•Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage

Home , Common drain, Common gate, Current source, Output impedance, Transconductance

Single Stage Amplifiers

Hassan Aboushady University of Paris VI

References

• B. Razavi, “Design of Analog CMOS Integrated Circuits”, McGraw-Hill, 2001.

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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.

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2 Approximation of a nonlinear system

Input-Output Characteristic of a nonlinear system ≈ α +α +α 2 + +α n ≤ ≤ y(t) 0 1x(t) 2 x (t) ... n x (t) x1 x x2

In a sufficiently narrow range: ≈ α +α y(t) 0 1x(t)

α where 0 can be considered the operating (bias) point and α1 the small signal gain

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Analog Design Octagon

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3 Single Stage Amplifiers

•Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage

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Common Source Stage with Resistive Load

= − Vout VDD RD I D µ C W 2 M1 in the saturation region: V = V − R n ox (V −V ) out DD D 2 L in TH µ nCox W 2 M1 in limit of saturation: V −V = V − R (V −V ) in1 TH DD D 2 L in1 TH ⎡ 2 ⎤ M1 in the = − µ W − − Vout Vout VDD RD nCox ⎢(Vin VTH )Vout ⎥ linear region: L ⎣ 2 ⎦ H. Aboushady University of Paris VI

4 Common Source Stage with Resistive Load

R V M1 in deep V = V on = DD linear region: out DD R + R W on D 1+ µ C R (V −V ) n ox L D in TH

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Common Source Stage with Resistive Load

M1 in the saturation region: µ C W V = V − R n ox (V −V )2 out DD D 2 L in TH Small signal gain: ∂V W A = out = −R µ C (V −V ) v ∂ D n ox in TH Vin L = − gm 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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5 Example 1

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

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

2 µ C W W ⎡ Vout ⎤ = n ox − 2 I = µ C (V −V )V − I D (Vin VTH ) D n ox ⎢ in1 TH out ⎥ 2 L L ⎣ 2 ⎦ ∂I W ∂I W g = D = µ C (V −V ) g = D = µ C V m ∂ n ox in TH m ∂ n ox out VGS L VGS L = VDD Vout + µ W − 1 nCox RD (Vin VTH ) H. Aboushady L University of Paris VI

Voltage Gain of a Common Source Stage

= − Av gm RD

= − µ W VRD Av 2 nCox I D L I D

= − µ W VRD Av 2 nCox L I D

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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6 Taking Channel Length Modulation into account

Calculating Av starting from the Large Signal Equations: µ C W V = V − R n ox (V −V )2 (1+ λV ) out DD D 2 L in TH out

∂V W A = out = −R µ C (V −V )(1+ λV ) v ∂ D n ox in TH out Vin L µ C W ∂V − R n ox (V −V )2 λ out D in TH ∂ 2 L Vin

= − − λ Av RD gm RD I D Av

λI =1/ r r R RD gm D O = − O D A = − Av gm v + λ r + R 1 RD I D O D

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Taking Channel Length Modulation into account

Calculating Av starting from the Small Signal model:

= − gmV1(rO // RD ) Vout V A = out = −g (r // R ) = v V m O D V1 Vin in

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7 Example 2

Assuming M1 biased in saturation, calculate the small signal voltage gain :

• I1 : Ideal current source Infinite Impedance = − Av gmrO

• Intrinsic gain of a transistor: This quantity represents the maximum voltage gain that can be achieved using a single device.

µ C W I = n ox (V −V )2 (1+ λV ) = I D1 2 L in TH out 1 • 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 gm (rO1 // rO2 ) • 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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8 CS with Source Degeneration Large Signal model: Small Signal model: ∂I ∂I ∂V G = D = D GS m ∂ ∂ ∂ Vin VGS Vin = − VGS Vin I D RS ∂V ∂I GS =1− D R ∂ ∂ S Vin Vin ∂ ⎛ ∂ ⎞ = I D − I D Gm ⎜1 RS ⎟ ∂ ∂ ⎜ ∂ ⎟ I D I D gmV1 VGS ⎝ Vin ⎠ g = G = = m ∂V m + = ()− GS Vin V1 gmV1RS Gm gm 1 RSGm g g G = m A = −G R G = m m + v m D m + 1 gm RS 1 gm RS g R A = − m D v + 1 gm RS H. Aboushady University of Paris VI

CS with Source Degeneration

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

ID is linearized at the cost of lower gain.

Small Signal model including body effect and channel length modulation:

= − − VX Iout gmV1 gmbVX rO

= − + − − Iout RS gm (Vin Iout RS ) gmb ( Iout RS ) rO I g r G = out = m O m + + + Vin RS [1 (gm gmb )RS ]rO

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9 With and Without Source Degeneration

g r G = m O m + + + 1 [1 (gm gmb )RS ]rO

= ≠ RS 0 RS 0

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Estimating Gain by Inspection

g R R A = − m D = − D v + + 1 gm RS 1/ gm RS

Resistance seen at the Drain Gain = − Total Resistance in the Source Path

Example:

R A = − D v + 1/ gm1 1/ gm2

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10 Output Resistance of Degenerated CS

= − V1 I X RS

The current flowing in r : − + O I X (gm gmb )V1 = + + I X (gm gmb )RS I X

= + + + VX rO[I X (gm gmb )RS I X ] I X RS

= VX = + + + Rout rO[1 (gm gmb )RS ] RS I X = + + + Rout [1 (gm gmb )rO ]RS rO

≈ + + = + + Rout (gm gmb )rO RS rO Rout [1 (gm gmb )RS ]rO

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Voltage Gain of Degenerated CS

The current through RS must equal that through RD: V I = I = out RD RS RD VS = − RS VS Vout RD

Vout The current through rO : I = − − (g V + g V ) rO m 1 mb BS RD V R R V I = − out −[g (V +V S ) + g V S ] V = I r − out R rO m in out mb out out rO O S RD RD RD RD = −Vout − + RS + RS − RS Vout rO [gm (Vin Vout ) gmbVout ]rO Vout RD RD RD RD

Vout = − gmrO RD V R + R + r + (g + g )R r H. Aboushady in D S O m mb S O University of Paris VI

11 Voltage Gain of Degenerated CS

Vout = − gmrO RD + + + + Vin RD RS rO (gm gmb )RS rO

+ + + Vout = − gmrO RD[RS rO (gm gmb )RS rO ] + + + + + + + Vin RS [1 (gm gmb )RS ]rO RD RS rO (gm gmb )RS rO

Vout = () Gm Rout // RD Vin

The output resistance = + + Rout [1 (gm gmb )RS ]rO of a degenerated CS stage: I g r The Transconductance G = out = m O of a degenerated CS stage: m + + + Vin RS [1 (gm gmb )RS ]rO

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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.

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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 µ C W V = n ox (V −V −V )2 R out 2 L in out TH S

To calculate gm : ∂V W ∂V ∂V out = µ C (V −V −V )(1− out − TH )R ∂ n ox in out TH ∂ ∂ S Vin L Vin Vin

Since, VTH = VTH 0 + γ ( 2ΦF +VSB − 2ΦF )

∂ TH ∂ TH ∂ V γ ∂ V V = V SB = SB ∂ ∂ ∂ Φ + ∂ Vin VSB Vin 2 2 F VSB Vin ∂ =η Vout ∂ V H. Aboushady in University of Paris VI

13 Source Follower Voltage Gain

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

g R Av = m S + + 1 (gm gmb )RS

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Source Follower Voltage Gain

Small Signal Equivalent Circuit

= []+ Vout gmV1 gmbVBS RS = []− − gm (Vin Vout ) gmbVout RS

V g R Av = out = m S + + Vin 1 (gm gmb )RS =η Since: gmb gm >> And for : gm RS 1

1 Av ≈ (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 V 1 I − g V − g V = 0 R = X = X m X mb X out + I X gm gmb

Body Effect decreases the output resistance of source followers.

↓ ↑ ↓ VX ⇒ VGS and VTH ↑ ⇒ I D

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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 R = // = out + gm gmb gm gmb

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15 Source Follower Thévenin Equivalent

1 g g A = mb = m v 1 1 + + gm gmb gm gmb

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Channel Length Modulation in M1 and M2

1 // r // r // R g O1 O2 L Av = mb 1 + 1 // rO1 // rO2 // RL gmb gm

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16 Source Follower Characteristics

+ High input impedance and Moderate output impedance - Nonlinearity - Limited voltage swing

α Example: VTH VSB

PMOS source follower with VSB=0

Without the source follower stage: > − VX VGS1 VTH1 µ < µ < p n ⇒ gmp gmn With the source follower stage: > + − > VX VGS 2 (VGS 3 VTH 3 ) ⇒ Routp Routn H. Aboushady University of Paris VI

Low Load Impedance: CS vs SF

Source Follower Amplifier Common Source Amplifier

R A ≈ −g R A ≈ L vCS m L vSF + RL 1/ gm

Assuming RL=1/gm

≈ ≈ − AvSF 1/ 2 AvCS 1

Source Followers are not necessarily efficient drivers.

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17 Single Stage Amplifiers

•Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage

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Common Gate Stage

Large Signal Behavior = − Vout VDD I D RD

Assuming M1 in saturation: µ C W V = V − n ox (V −V −V )2 R out DD 2 L b in TH D ∂V W ⎛ ∂V ⎞ out = −µ C (V −V −V )⎜−1− TH ⎟R ∂ n ox b in TH ⎜ ∂ ⎟ D Vin L ⎝ Vin ⎠ ∂V W out = µ C (V −V −V )()1+η R ∂ n ox b in TH D Vin L = +η Av gm (1 )RD

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18 Common Gate Stage Input Resistance

Same as Output Resistance of Source Follower:

1 R = in + gm gmb

Body Effect:

• increases Av • decreases Rin

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Common Gate Gain

Small Signal Signal Equivalent Circuit

V The current through is equal to : − out + = RS -Vout / RD V1 RS Vin 0 RD

The current through rO is equal to -Vout / RD - gmV1 - gmbV1 : ⎛ −V ⎞ V ⎜ out − − ⎟ − out + = rO ⎜ gmV1 gmbV1 ⎟ RS Vin Vout ⎝ RD ⎠ RD ⎡−V ⎛ R ⎞⎤ V out − + ⎜ S − ⎟ − out + = rO ⎢ (gm gmb )⎜Vout Vin ⎟⎥ RS Vin Vout ⎣ RD ⎝ RD ⎠⎦ RD

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19 Common Gate Gain

Common Gate Amplifier: (g + g )r +1 A = m mb O R vCG + + + + D RD RS rO (gm gmb )rO RS

Degenerated Common Source Amplifier: g r A = − m O R vCS + + + + D RD RS rO (gm gmb )rO RS

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Common Gate Stage Input Resistance

Since V1 = -VX : = + []− + VX RD I X rO I X (gm gmb )VX + VX = RD rO + + I X 1 (gm gmb )rO R 1 R ≈ D + in + + (gm gmb )rO (gm gmb ) • Assume R = 0 : D • Replace RD by ideal current source: 1 R = R = ∞ in + + in 1/ rO (gm gmb )

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

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20 Common Gate Stage Output Impedance

Similar to Output Impedance of a Degenerated Common Source Stage

= ()+ + + Rout [1 (gm gmb )rO ]RS rO // RD

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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

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 VTH1 ≥ + − Vb Vin VGS 2 VTH1

• M2 in saturation: − ≥ − − Vout VX Vb VX VTH 2 ≥ − + − Vout Vin VTH1 VGS 2 VTH 2

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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 (gm2 gmb2 )rO2 ]rO1 rO2 ≈ + Rout (gm2 gmb2 )rO2rO1 • 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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22 Cascode Stage Voltage Gain

= − Av Gm Rout ≈ Gm gm1

Ideal Current Source: ≈ + Rout (gm2 gmb2 )rO2rO1 ≈ + Av (gm2 gmb2 )rO2 gm1rO1

Cascode Current Source: ≈ Rout gm2rO2rO1 // gm3rO3rO4 ≈ () Av gm1 gm2rO2rO1 // gm3rO3rO4

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Shielding Property ∆ Assume VX is higher than VY by V. λ ≠ 0 Calculate the resulting difference between ID1 and ID2 (with ). µ C W I − I = n ox (V −V )2 (λV − λV ) D1 D2 2 L b TH DS1 DS 2 µ C W I − I = n ox (V −V )2 (λ ∆V ) D1 D2 2 L b TH DS

r ∆V ∆V = ∆V O1 ≈ PQ + + + + [1 (gm3 gmb3 )rO3 ]rO1 rO3 (gm3 gmb3 )rO3

µ C W λ ∆V I − I = n ox (V −V )2 D1 D2 b TH + 2 L (gm3 gmb3 )rO3

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23 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 (gm1 gmb1)rO1]RS rO1

Folded Cascode Stage:

M1 M2

RS rO1 // rO3

= + + + Rout [1 (gm2 gmb2 )rO2 ](rO1 // rO3 ) rO2

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