the design done, though? That is, how does one get back from the alternate parameter domain to the natural parameter domain?

2EBOXDD VCµV

PLW

=

{ }EBP,V

DQW ,IL

⎧ ⎫⎨ ⎬⎩ ⎭

47

How do we design an amplifier with a given architecture ?




5. Explore the resultant design space with the identified number of Degrees of Freedom


48

Design With the Basic Amplifier Structure


Consider basic op amp structure Alternate Parameter Domain

But, does the designer really have 2 degrees of freedom?

From circuit interconnection constraint we have

{ }EBP,V

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

VEB =-VSS-VT

49



Constraints : 1

Mathematical Degrees of Freedom: 2

Design Degrees of Freedom: MDoF – Constraints = 1

{ }EBP,V

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

VEB = -VSS-VT

50



Question: How can one meet two or more performance requirements with one design degree of freedom with this circuit?

Design Degrees of Freedom: 1

Luck or Can’t

{ }EBP,V

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

51

Design With the Basic Amplifier Structure Consider basic op amp structure Alternate Parameter Domain

What do you do if you can’t meet the performance requirements?

Look for a different architecture that either has more favorable performance characteristics or more design degrees of freedom.

{ }EBP,V

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

52

Design With the Basic Amplifier Structure Consider basic op amp structure

So, what performance can the designer really get with this circuit?

VEB = -VSS-VT

Designer really has no control of VEB with this circuit so

• Gain is fixed by the architecture • P can be used to determine GB

If gain is adequate, designer got “lucky” but GB can be engineered

{ }EBP,V

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

53

Design With the Basic Amplifier Structure Consider basic op amp structure

VEB = -VSS-VT

GB varies linearly with P ! GB is very costly !

If gain is adequate, designer got “lucky” but GB can be engineered

{ }EBP,V

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

54

Architectural Modification of the Basic Amplifier Structure

{ }XXEB V,VP,

• VEB used to determine the gain (P does not affect gain!)

Mathematical Degrees of Freedom : 3

EB XX SS TV V -V -V=

Circuit Constraints: 1

Effective Design Degrees of Freedom: 2

• P used to determine GB (but VEB does affect P needed for a given GB)

V0EB

2 1A = λ V

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

55

Consider the modified single-stage op amp

mL

gGBC

=

mV0

0

-gA =g

mV

L 0

-gA =sC +g

56

Design Space ExplorationQuestion: How does the GB of the modified

single-stage amplifier change with bias current?

GB increases linearly with IDQGB

IDQ

DQ

L EB

I2GB=C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

mL

gGBC

=

57



GB increases with the square root of IDQGB

IDQ

OXDQ

L

2µC WGB = IC L

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

58



GB independent of IDQGB

IDQ

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

59



GB decreases with the reciprocal of the square root of IDQGB

IDQ

OXLDQ

2µC W1 PGB=C LI

60

Design Space ExplorationQuestion: How does the GB of the modified single-

stage amplifier change with bias current?

GB decreases with the reciprocal of IDQGB

IDQ

LDQ

DD

OX

CILV

WPC

GB

32µ

=

61



⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

EB

DQ

L VI

CGB 2

32 OXDD

DQ L

C WPLV

GBI C

µ

=

21 OXLDQ

C WPGBC LI

µ=

Increases Linearly

Increases Quadratically

Decreases Linearly

Decreases Quadradicly

Independent of IDQ

It depends upon how the design space is explored !!!

OXDQ

L

2µC WGB = IC L

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

DD L EB

2 PGB=V C V⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

62

Design Space Exploration

Different trajectories through a design space

63

Design Space Exploration

Issue becomes more involved for amplifiers or circuits with more than onetransistor

Choice of design parameters can have major impact on insight into design

Size of parameter domain should agree with the number of degrees of freedom

Affects of any parameter on performance whether it be in the identified parameter domain or not is strongly dependent on how design space is explored

Small signal and natural parameter domains give little insight into design or performance

64

Single-Stage Low-Gain Op Amps

• Single-ended input

Basic single-stage op amp

65

Single-Stage Low-Gain Op Amps• Single-ended input

Observations:• This circuit often known as a common source amplifier• Gain in the 30dB to 45dB range• Inherently a transconductance amplifier since output impedance is high• Voltage gain is ratio of transconductance gain to output conductance• Critical to know degrees of freedom in design and know how to systematically explore design space• Alternative parameter domain much more useful for design than small-signal domain or natural domain• Performance of differential circuits will be obtained by inspection from those of the single-ended structures

lecture 2: basic op amp design - iowa state universityclass.ece.iastate.edu/ee435/previous course...

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