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6.976 High Speed Communication Circuits and … these parasitics from circuit layout M.H. Perrott...

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6.976 High Speed Communication Circuits and Systems Lecture 5 High Speed, Broadband Amplifiers Michael Perrott Massachusetts Institute of Technology Copyright © 2003 by Michael H. Perrott
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6.976High Speed Communication Circuits and Systems

Lecture 5High Speed, Broadband Amplifiers

Michael PerrottMassachusetts Institute of Technology

Copyright © 2003 by Michael H. Perrott

M.H. Perrott MIT OCW

Broadband Communication System

Example: high speed data link on a PC board

- We’ve now studied how to analyze the transmission line effects and package parasitics

- What’s next?

VLC1 RL

L1Delay = xCharacteristic Impedance = Zo

Transmission Line

Z1

VinC2

dieAdjoining pinsConnector

Controlled ImpedancePCB trace

package

On-ChipDrivingSource

M.H. Perrott MIT OCW

High Speed, Broadband Amplifiers

The first thing that you typically do to the input signal is amplify it

Function- Boosts signal levels to acceptable values- Provides reverse isolation

Key performance parameters- Gain, bandwidth, noise, linearity

VLC1 RL

L1Delay = xCharacteristic Impedance = Zo

Transmission Line

Z1

VinC2

dieAdjoining pinsConnector

Controlled ImpedancePCB trace

package

On-ChipDrivingSource

AmpVout

M.H. Perrott MIT OCW

Basics of MOS Large Signal Behavior (Qualitative)

S D

G

Cchannel = Cox(VGS-VT)

VGS

VDS=0

S D

GVGS

VD=∆V

S D

GVGS

VD>∆V

Triode

Pinch-off

SaturationVDS

ID

ID

ID

ID

Triode

Pinch-offSaturation

∆V

Overall I-V Characteristic

M.H. Perrott MIT OCW

Basics of MOS Large Signal Behavior (Quantitative)

S D

G

Cchannel = Cox(VGS-VT)

VGS

VDS=0

S D

GVGS

VD=∆V

S D

GVGS

VD>∆V

Triode

Pinch-off

Saturation

ID

ID

ID

ID = µnCoxWL

(VGS - VT - VDS/2)VDS

ID µnCoxWL

(VGS - VT)VDS

for VDS << VGS - VT

ID = µnCoxWL

12

(VGS-VT)2(1+λVDS)

(where λ corresponds tochannel length modulation)

∆V = VGS-VT

∆V =µnCoxW

2IDL

M.H. Perrott MIT OCW

Analysis of Amplifier Behavior

Typically focus on small signal behavior- Work with a linearized model such as hybrid-π- Thevenin modeling techniques allow fast and efficient

analysisTo do small signal analysis:

RS

RG

RD

vinvout

Vbias

ID 1) Solve for bias current Id2) Calculate small signal parameters (such as gm, ro)3) Solve for small signal response using transistor hybrid-π small signal model

Small Signal Analysis Steps

M.H. Perrott MIT OCW

MOS DC Small Signal Model

Assume transistor in saturation:

Thevenin modeling based on the above

RS

RG

RDRD

RS

RG

-gmbvsvgs

vs

rogmvgs

gm = µnCox(W/L)(VGS - VT)(1 + λVDS)

= 2µnCox(W/L)ID (assuming λVDS << 1)

Cox

2qεsNA

2 2|Φp| + VSB

γgm where γ =gmb =

In practice: gmb = gm/5 to gm/3

λID1ro =

ID

M.H. Perrott MIT OCW

Capacitors For MOS Device In Saturation

S D

GVGS

VD>∆V

ID

LDLD

overlap cap: Cov = WLDCox + WCfringe

B

CgcCcb

Cov

CjdbCjsb

Cov

Side View

gate to channel cap: Cgc = CoxW(L-2LD)

channel to bulk cap: Ccb - ignore in this class

S D

Top View

W

E

L

E

E

source to bulk cap: Cjsb = 1 + VSB ΦB

Cj(0)

1 + VSB ΦB

Cjsw(0)WE + (W + 2E)

junction bottom wall cap (per area)

junction sidewall cap (per length)

drain to bulk cap: Cjsd = 1 + VDB ΦB

Cj(0)

1 + VDB ΦB

Cjsw(0)WE + (W + 2E)

23

(make 2W for "4 sided" perimeter in some cases)

L

M.H. Perrott MIT OCW

MOS AC Small Signal Model (Device in Saturation)

RS

RG

RD

RD

RS

RG

-gmbvsvgs

vs

rogmvgs

ID

Csb

Cgs

CgdCdb

Cgs = Cgc + Cov = CoxW(L-2LD) + Cov23

Cgd = Cov

Csb = Cjsb (area + perimeter junction capacitance)

Cdb = Cjdb (area + perimeter junction capacitance)

M.H. Perrott MIT OCW

Wiring Parasitics

Capacitance- Gate: cap from poly to substrate and metal layers- Drain and source: cap from metal routing path to

substrate and other metal layersResistance- Gate: poly gate has resistance (reduced by silicide)- Drain and source: some resistance in diffusion region,

and from routing long metal linesInductance- Gate: poly gate has negligible inductance- Drain and source: becoming an issue for long wires

Extract these parasitics from circuit layout

M.H. Perrott MIT OCW

Frequency Performance of a CMOS Device

Two figures of merit in common use- ft : frequency for which current gain is unity- fmax : frequency for which power gain is unity

Common intuition about ft- Gain, bandwidth product is conserved

- We will see that MOS devices have an ft that shifts with bias

This effect strongly impacts high speed amplifier topology selection

We will focus on ft- Look at pages 70-72 of Tom Lee’s book for discussion on fmax

M.H. Perrott MIT OCW

Derivation of ft for MOS Device in Saturation

Assumption is that input is current, output of device is short circuited to a supply voltage- Note that voltage bias is required at gate

The calculated value of ft is a function of this bias voltage

-gmbvsvgs rogmvgs

ID+id

Csb

Cgs

CgdCdb

iinVbias

RLARGE

id

iin

M.H. Perrott MIT OCW

Derivation of ft for MOS Device in Saturation

-gmbvsvgs rogmvgs

ID+id

Csb

Cgs

CgdCdb

iinVbias

RLARGE

id

iin

M.H. Perrott MIT OCW

Derivation of ft for MOS Device in Saturation

1

idiin

fft

slope = -20 dB/dec

M.H. Perrott MIT OCW

Why is ft a Function of Voltage Bias?

ft is a ratio of gm to gate capacitance- gm is a function of gate bias, while gate cap is not (so long

as device remains biased)First order relationship between gm and gate bias:

- The larger the gate bias, the higher the value for ft

Alternately, ft is a function of current density

- So ft maximized at max current density (and minimum L)

M.H. Perrott MIT OCW

Speed of NMOS Versus PMOS Devices

NMOS devices have much higher mobility than PMOS devices (in current, non-strained, bulk CMOS processes)

- Intuition: NMOS devices provide approximately 2.5 x gmfor a given amount of capacitance and gate bias voltage

- Also: NMOS devices provide approximately 2.5 x Id for a given amount of capacitance and gate bias voltage

M.H. Perrott MIT OCW

Assumptions for High Speed Amplifier Analysis

Assume that amplifier is loaded by an identical amplifier and by fixed wiring capacitance

Intrinsic performance- Defined as the bandwidth achieved for a given gain when Cfixed is negligible- Amplifier approaches intrinsic performance as its device sizes (and current) are increased

In practice, optimal sizing (and power) of amplifier is roughly where Cin+Cout = Cfixed

Amp Amp

Cfixed

CinCin

Ctot = Cout+Cin+Cfixed

Cout

M.H. Perrott MIT OCW

The Miller Effect

Concerns impedances that connect from input to output of an amplifier

Input impedance:

Output impedance:

Amp

Zin

Zf

Av

ZL

VoutVin

Zout

M.H. Perrott MIT OCW

Example: The Impact of Capacitance in Feedback

Consider Cgd in the MOS device as Cf- Assume gain is negative

Impact on input capacitance:

Output impedance:

Amp

Zin

Av

ZL

VoutVin

Zout

Cf

M.H. Perrott MIT OCW

Amplifier Example – CMOS Inverter

Assume that we set Vbias such that the amplifier nominal output is such that NMOS and PMOS transistors are all in saturation- Note: this topology VERY sensitive to bias errors

Cfixed

Ctot = Cdb1+Cdb2 + Cgs3+Cgs4 + K(Cov3+Cov4) + Cfixed

M2

M1

M4

M3

Miller multiplication factor

Vbias

vin

(+Cov1+Cov2)

vout

M.H. Perrott MIT OCW

Transfer Function of CMOS Inverter

Low Bandwidth!

Cfixed

Ctot = Cdb1+Cdb2 + Cgs3+Cgs4 + K(Cov3+Cov4) + Cfixed

M2

M1

M4

M3

Miller multiplication factor

Vbias

vin

(+Cov1+Cov2)

vout

1

voutvin

f

slope = -20 dB/dec

(gm1+gm2)(ro1||ro2)

2πCtot(ro1||ro2)1 gm1+gm2

2πCtot

M.H. Perrott MIT OCW

Add Resistive Feedback

Bandwidth extended andless sensitivityto bias offset

Cfixed

Ctot = Cdb1+Cdb2 + Cgs3+Cgs4 + K(Cov3+Cov4) + CRf /2 + Cfixed

M2

M1

M4

M3

Miller multiplication factor

Vbias

vin

(+Cov1+Cov2)

voutRf

1

voutvin

f

slope = -20 dB/dec

(gm1+gm2)(ro1||ro2)

2πCtot(ro1||ro2)1 gm1+gm2

2πCtot

(gm1+gm2)Rf

2πCtotRf

1

M.H. Perrott MIT OCW

We Can Still Do Better

We are fundamentally looking for high gm to capacitance ratio to get the highest bandwidth- PMOS degrades this ratio- Gate bias voltage is constrained

Cfixed

Ctot = Cdb1+Cdb2 + Cgs3+Cgs4 + K(Cov3+Cov4) + CRf /2 + Cfixed

M2

M1

M4

M3

Miller multiplication factor

Vbias

vin

(+Cov1+Cov2)

voutRf

M.H. Perrott MIT OCW

Take PMOS Out of the Signal Path

Advantages- PMOS gate no longer loads the signal- NMOS device can be biased at a higher voltageIssue- PMOS is not an efficient current provider (Id/drain cap)

Drain cap close in value to Cgs- Signal path is loaded by cap of Rf and drain cap of PMOS

CL

M2

M1Vbias

vin

voutRf

Vbias2Ibias

CLM1Vbias

vin

voutRf

M.H. Perrott MIT OCW

Shunt-Series Amplifier

Use resistors to control the bias, gain, and input/output impedances- Improves accuracy over process and temp variations

Issues- Degeneration of M1 lowers slew rate for large signal

applications (such as limit amps)- There are better high speed approaches – the advantage

of this one is simply accuracy

Ibias

M1

Vbias

vin

voutRf

R1

Rs

RL

Rin Rout

M1

Vbias

vinvout

Rf

R1

Rs

RL

Rin Rout

M.H. Perrott MIT OCW

Shunt-Series Amplifier – Analysis Snapshot

From Chapter 8 of Tom Lee’s book (see pp 191-197):- Gain

- Input resistance

- Output resistance

M1

Vbias

vinvout

Rf

R1

Rs

RL

Rin Rout

vx

Same for Rs = RL!

M.H. Perrott MIT OCW

NMOS Load Amplifier

Gain set by the relative sizing of M1 and M2

CfixedM1Vbias

vin

vout

M2

gm2

1

1

voutvin

f

slope = -20 dB/dec

gm12πCtot

gm1

2πCtot

gm2

gm2Ctot = Cdb1+Csb2+Cgs2 + Cgs3+KCov3 + Cfixed

Miller multiplication factor(+Cov1)

M3

Vdd

Id

M.H. Perrott MIT OCW

Design of NMOS Load Amplifier

Size transistors for gain and speed- Choose minimum L for maximum speed- Choose ratio of W1 to W2 to achieve appropriate gain

Problem: VT of M2 lowers the bias voltage of the next stage (thus lowering its achievable ft)- Severely hampers performance when amplifier is cascaded- One person solved this issue by increasing Vdd of NMOS

load (see Sackinger et. al., “A 3-GHz 32-dB CMOS Limiting Amplifier for SONET OC-48 receivers”, JSSC, Dec 2000)

CfixedM1Vbias

vin

vout

M2

gm2

1

Ctot = Cdb1+Csb2+Cgs2 + Cgs3+KCov3 + Cfixed

Miller multiplication factor(+Cov1)

M3

Vdd

Id

M.H. Perrott MIT OCW

Resistor Loaded Amplifier (Unsilicided Poly)

This is the fastest non-enhanced amplifier I’ve found- Unsilicided poly is a pretty efficient current provider

(i..e, has a good current to capacitance ratio)- Output swing can go all the way up to Vdd

Allows following stage to achieve high ft- Linear settling behavior (in contrast to NMOS load)

M1

RL

vout

M2

Cfixed

Id

Vbias

vin

Ctot = Cdb1+CRL/2 + Cgs2+KCov2 + Cfixed

Miller multiplication factor(+Cov1)

1

voutvin

f

slope = -20 dB/dec

gm12πCtot

gm1RL

2πRLCtot

1

Vdd

M.H. Perrott MIT OCW

Implementation of Resistor Loaded Amplifier

Typically implement using differential pairs

Benefits- Self-biased- Common-mode rejection

Negative- More power than single-ended version

M6

M1 M2

M5

αIbias

Vin+

R1

Vin-

R2Vo+

Vo-

M7

M3 M4

Cfixed Cfixed

Ibias

Ibias/2

Vdd

M.H. Perrott MIT OCW

The Issue of Velocity Saturation

We classically assume that MOS current is calculated as

Which is really

- Vdsat,l corresponds to the saturation voltage at a given length, which we often refer to as ∆V

It may be shown that

- If Vgs-VT approaches LEsat in value, then the top equation is no longer valid

We say that the device is in velocity saturation

M.H. Perrott MIT OCW

Analytical Device Modeling in Velocity Saturation

If L small (as in modern devices), than velocity saturation will impact us for even moderate values of Vgs-VT

- Current increases linearly with Vgs-VT!Transconductance in velocity saturation:

- No longer a function of Vgs!

M.H. Perrott MIT OCW

Example: Current Versus Voltage for 0.18µ Device

0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

Vgs

(Volts)

I d (m

illiA

mps

)

Id versus V

gsM1

IdVgs

WL = 1.8µ

0.18µ

M.H. Perrott MIT OCW

0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Vgs

(Volts)

g m (

mill

iAm

ps/V

olts

)

gm

versus Vgs

Example: Gm Versus Voltage for 0.18µ Device

M1

IdVgs

WL = 1.8µ

0.18µ

M.H. Perrott MIT OCW0 100 200 300 400 500 600 700

0

0.2

0.4

0.6

0.8

1

Current Density (microAmps/micron)

Tra

nsco

nduc

tanc

e (m

illiA

mps

/Vol

ts)

Transconductance versus Current Density

Example: Gm Versus Current Density for 0.18µ Device

M1

IdVgs

WL = 1.8µ

0.18µ

M.H. Perrott MIT OCW

How Do We Design the Amplifier?

Highly inaccurate to assume square law behaviorWe will now introduce a numerical procedure based on the simulated gm curve of a transistor- A look at gm assuming square law device:

- Observe that if we keep the current density (Id/W) constant, then gm scales directly with W

This turns out to be true outside the square-law regime as well

- We can therefore relate gmof devices with different widths given that have the same current density

M.H. Perrott MIT OCW

A Numerical Design Procedure for Resistor Amp – Step 1

Two key equations- Set gain and swing (single-

ended)

Equate (1) and (2) through RM6

M1 M2

M5

αIbias

Vin+

R

Vin-

RVo+

Vo-

2Ibias

Ibias

Vdd

Can we relate this formula to a gm curve takenfrom a device of width Wo?

M.H. Perrott MIT OCW

We now know:

Substitute (2) into (1)

The above expression allows us to design the resistor loaded amp based on the gm curve of a representative transistor of width Wo!

A Numerical Design Procedure for Resistor Amp – Step 2

M.H. Perrott MIT OCW

Example: Design for Swing of 1 V, Gain of 1 and 2

Assume L=0.18µ, use previous gm plot (Wo=1.8µ)

0 100 200 300 400 500 600 7000

0.2

0.4

0.6

0.8

1

Current Density - Iden (microAmps/micron)

Tran

scon

duct

ance

(m

illiA

mps

/Vol

ts)

Transconductance versus Current Density

A=2 A=1

gm(wo=1.8µ,Iden)

For gain of 1, current density = 250 µA/µmFor gain of 2, current density = 115 µA/µm Note that current density reduced as gain increases!- ft effectively

decreased

M.H. Perrott MIT OCW

Example (Continued)

Knowledge of the current density allows us to design the amplifier- Recall- Free parameters are W, Ibias, and R (L assumed to be fixed)

Given Iden = 115 µA/µm (Swing = 1V, Gain = 2)- If we choose Ibias = 300 µA

Note that we could instead choose W or R, and then calculate the other parameters

M.H. Perrott MIT OCW

How Do We Choose Ibias For High Bandwidth?

As you increase Ibias, the size of transistors also increases to keep a constant current density- The size of Cin and Cout increases relative to Cfixed

To achieve high bandwidth, want to size the devices (i.e., choose the value for Ibias), such that - Cin+Cout roughly equal to Cfixed

Amp Amp

Cfixed

CinCin

Ctot = Cout+Cin+Cfixed

Cout


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