Single Stage Amplifiers
•Basic Concepts
•Common Source Stage
•Source Follower
•Common Gate Stage
•Cascode Stage
Hassan Aboushady
University of Paris VI
References
• B. Razavi, “Design of Analog CMOS Integrated Circuits”,
McGraw-Hill, 2001.
H. Aboushady
University of Paris VI
1
Single Stage Amplifiers
•Basic Concepts
•Common Source Stage
•Source Follower
•Common Gate Stage
•Cascode Stage
Hassan Aboushady
University of Paris VI
Basic Concepts I
• Amplification is an essential function in most analog circuits !
• Why do we amplify a signal ?
• The signal is too small to drive a load
• To overcome the noise of a subsequent stage
• Amplification plays a critical role in feedback systems
In this lecture:
• Low frequency behavior of single stage CMOS amplifiers:
• Common Source, Common Gate, Source Follower, ...
• Large and small signal analysis.
• We begin with a simple model and gradually add 2nd order effects
• Understand basic building blocks for more complex systems.
H. Aboushady
University of Paris VI
2
Approximation of a nonlinear system
Input-Output Characteristic of a nonlinear system
y (t ) ≈ α 0 + α1 x (t ) + α 2 x 2 (t ) + ... + α n x n (t )
x1 ≤ x ≤ x2
In a sufficiently narrow range:
y (t ) ≈ α 0 + α1 x(t )
where α0 can be considered
the operating (bias) point and
α1 the small signal gain
H. Aboushady
University of Paris VI
Analog Design Octagon
H. Aboushady
University of Paris VI
3
Single Stage Amplifiers
•Basic Concepts
•Common Source Stage
•Source Follower
•Common Gate Stage
•Cascode Stage
Hassan Aboushady
University of Paris VI
Common Source Stage with Resistive Load
Vout = VDD − RD I D
M1 in the saturation region:
Vout = VDD − RD
M1 in limit of saturation: Vin1 − VTH
M1 in the
linear region:
H. Aboushady
µ nCox W
(Vin − VTH ) 2
2 L
µC W
(Vin1 − VTH ) 2
= VDD − RD n ox
2 L
Vout = VDD − RD µ nCox
W
L
2
⎡
⎤
Vout
V
V
V
(
−
)
−
⎢ in TH out
⎥
2 ⎦
⎣
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4
Common Source Stage with Resistive Load
M1 in deep
linear region:
Vout = VDD
Ron
VDD
=
W
Ron + RD 1 + µ C
RD (Vin − VTH )
n ox
L
H. Aboushady
University of Paris VI
Common Source Stage with Resistive Load
M1 in the saturation region:
Vout = VDD − RD
µ nCox W
2
L
(Vin − VTH ) 2
Small signal gain:
Av =
∂Vout
W
= − RD µ nCox (Vin − VTH )
∂Vin
L
= − g m RD
Same relation can be derived
from the small signal equivalent
circuit
Small signal model for
the saturation region
To minimize nonlinearity, the gain equation must be a weak
function of signal dependent parameters such as gm !
H. Aboushady
University of Paris VI
5
Example 1
Sketch ID and gm of M1 as a function of the Vin:
• M1 in the saturation region:
ID =
µ nCox W
2
(Vin − VTH )
2 L
∂I
W
g m = D = µ nCox (Vin − VTH )
∂VGS
L
H. Aboushady
• M1 in the linear region:
2
⎤
W⎡
Vout
I D = µ nCox ⎢(Vin1 − VTH )Vout −
⎥
2 ⎦
L⎣
∂I
W
g m = D = µ nCox Vout
∂VGS
L
VDD
Vout =
W
1 + µ nCox RD (Vin − VTH )
L
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Voltage Gain of a Common Source Stage
Av = − g m RD
Av = − 2 µ nCox
V
W
I D RD
ID
L
Av = − 2 µ nCox
W VRD
L ID
How to increase Av ?
Trade-offs:
• Increase W/L
Greater device capacitances.
• Increase VRD
Limits Vout swing.
• Reduce ID
Greater Time Constant.
H. Aboushady
University of Paris VI
6
Taking Channel Length Modulation into account
Calculating Av starting from the Large Signal Equations:
Vout = VDD − RD
Av =
µ nCox W
2
L
(Vin − VTH ) 2 (1 + λVout )
∂Vout
W
= − RD µ nCox (Vin − VTH )(1 + λVout )
∂Vin
L
− RD
µ nCox W
2
L
(Vin − VTH ) 2 λ
∂Vout
∂Vin
Av = − RD g m − RD I D λ Av
Av = −
RD g m
1 + RD λ I D
λ I D = 1 / rO
Av = − g m
rO RD
rO + RD
H. Aboushady
University of Paris VI
Taking Channel Length Modulation into account
Calculating Av starting from the Small Signal model:
g mV1 (rO // RD ) = −Vout
V1 = Vin
H. Aboushady
Av =
Vout
= − g m (rO // RD )
Vin
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Example 2
Assuming M1 biased in saturation,
calculate the small signal voltage gain :
• I1 : Ideal current source
Infinite Impedance
Av = − g m rO
• Intrinsic gain of a transistor:
This quantity represents the maximum voltage
gain that can be achieved using a single device.
I D1 =
µ nCox W
2
L
(Vin − VTH ) 2 (1 + λVout ) = I1
• Constant Current:
As Vin increases, Vout must decrease such that the
product remains constant
H. Aboushady
University of Paris VI
CS Stage with Current-Source Load
• Both transistors operate in the saturation region:
Av = − g m (rO1 // rO 2 )
• The output impedance and the minimum
required VDS of M2 are less strongly coupled than
the value and voltage drop of a resistor.
VDS 2,min = VGS 2 − VTH 2
• This value can be reduced to a few hundred millivolts by
simply increasing the width of M2.
•If rO2 is not sufficiently high, the length and width of M2 can be
increased to achieve a smaller λ while maintaining the same
overdrive voltage.
•The penalty is the large capacitance introduced by M2 at the
output node.
•Increasing L2 while keeping W2 constant increases rO2 and
hence the voltage gain, but at the cost of higher |VDS2|
required to maintain M2 in saturation
H. Aboushady
University of Paris VI
8
CS with Source Degeneration
Large Signal model:
Small Signal model:
∂I
∂I ∂VGS
Gm = D = D
∂Vin ∂VGS ∂Vin
VGS = Vin − I D RS
∂VGS
∂I
= 1 − D RS
∂Vin
∂Vin
⎛
∂I ⎞
⎜⎜1 − RS D ⎟⎟
∂Vin ⎠
⎝
Gm = g m (1 − RS Gm )
Gm =
∂I D
∂VGS
Gm =
gm
1 + g m RS
gm =
∂I D
∂VGS
Gm =
Av = −Gm RD
Av = −
H. Aboushady
ID
g mV1
=
Vin V1 + g mV1RS
Gm =
g m RD
1 + g m RS
gm
1 + g m RS
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CS with Source Degeneration
Gm =
gm
1
=
1 + g m RS 1 / g m + RS
for
RS >> 1 / g m
Gm ≈ 1 / RS
ID is linearized at the cost of lower gain.
Small Signal model including body effect
and channel length modulation:
I out = g mV1 − g mbVX −
VX
rO
= g m (Vin − I out RS ) + g mb (− I out RS ) −
Gm =
H. Aboushady
I out RS
rO
I out
g m rO
=
Vin RS + [1 + ( g m + g mb ) RS ]rO
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9
With and Without Source Degeneration
Gm =
g m rO
1 + [1 + ( g m + g mb ) RS ]rO
RS = 0
H. Aboushady
RS ≠ 0
University of Paris VI
Estimating Gain by Inspection
Av = −
Gain = −
g m RD
RD
=−
1 + g m RS
1 / g m + RS
Resistance seen at the Drain
Total Resistance in the Source Path
Example:
Av = −
H. Aboushady
RD
1 / g m1 + 1 / g m 2
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10
Output Resistance of Degenerated CS
V1 = − I X RS
The current flowing in rO :
I X − ( g m + g mb )V1
= I X + ( g m + g mb ) RS I X
VX = rO [ I X + ( g m + g mb ) RS I X ] + I X RS
Rout =
VX
= rO [1 + ( g m + g mb ) RS ] + RS
IX
Rout = [1 + ( g m + g mb )rO ]RS + rO
Rout ≈ ( g m + g mb )rO RS + rO
Rout = [1 + ( g m + g mb ) RS ]rO
H. Aboushady
University of Paris VI
Voltage Gain of Degenerated CS
The current through RS must equal that through RD:
I R D = I RS =
Vout
RD
VS = −Vout
RS
RD
VS
Vout
− ( g mV1 + g mbVBS )
RD
R
V
R
V
Vout = I rO rO − out RS
I rO = − out − [ g m (Vin + Vout S ) + g mbVout S ]
RD
RD
RD
RD
R
R
R
V
Vout = − out rO − [ g m (Vin + Vout S ) + g mbVout S ]rO − Vout S
RD
RD
RD
RD
The current through rO :
H. Aboushady
I rO = −
Vout
g m rO RD
=−
Vin
RD + RS + rO + ( g m + g mb ) RS rO
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11
Voltage Gain of Degenerated CS
Vout
g m rO RD
=−
Vin
RD + RS + rO + ( g m + g mb ) RS rO
Vout
g m rO
RD [ RS + rO + ( g m + g mb ) RS rO ]
=−
Vin
RS + [1 + ( g m + g mb ) RS ]rO RD + RS + rO + ( g m + g mb ) RS rO
Vout
= Gm (Rout // RD )
Vin
The output resistance
of a degenerated CS stage:
Rout = [1 + ( g m + g mb ) RS ]rO
The Transconductance
of a degenerated CS stage:
Gm =
I out
g m rO
=
Vin RS + [1 + ( g m + g mb ) RS ]rO
H. Aboushady
University of Paris VI
General expression to calculate Av by inspection
Lemma:
Av = −Gm Rout
Gm : the transconductance of
the circuit when the output is
shorted to grounded.
Rout : the output resistance
of the circuit when the input
voltage is set to zero.
• For high voltage gain the output resistance must be high!
A “buffer” is needed to drive a low-impedance load.
The source follower can operate as a voltage buffer.
H. Aboushady
University of Paris VI
12
Single Stage Amplifiers
•Basic Concepts
•Common Source Stage
•Source Follower
•Common Gate Stage
•Cascode Stage
Hassan Aboushady
University of Paris VI
Source Follower (Common Drain)
Large Signal Behavior
M1 turns on in saturation:
Vout = I D RS
Vout =
µ nCox W
2
L
(Vin − Vout − VTH ) 2 RS
To calculate gm :
W
∂V
∂V
∂Vout
= µ nCox (Vin − Vout − VTH )(1 − out − TH ) RS
L
∂Vin ∂Vin
∂Vin
(
)
VTH = VTH 0 + γ 2ΦF + VSB − 2ΦF
∂ VSB
γ
∂ VTH ∂ VTH ∂ VSB
=
=
∂ Vin ∂ VSB ∂ Vin 2 2ΦF + VSB ∂ Vin
∂ Vout
=η
∂ Vin
H. Aboushady
Since,
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Source Follower Voltage Gain
W
∂V
∂V
∂Vout
= µ nCox (Vin − Vout − VTH )(1 − out − TH ) RS
L
∂Vin ∂Vin
∂Vin
W
∂V
∂V
∂Vout
= µ nCox (Vin − Vout − VTH )(1 − out − η out ) RS
L
∂Vin
∂Vin
∂Vin
W
µ nCox (Vin − Vout − VTH ) RS
∂Vout
L
=
∂Vin 1 + µ C W (V − V − V ) R (1 + η )
n ox
in
out
TH
S
L
We also have,
g m = µ nCox
Av =
W
(Vin − Vout − VTH )
L
g m RS
1 + ( g m + g mb ) RS
H. Aboushady
University of Paris VI
Source Follower Voltage Gain
Small Signal Equivalent Circuit
Vout = [g mV1 + g mbVBS ]RS
= [g m (Vin − Vout ) − g mbVout ]RS
Av =
Since:
And for :
Vout
g m RS
=
Vin 1 + ( g m + g mb ) RS
g mb = η g m
g m RS >> 1
Av ≈
H. Aboushady
1
(1 + η )
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14
Source Follower Output Resistance
Rout : the output resistance when the input voltage is set to zero.
V1 = VBS = −VX
Rout =
I X − g mVX − g mbVX = 0
VX
1
=
I X g m + g mb
Body Effect decreases the output resistance of source followers.
VX ↓ ⇒ VGS ↑ and VTH ↓
⇒ ID ↑
H. Aboushady
University of Paris VI
Source Follower body effect
Rout : the output resistance when the input voltage is set to zero.
Small Signal Model Simplification
Note that the value of the current source gmbVbs is linearly
proportional to the voltage across it.
Rout =
H. Aboushady
1
1
1
//
=
g m g mb g m + g mb
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15
Source Follower Thévenin Equivalent
Av =
1
g mb
1
1
+
g m g mb
=
gm
g m + g mb
H. Aboushady
University of Paris VI
Channel Length Modulation in M1 and M2
Av =
H. Aboushady
1
// rO1 // rO 2 // RL
g mb
1
1
// rO1 // rO 2 // RL +
g mb
gm
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Source Follower Characteristics
+ High input impedance and Moderate output impedance
- Nonlinearity
- Limited voltage swing
Example:
VTH α VSB
PMOS source follower with VSB=0
Without the source follower stage:
VX > VGS 1 − VTH 1
µ p < µ n ⇒ g mp < g mn
⇒ Routp > Routn
With the source follower stage:
VX > VGS 2 + (VGS 3 − VTH 3 )
H. Aboushady
University of Paris VI
Low Load Impedance: CS vs SF
Source Follower Amplifier
AvSF ≈
RL
RL + 1 / g m
Common Source Amplifier
AvCS ≈ − g m RL
Assuming RL=1/gm
AvSF ≈ 1 / 2
AvCS ≈ −1
Source Followers are not necessarily efficient drivers.
H. Aboushady
University of Paris VI
17
Single Stage Amplifiers
•Basic Concepts
•Common Source Stage
•Source Follower
•Common Gate Stage
•Cascode Stage
Hassan Aboushady
University of Paris VI
Common Gate Stage
Large Signal Behavior
Vout = VDD − I D RD
Assuming M1 in saturation:
Vout = VDD −
µ nCox W
(Vb − Vin − VTH ) 2 RD
2 L
⎛
∂V
W
∂Vout
= − µ n Cox (Vb − Vin − VTH )⎜⎜ − 1 − TH
L
∂Vin
∂Vin
⎝
⎞
⎟⎟ RD
⎠
W
∂Vout
= µ n Cox (Vb − Vin − VTH )(1 + η )RD
L
∂Vin
Av = g m (1 + η ) RD
H. Aboushady
University of Paris VI
18
Common Gate Stage Input Resistance
Same as Output Resistance of Source Follower:
Rin =
1
g m + g mb
Body Effect:
• increases Av
• decreases Rin
H. Aboushady
University of Paris VI
Common Gate Gain
Small Signal Signal Equivalent Circuit
The current through RS is equal to -Vout / RD :
V1 −
Vout
RS + Vin = 0
RD
The current through rO is equal to -Vout / RD - gmV1 - gmbV1 :
⎡ −V
rO ⎢ out
⎣ RD
H. Aboushady
⎞ V
⎛ −V
rO ⎜⎜ out − g mV1 − g mbV1 ⎟⎟ − out RS + Vin = Vout
⎠ RD
⎝ RD
⎛
⎞⎤ V
R
− ( g m + g mb )⎜⎜Vout S − Vin ⎟⎟⎥ − out RS + Vin = Vout
RD
⎝
⎠⎦ RD
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19
Common Gate Gain
Common Gate Amplifier:
AvCG =
( g m + g mb )rO + 1
RD
RD + RS + rO + ( g m + g mb )rO RS
Degenerated Common Source Amplifier:
AvCS = −
g m rO
RD
RD + RS + rO + ( g m + g mb )rO RS
H. Aboushady
University of Paris VI
Common Gate Stage Input Resistance
Since V1 = -VX :
VX = RD I X + rO [I X − ( g m + g mb )VX ]
VX
RD + rO
=
I X 1 + ( g m + g mb )rO
Rin ≈
RD
1
+
( g m + g mb )rO ( g m + g mb )
• Assume RD = 0 :
Rin =
1
1 / rO + ( g m + g mb )
• Replace RD by ideal current source:
Rin = ∞
Rin of a common gate stage is low only if RD is small.
H. Aboushady
University of Paris VI
20
Common Gate Stage Output Impedance
Similar to Output Impedance of a
Degenerated Common Source Stage
Rout = ([1 + ( g m + g mb )rO ]RS + rO ) // RD
H. Aboushady
University of Paris VI
Single Stage Amplifiers
•Basic Concepts
•Common Source Stage
•Source Follower
•Common Gate Stage
•Cascode Stage
Hassan Aboushady
University of Paris VI
21
Biasing of a Cascode Stage
The cascade of CS stage and a CG stage is called “cascode”.
M1 : the input device
M2 : the cascode device
Biasing conditions:
• M1 in saturation:
VX = Vb − VGS 2
Vb − VGS 2 ≥ Vin − VTH 1
Vb ≥ Vin + VGS 2 − VTH 1
• M2 in saturation:
Vout − VX ≥ Vb − VX − VTH 2
Vout ≥ Vin − VTH 1 + VGS 2 − VTH 2
H. Aboushady
University of Paris VI
Cascode Stage Characteristics
Large signal behavior:
As Vin goes from zero to VDD
For Vin < VTH M1 and M2 are OFF
Vout =VDD
Output Resistance:
• Same common source stage with
a degeneration resistor equal to rO1
Rout = [1 + ( g m 2 + g mb 2 )rO 2 ]rO1 + rO 2
Rout ≈ ( g m 2 + g mb 2 )rO 2 rO1
• M2 boosts the output impedance of M1
by a factor of gmr02
• Triple cascode Rout ↑↑
difficult biasing at low supply voltage.
H. Aboushady
University of Paris VI
22
Cascode Stage Voltage Gain
Av = −Gm Rout
Gm ≈ g m1
Ideal Current Source:
Rout ≈ ( g m 2 + g mb 2 )rO 2 rO1
Av ≈ ( g m 2 + g mb 2 )rO 2 g m1rO1
Cascode Current Source:
Rout ≈ g m 2 rO 2 rO1 // g m3rO 3rO 4
Av ≈ g m1 ( g m 2 rO 2 rO1 // g m 3rO 3 rO 4 )
H. Aboushady
University of Paris VI
Shielding Property
Assume VX is higher than VY by ∆V.
Calculate the resulting difference between ID1 and ID2 (with λ ≠ 0 ).
µ nCox
2
µC
= n ox
2
I D1 − I D 2 =
W
(Vb − VTH ) 2 (λVDS 1 − λVDS 2 )
L
I D1 − I D 2
W
(Vb − VTH ) 2 (λ ∆VDS )
L
∆VPQ = ∆V
rO1
[1 + ( g m 3 + g mb3 )rO 3 ]rO1 + rO 3
I D1 − I D 2 =
H. Aboushady
≈
∆V
( g m3 + g mb 3 )rO 3
µ nCox W
λ ∆V
(Vb − VTH ) 2
2 L
( g m3 + g mb 3 )rO 3
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23
Folded Cascode
Simple Folded
Cascode
Folded Cascode
with biasing
Folded Cascode
with NMOS input
Large Signal Characteristics:
H. Aboushady
University of Paris VI
Output Resistance of Folded Cascode
Degenerated Common Source Stage:
Rout = [1 + ( g m1 + g mb1 )rO1 ]RS + rO1
Folded Cascode Stage:
M1
M2
RS
rO1 // rO3
Rout = [1 + ( g m 2 + g mb 2 )rO 2 ](rO1 // rO 3 ) + rO 2
H. Aboushady
University of Paris VI
24