1

Chapter 5. Metal Oxide SiliconField-Effect Transistors (MOSFETs)

Transistor • Three terminal device

• Voltage between two terminals to control current flow in third terminal

• Versatile for many applications– Amplification

– Memory

– Logic

– voltage controlled current source

– switch

• Two popular types: – Bipolar Junction Transistor (BJT): used in power amplifier

– MOSFET: used in integrated circuits

Enhancement-type NMOS transistor•p-type material as substrate (i.e., body)•n-type material chemically bonded on body at source and drain → source and drain are electrically indistinguishable•Equivalent to having two diodes back to back → current cannot flow between source and drain•Typical dimensions:

o L = 0.1 to 3 μm,o W = 0.2 to 100 μmo tox = 2 to 50 nm

pn junction pn junction

• Four terminals shown: Source (S), Gate (G), Drain (D) and Body (B).

• Body is typically grounded (along with one of the other three terminals) and does not play any role.

• Gate is electrically insulated from the body by Silicon Oxide (SiO2)

• With no external voltages applied, normally there is no current between S and D.

• When certain voltage is applied at G, current flows from D to S. → The gate voltage controls the flow of current.

• With S and D grounded, apply positive voltage to G (vGS > 0).• Because G is electrically insulated to the body, in the channel, the gate voltage

attracts electrons from the body.• A thin layer of “induced n-type channel is formed between S and D.• Across the induced n-type channel, there is no pn junction between S and D.• The thickness of the induced n-type channel is proportional to vGS.• Now, between S and D there is continuous n-type material.• In the n-type region, there are excess electrons floating around (i.e., drift current

flowing in random directions).

Channel region

• With vGS > 0, now apply small vDS > 0.• Then, (diffusion) current starts flowing from D to S. There is no

current flowing into G, because of SiO2 insulator. • To form an induced n-type channel sufficient to support current

flow, vGS > Vt. Vt is called the threshold voltage.

• For small vDS, iD is a linear function of vDS.

• The slope is the inverse of the resistance between D and S.

• When vGS < Vt, the resistance is infinite

• Increase vDS with fixed vGS > Vt

• Voltage between G and S = vGS • Voltage between G and D = vGS - vDS

• n-channel is thickest at S, and thinnest at D.• As vDS increases, the resistance across the channel increases.

• When vGS - vDS = Vt, the channel depth at D is ≈ 0.• Channel is then, “pinched off.”• Increasing vDS beyond the point vDS = vGS - Vt has no effect on iD.• This region is called the saturation. vDSsat = vGS - Vt

• Device in saturation: region vDS ≥ vDSsat

• Device in triode region: vDS < vDSsat

The above curves exist for each fixed value of vGS.

The PMOS transistor works similarly, but with n-type body and p-type S and D.

• To establish a p-type channel between S and D for the PMOS transistor vGS < Vt where Vt < 0.

• The current flows from S to D when vDS < 0.• PMOS is not used by itself very often.

Complementary MOS (CMOS) Transistor• Combines NMOS and PMOS on single substrate• Most popular transistor for integrated circuit • Very dense structure, consumes low power.• Very powerful and versatile

Circuit Symbols for NMOS

Most popular one to use

1

small

is the internal resistance between D and S in triode mode.

: transconductance parameter, : mobility of electron in channel

DS

GS GS

DSD n GS t DS DS n GS t

vDv V

DS

n n OX n

vW Wi k v V v r k v V

L i L

r

k C

2Common value: 1.0 mA/V and 1 V n t

Wk V

L

Triode Region• vGS ≥ Vt: channel is induces between S and D.

• vDS ≤ vGS - Vt: channel is continuous (i.e., no pinch off).

Saturation Region• vGS ≥ Vt: channel is induces between S and D.

• vDS ≥ vGS - Vt: channel is pinched off.

• At the boundary of triode and saturation: vDS = vGS - Vt

21

2 D n GS t

Wi k v V

L

iD - vGS relationship in saturation

Large Signal Circuit Model NMOS in Saturation

Voltage Characteristics for NMOS Transistor

More Accurate Model• In saturation, slope in iD – vDS curve is not entirely flat.

• There is internal resistance, ro.

211

2: process parameter

D n GS t DS

Wi k v V v

L

Large Signal Model for NMOS with ro

1 11

2

constant2

: internal resistance between D and S in saturation mode

GS

nDo n GS t GS t

DS v

o

ki W Wr k v V v V

v L L

r

Circuit Symbols for PMOS

Most popular one to use

Nominal Current Directions and Voltage Polarity

Triode Region• vGS ≤ Vt: channel is induces between S and D where Vt < 0.

• vDS ≥ vGS - Vt: channel is continuous (i.e., no pinch off) where vDS < 0.

2

Or more accurately,

11

2Note: , , , < 0

D p GS t DS

D p GS t DS

GS t DS

Wi k v V v

L

Wi k v V v

Lv V v

Voltage Characteristics for PMOS Transistor

Skip All Sections for PMOS.

MOSFET Circuit with DC Inputs

2n

Assume 0.

Want to design the circuit such that

0.4 mA and 0.5 V.

Find and .

Given: 0.7 V, 100 A/V

1 m, 32 m,

o

D D

S D

t n OX

r

I V

R R

V k C

L W

2

26 6

2

Note that 0. Want to design 0.5. Thus the

transistor is in saturation mode.

1

21 32

400 10 100 102 1

0.25

0.5 since in saturation mode.

1.2 V

G D

D n OX GS t

GS t

GS t

GS t GS t

GS

V V

WI C V V

L

V V

V V

V V V V

V

1.2 ( 2.5)= = 3.25 k

0.4

2.5 0.5= = 5 k

0.4

S SSS

D

DD DD

D

V VR

I

V VR

I

2n

Example:

Assume 0.

Want to design the circuit such that 80 A.

Find and .

Given: 0.6 V, 200 A/V

0.8 m, 4 m,

o

D

D

t n OX

r

I

R V

V k C

L W

2

26 6

2

Note that . Transistor in saturation mode.

1

21 4

80 10 200 102 0.8

0.16

0.4 since in saturation mode.

1.0 V

1.0 V

3 1 = =

0.08

D G

D n OX GS t

GS t

GS t

GS t GS t

GS

G D

DD D

D

V V

WI C V V

L

V V

V V

V V V V

V

V V

V VR

I25 k

2

Example:

Want to design the circuit such that 0.1 V.

Given: 1 V, 1 mA/V

D

t n

V

WV k

L

-3

Note that 0.1 5 1 4.

Transistor in triode mode.

= 0.4 10 for small

5 0.1 = = 1.225 k

0.4

0.1 = 0.25 k

0.4

DS GS t

D n GS t DS DS

DD DD

D

DSDS

D

V V V

WI k V V V V

LV V

RI

Vr

I

MOSFET as Amplifier• Utilize saturation mode.• iD as function of vGS. • Transconductance amplifier

Often, we want amplifiers to be linear. But iD is a quadratic function of vGS. Use DC biasing technique. Shift small signal around a point that mimics linearity.

Common source amplifier

Q1 is too close to cut-off and Q2 is too close to triode boundary. We want the quiescent point to be in the middle of saturation region. For example, Q3 may be a reasonable point.

Boundary with triode region

Q3•

Quiescent point is determined by VGS and RD.

For linear amplifier, MOS's are in saturation mode.

Desire to operate at a certain - characteristics

Determine point of interest

- Set DC current

Biasing MOS Amplifier Circuit

i v

Q

I

- Set DC voltage

- Make AC signal small around point

D

DSV

Q

2

,

Biasing Scenario 1: Fix

1

2, , , and vary in temperature and by manufacturing procedure.

becomes unpredictable by simply fixing only.

GS

D n OX GS t

n OX t

D GS

V

WI C V V

LC W L V

I V

Biasing Scenario 2: Fix and add resistor in S

if is large enough.

is mostly determined by and .

is called the degeneration resistance.

G

G GS S D G GS S

GG S D D D G S

S

S

V

V V R I V V R

VV R I I I V R

R

R

Practical Methods of Biasing VG

2

Example:

Goal: 0.5 mA

Given: 1 V

1 mA/V

0

15 V

D

t

n

DD

I

V

Wk

L

V

2

2

2

1Rule of Thumb:

310 V and 5 V

15 10 = = 10 k

0.5

5 = = 10 k

0.5

1

21

0.5 12

1

2 V

2 5 7 V

D SR R DS DD

D S

DD DD

D

SS

D

D n GS t

GS t

GS t

GS

G GS S

V V V V

V V

V VR

I

VR

I

WI k V V

L

V V

V V

V

V V V

1 2

To get 7 V, select

= 8 M and = 7 MG

G G

V

R R

Biasing Scenario 3: D-G resistor

(since 0.)

GS DS DD D D G

DD GS D D

V V V R I I

V V R I

Biasing Scenario 4: Current Source

Determine to establish point. I Q

D2

1

211

1

1

Implementation of current source using second MOS (Goal: Set )

1

2

For each , solve the two equations above to get and .

DD SS GSD

D n GS t

D GS

I

V V VI

RW

I k V VL

I R V

222

2

2 22

1 1 1

2 1

1

2

Since is the same for both transistors,

/

/

If two transistors are identical, .

D n GS t

GS

D

D

D D

WI k V V

L

V

W LI

I W L

I I

MOS is in saturation mode.

MOS has been properly biased.

- characteristics is linearly approximated around

certain point.

Now apply small AC signal around

Applying AC at Point

i v

Q

Q

point.

The resulting circuit is a linear system.

DC and AC inputs can be applied separately by

superposition.

Q

2

2

2 2

2

For DC:

1

2

For total signal:

1

21 1

2 2

If is kept small,

1

2

Transco

D n GS t

D DD D D

GS GS gs

D n GS gs t

n GS t n GS t gs n gs

gs

D n GS t n GS t gs

D D d

WI k V V

LV V I R

v V v

Wi k V v V

LW W W

k V V k V V v k vL L L

v

W Wi k V V k V V v

L Li I i

nductance for small signal:

dm n GS t

gs

i Wg k V V

v L

D DD D D

DD D D d

D D d

d D d m gs D

dv m D

gs

v V i R

V R I i

V R i

v R i g v R

vA g R

v

Small Signal (or AC) Equivalent Models

λ = 0 λ ≠ 0

gm and ro are determined for each Q point.For analyzing small signal circuits, DC sources must be eliminated.

• Voltage source: short circuit• Current source: open circuit

Three Different Ways to Get

1.

2. 2

23.

To achieve a desired value, use , , and .

m

m n GS t

m D n

Dm

GS t

m GS D

g

Wg k V V

L

Wg I k

L

Ig

V V

Wg V I

L

2Example: Given 1.5 V, 0.25 mA/V , and 50 V.

Note: 1. The capacitor at the input blocks any DC through it.

2. point is set entirely by , , and 15V power supply

t n A

D G

WV k V

L

Q R R

(i.e., ).

1 3. = .

4. In small signal analysis, replace capacitors by short circuits.

DD

A

V

V

Given circuit → Small signal model

2 2 21 1 10.25 1.5 0.25 1.5

2 2 2Note that 0, and

15 10

Solve the above two equation together for and .

=1.06 mA and 4.4 V

0.25 4.4 1.5

D n GS t GS D

G GS D

D DD D D D

D D

D D

m n GS t

WI k V V V V

LI V V

V V I R I

I V

I V

Wg k V V

L

0.725 mA/V

50= 47 k

1.06

Ao

D

Vr

I

D L o

|| ||

R || R || r

0.725 10 || 10 || 47

3.3 V/V

1

1 3.3

=4.3 0.4310

2.33 M

o m gs D L o

ov m

i

i o i oi

G G i

i

G

ii

iin

i

v g v R R r

vA g

v

v v v vi

R R v

v

R

vv

vR

i

i

G

v

R

Source transform

GR

Make large such that .iG m gs

G

vR g v

R

Alternative Small Signal Models (T Models)

Separate DC and AC analysis

Discrete resistors as load

In complex ICs, MOS's are used together to

act as loads, current sources, and voltage references

as

Single Stage MOS Amplifier Circuits

well as amplifiers.

There are several popular configurations for single

stage MOS amplifier.

Basic Structure

sig sig sig sig

Source: and ( and may be from a separate source

or represent the output of precedeing amplifier stage.)

Output: ( may be an ac

Chatersistics of MOS Amplifiers

L L

v R v R

R R

sig

sig

tual resistor or represent

an input resistance of the succeeding amplifier stage.)

Circuit (or and ) independent parameters: , , , ,

Circuit (or and ) dependent parameters

L i o vo is m

L

R R R R A A G

R R

sigout 0

: , , , , ,

, L

in out v i vo v

i in oR R

R R A A G G

R R R R

Original Circuit

Equivalent Circuits

1

Very widely used.

: Large bypass capacitor to

eliminate output

associated with source .

: Large coupling capacitor

to bl

Common Source (CS) Configuration

S

C

C

I

C

sig

2

ock DC value

associated with .

: Large coupling capacitor to

block DC value associated

with .

C

D

O o d

v

C

v

v v v

sig sigsig

sig sig

out

0 (if large)

|| ||

|| || ||

|| ||

||

Gg in G gs i G

G

o m gs o D L

v m o D L vo m o D

in inv v m o D L

in in

o D

Ri R R v v v v R

R R

v g v r R R

A g r R R A g r R

R RG A g r R R

R R R R

R r R

CS Variation: Source Resistor SR

sigsig

sig

1

1 11 1

|| ||

1 1 1

||

1

Gi

G

m i i m igs i d

m S m SS S

m m

m D L m D L m Do i v vo

m S m S m S

m D LGv

G m S

Rv v

R R

g v v g vv v i i

g R g RR R

g g

g R R g R R g Rv v A A

g R g R g R

g R RRG

R R g R

1 2

G: Ground S: Input D: Output

, : Large coupling capacitors

Common Gate (CG) Configuration

C CC C

sigsig

sig sig

i1

in

1

1 1 Note: is seen only through S, i.e., not though G or D.

1

|| ||

||

inm m

ini

in m

m i

d m i

o d d D L m D L i

v m D L vo m D

Rg g

vRv v

R R g R

vi g v

R

i i i g v

v v i R R g R R v

A g R R A g R

sig

||

1m D L

v out o Dm

g R RG R R R

g R

1 2

D: Ground G: Input S: Output

Also called the source follower configuration

, : Large coupling capacitors

Common Drain (CD) Configuration

C CC C

sig sig sigsig sig

sig

||

1 ||

||

1 1 ||

||

1 ||

1 ||

in Gin G i

in G

L oo i

L om

L o ov vo

L o om m

L oGv

GL o

m

out om

R RR R v v v v

R R R R

R rv v

R rg

R r rA A

R r rg g

R rRG

R R R rg

R rg

CMOS is widely used in digital

logic circuits.

Matched NMOS ( : Driver) and

PMOS ( : Load) as signle device.

Tens of millions of CMOS's

CMOS Digital Logic Inverter

N

P

Q

Q

can be

put in a single integrated chip (IC).

Logic 0: 0 V Logic 1:

I I DD

DP DN

v v V

I I

When (logic high input), on and off

Intersection point: in triode and off

is small. logic low output (i.e., inversion)

is very small. very low power consumption

1

I DD N P

N P

O DSN

DSNn

n

v V Q Q

Q Q

v v

i

rW

kL

(part of steep slope low resistence)

DD tnn

V V

When 0 (logic low input), off and on

Intersection point: in triode and off

logic high output (i.e., inversion)

is very small. very low power consumption

1 (pa

I N P

P N

O DD

DSPp

p DD tpp

v Q Q

Q Q

v V

i

rW

k V VL

rt of steep slope, low resistence)

Two output voltages: 0 and

Static power consumption 0

and can be made very small.

: Pull-up transistor : Pull-down transistor

Hig

Salient Characteristics of CMOS Inverter

DD

DSP DSN

P N

V

r r

Q Q

h current driving capability through and

High speed operation (short charge and discharge cycle)

0 Infinite input resistance

Can drive many other inverters.

P N

G

Q Q

I

- Characteristics with Matched and pnN P n p

n p

WWi v Q Q k k

L L

15 2

81

3 28

1 3 2

8

1 3 2

8

IH DD t

IL DD t

H OH IH

DD t

L IL OL

DD t

V V V

V V V

NM V V

V V

NM V V

V V

• Next stage amplifier may present certain capacitance C.• As a result of C, vO does not change instantly.

• Thus tPHL and tPLH are observed.

• For every on-off cycle, QN and QP each consumes power equaling 0.5CVDD

2 where C is internal capacitance of transistors.• Dynamic power dissipation PD = f CVDD

2 where f is the frequency of on-off cycle (i.e., clocking frequency)