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Allen and Holberg - CMOS Analog Circuit Design

Design Strategy

The design process involves two distinct activities:

Architecture Design

Find an architecture already available and adapt it to present requirements

Create a new architecture that can meet requirements Component Design

Design transistor sizes

Design compensation network

If available architectures do not meet requirements, then an existing architecture must be modified, or a new one designed. Once a satisfactory architecture has been obtained, then devices and the compensation network must be designed.

Allen and Holberg - CMOS Analog Circuit Design

Op Amp Architecture

VDD

M3

M4

 

 

 

 

M6

 

M1

M2

 

 

-

+

Compensation

vOUT

IBias

 

 

 

M5

 

M7

 

Allen and Holberg - CMOS Analog Circuit Design

Compensation

In virtually all op amp applications, feedback will be applied around the amplifier. Therefore, stable performance requires that the amplifier be compensated. Essentially we desire that the loop gain be less than unity when the phase shift around the loop is greater than 135˚

β

IN

-

+

Σ A

OUT

=

A

 

1 + Aβ

IN

OUT

Goal: 1 + Aβ > 0

Rule of thumb: arg[Aβ] < 135˚ at mag[Aβ] = 1

Allen and Holberg - CMOS Analog Circuit Design

Graphical Illustration of Stability Requirements

β=1

|Aβ| (dB)

-6dB/oct

 

 

GB

0 dB

 

 

ω1

ω2

-12dB/oct

Frequency

 

180o

Arg[Aβ]

90o

0o

Frequency

Allen and Holberg - CMOS Analog Circuit Design

Step Response of Two-Pole System

Impact of placing ω2 at different locations:

 

 

stability

 

 

Date/Time run: 04/08/97

19:45:36

 

 

Temperature: 27.0

1.5V

 

 

 

 

 

 

3

 

 

 

 

2

 

 

 

1

 

 

 

1.0V

 

 

 

 

 

 

ω1 = 1000 rps

 

 

0.5V

 

Case 1: ω2 = 1 x 106 rps

 

 

 

 

 

 

 

 

Case 2: ω2 = 0.5 x 106 rps

 

 

 

 

Case 3: ω2 = 0.25 x 106 rps

 

 

0V

 

 

 

 

0s

5us

10us

15us

20us

v(5)

 

 

 

 

 

Time

 

 

Allen and Holberg - CMOS Analog Circuit Design

Types of Compensation

1.Miller - Use of a capacitor feeding back around a high-gain, inverting stage.

Miller capacitor only

Miller capacitor with an unity-gain buffer to block the forward path through the compensation capacitor. Can eliminate the RHP zero.

Miller with a nulling resistor. Similar to Miller but with an added series resistance to gain control over the RHP zero.

2.Self compensating - Load capacitor compensates the op amp (later).

3.Feedforward - Bypassing a positive gain amplifier resulting in phase lead. Gain can be less than unity.

Allen and Holberg - CMOS Analog Circuit Design

Miller Compensation

VDD

M3

M4

 

 

CM

Cc

M6

 

M1

M2

 

 

-

+

 

vOUT

IBias

C1

 

 

M5

 

M7

CL

Small-signal model

 

 

 

 

 

 

1

 

 

1

 

 

 

 

 

 

 

gds2+gds4

 

Cc

gds6+gds7

 

 

 

 

+

gm4v 1

 

 

 

 

 

+

gm1

-vin

 

CM

 

vin

v2

 

 

 

vout

2

 

v 1

1

C

1 gm6v2

CL

 

rds1

-

gm3

g m2 2

-

 

-

Simplified small-signal model

1

 

1

gds2+gds4

Cc

gds6+gds7

+

 

 

 

 

+

vin

gm1vin

v2

C1

C

vout

 

 

 

 

gm6v2

L

-

 

-

 

-

 

 

 

Allen and Holberg - CMOS Analog Circuit Design

Analysis

Vo(s) (gmI)(gmII)(RI)(RII)(1 - sCc/gmII)

Vin(s) = 1 + s[RI(C1 + Cc) + RII(CL + Cc) + gmIIRIRIICc] + s2RIRII[C1CL + Cc(C1+ CL)]

-1

p1 gmII RI RII Cc

-gmIICc

p2 C1CL + CLCc + C1Cc

-gmII p2 CL

gmII z1 = Cc

where

gmI = gm1 = gm2

gmII = gm6

1

1

RI = gds2+gds4

RII = gds6+gds7

Allen and Holberg - CMOS Analog Circuit Design

Miller Compensation

 

β=1

|Aβ| (dB)

-6dB/oct

Before compensation

After compensation

 

GB

 

 

ω1

ω2

-12dB/oct

Frequency

 

180o

Before compensation

Arg[Aβ]

90o

After compensation

Phase

0o

Frequency

Allen and Holberg - CMOS Analog Circuit Design

Conditions for Stability

• Unity-gainbandwith is given as:

 

 

 

 

 

 

 

1

 

gmI

 

 

GB = Av(0)·|p1| = ( gmIgmIIRIRII) · gmIIRIRIICc =

Cc

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

• The requirement for 45° phase margin is:

 

 

 

 

 

 

 

ω

 

 

 

ω

 

ω

 

 

Arg[Aß] = ±180° - tan-1

 

- tan-1

 

- tan-1

= 45°

 

|p1|

 

|p2|

 

z

 

 

Let ω = GB and assume that z 10GB, therefore we get,

 

 

 

GB

GB

 

GB

= 45°

 

 

 

±180° - tan-1

- tan-1

 

 

 

- tan-1

 

 

 

 

|p1|

|p2|

 

z

 

 

 

 

 

or

 

 

 

 

 

 

 

 

 

 

 

 

135° ≈ tan-1(Av(0)) + tan-1 GB

+ tan-1(0.1) = 90° + tan-1 GB

 

 

|p

2

|

 

 

 

 

 

 

|p

2

 

 

 

 

 

 

 

 

+ 5.7° |

39.3° ≈ tan-1 GB

GB

= 0.818 |p2 | 1.22GB

|p2|

|p2|

 

• The requirement for 60° phase margin:

| p2 | 2 . 2GB if z 10GB

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