Lecture 8: Ballistic FET I-V - Lunds tekniska högskola · 2016-02-09 Lecture 8, High Speed Devices 2016 4 L>>l0 L

Lecture 8: Ballistic FET I-V

2016-02-09 1Lecture 8, High Speed Devices 2016

Lecture 12: Ballistic FETs

2016-02-09 2Lecture 8, High Speed Devices 2016

Jena: 61-70

Diffusive Field Effect Transistor

2016-02-09 3Lecture 8, High Speed Devices 2016

Lg >> l

Mean free path much shorter than channel lengthDiffusive transport – drift and diffusion

Source DrainGate

E1

Source DrainGate

xxnqJ n

VGS

UCH=VGS –VT -VCS(x)

xUCtxq choxch

','

2110 L

xUxU CHCH

ID

n

cqn

1

µn=qt/m*

What happens when an electron travels through the channel without any scattering events?

Diffusive Field Effect Transistor

2016-02-09 4Lecture 8, High Speed Devices 2016

L>>l0 L<<l0

Mean free path much shorter than channel lengthDiffusive transport – drift and diffusion

Mean free path much longer than channel length Ballistic transport

Fully ballistic transport will give the upper limit of a transistor’s performance!How do we calculate the ballistic current flowing through a FET?

Source DrainGate

E1 E1

Top of the barrier limited transport

2016-02-09 5Lecture 8, High Speed Devices 2016

k

kEkv

Ekm

kE

1

21

2

*

2

E1

Ef,s

E1

Ef,d

qVds

E(0)

E

k

Ef,s E

k

Ef,d

We model the source/drain regions as electron reservoirsElectrons with positive k-values moves “right”Electrons with negative k-values moves “left”

t t

𝑑

𝑑𝑥𝐽 𝑥 = 0

Top of the barrier limited transport

2016-02-09 6Lecture 8, High Speed Devices 2016

k

kEkv

Ekm

kE

02

2

*

2

E1

Ef,s

E1

Ef,d

qVds

E(0)

E

k

Ef,s

Ef,d

E(0)

Injected from the source: Ef,s-E(0)Injected from the drain: Ef,d-E(0)

More electrons moving towards the source – net electron current!

E(0) is mainly set by changing the gate voltage!

2 minute excercise

2016-02-09 7Lecture 8, High Speed Devices 2016

E1

Ef,s

E1

Ef,d

qVds

E(0)

Vds

~ qEf,d

Net

Ele

ctro

n c

urr

ent

0

Sketch the net electron current as Vds is increased!For simplicity: Assume that E(0) are independent of Vds!

2D Ballistic FET – number of electrons

2016-02-09 8Lecture 8, High Speed Devices 2016

𝑛𝑠 =𝑚∗

2𝜋ℏ2

𝐸𝐶

∞

𝑓1 𝐸 − 𝑓2 𝐸 𝑑𝐸

𝑛𝑠 =𝑁2𝐷2[ℱ0 𝜂𝐹1 + ℱ0(𝜂𝐹2)]

𝑁2𝐷 =𝑚∗𝑘𝑇

𝜋ℏ2

𝑛𝑠 ≈ 𝐶𝐺(𝑉𝐺𝑆 − 𝑉𝑇)

Above threshold:

E1

Ef,s

E1

Ef,d

qVds

E(0)

𝜂𝐹𝑠 =𝐸𝑓𝑠 − 𝐸 0

𝑘𝑇

𝜂𝐹𝑑 =𝐸𝑓𝑠 − 𝐸 0 − 𝑞𝑉𝐷𝑆

𝑘𝑇

1

𝐶𝐺=1

𝐶𝑜𝑥+1

𝐶𝑞+1

𝐶𝐶

𝐶𝑞 = 𝑞2𝑚∗/(𝜋ℏ2)

2D Ballistic FET – number of electrons

2016-02-09 9Lecture 8, High Speed Devices 2016

𝑛𝑠 =𝑁2𝐷2[ℱ0 𝜂𝐹𝑠 + ℱ0(𝜂𝐹𝑑)]

𝑁2𝐷 =𝑚∗𝑘𝑇

𝜋ℏ2

𝑛𝑠 ≈ 𝐶𝐺(𝑉𝐺𝑆 − 𝑉𝑇)

VDS = 0V

E1

Ef,s

E1

Ef,d

1

𝐶𝐺=1

𝐶𝑜𝑥+1

𝐶𝑞+1

𝐶𝐶

E1

Ef,s

E1

Ef,d

qVds

E(0)

E(0)

𝑘𝑥+ 𝑘𝑥

−

𝑛𝑠 ≈ 𝐶𝐺′ (𝑉𝐺𝑆 − 𝑉𝑇)

VDS >> 0V

1

𝐶𝐺=1

𝐶𝑜𝑥+2

𝐶𝑞+1

𝐶𝐶

MOS-limit:Cox << Cq: 𝑛𝑠 = 𝑛𝑠

+ +𝑛𝑠− stays constant for

all VDS

𝑛𝑠+ increases with VDS. 𝜂𝐹𝑠 must increase, until VDS is large so that 𝑛𝑠

− ≈ 0.

2D Ballistic FET - Current I

2016-02-09 10Lecture 8, High Speed Devices 2016

𝜂𝐹𝑠 =𝐸𝑓𝑠 − 𝐸 0

𝑘𝐵𝑇

𝜂𝐹𝑑 =𝐸𝑓𝑠 − 𝐸 0 − 𝑞𝑉𝐷𝑆

𝑘𝐵𝑇

𝑱 =2𝑞

ℏ 2𝜋 2 𝑑𝑘𝑥𝑑𝑘𝑦𝛻𝒌 𝐸 𝒌 𝑓0 𝐸 𝒌 − 𝐸𝑓𝑠

𝐽→ : Current in x direction. (+kx)

𝛻𝑘𝑥𝐸 𝒌 =𝜕

𝜕𝑘𝑥

ℏ2

2𝑚∗𝑘𝑥2 + 𝑘𝑦

2 =ℏ2

𝑚∗𝑘𝑥 =

ℏ2

𝑚∗𝑘 cos 𝜃

𝐽→ =2𝑞ℏ

2𝜋 2𝑚∗

𝑘=0,𝜃=−𝜋2

𝑘=∞,𝜃=𝜋2

𝑘𝑐𝑜𝑠 𝜃 𝑘

1 + exp𝐸 𝑘 − 𝐸𝑓𝑠𝑘𝐵𝑇

𝑑𝑘𝑑𝜃

𝐸 − 𝐸1 =ℏ2𝑘2

2𝑚∗𝑘 =

2𝑚∗ 𝐸 − 𝐸1ℏ

𝑑𝐸 =ℏ2𝑘

𝑚∗𝑑𝑘 𝜂 =

𝐸

𝑘𝐵𝑇

2D Ballistic FET - current

2016-02-09 11Lecture 8, High Speed Devices 2016

𝜂𝐹𝑠 =𝐸𝑓𝑠 − 𝐸 0

𝑘𝐵𝑇

𝜂𝐹𝑑 =𝐸𝑓𝑠 − 𝐸 0 − 𝑞𝑉𝐷𝑆

𝑘𝐵𝑇

𝐽→ =2𝑞 2𝑚∗ 𝑘𝑇

32

2𝜋2ℏ2

0

∞𝜂

1 + 𝑒𝜂−𝜂𝐹𝑆𝑑𝜂

2

𝜋𝐹1/2 𝜂𝑓𝑠

𝐽→ =𝑞 2𝑚∗ 𝑘𝑇

32 𝜋

2𝜋2ℏ2𝐹1/2 𝜂𝑓𝑠 = 𝐽2𝐷

0 𝐹1/2 𝜂𝑓𝑠

𝐽→ ≈ 𝐽2𝐷04

3 𝜋𝜂𝑓𝑠

32 =𝑞 2𝑚∗

2𝜋2ℏ24

3𝐸𝑓𝑠 − 𝐸1

3/2Degenerate: Efs>>E1

𝐽← =𝑞 2𝑚∗ 𝑘𝑇

32 𝜋

2𝜋2ℏ2𝐹1/2 𝜂𝑓𝑑 = 𝐽2𝐷

0 𝐹1/2 𝜂𝑓𝑠 − 𝑣𝑑Drain to source current

2D Ballistic FET - Saturated Current

2016-02-09 12Lecture 8, High Speed Devices 2016

We need to determine 𝜂𝐹𝑠 to calculate ID

𝑞𝑛𝑠 ≈ 𝐶𝐺′ (𝑉𝐺𝑆 − 𝑉𝑇) In saturation: 𝐹0 𝜂𝐹𝑑 ≈ 0 and VGS>>VT

𝑞𝑛𝑠 =𝑞𝑁2𝐷2[𝐹0 𝜂𝐹𝑠 + 𝐹0 𝜂𝐹𝑑 ] ≈

𝑞𝑁2𝐷2𝜂𝐹𝑠 =

𝐶𝑞2

𝑘𝑇

𝑞𝜂𝐹𝑠

𝜂𝐹𝑠 =2𝐶𝐺′

𝐶𝑞

𝑉𝐺𝑆 − 𝑉𝑇𝑉𝑇ℎ

𝐼

𝑊= 𝐶𝐺′ 𝑉𝐺𝑆 − 𝑉𝑇

8ℏ

3𝑚∗ 𝑞𝜋𝐶𝐺′ 𝑉𝐺𝑆 − 𝑉𝑇

𝑞𝑛𝑠 < 𝑣𝑥 >

Since only +kx states are

occupied. 𝐶𝑞′ =𝐶𝑞

2should

be used for the calculation of 𝐶𝐺

′ .

𝐽 ≈ 𝐽→ ≈ 𝐽2𝐷04

3 𝜋𝜂𝑓𝑠

32

2D Ballistic FET - I(VDS,VGS)

2016-02-09 13Lecture 8, High Speed Devices 2016

General (approximate) solution for VGS>VT

𝑞𝑛𝑠 =𝑞𝑁2𝐷2[𝐹0 𝜂𝐹𝑠 + 𝐹0 𝜂𝐹𝑑 − 𝑣𝑑 ] = 𝐶𝐺(𝑉𝐺𝑆 − 𝑉𝑇)

𝐹0(𝜂) = ln(1 + 𝑒𝜂)

2𝐶𝐺𝐶𝑞

𝑉𝐺𝑆 − 𝑉𝑇𝑉𝑇ℎ

= 2𝜂𝐺 = ln 𝑒𝜂𝐺 = ln 1 + 𝑒𝜂𝐹𝑠 1 + 𝑒𝜂𝐹𝑠−𝑣𝑑

𝜂𝐹𝑠 = ln 1 + 𝑒𝑣𝑑 2 + 4𝑒𝑣𝑑 2𝑒𝜂𝐺 − 1 − 1 + 𝑒𝑣𝑑 − ln 2

Valid if 𝐶𝐺 ≈ 𝐶𝐺′ : CG<<Cq

Assuming MOS-limit: Cox << Cq

𝐽 = 𝐽2𝐷0 𝐹1

2𝜂𝑓𝑠 − 𝐹1

2𝜂𝑓𝑠 − 𝑣𝑑 ≈ 𝐽2𝐷

0 4/3 𝜋 𝜂𝐹𝑠1.5 − 𝜂𝐹𝑠 − 𝑣𝑑

1.5

Analytical but complicated expression for the ballistic current!

2D Ballistic FET - Current II

2016-02-09 14Lecture 8, High Speed Devices 2016

𝐶𝑞 = 0.025 𝐹/𝑚2

𝐶𝐺 = 0.01 𝐹/𝑚2

0 < VGS< 0.4 V

• Similar ”shape” as diffusivetransport : but 𝐼 ∝ (𝑉𝐺𝑆 −

Back Scattering

2016-02-09 15Lecture 8, High Speed Devices 2016

𝑇 ≈𝜆0𝐿𝐺 + 𝜆0

T

1-T

𝜆0 ≈ 2µ𝑛𝑘𝑇𝐿𝑞

1

𝑣𝑡ℎ𝑣𝑡ℎ =

2𝑘𝑇

𝑚∗𝜋LG

Non-degenerate limitSmall VDS

𝐼

𝑊= 𝐶𝐺′ 𝑉𝐺𝑆 − 𝑉𝑇

8ℏ

3𝑚∗ 𝑞𝜋𝐶𝐺′ 𝑉𝐺𝑆 − 𝑉𝑇

𝐼

𝑊≈𝜆0𝐿𝐺 + 𝜆0

𝐶𝐺′ 𝑉𝐺𝑆 − 𝑉𝑇

8ℏ

3𝑚∗ 𝑞𝜋𝐶𝐺′ 𝑉𝐺𝑆 − 𝑉𝑇

No scattering

Approximate effect of scattering!

This relates the mobility (long channel property) to the quasi-ballistic current!

2D Ballistic FET – current

2016-02-09 16Lecture 8, High Speed Devices 2016

Fully Ballistic

tw=10 nmtox=4 nm (er=14)m*=0.03

µn=3000 cm2/Vsvsat=107 cm/s

VGS-VT=0.5V

Experimental Lg=40 nmIn0.7Ga0.3As HEMTIDS~ 1 mA/µm(IEDM 2011)

(Lower IDS - Source/Drain resistance)

Below VT: Sub threshold current

2016-02-09 17Lecture 8, High Speed Devices 2016

𝐼𝐷 = 𝑞𝑁2𝐷2𝑊𝑣𝑇 𝐹1/2 𝜂𝐹𝑠 − 𝐹1/2 𝜂𝐹𝑠 − 𝑣𝑑 ≈ 𝑞𝑊

𝑁2𝐷2𝑣𝑇 𝐹1/2 𝜂𝐹𝑠

𝐼𝐷 = 𝑞𝑊𝑁2𝐷2𝑣𝑇𝑒

𝑞𝑘𝑇𝑉𝐺𝑆−𝑉𝑇

𝑆𝑆 =𝑑

𝑑𝑉𝐺𝑆log10 𝐼𝐷

−1

= 2.3𝑘𝑇

𝑞= 60𝑚𝑉/𝑑𝑒𝑐𝑎𝑑𝑒

Inverse subthreshold slope:Diffusive Transport gives the same subthreshold slope

𝐽→ =𝑞 2𝑚∗ 𝑘𝑇

32 𝜋

2𝜋2ℏ2𝐹1/2 𝜂𝑓𝑠 = 𝐽2𝐷

0 𝐹1/2 𝜂𝑓𝑠𝑁2𝐷 =

𝑚∗𝑘𝑇

𝜋ℏ2

𝑣𝑇 =2𝑘𝑇

𝜋𝑚∗Below threshold: 𝑛𝑠 ≈ 0, so that 𝜓𝑠 moves 1:1 with the applied VGS! (If no short channel effects)

If VDS large