VLSI Design - Unit I

8/7/2019 VLSI Design - Unit I

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Basic MOS Transistor

Unit I


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Basic MOS Transistor

Its just a repetition of what we had in the third

semester (Electron Devices) But

With simple additions needed exclusively (!?) for

VLSI design

S. B. Sivasubramaniyan MSEC, Chennai


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A Transistor,

Well it is a three terminal device

The third extra terminal is far more important than the 2 terminal

device (diode)

The 3rd terminal is helpful in creating a controlled source

The concept is the usage of voltage between the two terminals to

control the current flowing in the third terminal

This is the basis for amplifierdesign (?!)

The second most important application is that the current in the

third terminal can be made to vary from zero to maximum value

This makes the transistor to act as a switch



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An amplifier& a switch

We will discuss this later



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MOSFET

There are two types of MOSFETs

Enhancement type MOSFET and Depletion typeMOSFET



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MOSFET Enhancement type

There are 2 types of Enhancement type MOSFETs

n-type and p-type Accordingly it is referred as nMOS and pMOS



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Enhancement type - Structure

NMOS transistor is built on a p type substrate



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Due to this construction, MOSFET is also referred

to as IGFET Insulated Gate Field EffectTransistor

The oxide layer between the Gate and the

substrate makes the input current extremely

small of the order of femto Amperes



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Looking at the structure again, we see that the

body (substrate) forms pn junction with sourceas well as with the drain



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The pn junctions formed can be eliminated by

simply connecting source (S) with body (B)



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Enhancement type - Symbol

We had it already



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Enhancement type Operation

With no Gate voltage, VGS = 0 & Drain and Source

grounded



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The pn junctions formed prevent the current

conduction from source to drain The resistance offered is too high in the order of

tera ohm



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With Gate voltage applied (VGS = some value)



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The threshold value

Increasing VGS above a certain value

To effect the formation of channel betweensource and drain, VGS has to increased beyond a

certain amount called threshold value (Vt)

Note: Formation of channel here refers to

sufficient number of electrons getting

accumulated and remember no actual current

flow exists



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Applying a small voltage between Drain and

Source (VDS)



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Current flow

The Voltage between drain and the source (VDS)

drifts the electrons from the drain to the source For this to happen, channel has to be created

which is taken care by the Gate to Source voltage

(VGS) being greater than the threshold voltage

(Vt), often referred to as over-drive voltage oreffective voltage



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Current flow

Suppose that the voltage between drain and

source (VDS) is low in the order of, say, few tensof mV

This causes drain current (ID) to flow from Drain

to Source (opposite to the actual flow of

electrons)



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Current flow

Drain current ID is dependent on VDS, as is must

be the case An important thing to note here is that Gate to

Source Voltage (VGS) plays an important role in

deciding the magnitude of drain current flowing



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Current flow

The current ID is dependent on VDS, as is must

be the case An important thing to note here is that Gate to

Source Voltage (VGS) plays an important role in

deciding the magnitude of drain current flowing



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Current flow



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Current flow

We can conclude that, for current to flow

through the channel, First, the channel has to be induced,

accomplished by over drive voltage

Thus increasing the voltage above the threshold

voltage enhances the voltage, hence

enhancement type mosfet

Also, note that drain current is proportional to

over drive voltage and gate to source voltageS. B. Sivasubramaniyan MSEC, Chennai


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Problem

Calculate from the above figure used for

illustration, the slope offered by thecharacteristics for four values shown

Comment on the characteristics



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Enhancement type With increased VDS

Carefully note down what happens to MOSFETs

conduction Before that, let us do some work with respect to

voltages between the terminals of the device



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Enhancement type With increased VDS



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iD vDS characteristics



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iD vDS characteristics - Derivation

The derivation can be better understood and

carried out with the understanding of theconstruction, we had earlier

We know, vGS is applied between Gate and

Source, and this vGS should be greater Vt

Also vDS is applied between Drain and Source



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Considering operation in the triode region, we

have, vGS > Vt and vDS < vGS Vt

These are the voltage levels at the terminals

Before going into the actual derivation, let us

refresh our basics slightly and then concentrate

on the drain current expression



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Some basics before movingon

The capacitive effect of the MOSFET is due to the

gate oxide whose thickness is given by tox If the capacitance per unit gate area is given as

Cox, then

Where, is the permittivity of the silicon oxide


oxox

ox

Ct

I!

oxI

03.9oxI I 12 113.9 8.854 10 3.45 10 /F m ! !


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The oxide thickness is determined by the

process technology. Typical values (as far as, the scope of the

paper is concerned) are

32

10 , 3.45 10 /ox oxt nm C F m! !

23.45 / fF mQ!


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Charge = (Capacitance) (Voltage)


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Consider the gate oxide as shown


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Problem


For a process technology defined as

Determine, (i)

(ii) For a MOSFET with (W/L) = 8Qm/0.8Qm,determine vGS and vDSmin needed to operate the

transistor in the saturation region with a dc current ofID = 100 QA

(iii) For the same device, find vGS required to make it

work as a 1000 ; resistor for very small vDS

2

0.4 , 8 , 450 / . , 0.7ox n t L m t nm cm V s V V Q Q! ! ! !,o x nk d


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Solution


(i)ox

C11

9

(3.45)10

(8)1

0

! 3 24.32 10 /F m!

3 12 12 24.32 10 10 / 10F m ! 15 24.32 10 /F mQ!

n

k dn ox

CQ! 2 2450 / . 4.32 /cm V s fF mQ!

24.32 /fF mQ!

8 2 2450 10 / . 4.32 /m V s fF mQ Q!

8 2 15 2450 10 / . 4.32 10 /m V s F mQ Q!

ox

oxt

I!

6194 10 / . F V s! 194 / . F V sQ! 2194 /A VQ!


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Solution


(ii) For transistor operating in the saturation region, iD

is given by

Given,

To find,

For a transistor with,

Substituting, we get

21

2D n GS t

Wi k v V

Ld!

100D

i AQ!

min,GS DSv v 8

0.8

W m

L m

QQ

!

22 21 8

100 194 / 0.72

0.8

GS

mA A V v V

m

QQ Q

Q

!


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Solution


Solving, we get

We know,

(iii) MOSFET with very small, acts in the triode

region The expression for current is given by,

0.7 0.32GSv ! 1.02GSv V !

,minDS GS t v v V! ,min 1.02 0.7 0.32DSv V ! !

DSv

21

2D n GS t DS DS

Wi k v V v v

Ld!


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Solution


being small, the term can be omitted

The expression then becomes

Resistance (drain to source) is given as 1000 W, for

which the operating is to be determined

From the approximated expression for current, thedrain to source resistance is given as

DSv

D n GS t DSW

i k v V vL

d!

21

2DSv

GSv

DSDS

D

vr

i!

1

n GS t

Wk v V

L

!

d 2

11000

8194 / 0.7

0.8GS

mA V v V

m

QQ

Q

!

1.22GS

v V !


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Problem


For a 0.8 Qm process technology for whichand , find , and the over-

drive voltage , required to operate a

transistor having in saturation with

What is the minimum value of needed?

15oxt nm!2550 / .n cm V sQ ! oxC

nk d

OV GS tV V V! / 20W L ! 20 .

DI mA!

DSv


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Solution


oxC

11

9

(3.45)10

(15)10

! 3 22.3 10 /F m!

nk d n oxQ! 2 2550 / . 2.3 /cm V s fF mQ!

22.3 /fF mQ!

8 2 2550 10 / . 2.3 /m V s fF mQ Q!

8 2 15 2

550 10 / . 2.3 10 /m V s F mQ Q

!

ox

oxt

I!

6126.5 10 / . F V s! 126.5 / .F V sQ! 2

126.5 /A VQ!


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Solution


(ii) For transistor operating in the saturation region, iD

is given by

Given,

To find,

For a transistor with,

Substituting, we get

21

2D n GS t

Wi k v V

Ld!

0.2D

i mA!

OV GS tV v V! 20

W

L!

23 2 21

0.2 10 126.5 / 202

GS tA V v V VQ

!


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Solution


Solving, we get

We know, 0.3975GS tv V ! 0.3975OVv V !

,minDS GS t v v V! ,min 0.3975DSv V !


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Problem


Use the expression for operation in the triode region

to show that an n-channel MOSFET operated in

saturation with an overdrive voltage and

having a small across it behaves approximately

as a linear resistance ,

Calculate the value of obtained for a device

having & when operated

with an overdrive voltage of 0.5V

OV GS tV V V!

DSv

DSr

DSr2

100 /nk A VQd! / 10W L !

1 /DS n OV

Wr k V

L

d!


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Solution


MOSFET with very small, acts in the triode

region

The expression for current is given by,

, being small, the term can be omitted The expression then becomes

DSv

21

2D n GS t DS DS

Wi k v V v v

Ld!

DSv

21

2 DSv

D n GS t DSW

i k v V vL

d!


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Solution


From the approximated expression for current, the

drain to source resistance is given as

DSDS

D

vr

i!

1

n GS t

Wk v V

L

!d

2

1

100 / 10DS

GS t

r A V v V V Q

!

2DSr k ! ;

21

100 / 10DSOV

r A V V V Q ! 21

100 / 10 0.5DSr A V V Q !


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So far, everything look perfect,

But practically, something else happens

The MOSFET which we discussed, has promisedto be perfect with infinite input impedance and

also infinite output impedance

Input impedance is infinite (gate currentbeing

approximatelyzero), even though appearsimpractical can be considered to be infinite

But the case of output impedance is shaky



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Finite output resistance

But in practice, something else happens

The MOSFET which we discussed, has promisedto be perfect with infinite input impedance and

also infinite output impedance

Input impedance is infinite (gate currentbeing

approximatelyzero)

But the case of output impedance is shaky



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Output impedance

Increasing vDS causes the channel to be pinched

off This causes the current to attain the saturation

value

That is, iD is independent of the applied vDS

In practice, any increase in vDS about this

saturation value decreases the channel in such a

way that the pinch off pointmove towards

sourceS. B. Sivasubramaniyan MSEC, Chennai


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To make it clear



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Characteristics



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Accounting dependence of iD on vDS

We have to replace L by (L in the current

equation The current equation is given by

Now, a simple mathematical treatment,


21

2D n GS t

Wi k v V

Ld!

21

2D n GS t

Wi k v V

L Ld !

(

21 1

21

D n GS t

Wi k v V

LL

L

d ! (

21

12

D n GS t

W Li k v V

L L

( d !

, 1L

takingL

(=


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Now, assuming,

Substituting, in the current equation, we get,


DSL vE( DSL vPd ( !

21

12

DSD n GS t

vWi k v V

L L

Pd d !

21

12

D n DS GS t

Wi k v v V

LPd ! ,taking

L

PP

d!


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P is the process technology parameter with the

dimensions of (?!) In terms ofP, the drain current equation can be

given as


211

2D n GS t DS

Wi k v V v

LPd!

1

V


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Coming back to characteristics again,



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Finite output resistance

From the previous discussion, we infer that the

MOSFET has finite output resistance given by

Simplifying , we get

S B Sivasubramaniyan MSEC Chennai

1

D

o

DS

ir

v

x

! x

1

2

2

no GS t

k Wr v V

LP

d !

1

o

D

rIP

! Ao

D

Vr

I !

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VLSI Design - Unit I

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