inf3410 — fall 2014€¦ · inf3410 — fall 2014 book chapter 3: basic current mirrors and...

INF3410 — Fall 2014

Book Chapter 3: Basic Current Mirrors andSingle-Stage Amplifiers

ContentSimple Current MirrorCommon-Source AmplifierCommon-Drain Amplifier with active load / SourceFollowerCommon-Gate Amplifier with active loadSource-Degenerated Current MirrorsCascode Current MirrorsCascode Gain StageDiff Pair and Gain Stage

Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 2

Simple Current Mirror

When will Iout = Iin, Iout 6= Iin, Iout ≈ Iin, Iout ≈ xIin?Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 4

Diode Connected nMOS

Is the simplified small signal model very useful here?For example when you want to know the voltage whenyou apply a given current?


Current Mirror Small Signal Model

How do you describe the effect of different WL ratios of

the two transistors in terms of the small signal model?


Current Mirror Small Signal Model, Example3.1 (1/2)

A detour solution (shorter in the book):

Veff = Vin − Vt0 =

√

√

√

2

βIin

=

√

√

√

2

10μm/0.4μm∗ 190μA/V2 ∗ 100μA

= 205mV

Vin = 205mV + 570mV = 775mVBook Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 7


Continued detour solution (shorter in the book):

Iout =1

2βV2

eff

=10μm/0.4μm∗ 190μA/V2 ∗ 205mV2

2= 100μA



Wrong (!!!) approach:

gm = β∗ Veff = 10μm/0.4μm∗ 190μA/V2 ∗ 0.205V= 974μA/V

That was still correct but ...

Iout = Vin ∗ gm = 755μA

... is a wrong conclusion!!!



Continued correct detour solution:

rout =1

λIout

=L

λLIout

=0.4μm

0.16μm/V ∗ 100μA= 0.025V/μA = 25kΩ


Common-Source Amplifier with active load

Active: current mirror or FET with fixed gate voltage asloadPassive: a resistor to Vdd (normally lower outputimpedance or excessive voltage across resistor)

What are the conditions forthis circuit to work as anamplifier? When will it notwork?


Small Signal Model

AV =vout

vin= −gmR2 = −gm(rds1 ‖ rds2)

(Rin not well motivated in the book: makes only sensewith including parasitic capacitances. It is actuallyattributed to a non-ideal voltage source, i.e. not thetransistor.)Book Chapter 3: Basic Current Mirrors and Single-Stage

Amplifiers 13

Example 3.2 (1/3)

g2m1 = 2βIbias = 2∗ 190

μA

V2 ∗12μm

0.5μm∗ ID/bias

= 9120μA

V2 ∗ ID/bias

rds1/2 =L

λLID/bias=

0.5μm

0.16μmV

∗1

ID/bias

= 3.125V∗1

ID/bias


Example 3.2 (2/3)

A2V

= 152 = (gm1 ∗ (rds1 ‖ rds2))2 =

gm1 ∗rds1

2

2

= 9120μA

V2 ∗ ID/bias ∗ [3.125V]2 ∗

1

2ID/bias

2

= 22266μA∗1

ID/bias

ID/bias =22266μA

152= 99μA


Example 3.2 (3/3)

Veff = Vin/DC =

√

√

√

2ID

β

=

√

√

√

√

2∗ 99μA

190μAV2 ∗

12μm0.5μm

= 208.4mV

This Veff is the DC level (large signal value) of the input!For the Veff of the pFETs one would need to use adifferent μCox resp. β!Book Chapter 3: Basic Current Mirrors and Single-Stage

Amplifiers 16

Example 3.2 alternative

For Common-Source with active load where rds1 = rds2:

AV =Ai

2=gm1rds1

2≈

1

λVeff

Veff =1

λ15=

0.5μm

0.16μmV 15

= 208.4mV

ID =1

2βV2

eff= 99μA


Source Follower

What are the conditions forthis circuit to work? Wouldyou call it an amplifier?


Small Signal Model


Simplified Small Signal ModelThis circuit is used to follow avoltage but with an offset andsmall output resistance.

AV = gm1(1

gm1‖

1

gs1‖ rds1 ‖ rds2)

=gm1

gm1 + gs1 + 1rds1

+ 1rds2

≈ 1

(but actually somewhat smaller)Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 21

Common Gate Amplifier

Similar to common-sourceamplifier but with a lessthan infinite inputresistance. Good for e.gterminating resistance forinput from a (e.g. 50Ω)cable or current input(transresistance).


Small Signal Model


Simplified Small Signal Model and AV

AV is approximately thesame as for the comonsource amplifier.

(vout − vs1)gds1 + voutGL − vs1(gm1 + gs1) = 0

AV =vout

vs1=gm1 + gs1 + gds1

GL + gds1≈

gm1

GL + gds1= gm1(RL ‖ rds1)


Small Signal Model and rin (1/2)

iin = vin(gm1 + gs1)− (vout − vin)gds1

= vin(gm1 + gs1 + gds1)−vin(gm1 + gs1 + gds1)

GL + gds1gds1

= vin(gm1 + gs1 + gds1)

1−gds1

GL + gds1

= vin(gm1 + gs1 + gds1)GL

GL + gds1

= vin(gm1 + gs1 + gds1)1

1 +gds1GLBook Chapter 3: Basic Current Mirrors and Single-Stage

Amplifiers 26

Small Signal Model ans rin (2/2)

gin =iin

vin= (gm1 + gs1 + gds1)

1

1 +gds1GL

≈ gm11

1 +gds1GL

rin ≈1

gm1(1 +

gds1

GL)


A realistic voltage source at input

Since the input resistance is not infinit, one needs to becareful here: if considering a resistance RS betweenideal source and common-gate amplifier, e.g. a 50Ωcable or simply a realistic source with rout > 0, AV willappear smaller! That is to say when you first meaure vinwithout connecting your source to the amplifier, thenconnect the amp, and then measure vout, you will notsee the AV computed before, but A′

V(still called AV in

the book)!

vs1

vin=

rin

rin +RS⇒ A′

V= AV

rin

rin +RS

A′V≈

gm1(RL ‖ rds1)

1 +RSgm1+gs1+gds1

1+RL/rds1


Source-Degenerated Current Mirror

A first attempt to providehigher output resistance(i.e. a better currentsource). Note: requireshigher voltage at output towork at all!


Small Signal ModelImportant point: tocompute small signalrout the input currentcan be consideredconstant and thus thesmall signal inputcurrent source is anopen circuit, i.e.vgs = −vs.

rout = rds2 [1 +RS(gm2 + gs2 + gds2)] ≈ rds2 [1 +RS(gm2 + gs2)]


Cascode Current Mirror

A second attempt toprovide higher outputresistance (i.e. a bettercurrent source). Note: alsorequires higher voltage atoutput to work at all but isless dependent on (largesignal) current!


Small Signal routThe analysis is based on the source degeneratedcurrent mirror where former RS is replaced by rds2

rout = rds4 [1 + rds2(gm4 + gs4 + gds4)] ≈ rds4rds2gm4 (3.37)

VOUT!> 2Veff +Vtn (3.42)


Cascode Gain Stage

Offers large gain for a single stage (depends on qualityof current source!) and limits voltage accross the inputdrive transistor (e.g. avoiding short channel effects).Book Chapter 3: Basic Current Mirrors and Single-Stage

Amplifiers 36

Small Signal ModelThe small signal analysismakes use of the previouscommon source andcommon gate analysis,simply multiplying theprvious two voltage gains:

vout

vin=

vs2

vin

vout

vs2(3.54)

≈ −gm1gm2

∗(rds1 ‖ rin2)(rds2 ‖ RL)


Small Signal Model Folded Cascode

Similar to the telescopic cascode gain stage but noninverting and with an extra load resistor (currentsource) diminishing the gain:

vout

vin=vs2

vin

vout

vs2(3.54) ≈ gm1gm2(rds1 ‖ rin2 ‖ RL)(rds2 ‖ RL)


Assuming High Quality Current Mirrors

vout

vin≈ −gm1gm2(rds1 ‖ rin2)(rds2 ‖ RL) ≈ −

1

2g2mr2ds


Diff Pair

Realizes a differential inputfor most integratedamplifiers: immunity tonoise/offsets that affectboth inputs (e.g. pick-upnoise on twisted-pair cablesetc.)


Small signal model

The book uses the T-model. Here comes an alternativededuction based on the ’normal’ model.

g vm +x

vx

g vm -xrdsrds

i+ i-

i+/− = (v+/− − vx)gm − vx1

rdsi+ − i− = (v+ − v−)gm (3.69)


Small signal model with resistive loads oneach branch

g vm +x

vx

g vm -xrdsrds

i+ i-

vout+ vout-

RL

RL

i+/− = −vout+/−1

RL= (v+/− − vx)gm + (vout+/− − vx)

1

rdsBook Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 43

Small signal model with resistive loads oneach branch

g vm +x

vx

g vm -xrdsrds

i+ i-

vout+ vout-

RL

RL

−vout+/−

1

RL+

1

rds

= v+/−gm − vx

gm +1

rds

−(vout+ − vout−) =(v+ − v−)gm

1RL

+ 1rds

≈ (v+ − v−)gmRLBook Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 44

Differential Gain Stage, TranscunductanceAmp

The simple analysis: consider thatthe current mirror faithfully copiesthe current through the left branchand consider a voltage source atthe output, then the current intothat voltage source is exactly theprevious iout = i+ − i− . With novoltage source, the difference incurrent has to flow trough theoutput resistance ⇒ vout = ioutrout.With (3.69) and (3.78)vout = gmvinrout ⇒AV ≈ gm(rds2 ‖ rds4) which is (3.79).


More Careful Analysis for rout (1/4)

(different from the book)

-g vm1 2

v2

rds1rds2

-g vm4 1rds4

iout

i1 vout

i2

-g vm2 2

1/gm3

v1

iout = i2 − i1



-g vm1 2

v2

rds1rds2

-g vm4 1rds4

iout

i1 vout

i2

-g vm2 2

1/gm3

v1

i1 = −1

rds4vout − gm4v1

i2 =1

rds2(vout − v2)− gm2v2

≈1

rds2vout − gm2v2

i2 = (v2 − v1)1

rds1+ gm1v2

≈ v11

rds1+ gm1v2

i2 = v1gm3



-g vm1 2

v2

rds1rds2

-g vm4 1rds4

iout

i1 vout

i2

-g vm2 2

1/gm3

v1

Using 2. and 3. term on previous pagefor i2:

v2 ≈gm3

gm1v1

using 1. and 2. term on previous pagefor i2 and gm1 = gm2

vout1

rds2≈ v1

1

rds1+ gm3v1 +

gm3gm2

gm1v1

≈ 2gm3v1

v1 ≈ vout1

2gm3rds2Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 48


-g vm1 2

v2

rds1rds2

-g vm4 1rds4

iout

i1 vout

i2

-g vm2 2

1/gm3

v1

Finally substituting all into iout = i2 − i1using the simplest expression for i2(3rd on page (2/4))

iout = vout1

2rds2+ vout

1

2rds2+

1

rds4vout

iout = vout(1

rds2+

1

rds4)


inf3410 — fall 2014€¦ · inf3410 — fall 2014 book chapter 3: basic current mirrors and...

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