EE 330 Lecture 13 Devices in Semiconductor Processesclass.ece.iastate.edu/ee330/lectures/EE 330 Lect...

EE 330Lecture 13

Devices in Semiconductor Processes

• Diodes

• Capacitors

• MOSFETs

Flash Memory Cell

• Single FET with dual gate

• Electrically isolated floating gate is the storage element

• Electrons added to or removed from the floating gate shift the Vt of the cell to store a 1 or a 0

• Two types: NAND and NOR

• NAND – Better array efficiency, lower cost per die for mass-storage

• NOR – Faster read/write speeds for code storage and execution

Review from last lecture

Diodes (pn junctions)

Depletion region created that is ionized but void of carriers


pn Junctions

ID

VD

Anode

Cathode

Anode

CathodeCircuit Symbol


Rectifier Application: Simple Diode Model:

VD

ID

1K

VOUT

D1

VIN=VMsinωt

VIN

VM

t

VM

t

VIN

VOUT


pn Junctions

VI

0V0

0VAeJITnV

V

S

I

V

Diode Equation:(good enough for most applications)

JS= Sat Current Density (in the 1aA/u2 to 1fA/u2 range)

A= Junction Cross Section Area

VT=kT/q (k/q=1.381x10-23V•C/°K/1.6x10-19C=8.62x10-5V/°K)

n is approximately 1

Anode

Cathode

Note: IS=JsA


pn Junctions

0V0

0VAeJITnV

V

S I

V

Diode Equation:Anode

CathodeJS is strongly temperature dependent

G0 D

t t

-V V

V Vm

SXI(T) J T e Ae

With n=1, for V>0,

Typical values for key parameters: JSX=0.5A/μ2, VG0=1.17V, m=2.3


pn Junctions

VI

0V0

0VAeJITnV

V

S

I

V

Diode Equation:(good enough for most applications)

Anode

Cathode

IS=JsA

VD

ID

Simple Diode Model:

Often termed the “conducting” or “ON” state

Often termed the “nonconducting” or “OFF” state

VIN R

ID

VD

VOUT

O U TV ?

Consider again the basic rectifier circuit

• Previously considered sinusoidal excitation• Previously gave “qualitative” analysis• Rigorous analysis method is essential

Analysis of Nonlinear Circuits (Circuits with one or more nonlinear devices)

What analysis tools or methods can be used?

KCL ?

KVL?

Superposition?

Voltage Divider ?

Current Divider?

Thevenin and Norton Equivalent Circuits?

Nodal Analysis

Mesh Analysis

Two-Port Subcircuits

RIVV DDIN

RIV DOUT

1eII t

d

V

V

SD

VIN R

ID

VD

VOUT

1eRIV t

OU TIN

V

VV

SOUT

Consider again the basic rectifier circuit

Even the simplest diode circuit does not have a closed-form solution

when diode equation is used to model the diode !!

Due to the nonlinear nature of the diode equation

Simplifications are essential if analytical results are to be obtained

Lets study the diode equation a little further

Diode Characteristics

0

2000

4000

6000

8000

10000

0 0.2 0.4 0.6 0.8 1

Vd (volts)

Id (

am

ps)

Power Dissipation Becomes Destructive if Vd > 0.85V (actually less)

1eII t

d

V

V

Sd

Linear-Linear Axis


For two decades of current change, Vd is close to 0.6V


1E-12

1E-10

1E-08

1E-06

0.0001

0.01

1

100

10000

0 0.2 0.4 0.6 0.8 1

Vd (volts)

Id (

am

ps)

This is the most useful current range for many applications

1eII t

d

V

V

Sd

Linear-Log Axis


For two decades of current change, Vd is close to 0.6V

This is the most useful current range when conducting for many applications


0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)

1eII t

d

V

V

Sd



0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)

Widely Used Piecewise Linear Model

1eII t

d

V

V

Sd 0I0 .6 VV

0 .6 VV0I

dd

dd



0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)

Better model in “ON” state though often not needed

1eII t

d

V

V

Sd

Includes Diode “ON” resistance


0IRI0 .6 VV

0 .6 VV0I

dDdd

dd

1eII t

d

V

V

Sd

Piecewise Linear Model with Diode Resistance

Equivalent Circuit

Vd

Id

A C

(RD is rather small: often in the 20Ώ to 100Ώ range):

Vd

Id0.6V

A

A

C

C

Off State

On State

RD

if

if

The Ideal Diode

I

ID

VD

0

0

D D

D D

I 0 if V

V if I 0

IDVD I

0

0

D D

D D

I 0 if V

V if I 0

“OFF”

“ON”

The Ideal Diode

ID

VD

I

ID VD

“ON” “OFF”

ID >0 VD ≤0Valid for

Diode Models

ID

VD


0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)


0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)


0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)

Which model should be used?

The simplest model that will give acceptable results in the analysis of a circuit

0

0

D D

D D

I 0 if V

V if I 0

Diode Models

ID

VD

Diode Equation

1eII t

d

V

V

SD


0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)

Piecewise Linear

Models

0I0 .6 VV

0 .6 VV0I

dd

dd


0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)

0IRI0 .6 VV

0 .6 VV0I

dddd

dd


0

0.002

0.004

0.006

0.008

0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Vd (volts)

Id (

am

ps)

if

if

if

if

0

0

D D

D D

I 0 if V

V if I 0

Diode Models

Diode Equation

1eII t

d

V

V

SD

Piecewise Linear

Models

0I0 .6 VV

0 .6 VV0I

dd

dd

0IRI0 .6 VV

0 .6 VV0I

dddd

dd

When are the piecewise-linear models adequate?

When it doesn’t make much difference whether Vd=0.6V or Vd=0.7V is used

When is the ideal PWL model adequate?

When it doesn’t make much difference whether Vd=0V or Vd=0.7V is used

Example: Determine IOUT for the following circuit

10K

12V

IOUT

D1

Solution:

1. Assume PWL model with VD=0.6V, RD=0

2. Guess state of diode (ON)

3. Analyze circuit with model

4. Validate state of guess in step 2 (verify the “if” condition in model)

5. Assume PWL with VD=0.7V



8. Validate state of guess in step 6 (verify the “if” condition in model)

9. Show difference between results using these two models is small

10. If difference is not small, must use a different model

Strategy:

Solution:

10K

12V

IOUT

0.6V




1 14OUT

12V-0.6VI = .

10KmA

4. Validate state of guess in step 2

To validate state, must show ID>0

D OUTI =I =1.14mA>0

Solution:

10K

12V

IOUT

0.7V




1 13OUT

12V-0.7VI = .

10KmA



D OUTI =I =1.13mA>0

Solution:


OUT OUTI =1.14mA and I =1.13 mA are close

Thus, can conclude OUTI 1.14mA

Example: Determine IOUT for the following circuit

10K

0.8V

IOUT

D1

Solution:





5. Assume PWL with VD=0.7V






Strategy:

Solution:

10K

0.8V

IOUT

0.6V




20OUT

0.8 - 0.6VI =

10KA



D OUTI =I =20 A>0

Solution:

10K

0.8V

IOUT

0.7V




10OUT

0.8V-0.7VI =

10KA



D OUTI =I =10 A>0

Solution:


OUT OUTI =10 A and I =20 A are not close


Thus must use diode equation to model the device

10K

0.8V

IOUT

0.6V

VDD

OUT

0.8-VI =

10K

D

t

V

V

OUT SI =I e

Solve simultaneously, assume Vt=25mV, IS=1fA

Solving these two equations by iteration, obtain VD= 0.6148V and IOUT=18.60μA

Use of Piecewise Models for Nonlinear Devices when

Analyzing Electronic Circuits

Process:

1. Guess state of the device

2. Analyze circuit

3. Verify State

4. Repeat steps 1 to 3 if verification fails5. Verify model (if necessary)

Observations:

o Analysis generally simplified dramatically (particularly if piecewise model is linear)

o Approach applicable to wide variety of nonlinear deviceso Closed-form solutions give insight into performance of circuito Usually much faster than solving the nonlinear circuit directlyo Wrong guesses in the state of the device do not compromise solution

(verification will fail)

o Helps to guess right the first timeo Model is often not necessary with most nonlinear devices

Types of Diodes

Vd

Id Id

Vd

Id

VdVd

Id

Vd

Id

Vd

Id

Vd

Id

Vd

Id

pn junction diodes

Metal-semiconductor junction diodes

Signal or

Rectifier

Pin or

Photo

Light Emitting

LED

Laser Diode

Zener Varactor or

Varicap

Schottky Barrier

Vd

Id

Basic Devices and Device Models

• Resistor

• Diode

• Capacitor

• MOSFET

• BJT

Capacitors

• Types

– Parallel Plate

– Fringe

– Junction

Parallel Plate Capacitors

C

d

A1

A2

cond1

cond2

insulator

A = area of intersection of A1 & A2

d

AC

One (top) plate intentionally sized smaller to determine C

Parallel Plate Capacitors

ACC d

d

AεC

areaunit

CapC If d

d

εCd

where

Fringe Capacitors

d

C

d

AεC

A is the area where the two plates are parallel

Only a single layer is needed to make fringe capacitors

Fringe Capacitors

C

Capacitance

2

φV

φ

V1

ACC B

FBn

B

D

j o

for

ddepletion

region

C

Junction Capacitor

d

AC

Note: d is voltage dependent

-capacitance is voltage dependent

-usually parasitic caps

-varicaps or varactor diodes exploit

voltage dep. of C

d

p

n

VD

0.6VφB n ; 0.5

Capacitance

2

φV

φ

V1

ACC B

FBn

B

D

j o

for

Junction Capacitor

0.6VφB n ; 0.5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

-4 -3 -2 -1 0 1

j0

C

C A

VD

Voltage dependence is substantial

VD

Basic Devices and Device Models

• Resistor

• Diode

• Capacitor

• MOSFET

• BJT

n-Channel MOSFET

Poly

n-active

Gate oxide

p-sub

n-Channel MOSFET

LEFF

L

W

Source

DrainGate

Bulk

n-Channel MOSFET

Poly

n-active

Gate oxide

p-subdepletion region (electrically induced)

n-Channel MOSFET Operation and Model

VBS

VGS

VDS

Apply small VGS

(VDS and VBS assumed to be small)ID=0

IG=0

IB=0

Depletion region electrically induced in channel

IDIG

IB

Termed “cutoff” region of operation


VBS

VGS

VDS

Increase VGS

(VDS and VBS assumed to be small)ID=0

IG=0

IB=0Depletion region in channel becomes larger

IDIG

IB


VBS

VGS

VDS

ID=0

IG=0

IB=0

IDIG

IB

Model in Cutoff Region


VBS

VGS

VDS

Increase VGS more

IDRCH=VDS

IG=0

IB=0

Inversion layer forms in channel

IDIG

IB

(VDS and VBS small)

Inversion layer will support current flow from D to S

Channel behaves as thin-film resistor

Critical value of

VGS that creates

inversion layer

termed threshold

voltage, VT)

End of Lecture 13

EE 330 Lecture 13 Devices in Semiconductor Processesclass.ece.iastate.edu/ee330/lectures/EE 330 Lect...

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