cmos lecture5-1.pdf

8/10/2019 CMOS Lecture5-1.pdf

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Zou Zhige VLSI, EST 3

Last Class Exercise

Calculate the

Output voltage swing

(Vx, Vy, Vout=Vx-Vy)

low frequency Av

1in

out

2

DD

SS

b 3 4

( )//V m on opA g r r=

, , ,

, 1

,

in com thn x y DD od mp

in com GS P

od mn P

V V V V V

V V V

V V

< <

= +


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Introduction

We have learned current source:Current source can act as a large resistorwithout consuming excessivevoltage headroom.

MOS in saturation can act as current source.All the amplifier need current source or resistor for load current.

Current source has many applications:

current source load for common source amplifiers

tail current source for differential pairsbias currents for folded cascode amplifierDAC

Current count and so on.This Chapter we will learn current mirrorforbias elements and signalprocessing components.


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Table of Contents

Introduction

Basic Current Mirrors

Cascode Current Mirrors

Active Current Mirrors

Introduction

Large-signal analyses

Small-signal analyses Common response


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Current Source in Differential amplifier


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The simplest current source is as follows:

2 ,1

( )2 n oxout TGS

WI C V V

L=

2

1 2DD

GS

RV V

R R=

+

Is this current source

always constant ?


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The problem of Basic CM

Influence of: power supply, process, temperature

Vgs and Vth are not constant.

Even if the Vgs is precisely defined, the Id is not constant.

, Vth will change due to different process or temperature

We must seek other method of biasing MOS currentsources.

Current copy from a constant current reference!

Why?

But how to

copy?


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Copying Current

Current

Mirror


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Basic Current Mirror

Both in saturation and Neglecting the Lambda, We get:

It allows precise copying of the current with no dependence on processand temperature !

2 1( ) /( )out ref W W

I I

L L

=


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Current Source & Current Sink

current source

current sink

out1

REF

out2

DD


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Amplifier Bias Example

All the MOS should have the same channel length!


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Another Example

2

31

)/()/(

LWLWRgA LmV=

Current Transfer!

)||()||1

||( 3322

11 Lomo

m

omV Rrgrg

rgA =

Av=?


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Table of Contents

Introduction


Cascode Current Mirrors

Active Current Mirrors

Introduction

Large-signal analyses

Small-signal analyses Common response


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Drawback of Basic CM

Consider the channel length modulation effect:

Because the load is not constant, so:

We should reduce the channel length modulation. Or there isother method?

)1()(

2

1 2DSTHGSoxnOUT VVV

L

WCuI +=

)1(

)1(

1)/(

2)/(

1

2

1

2

DS

DS

D

D

V

V

LW

LW

I

I

+

+=

21 DSDS VV


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How about this current mirror ?


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Simple cascode amplifier

Large signal behavior (Vin fixed to VG1, Vout (VDS) sweeping from 0 to 3V)

Which one is better for the ideal current source?


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Cascode Current Mirror

Choose , let ,then

Output resistance:

Trade-off: accuracy and voltage headroom

3 3 2pout m o oR g r r=

YX VV =bV REFout II

2

3 3 2 3 3

1oy p p

m o o m o

rV V V

g r r g r = =

2oY rR =

How to get Vb ?


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To ensure:

We should have :

XGSb VVV += 3

YX VV =


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Add M0 for Vb

Choose proper dimension of M0 and M3 for VGS3=VGS0

If Then:3 1 2( ) /( ) ( ) /( )oW W W W

L L L L=

YX VV =


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Region of M2 and M3

M2 and M3 both are in triode

M2 in sat, and M3 in triode

M2 and M3 both are in sat

When Vout

if M2

triode first, but Vgs2

constant

so Vds2 and Id2,3

Vb constant, Vgs3 its impossible for M3 still in saturation.

Conclusion: M3 enter triode zone first, then M2


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Voltage Headroom 1

min 2 2 21y GS T odV V V V = =


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Voltage Headroom 2

min 3 2 3

1 3

2

p GS od TH

od od

od

V V V V V V

V

= = +

=

Notice: YX VV

REFout II So:


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Voltage Headroom 3

min 3 2

3 2 2 2

2

p od GS

od GS TH TH

od TH

V V V

V V V V

V V

= +

= + += +

To ensure:YX VV =

REFout II So:


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Low voltage Cascode CM

In Sub-micro process, the ro isvery small. Cascode is useful.

In sub-micro process, the voltage

supply is very low. So, we should reduce the voltage

headroom.

Here is A good example

out

1

2

3

4b

DD


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Low voltage Cascode CM

M1 in satu

M2 in satu

Range of Vb

The over-drive voltage of M2 is lower than Vth1

2 2 1GS TH TH V V V

2x b THV V V

2 2 1x T H b G S x T HV V V V V V + +

112 HTGSGSbA VVVVV =

1GSX VV =

out

1

2

3

4

b

DD

122 THXGSAGSb VVVVVV ++=


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Vb of Low voltage Cascode CM M1 at the edge of satu:

Vout reaches the lowest level (headroom)

min 4 2 1 4p b TH GS x TH THV V V V V V V = = +

4 4 3 4 3GS TH x TH od od V V V V V V = + = +

)( 112 THGSGSb VVVV +=

out

1

2

3

4b

DD

How to generate Vb?


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How to generate Vb?

The left figure:

I*R=VTH1

VGS2=VGS5

The right figure:

We need big (W/L)7 for VGS7=VTH1

VGS2=VGS5

)( 112 THGSGSb VVVV +=

Are the above result always correct?


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General Advantages of MOS Current Sources

Effective current gain

No dc loading of slave stages on the master stage

( Unlike the BJT multi-stage current mirrors)Current ratioMOS channel geometric ratio

Iout can be as small as several nA

The ratio is not constant any more


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Design Considerations

Work with integer ratios and unit devices as much as possible.Using a unit device of size 1

Keep mirror ratio (IOUT/IREF) reasonably small

Typically no larger than 1020 Typically, we'll only have one single reference current

generator on a chip

Can generate/distribute currents across chip in two different

ways Distribute gate voltage

Can cause big problems due to IR drop

Usually limited to local distribution

Distribute currents Have one global bias cell close to reference that sends currents into local

biasing sub-circuits

Disadvantage: Consumes additional current


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Class Exercise

Ignore the outputvoltage of IREF

All W/L, except M4:W/4L

Please calculate :

Iout And the min Vpand VDD

2

12

34

DD

REF

out

5

6

A

B


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Thanks!

cmos lecture5-1.pdf

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