Lecture 14

MOSFET I-V

CHARACTERISTICS

Outline

1. MOSFET: cross-section, layout, symbols

2. Qualitative operation

3. I-V characteristics

Key questions

• How can carrier inversion be exploited to make

a transistor?

• How does a MOSFET work?

• How does one construct a simple first-order

model for the current-voltage characteristics

of a MOSFET?

3

1. MOSFET: layout, cross-

section, platform

Shallow

trench

isolation

(STI)

4

Key elements:

• inversion layer under gate (depending on gate

voltage)

• heavily-doped regions reach underneath gate

=->inversion layer electrically connects source and

drain

• 4-terminal device: body voltage important

5

Circuit symbols

Two complementary devices:

• n-channel device (n-MOSFET) on p-substrate

– uses electron inversion layer

• p-channel device (p-MOSFET) on n-substrate

– uses hole inversion layer

6

Qualitative Operation

•Drain Current (Id: proportional to inversion charge and the

velocity that the charge travels from source to drain

•Velocity :proportional to electric field from drain to source

•Gate-Source Voltage (VGS controls amount of inversion charge

that carries the current

•Drain-Source Voltage (VDS: controls the electric field that

drifts the inversion charge from the source to drain

7

Want to understand the relationship between the drain

current in the MOSFET as a function of gate-to-source

voltage and drain-to-source voltage.

Initially consider source tied up to body (substrate)

8

Three regimes of operation:

9

• MOSFET:

-VGS<VT, with VDS≥0

• Inversion Charge=0

• VDS drops across drain depletion region

• ID=0

Linear or Triode regime:

10

Electrons drift from source to drain →electrical current!

GS N D

DS y D

V Q I

V E I

DS GS TV V V

Saturation Region VDS >VGS -VT

ID is independent of VDs: ID=Idsat

Electric field in channel cannot increase with VDs

11

GD,GS T T DS GS TV V V V V V V

Geometry of problem:

3. I-V Characteristics (Assume VSB=0)

12

General expression of channel current

Current can only flow in the y-direction, Total channel flux:

Drain current is equal to minus channel current:

Rewrite in terms of voltage at channel location y, V (y):

• If electric field is not too high (velocity saturation doesn’t occur):

• For QN(y), use charge-control relation at location y:

13

y N yI W Q y v y

D N yI W Q y v y

y n y n

dVv y E y

dy

N ox GS TQ y C V V y V

GS Tfor V V y V

All together the drain current is given by:

Solve by separating variables:

Integrate along the channel in the linear regime subject the

boundary conditions :

Then:

14

D n ox GS T

dV yI W C V V y V

dy

D y n ox GS TI d W C V V y V dV

0 0

DSVL

D n ox GS TI dy W C V V y V dV

Resulting in:

For small VDS:

15

0

02

DSV

L

D D n ox GS T

VI y I L W C V V V

D n ox GS T DS

WI C V V V

L

DS GS Tfor V V V

2

DSD n ox GS T DS

VWI C V V V

L

Key dependencies:

VDS↑→ ID↑ (higher lateral electric field)

VGS↑→ ID↑ (higher electron concentration)

L ↑→ID ↓ (lower lateral electric field)

W ↑→ID ↑ (wider conduction channel)

16

This is the linear or triode region:

It is linear if VDS <<VGS - VT

1. Equation only valid if VGS – V(y) ≥ VT at every y. Worst

point is y=L, where V(y) = VDS, hence, equation is valid if

Two important observations

17

DS GS TV V V

2. As VDS approaches VGS – VT, the rate of increase of

ID decreases.

As y increases down the channel, V(y) ↑, |QN(y)| ↓, and Ey(y)

↑ (fewer carriers moving faster)

inversion layer thins down from source to drain

Local ”channel overdrive” reduced closer to drain.

ID grows more slowly.

To understand why ID bends over, must understand first :

channel debiasing!

18

19

N ox GS TQ y C V V y V

Impact of VDS:

As VDS ↑, channel debiasing more prominent

=> ID rises more slowly with VDS20

Key conclusions • The MOSFET is a field-effect transistor:

– the amount of charge in the inversion layer is controlled by the field-effect action of the gate

– the charge in the inversion layer is mobile ⇒ conduction possible between source and drain

• In the linear regime:

– VGS ↑⇒ ID ↑: more electrons in the channel

– VDS ↑⇒ ID ↑: stronger field pulling electrons out of the source

• Channel debiasing: inversion layer ”thins down” from source to drain ⇒ current saturation as VDS approaches:

21DSsat GS TV V V

Drain current saturation

As VDS approaches

increase in Ey compensated by decrease in |QN|

⇒ ID saturates when |QN| equals 0 at drain end.

Value of drain saturation current:

22

DSsat GS TV V V

Dsat Dlin DS DSsat GS TI I V V V V

2DS GS T

DSDsat n ox GS T DS

V V V

VWI C V V V

L

21

2Dsat n ox GS T

WI C V V

L

Transfer characteristics in

saturationOutput Characteristics

23

What happens when VDS = VGS−VT ?

Charge control relation at drain end of channel:

No inversion layer at end of channel??!! ⇒ Pinchoff

24

0n ox GS DS TQ L C V V V

c DSsat GS TV L V V V

Key dependencies of IDsat

Drain current at pinchoff:

∝lateral electric field ∝VDSsat = VGS −VT

∝electron concentration ∝VGS−VT

25

2

Dsat GS TI V V

1Dsat

y

IL

L E

What happens when VDS > VGS−VT?

Depletion region separating pinchoff point and drain

widens (just like in reverse biased pn junction)

26

27

To first order, ID does not increase past pinchoff:

To second order, electrical channel length affected

(“channel length modulation”):

VDS ↑⇒Lchannel↓⇒ID ↑

Experimental finding:

Hence:

2

2D Dsat n ox GS T

WI I C V V

L

1 11D

LI

L L L L

DS DSsatL V V

DS Dsat

LV V

L

28

Improved model in saturation:

Also, experimental finding:

2

12

Dsat n ox GS T DS DSsat

WI C V V V V

L

1

L

2. Backgate characteristics There is a fourth terminal in a MOSFET: the body.

What does the body do?

29

30

Body contact allows application of bias to body with respect to

inversion layer, VBS .

Only interested in VBS<0 (pn diode in reverse bias).Interested in

effect on inversion layer⇒ examine for VGS >VT (keep VGS constant).

Application of VBS< 0 increases potential buildup across

semiconductor:

Depletion region must widen to produce required extra field:

2 2p p BSV

31

32

Consequences of application of VBS< 0:

Application of VBS < 0 with constant VGS reduces

electron concentration in inversion layer ⇒VT ↑

max

2 2

since constant, unchanged

unchanged

unchanged

unchanged, but

inversion layer charge is reduced!

p p BS

B d

GS ox

ox

S G

S n B B n

V

Q x

V V

E

Q Q

Q Q Q Q Q

33

How does VT change with VBS?

In VT formula change −2φp to −2φp −VBS:

In MOSFETs, interested in VT between gate and source:

Then:

And:

In the context of the MOSFET, VT is always defined in

terms of gate-to-source voltage.

1

2 2 2GB

T BS FB p BS s a p BS

ox

V V V V qN VC

GB GS

GB GS BS T T BSV V V V V V

GS GB

T T BSV V V

1

2 2 2GS

T BS FB p s a p BS T BS

ox

V V V qN V V VC

34

Then :

Define backgate effect parameter [units: V1/2]:

Define Zero-bias threshold voltage

12 s a

ox

qNC

0To T BSV V V

2 2T BS To p BS pV V V V

Key conclusions• MOSFET in saturation (VDS≥ VDSsat): pinchoff point at drain end

of channel

– electron concentration small, but

– electrons move very fast;

– pinchoff point does not represent a barrier to electron flow

• In saturation, ID saturates:

35

•But due to channel length modulation, IDsat increases slightly

with VDS

•Application of back bias shifts VT (backgate effect)

2

2Dsat n ox GS T

WI C V V

L

36

Example: MOSFET as a voltage controlled resistor

The circuit below shows an n-channel MOSFET that is used as

voltage-controlled resistor.

(a) Find the sheet resistance of the MOSFET over the range

VGS=1.5 V to VGS=4V using un= 215 cm2V-1S-1, Cox=2.3fF/um2

and Vtn =1 V.

Fig. an n-channel MOSFET used a voltage controlled resistor

37

For a particular value of VGS, ID is a linear function of VDS and

the circuit model for the MOSFET is a resistor. Now we relate

R to the sheet resistance

1

D n ox GS Tn DS DS

WI C V V V V

L R

2 1 1 7 2

1 1/

as a function of is

1 1 20

1 1215 2.3 10 /

n ox GS Tnn ox GS Tn

GS

GS GS

LR R L W

W C V V WC V V

L

R V

k VR

V V V Vcm V s F cm

Note that for a gate-source voltage VGS>VTn+0.1V=1.1V, the

MOSFET operates in the triode region. Since the drain-source

voltage is small

SOLUTION

38

The plot of sheet resistance as a

function of VGS with very small VDS

39

(b) For a particular application, we need to control the resistor between 200 Ω

and 1 kΩ for VGS=1.5V to 4 V. How wide should the MOSFET be if the channel

length L=1.5 um?

Solution:

We already solved for the sheet resistance in part (a), so we can find the

range of sheet resistances for VGS=1.5V to 4V

Solving for (W/L) to obtain Rmin=200 Ω and Rmax=1 kΩ yields

(W/L)=33.3 is adopted so the width of the MOSFET should be

min max

20 206666.7 and 40

4 1 1.5 1

k V k VR R k

V V V V

min max

6666.7 4000033.3 and 40

200 1000

W W

L L

1.5 33.3 50W m m

40

(c) Design the layout for

this MOS resistor so it

occupies a minimum area.

The length of the

source/drain diffusions is

Ldiff=6um with contact that

are 2um × 2um.

Solution:

Given the high ratio of

width to length (W/L=33.3),

it is desirable to fold the

MOSFET. Since the

diffusions are 6 um long,

the total length is

3 2

3 6um+2 1.5um=21um

T diffL L L

Fig (a) layout of folded n-channel MSOFET

(b) Equivalent schematic circuit

41

Example: Measuring the backgate effect parameter

The test circuit below can be used to find an experimental value for the

backgate parameter γn. Note that a negative voltage VBS is applied from

the bulk to the source of the MOSFET. The circuit varies VGS

continuously from 0 to 5 V, for VBS=0 VBS=-5V. The drain-source voltage

is VDS=100mV.

(a) From the drain current measurements plotted below, find the backgate

effect parameter. The device parameters are un= 215 cm2V-1S-1,

Cox=2.3fF/um2 , Vt0 =1 V and Na=1017 cm-3.

Fig. circuit to find the

backgate effect parameter

42

Solution:

Since the drain voltage is small, the MOSFET operates in its triode region

once VGS exceeds the threshold voltage. The drain current is linear with VGS

The threshold voltage is

From the graph, we have

D n ox GS Tn BS DS

WI C V V V V

L

2 2T BS To p BS pV V V V

1/22 1 0.84 5 0.84 0.67n nV V V V V V

Homework 15

Consider an n-channel MOSFET with the following

parameters: un/Cox=0.18 mA/V2, W/L =8, and VT

=0.4 V. Determine the drain current ID for

(a) VGS =0.8 V, VDS = 0.2 V;

(b) VGS = 0.8 V, VDS = 1.2 V;

(c) VGS = 0.8 V, VDS = 2.5 V;

(d) VGS = 1.2 V, VDS = 2.5 V.