t vr t vr j t - Cornell University · s s out in v j R C A A v Av 1 1 vs Rs vin The amplifier input voltage, and the output voltage, will now begin to drop-off at a much lower frequency!!

1

ECE 315 – Spring 2005 – Farhan Rana – Cornell University

Lecture 19

High Frequency Analysis of FET Circuits

In this lecture you will learn:

• High Frequency Analysis of FET Circuits• Miller Effect and the Miller Capacitance


Phasor Analysis: Complex Impedances

R

C

tjevtv Re

tjeiti Re

+

-

tjrr evtv Re

Capacitive Impedance/Admittance:

CjYZ

11

Inductance Impedance/Admittance:

LjY

Z

1

L~

2


Phasor Analysis: Calculations with Impedances

R

C

v

i

+

-

rv cv

+ -

vRCj

v

CjR

Cjvc

1

11

1

vRCj

RCjv

CjR

Rvr

11

One can compute the voltage phasors using the impedances:


Phasor Analysis: Bode Plots

R

C

v

i

+

-

rv cv

+ -

HRCjv

vc

1

1

210log10 H

RC1

~3 dB

Slope = -20 dB/decade

A transfer function with a pole at frequency 1/RC

3dB break point

3


Poles and Zeroes of Transfer Functions

R

C

v

i

+

-

rv cv

+ -

1

1cv

v j RC

1

rv j RC

v j RC

Pole at the complex frequency j

RC

Pole at the complex frequency j

RC

Zero at the complex frequency 0

With a little abuse of language we will say a pole at the frequency 1

RC

With a little abuse of language we will say pole at the frequency and a zero at 0 frequency

1

RC


Laplace Transform Calculations: Language Differences to Note

R

C

v s

i s

+

-

rv s cv s

+ -

1

11 1c

sCv s v s v ssRCR

sC

1 1r

R sRCv s v s v s

sRCRsC

One can compute the voltages using Laplace transforms as well:

s j

Pole at 1

RC

Pole at 1

RC

Zero at 0

4


Phasor Analysis: Bode Plots

R

C

v

i

+

-

rv cv

+ -

HRCj

RCjvvr

1

210log10 H

RC1

~3 dB

Slope = +20 dB/decade

A transfer function with a zero at zero frequency and a pole at frequency 1/RC


The Miller Effect and the Miller CapacitanceJohn A. Miller (1920)

Consider a voltage amplifier:

sv

sR

out in sv Av Av inv

Consider now a capacitor sitting at the input of an amplifier:

1

out in

ss

v Av

Av

j R C

sv

sR inv

C

Input voltage to the amplifier decreases, and so does the output voltage of the amplifier, at high frequencies (but not too bad…..)

|A| >> 1A < 0

1

in sin s

s in s

Z vv v

R Z j R C

1inZ

j C

Open circuit voltage gain

5


Consider now a capacitor straddling the input and the output of an amplifier:

out inv Av tv

inv

The Miller Effect and the Miller Capacitance

C ti

1

1

1

t t out t t

t

tin

t

i j C v v j C v Av

j C A v

vZ

i j C A

|A| >> 1A < 0

The capacitance, as seen from the input end, is effectively very large!


Capacitor straddling the input and the output of an amplifier:

ss

inout

vACRj

A

Avv

11

sv

sR

inv

The amplifier input voltage, and the output voltage, will now begin to drop-off at a much lower frequency!! This is the Miller effect and the capacitance positioned this way is called the Miller capacitance

The Miller Effect and the Miller Capacitance

C

1 1

in sin s

s in s

Z vv v

R Z j R C A

1

1inZj C A

The pole has shifted to a much lower frequency!

|A| >> 1A < 0

6


The Miller Effect and the Miller Capacitance: Summary

ss

inout

vACRj

A

Avv

11

In both cases:

A A

1

1inZj C A

Lesson:

A capacitor C straddling the input-output terminals of an amplifier of gain A = -|A| behaves like a very large capacitor C(1-A) sitting between the input and the ground terminals.

This capacitor reduces the gain bandwidth of the circuit

inin s

s in

Zv v

R Z


NFET: Capacitances

P-Si Substrate (or Bulk)

Gate

+_

_

N-Si

GSV

+_

DSV

SBV

+

SourceDrain

y0y Ly

N-Si

Csb Cdb

Cgs Cgd

7


NFET: High Frequency Small Signal Model

+gsv-

Gate

+bsv-Source

Base

Draindi

gsmvg bsmbvg+dsv-

oggsC

dbCsbC

gigdC

gbC

FET high frequency model:


NFET: High Frequency Small Signal Model

+gsv-

Gate

+bsv-

Source Base

Draindi

gsmvg bsmbvg+dsv

-

oggsC

gigdC

Our simplified high frequency model:

8


NFET Gate ChargeThe total gate charge QTG (units: Coulombs) consists of the image charge due to:

1) The total inversion layer charge QTN

2) The total depletion layer charge QTB

TG TN TBQ Q Q

The inversion layer charge QN (units: Coulombs/unit area) is a function of position “y” in the device:

N ox GS TN CSQ y C V V V y

The total inversion layer charge QTN (units: Coulombs) can be found by integration over the entire area of the FET:

0

0

L

TN N

L

ox GS TN CS

Q W Q y dy

WC V V V y dy


NFET Gate Charge

0

0

DS

L

TN ox GS TN CS

V

ox GS TN CS CSCS

Q WC V V V y dy

dyWC V V V dV

dV

Recall that (Lecture 10):

D N n y

CSn ox GS TN CS

n ox GS TN CS

CS D

I WQ y E y

dV yW C V V V y

dy

W C V V V ydy

dV I

2 2

0

2 3 3

3

DSVn

TN ox GS TN CS CSD

nox GS TN DS GS TN

D

Q WC V V V dVI

WC V V V V VI

9


Drain

Gate

SiO2

Ly

L

The channel has been “pinched off”

Channel potential: TNGSCS VVyV

Channel potential: DSCS VyV

For :DS GS TNV V V

Drain

Gate

SiO2

Ly

Channel potential: DSCS VyV

Linear Region:

For :DS GS TNV V V Saturation Region:

NFET: Inversion Layer Charge in Linear and Saturation


NFET Gate Charge

Inversion layer charge QN :

2 3 3

3n

TN ox GS TN DS GS TND

Q WC V V V V VI

In saturation (VDS > VGS - VTN) but the inversion charge remains fixed at the value when VDS = VGS – VTN because of pinch-off:

2

3TN ox GS TNQ WLC V V

Expression only works in the linear region

, ,

2

3GD GB GD GB

TG TN TB

TG TNgs ox

GS GSV V V V

Q Q Q

Q QC WLC

V V

Therefore, in saturation (since the depletion region thickness and charge do not change with the gate voltage), we get:

10


NFET Capacitances in Saturation: A Simple Model

, , ,

0

GS GB GS GB GS GB

TG TG TNgd

GD DS DSV V V V V V

Q Q QC

V V V

In saturation, because of pinch off, the inversion layer charge QTN is not affected by the drain voltage:


Capacitances

In Saturation:

,

2 2

3 3GS SB

Ggs ox ov p ox

GS V V

QC WLC WC WC WLC

V

,

0 0

GS SB

Ggd ov p

GD V V

QC WC WC

V

Overlap capacitances

Parasitic capacitances

11


NFET: High Frequency Small Signal Model for Saturation Region

+gsv-

Gate

+bsv-

Source Base

Draindi

gsmvg bsmbvg+dsv

-

oggsC

gigdC

Our simplified high frequency model:

2

3gs ox ov pC WLC WC WC

gd ov pC WC WC


+

INV

DDV

R

tvV outOUT

tvs

+

The Common Source Amplifier

+ inv

-

di

gsmvg+ outv

-oggsC

gigdC

R sv

A high frequency small signal model can be built as follows:

+ gsv-

sR

sR

tjss evtv Re

12


The Common Source Amplifier: Open Circuit Voltage Gain

KCL at (1) gives:

gdinoutinmoutod Cjvvvgvgi

Also:

Riv dout

The above two give:

gdo

gdm

in

outv RCjRg

RCjRg

v

vA

1

+ inv

-

di

gsmvg+ outv

-oggsC

gigdC

R+ gsv-

(1)

Need to find:

in

outv v

vA and

s

out

vv

H



ogd

m

gd

omgdo

gdm

in

outv rRCj

g

Cj

rRgRCjRg

RCjRg

vv

A||1

1||

1

210log10 vA

gdo CrR ||

1


210 ||log10 om rRg

gd

m

Cg

A relatively largefrequencyThese poles might not limit the frequency performance…….PTO

21010log 1 0

13



ogd

m

gd

omgdo

gdm

in

outv rRCj

g

Cj

rRgRCjRg

RCjRg

v

vA

||1

1||

1

vA

gdo CrR ||

1

gd

m

Cg

0

2


+ inv

-

di

gsmvg+ outv

-oggsC

gigdC

R sv+ gsv-

sR

The Common Source Amplifier: Input-Output Response

So far we have found the open circuit voltage gain:

ogd

m

gd

omgdo

gdm

in

outv rRCj

g

Cj

rRgRCjRg

RCjRg

v

vA

||1

1

||1

What about:

??out

s

v

v

14


The Common Source Amplifier: Input Impedance

+ inv

-

di

gsmvg+ outv

-oggsC

ti gdC

R tv + gsv-

(2)

KCL at (2) gives:

1

1

1

t gs t gd t out

gs gd v t

tin

t gs gd v

i j C v j C v v

j C j C A v

vZ

i j C j C A

vin

out Av

v

vgdgsin ACCj

Z

1

1The input impedance is:

gi

+

-



+ inv

-

di

gsmvg+ outv

-oggsC

gigdC

R sv+ gsv-

sR

svgdgsins

in

s

in

RACCjZRZ

vv

111

Looking in from the input terminal the capacitance Cgd seems to be magnified by a factor proportional to the open circuit voltage gain of the amplifier!!

Input voltage divider Miller Effect!!

15


The Common Source Amplifier: Total Gain

+ inv

-

di

gsmvg+ outv

-oggsC

gigdC

R sv+ gsv-

sR

Finally:

Where:

svgdgs

v

ins

inv

s

in

in

out

s

out

RACCjA

ZRZ

Avv

vv

vv

H

11

ogd

m

gd

omin

outv rRCj

g

Cj

rRgv

vA

||1

1||


The Common Source Amplifier: Poles and Zeros of the Total Gain

21

3

11

10

11

jj

jH

RACCjA

Hsvgdgs

v

A little tedious algebra can show that the transfer function above has two poles and a zero

1) Assuming gmRS >> 1, these poles are:

om rRgH ||0

gd

m

Cg

3

1

ogdgs

m

rRCCg

||11

1

somgdgs RrRgCC ||

11

2

This pole will likely determine the smallest frequency at which the total gain rolls over

2) Assuming gmRS << 1, these poles are:

gssCR11

1

This pole will likely determine the smallest frequency at which the total gain rolls over

ogd rRC ||11

2

16


The Common Source Amplifier: The Miller Approximation

0 ||

1m gdout

v v m oin o gd

g R j RCvA A g R r

v g R j RC

Miller approximation

If the poles in Av() are at high enough frequencies such that the lowest poles of the full transfer function are determined by other poles (which would be the case if gmRS >> 1 ) then one may approximate the open circuit gain Av() by its low frequency value:

And then:

1 1

||

1 1 ||

out v

s gs gd v s

m o

gs gd m o s

v AH

v j C C A R

g R r

j C C g R r R

Looking in from the input terminal the capacitance Cgd seems to be magnified by a factor proportional to the low frequency open circuit voltage gain of the amplifier!!

And now the single pole is at: somgdgs RrRgCC ||1

11

1 1

out v

s gs gd v s

v AH

v j C C A R


+ inv

-

di

gsmvg+ outv

-oggsC

gigdC

R sv+ gsv-

sR

The Common Source Amplifier: Total Gain Under the Miller Approx

+ inv

-

di

gsmvg+ outv

-og

gsC

gi omgd rRgC ||1

R sv+

gsv-

sR

Miller approximation is equivalent to saying that the following circuit will describe the input-output response at not too high frequencies………

17



+ inv

-

di

gsmvg+ outv

-og

gsC

gi omgd rRgC ||1

R sv+

gsv-

sR

Looking in from the input terminal the capacitance Cgd seems to be magnified by a factor proportional to the low frequency open circuit voltage gain of the amplifier!!

omgdgsin rRgCCj

Z||1

1

Miller approximation is equivalent to saying that the following circuit will describe the input-output response at not too high frequencies………


210log10 H

-20 dB/dec

210 ||log10 om rRg

The Common Source Amplifier: Total Gain Under the Miller Approx

Hv

v

s

out

||1

1

omgdgss rRgCCR Can be large

+ inv

-

di

gsmvg+ outv

-og

gsC

gi omgd rRgC ||1

R sv+

gsv-

sR

18


gsmvg oggsC

gdC

R+ gsv-

sR

The Common Source Amplifier: Output Impedance

This gives:

ti(1)

tv

KCL at (2) gives:

s

gsgsgsgdgst R

vvCjCjvv

(2)

t

sgdgs

sgdgs v

RCCj

RCjv

1

Need to find:

outZ


gsmvg oggsC

gdC

R+ gsv-

sR

The Common Source Amplifier: Output Impedance

KCL at (1) gives:

1||11

||

sgdgs

sgdgsm

oout

gdgstgsmo

tt

RCCj

RCjCjg

rRZ

CjvvvgrR

vi

ti(1)

tv

(2)

t

sgdgs

sgdgs v

RCCj

RCjv

1

Not a bad approximation even at moderately high frequencies

19


+ inv

-

outi

gsmvg oggsC

ing ii gdC

R si+ gsv-

sR

The Common Source Amplifier: Short Circuit Current Gain

Need to find:

in

out

ii

gdgs

m

inmingdmin

in

in

out

in

out

CCjg

ZgZCjgiv

v

i

i

i

gdm

in

out

outinmgdin

Cjgv

i

ivgCjv

00

Start from KCL at (1):

(1)

gdgsin CCj

Z

1


Short Circuit Current Gain and the Transition Frequency (fT or T)

For most transistors, the short circuit current gain falls of with frequency with a -20 dB/dec slope (at high enough frequencies)

The frequency at which the short circuit current gain equals unity is called the transition frequency (fT or T)

The transition frequency expresses the maximum useful frequency of operation of the transistor

2

10log10

in

out

i

i

0


T

???

20


The Common Source Amplifier: The Transition Frequency

gdgs

m

in

out

CCjg

ii

2

10log10

in

out

i

i

0


T

gs

m

gdgs

mT C

gCC

g

The transition frequency is:


The Common Source Amplifier: The Transition Frequency

gs

mT C

g

The FET transition frequency is:

This is the highest frequency at which the transistor is still usefulQ: What is its physical significance?

0y Ly

Gate

yESource Drain

VCS=VDSVCS=VGS-VTN

VCS=0

oxgs

TNGSoxnm

WLCC

VVCLW

g

32

22

3

L

VVCg TNGS

ngs

mT

Therefore the transition frequency is:

t

nynTNGS

nTNGS

nT Lv

L

E

LL

VV

L

VV

1

2

Electron transit time through the FET

Therefore, the electron transit time sets the maximum frequency of operation of the FET!!

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t vr t vr j t - Cornell University · s s out in v j R C A A v Av 1 1 vs Rs vin The amplifier input voltage, and the output voltage, will now begin to drop-off at a much lower frequency!!