INF3410 — Fall 2015
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
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 3
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?
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 5
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?
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 6
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
Current Mirror Small Signal Model, Example3.1 (2/4)
Continued detour solution (shorter in the book):
Iout =1
2βV2
eff
=10μm/0.4μm∗ 190μA/V2 ∗ 205mV2
2= 100μA
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 8
Current Mirror Small Signal Model, Example3.1 (3/4)
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!!!
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 9
Current Mirror Small Signal Model, Example3.1 (4/4)
Continued correct detour solution:
rout =1
λIout
=L
λLIout
=0.4μm
0.16μm/V ∗ 100μA= 0.025V/μA = 25kΩ
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 10
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 11
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?
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 12
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
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 14
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
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 15
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
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 17
Exam 2013, task 2 and 3
We’ll have a look at the circuits in B) and D) in the tasks2 and 3. Circuits A) and C) you do not yet understand,but maybe we can try to get an first intuitive feeling forthem as well ...
vout
vout
vout
-
+
AV
vbias1
vbias2
vout
vin
vin
vin
vin
vbias2
vbias1
vbias1
A) B) C) D)
Q1
Q2
Q1 Q1 Q1
Q2
Q2
Q2Q3Q3
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 18
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 19
Source Follower
What are the conditions forthis circuit to work? Wouldyou call it an amplifier?
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 20
Small Signal Model
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 21
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 22
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 23
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).
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 24
Small Signal Model
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 25
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)
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 26
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 27
Small Signal Model and rin (2/2)
gin =iin
vin= (gm1 + gs1 + gds1)
1
1 +gds1GL
≈ gm11
1 +gds1GL
rin ≈1
gm1(1 +
gds1
GL)
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 28
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
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 29
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 30
Source-Degenerated Current Mirror
A first attempt to providehigher output resistance(i.e. a better currentsource). Note: requireshigher voltage at output towork at all!
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 31
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)]
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 32
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 33
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!
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 34
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)
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 35
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 36
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 37
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)
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 38
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)
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 39
Assuming High Quality Current Mirrors
vout
vin≈ −gm1gm2(rds1 ‖ rin2)(rds2 ‖ RL) ≈ −
1
2g2mr2ds
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 40
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 41
Diff Pair
Realizes a differential inputfor most integratedamplifiers: immunity tonoise/offsets that affectboth inputs (e.g. pick-upnoise on twisted-pair cablesetc.)
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 42
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)
Book 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
i+/− = −vout+/−1
RL= (v+/− − vx)gm + (vout+/− − vx)
1
rdsBook Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 44
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 45
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).
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 46
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
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 47
More Careful Analysis for rout (2/4)
-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
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 48
More Careful Analysis for rout (3/4)
-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 49
More Careful Analysis for rout (4/4)
-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)
Book Chapter 3: Basic Current Mirrors and Single-StageAmplifiers 50