B 20090222 Single-Stage ICs

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1

ELECTRONICS II



2

Single-Stage IC Amplifier

ELECTRONICS II

• Introduction!



3

Single-Stage Integrated Circuit Amplifier

• Introduction :

– We have studied discrete–circuit am lifier

configurations. – The next domain is integrated–circuit amplifiers.

• There is a difference in IC design philosophy.

– Circuits combine MOS and bipolar transistors in a

– Chip-area considerations dictate that while resistors are

to be avoided, constant current sources are readily

available.

ELECTRONICS II

• …contd!



4Integrated Circuit Design Philosophy

• Large capacitors are components external to the IC chip but should

be used cautiously to keep number of chip terminals small to avoid

high cost.

• As of 2003, CMOS process technologies are capable of producingdevices with a 0.1 µm channel length. Advantages?

– These re uire overdrive volta es of onl 0.2 volts or so.

– However, bipolar circuits can still provide much higher output currents.

– Due to reliability under severe environmental conditions, the bipolar circuits are referred for a lications in the automotive industr .

• CMOS is the most popular technology for the implementation of

digital systems due to :

•

• Ease of fabrication and

• Low power dissipation

ELECTRONICS II

• Typical values of MOSFET parameters!



5Typical Values of BJT/MOSFET Parameters

• To pack more transistors on a chip, the trend is to reduce the

minimum allowable channel length.

• Magnitude of threshold voltage V has been decreasing with

decreasing length. Additionally VDD has been reduced from 5 voltsto 1.8 volts for newer technologies to keep power dissipation as low

as ossible.

• With submicron technologies, channel length modulation effect is

very pronounced. As a result V’A has been steadily decreasing= ’

short-channel MOSFETs exhibit low output resistance.

– Because r 0 = VA /ID

• gs gd. – Shorter devices exhibit much higher operating speeds and wider

amplifier bandwidths.

ELECTRONICS II

– : T or . µm rans s or can e as g as z.

• …contd!



6Typical Values of BJT/MOSFET Parameters

,

integrated circuit fabrication process has been the lack of

pnp transistors of a quality equal to that of npn devices.

• β is lower for pnp transistors

– And pnp transistors have much larger forward transit time τF . – F -

capacitance Cde and hence the transistor speed of operation.

ELECTRONICS II

• Comparison of important characteristics!



7Comparison of Important Characteristics

• MOSFET• Induce a channel vGS > vt where

• BJT• Forward-bias EBJ vBE >vBEon

t . .

• vDS > vGS – vt

• i – v characteristics

BEon .

• Reverse-bias CBJ• i – v characteristics

G

• Input resistance (CS) is infinite

• Transconductance

B C

• Input resistance (CE) r Π = β /gm

• Transconductance gm=Ic /VT

( )t GS

Dm

V V

I g

−=

2

gs

d m

v

ig = ( )t GSn V V

L

W k −′= or

( ) ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +−′= A

DSt GSn D

V

V V V

L

W k I 1

2

1 2

where 1/VA = λ process-technology parameter

ELECTRONICS II

• Low/High Frequency Models!



8Low/High Frequency Models

i d

i g = 0

v o+

_

g mv +

_ r ov Π r Π r o

v gs

+

_

g mv gs

CgdCμr x

i s

Vgs

gmVgsr o

Cgs+

_

VΠ

gmVΠ

r o

CΠ+

_

r Π

• The frequency at which magnitude of hfe drops to unity is called “unity-gain” bandwidth wT.

T

- .

• f T is 10 to 20 GHz for npn and 5 to 15 GHz for NMOS.

• The high-frequency response of IC amplifiers is limited by the transistor internal

capacitances, mainly Cgs and Cgd in the MOSFET and CΠ and Cμ in the BJT.

ELECTRONICS II

• The IC Biasing & MOSFET Current Mirror/Source!



9

Biasing Using a Current Source

• Consider the following circuit:

Vcc

Rc 8 k

+• The BJT can be biased using a

vOβ = 100

RB 100 k

RL 5 k

.

• It has the advantage that the emitter

current is independent of β and RB.

• Thus RB can be made large, enabling

- 10 V

_an increase in the input resistance at

the base without adversely affecting

bias stability.

• Implementation of constant current source?

ELECTRONICS II



10

Biasing Using a Current Source : The Circuit

• Q1 and Q2 are matched transistors.

• Assuming high β, and small value of base current, the current through Iref

V

RI

Iref Iref = {VCC – (- VEE) – VBE} / R

• Since VBE is same so collector currents

of Q1 and Q2 will be the same.

V

+ V

Q1 Q2

_

- VEE

• So I = Iref = {VCC + VEE – VBE} / R

• This is known as a current mirror.

ELECTRONICS II

• Implementation!



11Biasing using a current source : implementation

Vcc

Rc

= 100

8 k

+

vO

1 mA

RB 100 k

_

RL 5 k

-

R

Vcc

IIref

V

+_ VBE

Q1 Q2

ELECTRONICS II -

VEE

• The IC Biasing & MOSFET Current Mirror/Source!



12

• Biasing in IC design is based on the use of constant–current sources.

The IC Biasing & MOSFET Current Mirror/Source

• Now: ( )21

12

1t GSn D V V

L

W k I −⎟

⎠

⎞⎜⎝

⎛ ′=

• Also: ( )V V GS DD −

VDD

Rref D1

IRef Io

I

• And: ( )2

2

22

1t GSn Do V V

L

W k I I −⎟

⎠

⎞⎜⎝

⎛ ′==

G S

G

SVGS

+Q1Q2

• Then:2

⎟ ⎞

⎜⎛

⎟ ⎠

⎜⎝ =

L

W

L

I

I

ref

o

-• For identical transistors:

ref o I I =

• Because the circuit replicates or mirrors the reference current, it is given the name of

• A constant dc current (called reference current) is generated at one location and then

replicated at various other locations through a process called current steering.

•

current mirror.

ELECTRONICS II • MOS Current –Steering Circuits!

current.



13

• The circuit:

MOS Current –Steering CircuitsVDD +

V

VDD

Current Sink

I5

Q5R

IRef I2

I4

Q4-

VGS1

+Q1 Q2

Q3

-- VSS

1

current Iref .

• Q1, Q2 & Q3 form a two-

port current- mirror 2ref 2

II

⎟ ⎠

⎞⎜⎝

⎛

=L

W

and 3

ref 3II

⎟ ⎠

⎞⎜⎝

⎛

=L

W

where:

1⎟ ⎠⎜⎝ L

1⎟ ⎠⎜⎝ L

• For saturation region operation, the voltages at the drains of Q2 and Q3 are constrained as

ELECTRONICS II

vD2, vD3 = - vSS + vGS1 - vtn

• The CS Circuit With Active Load!



14The CS Circuit with Active LoadvDD

• e mos as c amp er s:- I

vvo

• The drain resistance RD has been replaced by a constant – current source I.

•

transistor, it is called an active-load and the amplifier is said to be active-

loaded.

• The small si nal anal sis of the am lifier is:-

• For the CS amplifier,

• Ri = α , Avo = - gmr o and Ro = r 0

v gs

+

_ g mv gs

+

_ ro

• e magn u e o vo s e max mum ga n ava a efrom a CS amplifier and is called the intrinsic gain,

• Ao = gmr o

ELECTRONICS II

• The active-loaded CE Circuit!



15

• The circuit is :-

The Active-Loaded CE Circuit

vCC

I

vi

vo

• To keep analysis simple, the bias network is not shown.

• The small signal analysis of the amplifier is:-

• For the CE amplifier,

• Ri = r Π

• Avo = - gmr o and

• Ro = r 0

vor or Π

+

_ vΠvi

gmvΠ

• The magnitude of Avo is the maximum gainavailable from a CE amplifier and is called

the intrinsic gain,

ELECTRONICS II

• o = gmr o

• High-Frequency Response of the CS & CE Amplifier !



16The MOSFET internal capacitances

• gs , gd , gb , sb, db

• It can be shown that :

Cgs = Cgd =1/2 WLCox (triode region)

Cgs = 2/3 WLCox and Cgd = 0 (saturation region)

gs gd gb ox -

• Another small capacitance that should be added to Cgs and Cgd is the capacitance that results from the

fact that the source and drain diffusions extend slightly under the gate oxide.

• If overlap length is denoted by Lov, the overlap capacitance Cov = WLovCox (typically Lov = 0.05 to 0.1 L)

• The unction ca acitances are iven b :

sboC dboC

o

SB

sb

V

V +1

an

o

DB

db

V

V +

=

1

Where Csbo = value of Csb at zero-body source bias, VSB is equal to magnitude of

reverse bias voltage. Vo is equal to junction built in voltage of 0.6 to 0.8 volts.

• …contd.

ELECTRONICS II



17The MOSFET internal capacitances …contd

• So the hi h fre uenc res onse of the MOSFET am lifier can be redicted bthe model:

G DCgd

Vgs

gmVgsr o Cdb

Cgs

+

Csb

mb bs

vbs

_

S B+

• Due to its complexity, such models are limited to computer simulation.

•

ELECTRONICS II

… .



18The MOSFET internal capacitances …contd(Source and body connected)

D

Vgs

gd

gmVgsr o CdbgmbVbs

Cgs

+

S B

Csb

_

G DCgd

+

Vgs

gmVgsr o

CgsCdb

_

ELECTRONICS II

S



19The MOSFET internal capacitances …contd(Source and body connected: with Cdb neglected)

DCgd

VgmVgs

r Cgs

+

S

_

• This is the commonly used high frequency model of the MOSFET.

ELECTRONICS II



20

The MOSFET internal capacitances …contd

V o /v in (dB)

LF band HF band

Gain falls due to

C C! , C C2 ,C S

Gain falls due to

C gs , C gd

f L f H f (Hz)

ELECTRONICS II



21

We cannot direct the wind….

but we can adjust the sails.

(Anonymous)

ELECTRONICS II



22Common-Source Amplifier : High Frequency Response

+ VDD+ VDD

• e c rcu s:

D

RD

D

RD

CC2

G

S

IRG

sig

vsig

RL

G

S

IRG

sig

vsig

RL

C1

- VSS- VSS

S

• The MOSFET is replaced by its high-

frequency model to determine the gain

or the transfer function at f H.3 dB

• At these frequencies (high) CC1, CC2,and CS will be behaving as perfect

short circuits.f L f H

ELECTRONICS II

• ..contd.



23

• So the circuit becomes:

Common-Source Amplifier : High Frequency Response…contd

v o

G DRsigC gd

R L

v gs

_

g mv gsr

o

g mv gsR

D

vsig RG

C gs

R / L

gd

G DR’sig=RG // Rsig C gd

I ’ L

v ov gs

+

_

g mv gs R / Lg mv gs

Vsig{ RG /(RG + Rsig) }C gs

I gd

ELECTRONICS II

SThevenins



24

• So KCL at o/p node is: Id

= gm

V s

+ I’L


• Or I’L = Igd - gmV gs

• At and near f H Igd << gmV gs

• So I’L = - gmV gsv o

G D

+

RG // Rsig C gd

i

I ’ L

• And vO = - gmV gs R’L

• Because I = sCV

v gs

_


Vsig= RG /(RG + Rsig) C gs

• gd gd gs - o

• Or Igd = sCgd (vgs - (- gmV gs R’L)

• And Igd = sCgd (1 + gmR’L) V gs

• Now consider:

G DRG // Rsig C gd

I ’ L

X

G D

igd

I ’ L

o

v gs

+

_


Vsig= RG /(RG + Rsig)C gs

igdX’

o

R / Lv gs

+

_

g mv gs

C gs

C eq

ELECTRONICS II

S



25

• Now Igd = sCgd (1 + gmR’L) V gs = sCeqVgs


• Which means Ceq = Cgd (1 + gmR’L)

• So output voltage across Cgs :

R

And :⎟⎟ ⎞

⎜⎜⎛

== oF s

vT

1where wo=1/CinR’Sig

F sig

sigG

gs v

R R

v

+

= ⎠⎝

+ow

3 dB or corner freq

• Hence:Here Cin = Cgs+ Ceq =Cgs+Cgd (1 + gmR’L)

’ ⎞⎛

Sig Sig G

⎟⎟⎟

⎠

⎜⎜⎜

⎝

+ow

s1

1( )sig

sigG

Ggs v

R Rv

+=

v G D

igd

I ’ L

R / Lv gs+

_

g mv gs

C gs

C eq

ELECTRONICS II •

…contd!



26Common-Source Amplifier : High Frequency Response…contd

⎟⎟⎟

⎜⎜⎜

⎝ +

ow

s1

1( )sig

sigG

Ggs v

R R

Rv

+=

• Hence with:

• We know vO= - gmvgs R’L or vgs = - vO / gmR’L

• So vgs = - vO / gmR’L⎟⎟⎟⎟ ⎞

⎜⎜⎜⎜

⎝

⎛

+ow

s1

1( )sig

sigG

G v R R

R

+=

( ) LmGo

Rg

Rv

'−=⎟⎟ ⎞

⎜⎜⎛

s

1

• So hi-freq gain of CS amplifier:

sigGsig ⎠⎝

+ow

Or:⎟⎟⎟ ⎞

⎜⎜⎜⎛

= M o

s

A

v

v

And: ⎟ ⎞

⎜⎛

=1where: ( Lm

G M Rg

R A '−=

⎠⎝ H

s g

w ⎠⎝ sigin RC '2π

• Notice that in the expression for Ceq = Cgd (1 + gmR’L), the factor (1 + gmR’L) is known as

Miller multi lier and multi lication of C b 1 + R’ is known as Miller effect.

sigG

+

ELECTRONICS II

• Example!



27Example : High Frequency Response• The circuit is:

+ VDD+ VDD

DD

RD = 15 k ohms

CC2

Cgs = 1 pF

Cgd = 0.4 pF

gm = 1 mA/V

• Find AM, f H of the CS amplifier.

G

S

I

vsig

G

S

I

RG = 4.7 M ohms

Rsig = 100 K ohm

vsig

C1

RL = 15 k ohms

o

- VSS- VSS

S

• Solution:( ) Lm

sigG

G M Rg

R R

R A '

+−= R’L =ro // RD // RL = 150 //15 //15 = 7.14 k ohms

gm R’L =1 X 7.14 = 7.14 V/V So Am = -7 V/V

• Now Ceq =(1 + gm R’L ) Cgd = 3.26 pF

• And Cin = Cgs + Ceq = 1 + 3.26 = 4.26 pF

ELECTRONICS II

• Finally f H = 1/2ΠCin (Rsig // RG ) = 382 k Hz.

• QED?



28Active-Loaded Common Source : High Frequency Response

• High-frequency equivalent-circuit model of CS amplifier is:

v o

v gs

+g mv gs

r og mv gs

R Lvsig

Rsig

C

gd

C _

L

And when Rsig is relatively large and CL is relatively small,

igd I ’ L

’

v o

v gs

+

_

g mv gs

C eq

R / L

eq m L gd

⎟ ⎞

⎜⎛

=1

Rg A '−=

gs

ELECTRONICS II

• QED!

⎠⎝ sigin RC π 2



29The CG Amplifier With Active Load

• Consider the following circuit:

• Body transconductance gmb = χ gm where χ = 0.1 to 0.2

• Note vbs gives rise to a drain current signal gmb vbs

• And since both gate and body are grounded, the two

+ VDD

I

voltages are equal.

• For analysis, consider the following circuit:

vo

vgsvbs

B

ioRS

vsig

vi

+vo

r o _

RS

+

RL

iii

• Because r o connects output to input node so Rin depends on RL and Rout on Rs.

sig

vi

_

= gm gmb vi

ELECTRONICS II

• …contd!



30The CG Amplifier With Active Load…contd

=

• Now vo = ii RL

vo

o

vo

o

i iro

• At source node ii =(gm+gmb) vi + iro

• Now Liioi

ro

Rivvvi

−=

−=

+

RL

o

ii i

+

RL

o

ii i

oo r r

• So Liiimbmi

Rivvggi −++= )(S

vsig vi

_

i=(gm+gmb) vi

Svsig vi

_

o

• Or

[ ] iombm Lioi vr gg Rir i1)( ++=+

• And[ ]

Lo

iombmi

Rr

vr ggi

+

++=

)(1

• Finally

[ ]ombm

Lo

i

iin

r gg

Rr

i

v R

)(1 +++==

ELECTRONICS II

• …contd!




=

• For voltage gain: vo = ioRL = iiRL and vi = ii Rin

vo

o

vo

o

i iro

• So Av = vo /vin = RL /Rin

• For Avo?

• ir o + vi = vo (because RL = α so i flows from top to bottom)

+

RL

o

ii i

+

RL

o

ii i

• And i = (gm +gmb)vi

• So vo = (gm +gmb)vir o+ vi

• Or A = 1+ + r vsig

Svsig vi

_

i=(gm+gmb) vi

S

vi

_

• Now

Loi Rr v R +==ombmi r gg ++

• Therefore Lo

in

Rr R

+=

vo

• Hence

Lvo

Lv

r

R A

R A

+==

ELECTRONICS II

• …contd!

on



32

problem is not its solutionu w

finding the solution.(Anonymous)

ELECTRONICS II




• To determine output resistance:

• There are two output resistances:- Ro when vi = 0 and Rout when vsig = 0.

• So Ro = r o

• For Rout

[ ] vr vggiv ombm x x +++= )(ix

• And

vx

r o

• Also v = ix Rs

[ ] sombmoout Rr ggr R )(1 +++=RS

+

ix

•

svooout R Ar R +=

V=ix

RS

_

ELECTRONICS II

• CB Amplifier with active-load!

34



34

The CB Amplifier With Active Load

• Similar treatment as CG but with finite β. Additionally, the base

conducts si nal current contrar to the behaviour of a MOSFET CG

amplifier.

ELECTRONICS II

• Emitter degeneration resistance!

35



35The CE Amplifier With Emitter Degeneration Resistance

• Emitter degeneration is more useful in CE amplifier than source degeneration in the CS

amplifier.

– Because emitter degeneration increases the input resistance of the CE amplifier.

– Incidentally the input resistance of CS is practically infinite to begin with.

• The circuit:

vCC

α i i - (V i – ir e ) /R e - α i

vo

I

vo

α i r o R L

i/( β +1)

Re

R

i

i

(V i – ir e ) i - (V i – ir e ) /R e

• Re is usually in the range of 1 to 5 times r e.

• The analysis?

in

(V i – ir e )/ R e

ELECTRONICS II

• …contd!

• vo = (V i – ir e ) – r o {i - (V i – ir e ) / R e}

36Th CE A lifi Wi h E i D i R i d



36The CE Amplifier With Emitter Degeneration Resistance…contd

• So vo = (V i – ir e ) – r o {i - (V i – ir e )/ R e} and vo = [ i - (V i – ir e ) /R e – α i ]RL

• And solving these two equations simultaneously gives us R in = V i /[i/( β + 1)]

• Or:

+ Lo

Rr

e

eein

RR

1)R(1)r(R

++

+++= Lor

β β

• Can be simplified to:

L

r

R1

11)R(1)r(R eein

++++= β β

• Hence inclusion of Re :

– Reduces effective transconductance by a factor (1 + gm RL)

– Increases its output resistance by the same factor

– .

– Increases amplifier bandwidth and finally the emitter

degeneration resistance Re increases the linearity of the

amplifier.

ELECTRONICS II

• The CS Amplifier!

37



37The CS Amplifier With Source Degeneration Resistance

• The source degeneration resistanceintroduces negative feedback.

• Broadens the bandwidth.

• More control over the amplifier.

ELECTRONICS II

• The Emitter Follower!

38



38The Emitter Follower

• The common–collector is also called emitter-follower because the voltage at

the emitter follows very closely the voltage at the input.

• The emitter-follower suitable for IC fabrication is:

Rsig

vCC

• Low frequency gain, input resistance and output

vsig

RL

vo

I

resistance is identical to capacitively coupled

version studied earlier.

• The high-frequency circuit is:

+

RsigC μ r xB B /

C

_

g mv Π r ovsig

r ΠC Π

v Π

R Lv

o

_

ELECTRONICS II

• Simplified circuit!

39Th E itt F ll td



39The Emitter Follower…contd

RC μ r B B /

+g mv Π r ovsig

r ΠC Π

v Π

• The simplified circuit is:

_

R L v o+

_

+v

R /sig

_

sig

r ΠC Π

μ v Π

/ v o

+

• Here R /si = ? and R /

L = ?

L

_

ELECTRONICS II

• …contd!

40Th E itt F ll td



40The Emitter Follower…contd

• The sim lified circuit is: R /

+g mv Π vsig

r ΠC Π

C μ v Π• The resistance seen by Cμ is :

_

R L / v o

+

_ ( ) Lsig Rr R R

/ / 1 // ++= β π μ

• And resistance seen by CΠ is:

Lsig R R R

/ / +=

• The high-frequency is:

e

Lsig

r

R

r

R1 ++

π

( )π π μ μ π RC RC f H

+=

2

1

ELECTRONICS II

• Home Assignment!

41



41

The Emitter Follower

Home Assi nment

Ex : 6.35 (pp 641)

ELECTRONICS II

• …Some useful transistor pairings!

42Some Useful Transistor Pairings



Some Useful Transistor Pairings

•performance is achieved through maximizing the advantages and minimizing

the shortcomings of each of the two individual configurations.

• In such cases the transistor air is considered as a com ound device and the

• The Darlington Configuration:

resulting amplifier is considered as a single stage.

C

Bβ = β1β2

E

ELECTRONICS II

• …contd!

43Example : Transistor Pairings



Example : Transistor Pairings

in 1 2 , E , sig E2 .

VCC

• Solution: Rin = (β1 + 1) {r e1 + (β2 + 1) (r e2 + RE)} ????

Rsig

Q1

Q

Rin

vo

vsig

RE

α 1i e1

r e1

ii

α 2 i evi

-VEE

r e2

e1

ie2

2

Rin

R E

• Final remark: R when load is infinite R when v is 0 A A ertain to am lifier ro er.

ELECTRONICS II

• Summary!

44Single-stage IC Amplifiers : Summary



•

Single-stage IC Amplifiers : Summary

small-area MOS transistors.

• For minimization of chip area, large-valued resistors and capacitors are

virtuall absent.

• Biasing in integrated circuits utilizes current sources.

– An accurate and stable reference current is generated and then replicated to provide

bias currents for various amplifier stages.

– The heart of the current-steering circuitry utilized to perform this function is the

current-mirror.

• The high-frequency response of IC amplifiers is limited by the transistor n erna capac ances, ma n y gs an gd n e an Π an μ n e

BJT.

• IC amplifiers employ constant-current sources in place of the resistances

D C .

– These active loads enable the realization of reasonably large voltage gains while using

low voltages supplies (as low as 1 V)

• …contd

ELECTRONICS II

45……contd



……contd

amplifier is equal to the intrinsic gain of the transistor: Ao =

gmr o

– for a BJT, it is 2000 to 4000 V/V.

– for a MOSFET, it is 20 to 100 V/V.

• CE am lifier has low in ut resistance and

• CS amplifier has an infinite input resistance

• Including a small resistance in the source (emitter) of a

amp er prov es e es gner w a oo o e ec

some performance improvements e.g. wider bandwidth in

return for gain reduction (a trade-off characteristic of

negative feedback).• Note some of the ICs dissipate as much as100 watts.

ELECTRONICS II

• End of the part.

46



Choose a job you love,and you will not have to work a

.(Confucius)

ELECTRONICS II

Documents

B 20090222 Single-Stage ICs