EE338L CMOS Analog Integrated Circuit Design
Lecture 6, Single-Stage Amplifiers (3)
Cascode Amplifiers
We will cover different cascode amplifiers, including
1.
2.
3.
4.
Simple cascode amplifier
Multi-level cascode amplifier
Gain boosted cascode amplfier
Folded cascode amplifier
1. Simple cascode amplifier
Vdd
IB
vout
M2
VB
M1
vin
Large signal behavior (Vin fixed to VG1, Vout (VDS) sweeping from 0 to 3V)
ID,SIMPLE
ID
ID,CASCODE
ID,SIMPLE
VG1
I
II
III
VG2
M1A
M2B
ID,CASCODE
VG1
M1B
VDS
Region I: M1B and M2B both in triode; Region II, M1B in saturation, M2B in
triode; Region III, M1B and M2B both in saturation
S. Yan, EE338L
1
Lecture 6
Small signal analysis
We will calculate small signal
i)
output resistance,
ii)
transconductance (when output is shorted to a fixed DC voltage),
iii)
DC voltage gain (when the output is open).
i) Output resistance
We have derived earlier,
rout  ( g m 2  g mb2 )rds1rds 2  rds1  rds 2 s
 g m 2 (1  2 )rds1rds 2  rds1  rds 2  g m 2 (1  2 )rds 2  1rds1  rds 2
ii) Transconductance (when output is shorted to a fixed DC voltage or AC
ground)
Short the output to an AC ground, and draw the small signal diagram as shown
above. According to KCL we can list the following equations,
 iout  i21  i22  i23  i11  i12
(1)
i11  g m1vgs1  g m1vin
(2a)
i12  gds1vs 2
(2b)
where
i21=gm2vgs2 i23=gmb2vbs2
vg2
vout=0
iout
i22
vgs2
vb2
gds2
vbs2
vs2
vin
vgs1
vbs1=0
i12
i11=gm1vgs1
= gm1vin
gmb1vbs1
=gmb10=0
i21  g m 2vgs 2  g m 2 (vg 2  vs 2 )   g m 2vs 2
S. Yan, EE338L
vb1
gds1
2
(2c)
Lecture 6
i22  gds 2vds 2  gds 2 (vd 2  vs 2 )   gds 2vs 2
(2d)
i23  gmb2vbs 2  gmb2 (vb 2  vs 2 )   gmb2vs 2
(2e)
From Eq. (1), we have
iout  (i21  i22  i23 )
(3a)
Substitute Eqs. (2c)-(2e) into Eq. (3a), we have,
iout  (i21  i22  i23 )  ( g m 2  gmb2  g ds 2 )vs 2
(3b)
From Eq. (1), we have
i21  i22  i23  i11  i12
(4a)
Substitute Eqs. (2a)-(2e) into Eq. (4a),
 ( g m 2  g mb2  gds 2 )vs 2  gm1vin  g ds1vs 2
(4b)
Solving Eq. (4b), we get
vs 2  
g m1
vin
g m 2  g mb 2 g ds 2  g ds1
(5)
Substitute Eq. (5) into Eq. (3b),
iout  ( g m 2  g mb2  g ds 2 )vs 2   g m1
g m 2  g mb2  g ds 2
vin
g m 2  g mb2  g ds 2  g ds1


g ds1
vin
  g m1 1 
 g m 2  g mb2  g ds 2  g ds1 
(6)
Thus the transconductance of the cascode amplifier is
Gm 
iout
g m 2  g mb 2  g ds 2
  g m1
vin
g m 2  g mb 2  g ds 2  g ds1


g ds1

  g m1 1 
 g m 2  g mb 2  g ds 2  g ds1 
  g m1
(7)
Observation: Compared with a single-transistor common source amplifier with a
transconductance of |Gm|=gm1 (Note that Gm is the transcoducance of the
amplifier, and gm is the transcoducance of the transistor), the transcoductance
of cascode amplifier is slightly less, whose transconductance Gm is given by


g ds1
  (90% to 99%) g m1 .
Gm   g m1 1 
 g m 2  g mb2  g ds 2  g ds1 
S. Yan, EE338L
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Lecture 6
iii) DC voltage gain (when the output is open)
i21=gm2vgs2 i23=gmb2vbs2
vg2
vout
i22
vb2
vgs2
gds2
vbs2
vs2
vin
vb1
gds1
vgs1
vbs1=0
i12
i11=gm1vgs1
= gm1vin
gmb1vbs1
=gmb10=0
With the output node open, according to KCL we can list the following equations,
0  i21  i22  i23  i11  i12
(8)
where i11, i12, i21, i22, and i23 are expressed by
i11  g m1vgs1  g m1vin
(9a)
i12  gds1vs 2
(9b)
i21  g m 2vgs 2  g m 2 (vg 2  vs 2 )   g m 2vs 2
(9c)
i22  gds2vds 2  gds2 (vd 2  vs 2 )  gds2 (vout  vs 2 )
(9d)
i23  gmb2vbs 2  gmb2 (vb 2  vs 2 )   gmb2vs 2
(9e)
From Eq. (8), we have
0  i11  i12 ,
(10)
Substitute Eqs. (9a) and (9b) into Eq. (10), we have
0  i11  i12  gm1vin  gds1vs 2
(11)
Rearrange the above equation,
vs 2  
g m1
vin
g ds1
(12)
From Eq. (8), we have
0  i21  i22  i23
(13)
Substitute Eqs. (9c) and (9e) into Eq. (13),
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Lecture 6
 ( gm2  gmb2  gds 2 )vs 2  gds 2vout  0
(14)
Rearrange Eq. (14), and substitute vs2 with Eq. (12)
 g  g mb2 
g  g  g mb2 
vout   m 2
 1vs 2   m1  m 2
 1vin
g ds 2
g ds1 
g ds 2



(15a)
Or
vout
g  g  g mb2 
  m1  m 2
 1
vin
g ds1 
g ds 2

  g m1rds1[( g m 2  g mb2 )rds 2  1]
Av 
(15b)
Note that, gm 2  gmb2  gm 2  2 gm 2  gm2 (1  2 ) , Eq. (15b) can be written as
Av 
vout
g  g (1  2 ) 
  m1  m 2
 1
vin
g ds1 
g ds 2

(15c)
  g m1rds1[ g m 2 (1  2 )rds 2  1]
Observation: Assuming the load of the casode amplifier is an ideal current
source, the voltage gain of the cascode amplifier is improved compared with
single transistor common source amplifier.
Av,cascode_ amp  A2 single_ transistor_ amp
S. Yan, EE338L
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Lecture 6
2. Multi-level cascode amplifier
vout
Large signal behavior
M3
M2
M1
vin
VB2
VB1
Small signal analysis
i) Output resistance
rout  g m3 (1  3 )g m 2 (1   2 )rds 2  1rds1  rds 2  1rds3  rds3
ii) Transconductance


g ds1
 g m1
Gm  1 
 g m 2  g mb2  g ds 2  g ds1 
iii) Voltage gain
Av  g m1rds1[ g m 2 (1   2 )rds 2  1][ g m3 (1  3 )rds3  1]
Observation: Assuming the load of the amplifier is an ideal current source, the
voltage gain of the three-transistor multi-level cascode amplifier is much
improved compared with single transistor common source amplifier.
Av,cascode_ amp  A3single_ transistor_ amp
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Lecture 6
3. Gain boosted cascode amplifier
Vdd
IB
vout
M2
A
M1
vin
VB
Large signal behavior (Vin fixed to VG1, Vout (VDS) sweeping from 0 to 3V)
ID,SIMPLE
ID
ID,CASCODE
ID,CASBOOSTED
VCAS
M2C
ID,CASBOOSTED
A
I
II
III
VG1
M1C
VDS
Region I: M1C and M2C both in triode; Region II, M1C in saturation, M2C in
triode; Region III, M1C and M2C both in saturation
S. Yan, EE338L
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Lecture 6
ID
ID,SIMPLE
ID,CASCODE
ID,CASBOOSTED
VDS
Zoomed-in view of drain currents vs. VDS of simple amplifier, cascode amplifier,
and gain boosted cascode amplifier
Small signal analysis
We try to obtain,
i) Output resistance,
ii) Transconductance (when output is shorted to a fixed DC voltage),
iii) DC voltage gain (when the output is open).
of the gain boosted cascode amplifier.
i) Derive the small signal output impedance, rout.
a) Set input voltage to zero (or short vin to ground).
b) Draw the small signal diagram as shown below.
c) Apply itst at the output node.
d) Calculate output voltage vtst.
Note that the current through gds1, i12, equals to itst,
S. Yan, EE338L
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Lecture 6
i22=gmb2vbs2
vg2
A
vtst(vout)
i23= itst- i21- i22
i21=gm2vgs2
vb2
gds2
vgs2
vs2
itst
i12=itst
gds1
vin
vgs1=0
i11=gm1vgs1=0
Small signal equivalent circuit diagram for calculating output resistance
vs 2  i12 / g ds1  itst / g ds1 .
(1)
Note that, vs2 applies to the negative input of the amplifier A. The output
voltage of the amplifier A is,
v g 2   Avs 2 .
(2)
The vgs of M2, vgs2, is given by,
vgs 2  vg 2  vs 2  ( A  1)vs 2
(3)
Thus, i21 is given by,
i21  g m 2vgs 2   g m 2 ( A  1)vs 2 .
(4)
The vbs of M2, vbs2, is
vbs 2  vb 2  vs 2  vs 2 ,
(5)
Thus, i22 is given by,
i22  g mb2vbs 2   g mb2vs 2 .
(6)
According to KCL, i21+i22+i23 = itst. We have,
i23  itst  i21  i22  itst   gm 2 ( A  1)vs 2    gmb2vs 2   itst  gm2 ( A  1)  gmb2 vs 2
(7)
Thus the drain-source voltage of M2, vds2, is given by,

1 
vds 2  i23 / g ds 2  itst 1  g m 2 ( A  1)  g mb2 

g ds1 

(8)
Note that, the voltage at the output node, vtst, is given by,
S. Yan, EE338L
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Lecture 6
vtst  vds1  vds 2  vs 2  vds 2 
itst

g ds1
itst  g m 2 ( A  1)  g mb 2 
itst
g ds1
g ds 2
(9)
 g ( A  1)  g mb 2
1
1 
  m2


itst
g ds1 g ds 2
g ds1 g ds 2 

Thus, the output impedance (resistance) is given by,
rout 
vtst g m 2 ( A  1)  g mb 2
1
1



 g m 2 ( A  1)  g mb 2 rds1rds 2  rds1  rds 2
itst
g ds1 g ds 2
g ds1 g ds 2
(10)
ii) Transconductance (when output is shorted to a fixed DC voltage)
Short the output node of the small signal equivalent circuit to ground, we can
draw Fig. 3.
iout
i21=gm2vgs2 i22=gmb2vbs2 i23=vs2gds2
vg2
A
Vout=0
vb2
gds2
vgs2
vs2
gds1
vin
vgs1
i11=gm1vgs1
i12=vs2gds1
Fig. 3. Small signal diagram to calculate transconductance Gm
From Fig. 3, we can list the following equation,
 iout  i21  i22  i23  i11  i12
(11)
Copy Eqs. (4), and (6) below for easy reference,
i21  g m 2vgs 2   g m 2 ( A  1)vs 2 .
(12)
i22  g mb2vbs 2   g mb2vs 2 .
(13)
From Fig. 3, we can list equations for i11, i12, and i23,
i11  g m1vgs1  g m1vin ,
(14)
i12  gds1vs 2 ,
(15)
S. Yan, EE338L
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Lecture 6
i23   gds2vs 2 ,
(16)
Substitute Eqs. (12)-(16) into Eq. (11), we have,
 iout  gm 2 ( A  1)  gmb2  gds 2 vs 2  gm1vin  gds1vs 2
(17)
Solving Eq. (17), we obtain,
g m1

v


vin
s
2

g m 2 ( A  1)  g mb 2  g ds 2  g ds1


g m 2 ( A  1)  g mb 2  g ds 2
i   g v  g v    g
vin
out
m1 in
ds1 s 2
m1

g m 2 ( A  1)  g mb 2  g ds 2  g ds1
(18)
Thus
Gm 
iout
g m 2 ( A  1)  g mb2  g ds 2
  g m1
vin
g m 2 ( A  1)  g mb2  g ds 2  g ds1
Note that,
g m 2 ( A  1)  g mb 2  g ds 2

g m 2 ( A  1)  g mb 2  g ds 2  g ds1 1 
(19)
1
 1 , as
g ds1
g m 2 ( A  1)  g mb 2  g ds 2
gm 2 ( A  1)  g mb2  gds 2  gds1 , thus Eq. (19) can be re-written as,
Gm 
iout
g m 2 ( A  1)  g mb 2  g ds 2
  g m1
vin
g m 2 ( A  1)  g mb 2  g ds 2  g ds1


g ds1
   g m1
  g m1 1 
 g m 2 ( A  1)  g mb 2  g ds 2  g ds1 
,
(20)
iii) DC voltage gain (when the output is open).
Small signal voltage gain, Av = Gm rout.
Multiplying Eq. (10) and Eq. (20),

g m 2 ( A  1)  g mb 2  g ds 2  g m 2 ( A  1)  g mb 2
1
1 


Av  Gm rout    g m1


g m 2 ( A  1)  g mb 2  g ds 2  g ds1 
g ds1 g ds 2
g ds1 g ds 2 


g m 2 ( A  1)  g mb 2  g ds 2  g m 2 ( A  1)  g mb 2  g ds 2  g ds1 


   g m1
g m 2 ( A  1)  g mb 2  g ds 2  g ds1 
g ds1 g ds 2


g ( A  1)  g mb 2  g ds 2
  g m1 m 2
g ds 2 g ds1
(21)
Eq. (21) can also be written as,
S. Yan, EE338L
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Lecture 6
Av  Gm rout   g m1
g m 2 ( A  1)  g mb 2  g ds 2
g ds 2 g ds1
(22)
  g m1g m 2 ( A  1)  g mb 2 rds1rds 2  rds1
4. Folded-cascode amplifier
Basic folded-cascode amplifier:
VDD
M1
VDD
RD
vin
RD
vout
M1
vin
M2
M2
VCAS
I1
(a)
vout
VB
(b)
VCAS
I1
Fig. 1
Folded-cascode circuit with proper biasing with the source terminal of M1 at
VDD (a) and at a suitable bias voltage at VB.
Why choose folded-cascode amplifier instead of telescopic configuration?
 More freedom to choose the DC input voltage at vin (such as Fig. 1(a)).
 Higher voltage swing.
 Convenience in shorting the input and the output in feedback configurations.
Large signal behavior
Fig. 2 Large-signal characteristics of folded cascode
In Fig. 2, I1 is the current flowing through M3 and is equal to the sum of ID1 and
ID2, VTH1=VT1.
S. Yan, EE338L
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Lecture 6




Vin > VDD-|VT1|, M1 is off and M2 carries all of I1, yielding Vout=VDD-I1RD.
For Vin<VDD-|VT1|, M1 turns on in saturation.
As Vin drops, ID2 decreases further, falling to zero if ID1=I1 (Vin=Vin1).
If Vin<Vin1, M1 enters triode.
Small signal behavior
Vdd
IB
vout
M1
VB
M2
vin
VG2
M3
VG3
i21=gm2vgs2= -gm2vs2
i22=gmb2vbs2= -gmb2vs2
vg2=vb2=0
vsg1
vin(vg1)
vout
gds2 vgs2
gds1
vs2
i11=gm1vsg1= -gm1vin
gds3
i23=gds2(vout-vs2)
i12=-gds1vs2
Fig. 3 Small signal diagram of the folded-cascode amplifier
i) Output resistance
rout  g m 2 (1   2 )rds 2  1(rds1 || rds 3 )  rds 2
S. Yan, EE338L
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Lecture 6
ii) Transconductance (when output is shorted to a fixed DC voltage)
Gm   g m1
g m 2  g mb 2  g ds 2
g m 2  g mb2  g ds 2  g ds1  g ds3


g ds1  g ds3

  g m1 1 
g

g

g

g

g
m2
mb 2
ds 2
ds1
ds 3 

  g m1
iii) DC voltage gain (when the output is open)
Av  Gm rout   g m1{[ g m 2 (1   2 )rds 2  1]( rds1 || rds3 )  rds 2 )
Example.
VDD
IB1
vout
M2
vin
M3
M1
Fig. 1 Example circuit 1
In the above Figure, the small-signal parameters of M1 to M3 are shown in the
following table:
Mi
Gate-to-Source
transconductance
Bulk
transconductance
Drain-to-Source
transconductance
gmi
gmbi
gdsi=1/rdsi
1) Draw the small-signal diagram.
2) Derive the low frequency small signal voltage gain
v out
.
v in
3) Derive the small signal output resistance.
S. Yan, EE338L
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Lecture 6
Solution:
1) The small signal diagram is drawn as below.
i32=gds3vg2
vg2
i21=gm2vgs2
i22=gmb2vbs2=-gmb2vs2
vout
gds3 vgs2
gds2
vb2=0
vs2
i31=gm3vgs3=gm3vs2
vin
vgs1
vgs3
gds1
i11=gm1vgs1=gm1vin
i23=gds2(vout-vs2)
i12=gds1vs2
Fig. 2 Small-signal diagram
2) and 3)
From Fig. 2, we can list the following equation,
v gs 2  v g 2  v s 2  g m 3v s 2 / g ds 3  v s 2  (g m 3 / g ds 3  1)v s 2 .
(1)
Therefore, it is equivalent to the gain boosted cascode amplifier explained
g
in the previous section with A  m 3
. Thus, the voltage gain and output
g ds 3
resistance is calculated by substituting A in the previous results for the gain
boosted cascode amplifier:
Av  g m1 g m 2 (g m3 rds 3  1)  g mb 2 rds1rds 2  rds1 
(2)
rout  g m 2 (g m3 rds 3  1)  g mb 2 rds1rds 2  rds1  rds 2
(3)
v out
and small signal output
v in
resistance from scratch by listing nodal equations.
You can also derive small signal voltage gain
S. Yan, EE338L
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Lecture 6
Example:
The small-signal parameters of M1 to M3 in the circuit are listed in the following
table:
Gate-to-Source
transconductance
Bulk
transconductance
Drain-to-Source
transconductance
gm1
gmb1
gds1=1/rds1
M1
VDD
IB2
RD
vout
VB
A
M1
Iin
IB1
Fig. 1
1) Draw the small-signal diagram.
2) What is the input impedance?
3) What is the transimpedance gain
v out
?
i in
Solution:
1)
i11=gm1vgs1 i12=gmb1vbs1=-gmb1vs1
A
iin
vg1
vgs1
gds1
vout
vb1=0
RD
vs1
i13=gds1(vout-vs1)
iin
Fig. 2 Small-signal diagram
2)
S. Yan, EE338L
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Lecture 6
According to KCL, we have,
i tst  (i11  i12  i13 )
(1)
From previous discussions, we can also have,
v gs1  ( A  1)v tst
(2)
i11=gm1vgs1 i12=gmb1vbs1=-gmb1vtst
itst
vg1
A
vgs1
gds1
vout
vb1=0
RD
vs1=vtst
itst
i13=gds1(vout-vtst)
vtst
Fig. 3 Small-signal diagram to calculate the input impedance
From Fig. 3 and (2), we have
i11  g m1v gs1  g m1( A  1)v tst
(3)
i12  g mb 1v tst
(4)
i13  gds1(v out  v tst )  gds1(i tstRD  v tst )
(5)
Substitute Eqs. (3)-(5) into Eq. (1), we have
i tst  g m1 ( A  1)v tst  g mb 1v tst  g ds1RD i tst  g ds1v tst  0
(6a)
Simplify Eq. (6a), we have
i tst (1  g ds1RD )  [g m1 ( A  1)  g mb 1  g ds1 ]v tst
(6b)
Thus
rin 
v tst
1  g ds1RD

i tst g m1 ( A  1)  g mb 1  g ds1
(7)
3) The transimpedance is given by
v out
 RD
i in
S. Yan, EE338L
(8)
17
Lecture 6