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EE130/230A Discussion 12 Peng Zheng 1

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EE130/230A Discussion 12. Peng Zheng. Velocity Saturation. Velocity saturation limits I Dsat in sub-micron MOSFETS Simple model: E sat is the electric field at velocity saturation:. for e < e sat. for e  e sat. EE130/230A Fall 2013. Lecture 22, Slide 2. - PowerPoint PPT Presentation

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Page 1: EE130/230A Discussion 12

1

EE130/230A Discussion 12

Peng Zheng

Page 2: EE130/230A Discussion 12

Velocity SaturationVelocity saturation limits IDsat in sub-micron MOSFETS

Simple model:

Esat is the electric field at velocity saturation:

sat

sat

vv

v

1

sat

satv2

for sat

for < sat

Siin holesfor cm/s 106

Siin sonfor electr cm/s 1086

6

satv

Lecture 22, Slide 2EE130/230A Fall 2013

Page 3: EE130/230A Discussion 12

Drain Saturation Voltage, VDsat

• If satL >> VGS-VT then the MOSFET is considered “long-channel”. This condition can be satisfied when– L is large, or – VGS is close to VT

LVVm

V satTGSDsat 11

satsat

v2

Lecture 22, Slide 3EE130/230A Fall 2013

Current saturation is limited by 2 mechanisms:1. Pinch-off2. Velocity saturation

Page 4: EE130/230A Discussion 12

Example: Drain Saturation VoltageQuestion: For VGS = 1.8 V, find VDsat for an NMOSFET with Toxe = 3 nm, VT = 0.25 V, and WT = 45 nm, if L = (a) 10 m, (b) 1 m, (c) 0.1 m (d) 0.05 m

Solution: From VGS , VT and Toxe, eff is 200 cm2V-1s-1. Esat= 2vsat / eff = 8 104 V/cm m = 1 + 3Toxe/WT = 1.2

11

LVVmV

satTGSDsat

Lecture 22, Slide 4EE130/230A Fall 2013

Page 5: EE130/230A Discussion 12

(a) L = 10 m: VDsat= (1/1.3V + 1/80V)-1 = 1.3 V

(b) L = 1 m: VDsat= (1/1.3V + 1/8V)-1 = 1.1 V

(c) L = 0.1 m: VDsat= (1/1.3V + 1/.8V)-1 = 0.5 V

(d) L = 0.05 m: VDsat= (1/1.3V + 1/.4V)-1 = 0.3 V

Lecture 22, Slide 5EE130/230A Fall 2013

Page 6: EE130/230A Discussion 12

MOSFET I-V with Velocity Saturation

LV

VVmVVCL

W

I

sat

DS

DSDSTGSeffoxe

D

1

2

LV

I channellongI

sat

DS

DD

1

In the linear region:

Lecture 22, Slide 6EE130/230A Fall 2013

Page 7: EE130/230A Discussion 12

IDsat with Velocity SaturationSubstituting VDsat for VDS in the linear-region ID equation gives

For very short L:

• IDsat is proportional to VGS–VT rather than (VGS – VT)2

• IDsat is not dependent on L

LmVV

I channellong

LmVV

VVCmLW

I

sat

TGS

Dsat

sat

TGS

TGSeffoxe

Dsat

11

22

mVVL TGSsat /

TGSoxesatTGSeffoxesatDsat VVCWvVVCWI 2

Lecture 22, Slide 7EE130/230A Fall 2013

Page 8: EE130/230A Discussion 12

Short- vs. Long-Channel NMOSFET

Short-channel NMOSFET:• IDsat is proportional to VGS-VTn rather than (VGS-VTn)2

• VDsat is lower than for long-channel MOSFET• Channel-length modulation is apparent

Lecture 22, Slide 8EE130/230A Fall 2013 C. C. Hu, Modern Semiconductor Devices for Integrated Circuits, Figure 6-23

Page 9: EE130/230A Discussion 12

The Short Channel Effect (SCE)

• |VT| decreases with L– Effect is exacerbated by high values of |VDS|

• This effect is undesirable (i.e. we want to minimize it!) because circuit designers would like VT to be invariant with transistor dimensions and bias condition

“VT roll-off”

Lecture 22, Slide 9EE130/230A Fall 2013

R. F. Pierret, Semiconductor Device Fundamentals, Fig. 19.3

Page 10: EE130/230A Discussion 12

depletionchargesupportedby gate(simplifiedanalysis) n+ n+

VG

p depletion region

Large L:S D

Small L:DS

Depletion charge supported by S/D

Depletion charge supported by S/D

The smaller L is, the greater the percentage of depletion charge balanced by the S/D pn junctions:

rj

Lecture 22, Slide 10EE130/230A Fall 2013

Page 11: EE130/230A Discussion 12

VT Roll-Off: First-Order Model

121)(

j

Tj

oxe

TATchannellongTT r

WLr

CWqNVVV

Minimize VT by• reducing Toxe

• reducing rj

• increasing NA

(trade-offs: degraded eff, m)Þ MOSFET vertical dimensions should be scaled along with horizontal dimensions!

Lecture 22, Slide 11EE130/230A Fall 2013

R. F. Pierret, Semiconductor Device Fundamentals, Fig. 19.4

Page 12: EE130/230A Discussion 12

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Questions?

1. CMOS fabrication process / technology?2. The MOSFET design project?

Good luck to Quiz#5!