Lecture 12: FET AC Properties - LTH · Lecture 12: FET AC Properties 2016-02-22 Lecture 11, High...

Lecture 12: FET AC Properties

2016-02-22 1Lecture 11, High Speed Devices 2016

Lecture 12: FET AC Properties


• Quasi-static operation

• Diffusive and Ballistic FETs

• y-parameters

• Hybrid p-model

• Non-quasi Static effects

Reading guide for chapter 6: 371-386, 394-421,. Omit any specific MESFET related topics. Non-quasi static effects briefly. Note that Liu only covers the diffusive FETs.

Time varying voltages


The transistor is charge neutral:QG(t)+QCH(t)=0

The channel charge is divided intotwo parts: ”Source” and ”Drain” charge: QCH=QS+QD

ig(t)

ID+id(t)

When Vgs changes (reasonable slowly):• ID (conduction current) changes• The amount of channel charge changes -> charges has to be supplied through id and

is! ig supplies the gate charge!

+

id(t)=dQd/dt – need to determineQD(vg,vd,vs).

Quasi-Static Approximation


VT

vgs(t)ID

Quasi-static approximation: the charge distribution in the channel reactsinstantiounsly to any changes in vDS, vGS.

t=t1

t=t2

t=t3

txsatD

ch

QStrv

L

v

L

v

L

I

Q,,,

x

n(x)

x=Lg

Valid for time scales longer than Given by the time it takes for the electrons to propagate from source to drain

𝑓 =1

2𝜋𝜏𝑡𝑟≈ 5 𝑇𝐻𝑧 for a ballistic fet with Lg=60 nm.

VDS>VDS,sat

t=t1

t=t2

t=t3

Ex – let vGS change from below VT to abiove

Charges and currents


𝑖𝑔 =𝑑

𝑑𝑡𝑄𝐺(𝑡) 𝑖𝐷(𝑡) = 𝐼𝐷 𝑡 +

𝑑

𝑑𝑡𝑄𝐷 𝑡 𝑖𝑆 = −𝐼𝐷 𝑡 +

𝑑

𝑑𝑡𝑄𝑆 𝑡

iD : total instantaneous drain current

ID: Quasi-static drain current

d/dt: capacitive/charging current from changein channel/gate charge storage. Flows onlywhen the voltage changes!

Gate Current Drain Current Source Current

𝑄𝐺 = 𝑄𝑆 + 𝑄𝐷The device is charge neutral:

(From previous lectures!)

Charging Currents


𝜕𝑄

𝜕𝑡+𝜕𝐽

𝜕𝑥= 0

𝜕𝑄𝑆𝜕𝑡+𝜕𝑄𝐷𝜕𝑡+𝜕𝑄𝐺𝜕𝑡

= 0

𝜕𝑄𝑆𝜕𝑡=𝜕𝑉𝐷𝜕𝑡

𝜕𝑄𝑆𝜕𝑉𝐷

+𝜕𝑉𝐺𝜕𝑡

𝜕𝑄𝑆𝜕𝑉𝐺

+𝜕𝑉𝑆𝜕𝑡

𝜕𝑄𝑆𝜕𝑉𝑆

= 𝑖𝑠

𝜕𝑄𝐷𝜕𝑡

=𝜕𝑉𝐷𝜕𝑡

𝜕𝑄𝐷𝜕𝑉𝐷


𝜕𝑄𝐷𝜕𝑉𝐺


𝜕𝑄𝐷𝜕𝑉𝑆

= 𝑖𝑑

𝜕𝑄𝐺𝜕𝑡

=𝜕𝑉𝐷𝜕𝑡

𝜕𝑄𝐺𝜕𝑉𝐷


𝜕𝑄𝐺𝜕𝑉𝐺


𝜕𝑄𝐺𝜕𝑉𝑆

= 𝑖𝑔

VD

VG

Charging currents:

𝑖𝐺 =𝜕𝑄𝐺𝜕𝑡

𝑖𝑆 =𝜕𝑄𝑠𝜕𝑡

𝑖𝐷 =𝜕𝑄𝐷𝜕𝑡

VS

QS QD

QG

id

ig

is

𝐶𝑖𝑖 =𝜕𝑄𝑖𝜕𝑉𝑖

𝐶𝑗𝑘 = −𝜕𝑄𝑗

𝜕𝑉𝑘

Definition of transcapacitances

Quasi-static:Charge neutral deviceKCL also holds

This makes most capacitances positive

Charging Currents - capacitances


𝜕𝑄

𝜕𝑡+𝜕𝐽

𝜕𝑥= 0

VD

VG

VS

𝑖𝑠𝑖𝑑𝑖𝑔

=

𝐶𝑆𝑆 −𝐶𝑆𝐷 −𝐶𝑆𝐺−𝐶𝐷𝑆 𝑪𝑫𝑫 −𝑪𝑫𝑮−𝐶𝐺𝑆 −𝑪𝑮𝑫 𝑪𝑮𝑮

𝑑𝑣𝑠𝑑𝑡𝑑𝑣𝑑𝑑𝑡𝑑𝑣𝑔

𝑑𝑡

If all voltages are equal: 𝑖𝑠 = 𝑖𝑑 = 𝑖𝑔 = 0

If one voltage is applied: 𝑖𝑠 + 𝑖𝑑 + 𝑖𝑔 = 0

Rows/ columns sum to zero

𝑖𝑔 = 𝐶𝐺𝐺𝑑𝑣𝑔

𝑑𝑡

𝑖𝑑 = −𝐶𝐷𝐺𝑑𝑣𝑔

𝑑𝑡

𝑖𝑠 = −𝐶𝑆𝐺𝑑𝑣𝑔

𝑑𝑡

Ex: Voltage is applied to the gate terminal only

CGG

ig

VGSVGS

isid

CSG CDG

ig

Charging Current / total Current


𝑖𝑠𝑖𝑑𝑖𝑔

=

𝐶𝑆𝑆 −𝐶𝑆𝐷 −𝐶𝑆𝐺−𝐶𝐷𝑆 𝑪𝑫𝑫 −𝑪𝑫𝑮−𝐶𝐺𝑆 −𝑪𝑮𝑫 𝑪𝑮𝑮

𝑑𝑣𝑠𝑑𝑡𝑑𝑣𝑑𝑑𝑡𝑑𝑣𝑔

𝑑𝑡

𝑖𝑔𝑖𝑑=

𝐶𝐺𝐺 −𝐶𝐺𝐷−𝐶𝐷𝐺 𝐶𝐷𝐷

𝜕𝑣𝐺𝑆𝜕𝑡𝜕𝑣𝐷𝑆𝜕𝑡

𝑖𝐷 = 𝐼𝐷 𝑣𝐺𝑆, 𝑣𝐷𝑆 + 𝐶𝐷𝐷 𝑣𝐺𝑆, 𝑣𝐷𝑆𝜕𝑣𝐺𝑆𝜕𝑡

− 𝐶𝐷𝐺 𝑣𝐺𝑆, 𝑣𝐷𝑆𝜕𝑣𝐷𝑆𝜕𝑡

𝑖𝐺 = 𝐶𝐺𝐺 𝑣𝐺𝑆, 𝑣𝐷𝑆𝜕𝑣𝐺𝑆𝜕𝑡

− 𝐶𝐺𝐷 𝑣𝐺𝑆, 𝑣𝐷𝑆𝜕𝑣𝐷𝑆𝜕𝑡

Large Signal, Quasi-static FET model

We need to calculate– QG(VGS,VDS) and QD(VGS,VDS)

Common Source

Diffusive FET Gate Charge: QG


Negative channel charges divides onto drain and source chargesPositive charge on gate terminal to make device neutral

1

1

3

2,,11),0(,)(

2'

0 0

2'

Tgsox

L L

dgsCHCHoxG VvWLCvvvL

xtudxtxuWCtQ

Total Gate charge:

-

+++ ++ ++ +

++ ++

- -- - -- - -- ---

QG

QDQS

More positive gate bias – more positive charge on the gateCharge neutral device: More negative charge in the channel

𝑢𝑐ℎ 𝑥, 𝑡 = 𝑢𝑐ℎ 0, 𝑡 1 −𝑥

𝐿1 − 𝛼2 𝛼 =

1 −𝑣𝐷𝑆𝑉𝐷𝑆,𝑠𝑎𝑡

, 𝑉𝐷𝑆 < 𝑉𝐷𝑆,𝑠𝑎𝑡

0

𝑉𝐷𝑆,𝑠𝑎𝑡 = (𝑉𝐺𝑆 − 𝑉𝑇)

𝑛 𝑥, 𝑡 = 𝑊𝐶𝐺 𝑣𝐺𝑆 𝑡 − 𝑉𝑇 1 −𝑥

𝐿1 − 𝛼2

Ballistic FET Gate Charge: QG


EFS

LG

A real ballistic FET has a complicated band profileCrude approximation : constant potential until drain reservior

𝑛 𝑥 = 𝑛 0 = 𝐶𝐺′ (𝑣𝐺𝑆 − 𝑉𝑇)

𝑄𝐺 = 𝑊𝐿𝑔𝐶𝐺′ (𝑣𝐺𝑆 − 𝑉𝑇)

𝑄𝐺 =

0

𝐿𝐺

𝑛 𝑥 𝑑𝑥

Simplest possible assumption:

𝐶𝐺′ =

1

𝐶𝑜𝑥+2

𝐶𝑞+1

𝐶𝑐

−1



EFS𝐼

𝑊= 𝐶𝐺

′ 𝑣𝐺𝑆 − 𝑉𝑇8ℏ

3𝑚∗ 𝜋𝑞𝐶𝐺′ 𝑣𝐺𝑆 − 𝑉𝑇

𝐼

𝑊= 𝑞𝑛 0 𝑣(0)

Δ𝐸𝑘𝑖𝑛 ≈ 𝑞𝑣𝐷𝑆𝑥

𝐿𝐺

2

LG

A real ballistic FET has a complicated band profileSimple approximation : quadratic drop between top-of-the-barrier and the drain

LG

Carriers gain extra kinetic energySince J(x) = constant : n(x) decreases

𝑚𝑣 𝑥 2

2= 𝐸0 + Δ𝐸𝑘𝑖𝑛

x

E



𝐼

𝑊= 𝑞𝑛 0 𝑣(0)

Δ𝐸𝑘𝑖𝑛 = 𝑞𝑣𝐷𝑆𝑥

𝐿𝐺

2

x=LG

𝑣 𝑥 = 𝑣 0 2 +2𝑞𝑣𝐷𝑆𝑚∗

𝑥

𝐿𝐺

2

𝑛 𝑥 = 𝑛 0𝑣 0

𝑣 0 2 +2𝑞𝑣𝐷𝑆𝑚∗

𝑥𝐿𝐺

2

x=0

𝑄𝐺 =𝑊𝐿𝐺𝐶𝐺

′ 𝑉𝐺𝑆−𝑉𝑇 𝑣 0 𝑚∗

2𝑞𝑉𝐷𝑆sinh−1

2𝑞𝑉𝐷𝑆

𝑚∗𝑣 0

𝑣 0 =8ℏ

3𝑚∗ 𝑞𝜋𝐶𝐺′ 𝑣𝐺𝑆 − 𝑉𝑇 𝑛 0 = 𝐶𝐺

′ 𝑣𝐺𝑆 − 𝑉𝑇

𝑄𝐺 =

0

𝐿𝐺

𝑛 𝑥 𝑑𝑥 Total Gate Charge

Total channel / gate charge in saturation


Lg=50 nm

Ballistic - flat

Ballistic -quadratic

VDS=0.5V

VDS=0.6V

tox= 5nmtw= 7 nmm*=0.03 m0

er=eox=14

QG decreases with increasing VDS

for the ballistic FET in saturation

QG for a ballistic FET in saturation



VDS=0.5V

VDS=0.6V

QG decreased with increasing VDS since:

A) Less back injected drain chargeB) More back injected drain chargeC) Larger carrier velocity towards the drainD) Smaller carrier velocity towards the drainE) ???

0A B C D E

nano.participoll.com

Transcapacitances: CGD and CGG


2

2'

1

41

3

2

ox

G

Ggg WLC

V

QC

2

2'

1

2

3

2

ox

D

Ggd WLC

V

QC

𝐶𝑥𝑦 = 𝛿𝑥𝑦𝜕𝑄𝑥𝜕𝑉𝑦

𝛿𝑥𝑦:= 1 if x=y= -1 if x≠y

Diffusive FET:

𝐶𝑔𝑔 = 𝑊𝐿𝐺𝐶𝐺′

How much does the total channel charge change as VG and VD is varied

𝐶𝑔𝑑 = 0

Ballistic FET – constant potential(Saturation)

Ballistic FET – quadratic potential(Saturation)

𝐶𝑔𝑔 =𝜕𝑄𝐺𝜕𝑉𝐺

= ⋯

𝐶𝑔𝑑 = −𝜕𝑄𝐺𝜕𝑉𝐷

> 0…

Transcapacitances in saturation: CGG


Diffusive


VDS=0.5V

VDS=0.6V

Transcapacitances CGD


CGG

CGDBallistic -quadratic

VDS=0.5V

VDS=0.6V

Diffusive

Drain Charges – Ballistic FET


Negative channel charges divides onto drain and source charges

-

+++ ++ ++ +

++ ++

- -- - -- - -- ---

QG

QDQS𝑄𝐺 + 𝑄𝑆 + 𝑄𝐷 = 0

QD: When QG changes – how much of that charge is ‘supplied’ from the drain?

QS – charge directly supplied from the source

QD – charge directly supplied from the drain

𝑛→ 𝑥 =𝐽→

𝑣+2(0) +

2𝑞𝑣𝐷𝑆𝑚∗

𝑥𝐿𝐺

2

𝑛← 𝑥 ≈𝐽←

𝑣−2 0 +

2𝑞𝑣𝐷𝑆𝑚∗

𝑥𝐿𝐺

2

≈ 0

𝑄𝑆 ∶ 𝐽→ 𝑄𝐷 ∶ 𝐽

←

Drain Charges – Diffusive FET


Negative channel charges divides onto drain and source charges

-

+++ ++ ++ +

++ ++

- -- - -- - -- ---

QG

QDQS

2

23'

0

'

115

48126)(

Tgsox

L

CHox

D VvWLCdxxxUL

WCtQ

t

QIti D

Dd

Definition of drain charge:

=0QD/QS=40/60

0 0.5 1 1.5 2-1

-0.5

0

0.5

1

1.5

VDS

(V)

Qx

QG

QS

QD

VT=-0.5VVg=0.7V

Vg=0.3V

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

1.5

VGS

(V)

Qx

QG

QD

QS

Vds=0.5V

Vds=2.0V

In saturation:

𝑄𝐺 = 𝑄𝑆 + 𝑄𝐷

QD: When QG changes – how much of that chare is ‘supplied’ from the drain?

Transcapacitances – Diffusive FET


2

2'

1

41

3

2

ox

G

Ggg WLC

V

QC

2

2'

1

2

3

2

ox

D

Ggd WLC

V

QC

3

32'

1

311142

15

2

ox

G

Ddg WLC

V

QC

3

32'

1

398

15

2

ox

D

Ddd WLC

V

QC

Note that Cdg ≠ Cgd

𝐶𝑥𝑦 = 𝛿𝑥𝑦𝜕𝑄𝑥𝜕𝑉𝑦

𝛿𝑥𝑦:= 1 if x=y= -1 if x≠y

Transcapacitances – Ballistic FET in saturation


𝑄𝐺 =𝑊𝐿𝐺𝐶𝐺

′ 𝑉𝐺𝑆−𝑉𝑇 𝑣 0 𝑚∗

2𝑞𝑉𝐷𝑆sinh−1

2𝑞𝑉𝐷𝑆

𝑚∗𝑣 0

𝑄𝑆 + 𝑄𝐺 = 0 𝑄𝐷 ≈ 0

𝐶𝐺𝐺 =𝜕𝑄𝐺𝜕𝑉𝐺

= 𝑊𝐿𝐺𝐶𝐺′ …

𝐶𝐺𝐷 = −𝜕𝑄𝐺𝜕𝑉𝐷

= 𝑊𝐿𝐺𝐶𝐺′ …

𝐶𝐷𝐺 = −𝜕𝑄𝐷𝜕𝑉𝐺

≈ 0

CGG

CGD

𝐶𝐷𝐷 =𝜕𝑄𝐷𝜕𝑉𝐷

≈ 0

Short channel effects can lead to negative CGD!

CDD,CGD

2 minutes excercise


2

2'

1

2

3

2

ox

D

Ggd WLC

V

QC

3

32'

1

311142

15

2

ox

G

Ddg WLC

V

QC

What is the physical reason that in saturation Cgd=0 and Cdg > 0 for an ideal, diffusive FET?

V

I

=0

A) The drain voltage does not influence the total channel charge after pinch-off

B) The gate voltage does not influence the total channel charge after pinch-off

C) QS+QD+QG=0 does not hold in saturation.D) This is due to approximations. Cij must

always be Cji.E) ???

0A B C D E

nano.participoll.com

Large Signal Model – common source


Common source:

dt

dvvvC

dt

dvvvC

dt

tvtvtvdQti ds

dsgsgd

gs

dsgsgg

sdgg

g ,,)(),(),(

dt

dvvvC

dt

dvvvCI

dt

tvtvtvdQtvtvIti

dsdsgsdd

gs

dsgsdgD

sdgD

dsgsDd

,,

)(),(),()(),(

dt

dv

dt

dv

dt

dv

dt

dv

dt

dv gsgdsds 0

vG

Cgg

𝐶𝑔𝑑𝑑𝑣𝐷𝑆𝑑𝑡

Cdd

𝐶𝑑𝑔𝑑𝑣𝐺𝑆𝑑𝑡

𝐼𝐷(𝑉𝐺𝑆, 𝑉𝐷𝑆)

Gate

Source

Drain

Small signal DC Model– Common Source


0gi

ds

VDS

Dgs

VGS

DDSGSD

dsDSgsGSDD

vV

Iv

V

IVVI

vVvVIi

GSDS

,

,

gmgd

satoxm

TGSnG

m

vWCg

VVL

WCg

'

1

),(

'

dsgsd

TGSnox

d

VVfg

VVL

WCg

Long channel

Velocity Saturation

vGS 𝑔𝑚𝑉𝐺𝑆

Gate

Source

Drain

𝑔𝑑

𝑑

𝑑𝑡→ 0

𝑔𝑚 =3

2𝐶𝐺′𝑊

8ℏ

3𝑚∗ 𝑞𝜋𝐶𝐺′ (𝑉𝐺𝑆 − 𝑉𝑇) 𝑔𝑑 > 0 Ballistic

Quasi-static Common Source Small Signal y-parameters


dt

dvC

dt

dvCti ds

gd

gs

ggg

dt

dvC

dt

dvCvgvgti ds

dd

gs

dgdsdgsmd tj

dd

tj

dds

eii

evv

~

~

tj

gg

tj

gsgs

eii

evv

~

~

ds

gs

QSddQSdg

QSgdQSgg

d

g

v

v

yy

yy

i

i~

~

~

~

,,

,,

dddQSdd

dgmQSdg

gdQSgd

ggQSgg

Cjgy

Cjgy

Cjy

Cjy

,

,

,

,

Small Signal Hybrid-p model


Cgd

Cgs Csdgd

Gate

jCmvgs

gmvgs

Drain

Source

Cm=Cdg-Cgd

ds

gs

QSddQSdg

QSgdQSgg

d

g

v

v

yy

yy

i

i~

~

~

~

,,

,,

dddQSdd

dgmQSdg

gdQSgd

ggQSgg

Cjgy

Cjgy

Cjy

Cjy

,

,

,

,

+

-

i1

y11+y12

(y21-y12)v2

-y12

y22+y12

v2

+

-

i2

Cgg-Cgd=Cgs

-Cgd+Cdd=Csd+

Small Signal Hybrid-p model


Cgd

Cgs Csdgd

Gate

jCmvgs

gmvgs

Drain

Source

Cm=Cdg-Cgd

General FET Model

Cgd

Cgs Csdgd

Gate

𝑔𝑚 − 𝑗𝜔𝐶𝑔𝑑 𝑣𝑔𝑠

Ballistic FET in saturation

Cgsgd

Gate

𝑔𝑚 + 𝑗𝜔𝐶𝑑𝑔 𝑣𝑔𝑠

Diffusive FET in saturation

Non-Quasi-Static


VT

vgs(t)

ID

For very short timescales:

Charges has to be supplied from the source – transient involves a charge-front –more complex math!

=0

)(,,

2

,

TGSnxsatD

ch

QStrVV

L

v

L

v

L

I

Q

txut

txux

ux

chchchn ,,

x

n(x)

x=Lg

𝑡 = ∞ 𝑠

t=t0 t0

t1

t2

t3

t4

t1 t2 t3 t4

Diffusive Transistor: Continuity and drift equation

Small signal non-quasi static


txut

txux

ux

chchchn ,,

tj

chCHch

tj

chCHch

exixItxi

exuxUtxu

~,

~,

Assume small signal sinusodial perturbation:

D

oxn

chCHD

CH

ch

I

WCD

iUIjDdU

id

'

2

2~

,

~

2/3

3/12

2/3

3/113

2ˆ3

2ˆ~CHCHCHCHCHch UjDIUCUjDIUCUi

Î is the modified Bessel function

Express Î as a series expansion, and solve for uch(x) from boundary conditions(very tedious algebra, see pp. 481-489)

Non QS: Channel Resistance


10

130,

10

130,

/1

/

.../1

/

j

jCjCjy

j

jCjCjy

gdgdNQSgd

gdggNQSgg

After plenty of algebra, one obtains the NQS y-parameters

20

L

VV TGSn

Essentially all corrections are zero/small if << 0

Cgs

rch

Gate

Source

2

23

, 21110

121015311Re

mNQSgs

chgy

r

rch important when gm is small, i.e. Not fully linear region / small Vgs

Or at very high frequencies!

Cgs

GateDrain

Source

rch

rgd

d

sd

g

C

03

2

1)1(

31

15

4

10 )/(1 j

gm

𝑔𝑚 →𝑔𝑚

1 +𝑗𝜔𝜔0𝜏1

Summary – hybrid p


Cgs

GateDrain

Source

rch

rgd

1vgm

Cgd

Cgs Csdgd

Gate Drain

Source

)1( mm Cjg

Quasi-Static AC DC

Non Quasi-Static AC

vGS 𝑔𝑚𝑉𝐺𝑆

Gate

Source

Drain

𝑔𝑑

+v1

-

Ignoring Cm

Lecture 12: FET AC Properties - LTH · Lecture 12: FET AC Properties 2016-02-22 Lecture 11, High...

Documents

Transcript of Lecture 12: FET AC Properties - LTH · Lecture 12: FET AC Properties 2016-02-22 Lecture 11, High...