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

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 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.

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 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

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Analog Design Octagon

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 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

µ nCox W
M1 in the saturation region: Vout = VDD − RD (Vin − VTH ) 2
2 L
µC W
M1 in limit of saturation: Vin1 − VTH = VDD − RD n ox (Vin1 − VTH ) 2
2 L

M1 in the W ⎡ 2
Vout ⎤
Vout = VDD − RD µ nCox (V − V )V
⎢ in TH out − ⎥
linear region: L 2 ⎦
⎣
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 Common Source Stage with Resistive Load

M1 in deep Ron VDD

Vout = VDD =
linear region: Ron + RD 1 + µ C W
n ox RD (Vin − VTH )
L
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Common Source Stage with Resistive Load

M1 in the saturation region:
µ nCox W
Vout = VDD − RD (Vin − VTH ) 2
2 L
Small signal gain:
∂Vout W
Av = = − RD µ nCox (Vin − VTH )
∂Vin L
= − g m RD
Same relation can be derived Small signal model for
the saturation region
from the small signal equivalent
circuit

To minimize nonlinearity, the gain equation must be a weak

function of signal dependent parameters such as gm !

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 Example 1
Sketch ID and gm of M1 as a function of the Vin:

• M1 in the saturation region: • M1 in the linear region:

µ nCox W W⎡ Vout2
⎤
ID = (Vin − VTH ) 2 I D = µ nCox ⎢(Vin1 − VTH )Vout − ⎥
2 L L⎣ 2 ⎦
∂I W ∂I W
g m = D = µ nCox (Vin − VTH ) g m = D = µ nCox Vout
∂VGS L ∂VGS L
VDD
Vout =
W
1 + µ nCox RD (Vin − VTH )
H. Aboushady L University of Paris VI

Voltage Gain of a Common Source Stage

Av = − g m RD
W V
Av = − 2 µ nCox I D RD
L ID
W VRD
Av = − 2 µ nCox
L ID

How to increase Av ?

Trade-offs:

• Increase W/L Greater device capacitances.

• Increase VRD Limits Vout swing.

• Reduce ID Greater Time Constant.

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 Taking Channel Length Modulation into account
Calculating Av starting from the Large Signal Equations:
µ nCox W
Vout = VDD − RD (Vin − VTH ) 2 (1 + λVout )
2 L

∂Vout W
Av = = − RD µ nCox (Vin − VTH )(1 + λVout )
∂Vin L
µ nCox W ∂Vout
− RD (Vin − VTH ) 2 λ
2 L ∂Vin

Av = − RD g m − RD I D λ Av

RD g m λ I D = 1 / rO rO RD
Av = − Av = − g m
1 + RD λ I D 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 Vout

Av = = − g m (rO // RD )
V1 = Vin 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.

µ nCox W
I D1 = (Vin − VTH ) 2 (1 + λVout ) = I1
2 L
• Constant Current:
As Vin increases, Vout must decrease such that the
product remains constant
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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
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 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 D ∂I ⎞
Gm = ⎜⎜1 − RS D ⎟⎟ ∂I D
∂VGS ∂Vin ⎠ ID g mV1
⎝ gm = Gm = =
∂VGS Vin V1 + g mV1RS
Gm = g m (1 − RS Gm )

gm gm
Gm = Av = −Gm RD Gm =
1 + g m RS 1 + g m RS
g m RD
Av = −
1 + g m RS
H. Aboushady University of Paris VI

CS with Source Degeneration

gm 1
Gm = = for RS >> 1 / g m Gm ≈ 1 / RS
1 + g m RS 1 / g m + RS

ID is linearized at the cost of lower gain.

Small Signal model including body effect

and channel length modulation:
VX
I out = g mV1 − g mbVX −
rO
I out RS
= g m (Vin − I out RS ) + g mb (− I out RS ) −
rO

I out g m rO
Gm = =
Vin RS + [1 + ( g m + g mb ) RS ]rO

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 With and Without Source Degeneration

g m rO
Gm =
1 + [1 + ( g m + g mb ) RS ]rO

RS = 0 RS ≠ 0

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Estimating Gain by Inspection

g m RD RD
Av = − =−
1 + g m RS 1 / g m + RS

Resistance seen at the Drain

Gain = −
Total Resistance in the Source Path

Example:

RD
Av = −
1 / g m1 + 1 / g m 2

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 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

VX
Rout = = 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:
Vout
I R D = I RS =
RD
VS
RS
VS = −Vout
RD

The current through rO :

Vout
I rO = −− ( g mV1 + g mbVBS )
RD
V R R V
I rO = − out − [ g m (Vin + Vout S ) + g mbVout S ] Vout = I rO rO − out RS
RD RD RD RD
V R R R
Vout = − out rO − [ g m (Vin + Vout S ) + g mbVout S ]rO − Vout S
RD RD RD RD

Vout g m rO RD
=−
H. Aboushady
Vin RD + RS + rO + ( g m + g mb ) RS rO University of Paris VI

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 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 Rout = [1 + ( g m + g mb ) RS ]rO

of a degenerated CS stage:

The Transconductance I out g m rO

Gm = =
of a degenerated CS stage: 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 Rout : the output resistance

the circuit when the output is of the circuit when the input
shorted to grounded. 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

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 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
µ nCox W
Vout = (Vin − Vout − VTH ) 2 RS
2 L
To calculate gm :
∂Vout W ∂V ∂V
= µ nCox (Vin − Vout − VTH )(1 − out − TH ) RS
∂Vin L ∂Vin ∂Vin
Since, (
VTH = VTH 0 + γ 2ΦF + VSB − 2ΦF )
∂ VTH ∂ VTH ∂ VSB γ ∂ VSB
= =
∂ Vin ∂ VSB ∂ Vin 2 2ΦF + VSB ∂ Vin
∂ Vout
=η
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∂ Vin University of Paris VI

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 Source Follower Voltage Gain

∂Vout W ∂V ∂V
= µ nCox (Vin − Vout − VTH )(1 − out − TH ) RS
∂Vin L ∂Vin ∂Vin
∂Vout W ∂V ∂V
= µ nCox (Vin − Vout − VTH )(1 − out − η out ) RS
∂Vin L ∂Vin ∂Vin
W
∂Vout
µ nCox (Vin − Vout − VTH ) RS
= L
∂Vin 1 + µ C W (V − V − V ) R (1 + η )
n ox in out TH S
L
W
We also have, g m = µ nCox (Vin − Vout − VTH )
L

g m RS
Av =
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

Vout g m RS
Av = =
Vin 1 + ( g m + g mb ) RS

Since: g mb = η g m

And for : g m RS >> 1

1
Av ≈
(1 + η )

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 Source Follower Output Resistance
Rout : the output resistance when the input voltage is set to zero.

V1 = VBS = −VX
VX 1
I X − g mVX − g mbVX = 0 Rout = =
I X g m + g mb

Body Effect decreases the output resistance of source followers.

VX ↓ ⇒ VGS ↑ and VTH ↓

⇒ ID ↑

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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.

1 1 1
Rout = // =
g m g mb g m + g mb

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 Source Follower Thévenin Equivalent

1
g mb gm
Av = =
1 1 g m + g mb
+
g m g mb

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Channel Length Modulation in M1 and M2

1
// rO1 // rO 2 // RL
g mb
Av =
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 With the source follower stage:

⇒ Routp > Routn VX > VGS 2 + (VGS 3 − VTH 3 )
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Low Load Impedance: CS vs SF

Source Follower Amplifier Common Source Amplifier

RL AvCS ≈ − g m RL
AvSF ≈
RL + 1 / g m
Assuming RL=1/gm

AvSF ≈ 1 / 2 AvCS ≈ −1

Source Followers are not necessarily efficient drivers.

H. Aboushady University of Paris VI

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 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:
µ nCox W
Vout = VDD − (Vb − Vin − VTH ) 2 RD
2 L
∂Vout W ⎛ ∂V ⎞
= − µ n Cox (Vb − Vin − VTH )⎜⎜ − 1 − TH ⎟⎟ RD
∂Vin L ⎝ ∂Vin ⎠
∂Vout W
= µ n Cox (Vb − Vin − VTH )(1 + η )RD
∂Vin L
Av = g m (1 + η ) RD
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 Common Gate Stage Input Resistance

Same as Output Resistance of Source Follower:

1
Rin =
g m + g mb

Body Effect:
• increases Av
• decreases Rin

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Common Gate Gain

Small Signal Signal Equivalent Circuit

Vout
The current through RS is equal to -Vout / RD : V1 − RS + Vin = 0
RD
The current through rO is equal to -Vout / RD - gmV1 - gmbV1 :

⎛ −V ⎞ V
rO ⎜⎜ out − g mV1 − g mbV1 ⎟⎟ − out RS + Vin = Vout
⎝ RD ⎠ RD
⎡ −V ⎛ R ⎞⎤ V
rO ⎢ out − ( g m + g mb )⎜⎜Vout S − Vin ⎟⎟⎥ − out RS + Vin = Vout
⎣ RD ⎝ RD ⎠⎦ RD
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 Common Gate Gain
Common Gate Amplifier:
( g m + g mb )rO + 1
AvCG = RD
RD + RS + rO + ( g m + g mb )rO RS

Degenerated Common Source Amplifier:

g m rO
AvCS = − 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
RD 1
Rin ≈ +
( g m + g mb )rO ( g m + g mb )
• Assume RD = 0 : • Replace RD by ideal current source:
1
Rin = Rin = ∞
1 / rO + ( g m + g mb )

Rin of a common gate stage is low only if RD is small.

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 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

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 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.
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 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 W
I D1 − I D 2 = (Vb − VTH ) 2 (λVDS 1 − λVDS 2 )
2 L
µC W
I D1 − I D 2 = n ox (Vb − VTH ) 2 (λ ∆VDS )
2 L

rO1 ∆V
∆VPQ = ∆V ≈
[1 + ( g m 3 + g mb3 )rO 3 ]rO1 + rO 3 ( g m3 + g mb 3 )rO 3

µ nCox W λ ∆V
I D1 − I D 2 = (Vb − VTH ) 2
2 L ( g m3 + g mb 3 )rO 3

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 Folded Cascode

Simple Folded Folded Cascode Folded Cascode

Cascode with biasing with NMOS input

Large Signal Characteristics:

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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

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