Download - MOSFET: Transfer Function & Biasing
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4. Active Loads and IC MOS Amplifiers
ECE 102, Winter 2011, F. Najmabadi
Reading: Sedra & Smith: Chapter 7 (MOS portions)
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Progress towards an IC relevant amplifier
Resistors and capacitors take a lot of space on ICs:o Minimize (i.e., very few) R & C and small sizes (e.g., nF or smaller)
o Get rid of coupling capacitors by direct coupling between stages (makes biasing design complicated)
o Replace Rs with a current sourceo Still need to get rid of Cs!o What to do about RD?
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Current mirrors are the principle method for biasing in ICs
Identical MOS: Same k’n and Vt
Q1 is always in saturationVDS1 = VGS > VGS – Vt
Q2 has to be in saturation for current mirror to workVDS2 > VGS – Vt
refo ILWLWI
1
2
)/()/(
=
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Small signal model of Current Mirrors
Small signal response of an ideal current source is an open circuit!
However, Current mirrors are NOT ideal current sources as we used λvDS << 1 in our “bias” analysis
Bias Model Small Signal Model
Real Circuit
refo ILWLWI
1
2
)/()/(
=
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Small signal model of Current Mirrors
0
in flows :D1at KCL
111
11
=⇒−== gsogsmgsD
ogsm
vrvgvvrvg
Real Circuit Small Signal Model
2orR =circuitopen is sourcecurrent 0 2 gsmgs vgv ⇒=
Current sourcebecomes open circuit
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But what happens if we replace Irefcurrent source with “practical” elements?
Not a Practical Circuit Practical Circuit
Would practical elements that fix Iref change the small signal response?
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Small signal response of a practical current mirror
Practical Circuit
0 )||(
||in flows :D1at KCL
111
11
=⇒−== gsDogsmgsD
Dogsm
vRrvgvvRrvg
2orR =circuitopen is sourcecurrent 0 2 gsmgs vgv ⇒=
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Generalized Small signal Model ofCurrent Mirrors
Any circuit that “fixes Iref”
V ..
)1()( )/('21
saturationin Q1
.
G1
12
111
11
1
→⇒=⇒==⇒
+−=
⇒=
==
GGGSGS
DStnGSnD
GSDS
refD
vconstvconstvv
vVvLWki
vvconstii
λ
Small Signal Model
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Summary of Current Mirrors
Bias Model Small Signal Model
refo ILWLWI
1
2
)/()/(
=
Bias current goes through this leg
Signal current goes through this leg(∞ capacitor)
It is sufficient to consideronly Q2 in circuit calculations
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PMOS Version:
Summary of Current Mirrors
NMOS Version:
“Intuitive” Model
It is sufficient to consideronly Q2 in circuit calculations
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Biasing a CS Stage: Can we place a current mirror in the source circuit?
Typical bias of a discrete CS amplifier
Placing a current mirror in the source circuit will not Work!
Current mirror Bias works fine!
o A large R in the source circuit for small signal reducing the gain by (1+ gm1 rO2)
Bias Small Signal
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We need to Bias a CS Stage by placing a current mirror in the drain circuit!
Signal
Current mirror provides RD !
Current mirror sets ID = Io
However, a precise bias voltage should be applied to the Gate (corresponding to the ID set by the current source)
Several ways to do this
Bias
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Basic gain cell in IC
NMOS Version:
PMOS Version:)||( 211 oomv rrgA −=
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Bias point of CS amp with current mirror
||22
2
tpSGOV
GDDSG
VVVVVV−=
−=
222 2 )/('
21
OVVLWkI pD =
222
111
111
//
/2
DAo
DAo
OVDm
IVrIVr
VIg
===
2/1
1
11
21
)/('2
=
=
LWkI V
II
n
DOV
DD
Ignore Channel Width Modulation,o Fast and relatively accurate
method to find gm1 , ro1 and ro2o Cannot find VDS1 and VDS2
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Bias point of CS amp with current mirror
Include Channel Width Modulation,o Lengthy Analysiso Gives VDS1 and VDS2o See S&S Example 7.2 (pp500-504)o Can gain insight with load-line analysis
iD2
vDS2
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Bias point of CS amp with current mirror
12 DSSDDD vvV +=
Setting Vss = 0 (For simplicity), the load line (or load curve!) equation for Q1 is (note iD1 = iD2)
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Biasing CS amp with current mirror allows a very large RD without increasing VDD
Q2 in triodeQ2 in saturation
Load line for a discrete resistor of value ro2
Needed VDD
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Biasing a Source Follower in ICs
Current mirror Bias works fine! Small signal OK
Common-Drain (Source Follower) stages are biased with current mirror in the source circuit (as above)
Common-Gate stages are biased with current mirror in the drain circuit similar to CS amplifier.
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PMOS version of Basic gain cell in IC
NMOS CS Amp PMOS CS Amp NMOS CD Amp PMOS CD Amp
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Implementation of CS and Follower configurations on IC
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Cascode Amplifiers and Current Mirrors
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Cascode amplifier is a two-stage, CS-CG configuration
Cascode Configuration
CG stage
CS stage
signalbias
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Cascode amplifier is a two-stage, CS-CG configuration
Cascode Configuration
CG stage
CS stage
Small Signal Model
Small Signal Model configured as two-stage amplifier
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Open-Loop gain of a Cascode amplifier
122 )1( vrgv omo ⋅+=⇒
22112211 )1( omomomomi
ovo rgrgrgrg
vvA ⋅⋅⋅−≈⋅+⋅⋅−==
Node Voltage Method:
0122
1 =⋅−− vg
rvv
mo
oNode vo:
0011
1 =+⋅+ imo
vgrv
Node v1: iom vrgv 111 ⋅−=⇒
By KCL around Q2
Open Loop Gain (RL →∞, io = 0):
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Output Resistance of a Cascode amplifier
1122 )1( oomoo rrgrR +⋅+=
RRgr mo ++ )1(
21221 oomooo rrgrrR ++=
Exercise: Compute Ro from the small signal circuit of the previous slide(Attach a voltage source vx to the output and compute ix, see S&S pp 509-510)
R
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Amplifier models*Voltage Amplifier
Current Amplifier
Transconductance Amplifier
0for == oivoo ivAv
Avo : Open-Loop voltage gain
0for == oimo vvGi
Gm : Short-circuit transconductance
0for == oiiso viAi
Ais : Short-circuit current gain
Ideal Amplifier
Ri →∞Ro = 0
Ri →∞Ro →∞
Ri = 0 Ro →∞
* See S&S pp 26-27
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Amplifier modelsVoltage Amplifier
Current Amplifier
Transconductance Amplifier
isiovo
imis
omvo
ARRARGARGA
)/(===
From the point of view of circuit analysis: amplifier models are identical (The output stages are
Thevenin/Norton Equivalent)
omvo
oimivoo
RGARvGvAvi
====
0For 0
Avo , Ais , and Gm are related to each other
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Amplifier models*Voltage Amplifier
Transconductance Amplifier
0for == oivoo ivAv
0for == oimo vvGi
Alternate method to compute Avo:1. Set RL = 0 (short output)2. Compute Gm = io/vi3. Avo = Gm Ro
omvo RGA =
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Example: CS amplifier gain from Gm
gsmo vgi ⋅−=
oo rR =
omomvo rgRGA −==
gsmimo vGvGi ==
gsmgsm vgvG −=
mm gG −=
CS Amplifier
Alternate method to compute Avo:1. Set RL = 0 (short output)2. Compute Gm = io/vi3. Avo = Gm Ro
Short circuit
Transconductance Amplifier
Gm = − gm because current source is flipped
0
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Cascode amplifier gain from Gm
1
11
1
111
oim
ogsmo r
vvgrvvgi −⋅−=−⋅−=
21221
2211 ) 1(
oomoo
omom
i
om rrgrr
rgrgviG
+++
−==
21221 oomooo rrgrrR ++=22112211 )1( omomomomomvo rgrgrgrgRGA −≈+−==
121 and vvvv gsigs −==
Node equation at v1
imoooo
moo
immoo
ogsmgsm
o
vgrrrr
grrv
vgvgrr
rvvgvg
rv
22121
1211
11221
2
12211
1
1
11
0
++−=
−=
++
=+−+
KCL at D1
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Insight into operation of Cascode amplifier
For simplicity assume ro1 = ro2 = ro and gm1 = gm2 = gm
CG section has kept Gm the same (i.e., since Gm = io/vi , same current is passed through) but has increased Ro substantially resulting in a large increase in overall open-loop voltage gain!
CG section acted as a current buffer!
Cascode amplifier needs a large load!
mom
omom
oomoo
omomm g
rgrgrg
rrgrrrgrgG −=−≈
+++
−= 221221
2211 )() 1.(
221221 )2( omomooomooo rgrgrrrgrrR ≈+=++=Cascode amplifier:
oomm rRgG =−= and 1CS amplifier (1st
stage of cascode):
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Distribution of the gain in a Cascode Amp.
)||( 222 Lomv RrgA =
22222
221 1
1 om
L
mom
LoiL rg
Rgrg
RrRR +≈+
+==
)||( 2111 iomv RrgA −=
21 vvv AAA =
CG stages “reduces” the load seen by the CS stage by gm2ro2
CG StageCS Stage
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Cascode Amplifier needs a large load
)||( 222 Lomv RrgA =22222
221 1
1 om
L
mom
LoiL rg
Rgrg
RrRR +≈+
+==)||( 2111 iomv RrgA −=
For simplicity assume ro1 = ro2 = ro and gm1 = gm2 = gm
∞ ∞ − gmro gmro − (gmro)2
(gmro) ro = Ro ro − 0.5gmro gmro − 0.5(gmro)2
ro 2/gm − 2 0.5 gmro − gmro
RL Ri2 = RL1 Av1(CS) Av2 (CG) Av= Av1 Av2
Max. Gain
Same gainas a CS Amp.
PracticalGain
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Cascode amplifier is a two-stage, CS-CG configuration
Cascode Configuration
CG stage
CS stage
Small Signal Model
Small Signal Model configured as two-stage amplifier
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Cascode Amplifier needs a large load
)||( 222 Lomv RrgA =22222
221 1
1 om
L
mom
LoiL rg
Rgrg
RrRR +≈+
+==)||( 2111 iomv RrgA −=
For simplicity assume ro1 = ro2 = ro and gm1 = gm2 = gm
∞ ∞ − gmro gmro − (gmro)2
(gmro) ro = Ro ro − 0.5gmro gmro − 0.5(gmro)2
ro 2/gm − 2 0.5 gmro − gmro
RL Ri2 = RL1 Av1(CS) Av2 (CG) Av= Av1 Av2
Max. Gain
Same gainas a CS Amp.
PracticalGain
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Cascode amplifier needs a large load to get a high gain
RL = ro3
Av ≈ − gmro
Gain did not increase compared to a CS amplifier.
This is still a useful circuit because of its high gain-bandwidth (we see this later).
To get a high gain, Av = − 0.5(gmro)2 , we need to increase the small-signal resistance of the current mirror to ≈ (gmro) roo Cascode current mirror
Current Mirror
Cascode
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Cascode Current mirror Identical MOS: Same k’n and Vt ,
and
Usually: (W/L)1 = (W/L)3 and (W/L)2 = (W/L)4
1
2
3
4
)/()/(
)/()/(
LWLW
LWLW
=
Q1 and Q3 are always in saturation Q2 and Q4 have to be in saturation
for current mirror to work VDS2 > VGS – Vt
VDS4 > VGS – Vt
Straight forward to show
refo ILWLWI
1
2
)/()/(
=
Exercise: For VSS = 0, show that a single current mirror (no cascoding) works only if VD2 > VOV and a cascode current mirror requires VD4 > Vt + 2VOV
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Small signal resistance of a cascode current mirror is quite large
Equivalentcircuit
Small-signalcircuit
2244 )1( oomoo rrgrR +⋅+=
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Cascode amplifier with a cascode current mirror
Cascodeamplifier
Cascodecurrent mirror
Signal
Bias
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Cascode amplifier with a cascode current mirror
A high gain, Av ≈ − 0.5(gmro)2 , high gain-bandwidth circuit.
Draw-back: Low voltage headroom because 4 MOS should be in active for a given VDD
Cascodeamplifier
Cascodecurrent mirror
4433 )1( oomo rrgr +⋅+
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Folded Cascode increases voltage overhead
PMOS CG stageNMOS CS stageBiased with I1 – I2
Exercise: Draw the small-signal circuit of a folded cascode and show that it is exactly the same as a regular cascode.