Module 2.3: PN Junctions Prof. Ali M. Niknejad Prof. …rfic.eecs.berkeley.edu/105/pdf/module2-3_pn.pdfEE 105Fall 2016 Prof. A. M. Niknejad 9 PN Junctions: Overview l The most important

Department of EECS University of California, Berkeley

Prof. Ali M. NiknejadProf. Rikky Muller

Module 2.3: PN Junctions

EE 105 Fall 2016 Prof. A. M. Niknejad

2

What is a pn-junction?

n-typep-type

NDNA

)(0 xpaNp =0

d

i

Nnp2

0 =

a

i

Nnn2

0 =

Transition Region

dNn =00px- 0nx

AKA: diode


3

Module 2.3 Outlinel Part 1: Carrier concentration variation and

potential– Show that any time there’s a variation in carrier

concentration, then at thermal equilibrium there must be a variation in potential

l Part 2: Apply this to a pn-junction at thermal equilibrium– Extend result to a reverse biased junction

l Part 3: Look at a forward biased pn-junction

University of California, Berkeley


4

Carrier Concentration and Potential

l In thermal equilibrium, there are no external fields and we thus expect the electron and hole current densities to be zero:


dxdnqDEqnJ o

nnn +== 000 µ

dxdn

kTqEn

Ddxdn

oon

no 00

fµ÷øö

çèæ=÷÷

ø

öççè

æ-=

0

0

00 n

dnVndn

qkTd th

o =÷÷ø

öççè

æ=f


5

Carrier Concentration and Potential (2)

l We have an equation relating the potential to the carrier concentration

l If we integrate the above equation we have

l We define the potential reference to be intrinsic Si:


)()(ln)()(00

0000 xn

xnVxx th=-ff

inxnx == )(0)( 0000f

0

0

00 n

dnVndn

qkTd th

o =÷÷ø

öççè

æ=f


6

Carrier Concentration Versus Potential

l The carrier concentration is thus a function of potential

l Check that for zero potential, we have intrinsic carrier concentration (reference).

l If we do a similar calculation for holes, we arrive at a similar equation

l Note that the law of mass action is upheld


thVxienxn /)(

00)( f=

thVxienxp /)(

00)( f-=

2/)(/)(200

00)()( iVxVx

i neenxpxn thth == - ff


7

The Doping Changes Potentiall Due to the log nature of the potential, the potential changes

linearly for exponential increase in doping:

l Quick calculation aid: For a p-type concentration of 1016

cm-3, the potential is -360 mVl N-type materials have a positive potential with respect to

intrinsic Si


100

0

0

0

00 10

)(log10lnmV26)()(lnmV26

)()(ln)( xn

xnxn

xnxnVx

iith »==f

100

0 10)(logmV60)( xnx »f

100

0 10)(logmV60)( xpx -»f



Revesed Biased PN Junctions


9

PN Junctions: Overviewl The most important device is a junction

between a p-type region and an n-type regionl When the junction is first formed, due to the

concentration gradient, mobile charges transfer near junction

l Electrons leave n-type region and holes leave p-type region

l These mobile carriers become minority carriers in new region (can’t penetrate far due to recombination)

l Due to charge transfer, a voltage difference occurs between regions

l This creates a field at the junction that causes drift currents to oppose the diffusion current

l In thermal equilibrium, drift current and diffusion must balance

n-type

p-type

ND

NA

− − − − − −

+ + + + ++ + + + ++ + + + +

− − − − − −− − − − − −

−V+


10

PN Junction Currentsl Consider the PN junction in thermal equilibriuml Again, the currents have to be zero, so we have

dxdnqDEqnJ o

nnn +== 000 µ

dxdnqDEqn o

nn -=00µ

dxdn

nqkT

ndxdnD

En

on

0

000

1-=

-=

µ

dxdp

pqkT

ndxdpD

Ep

op

0

000

1-==

µ


11

PN Junction Fields

n-typep-type

NDNA

)(0 xpaNp =0

d

i

Nnp2

0 =diffJ

0E

a

i

Nnn2

0 =

Transition Region

diffJ

dNn =0

– – + +

0E

0px- 0nx


12

Total Charge in Transition Regionl To solve for the electric fields, we need to write

down the charge density in the transition region:

l In the p-side of the junction, there are very few electrons and only acceptors:

l Since the hole concentration is decreasing on the p-side, the net charge is negative:

)()( 000 ad NNnpqx -+-=r

)()( 00 aNpqx -»r

0)(0 <xr0pNa >

00 <<- xxp


13

Charge on N-Sidel Analogous to the p-side, the charge on the n-side is

given by:

l The net charge here is positive since:

)()( 00 dNnqx +-»r 00 nxx <<

0)(0 >xr0nNd >

a

i

Nnn2

0 =

Transition Region

diffJ

dNn =0

– – + +

0E


14

“Exact” Solution for Fieldsl Given the above approximations, we now have an

expression for the charge density

l We also have the following result from electrostatics

l Notice that the potential appears on both sides of the equation… difficult problem to solve

l A much simpler way to solve the problem…

îíì

<<-<<--

@-

0/)(

/)(

0 0)(0)(

)(0

0

nVx

id

poaVx

i

xxenNqxxNenq

xth

th

f

f

r

s

xdxd

dxdE

erf )(0

2

20 =-=


15

Depletion Approximationl Let’s assume that the transition region is

completely depleted of free carriers (only immobile dopants exist)

l Then the charge density is given by

l The solution for electric field is now easyîíì

<<+<<--

@0

0 00

)(nd

poa

xxqNxxqN

xr

s

xdxdE

er )(00 =

)(')'()( 000

00

p

x

xs

xEdxxxEp

-+= ò- er

Field zero outsidetransition region


16

Depletion Approximation (2)l Since charge density is a constant

l If we start from the n-side we get the following result

)(')'()(0

00 po

s

ax

xs

xxqNdxxxEp

+-== ò- eer

)()()(')'()( 0000

000 xExxqNxEdxxxE n

s

dx

xs

nn +-=+= ò eer

)()( 00 xxqNxE ns

d --=e

Field zero outsidetransition region


17

Plot of Fields In Depletion Region

l E-Field zero outside of depletion regionl Note the asymmetrical depletion widthsl Which region has higher doping?l Slope of E-Field larger in n-region. Why?l Peak E-Field at junction. Why continuous?

n-typep-type

NDNA

– – – – –– – – – –– – – – –– – – – –

+ + + + +

+ + + + +

+ + + + +

+ + + + +

DepletionRegion

)()( 00 xxqNxE ns

d --=e

)()(0 pos

a xxqNxE +-=e


18

Continuity of E-Field Across Junction

l Recall that E-Field diverges on charge. For a sheet charge at the interface, the E-field could be discontinuous

l In our case, the depletion region is only populated by a background density of fixed charges so the E-Field is continuous

l What does this imply?

l Total fixed charge in n-region equals fixed charge in p-region! Somewhat obvious result.

)0()0(00

==-=-== xExqNxqNxE pno

s

dpo

s

an

ee

nodpoa xqNxqN =


19

Potential Across Junctionl From our earlier calculation we know that the

potential in the n-region is higher than p-regionl The potential has to smoothly transition from high

to low in crossing the junctionl Physically, the potential difference is due to the

charge transfer that occurs due to the concentration gradient

l Let’s integrate the field to get the potential:

ò- ++-=x

x pos

apo

p

dxxxqNxx0

')'()()(e

ffx

x

pos

ap

p

xxxqNx0

'2')(2

-÷÷ø

öççè

æ++=

eff


20

Potential Across Junctionl We arrive at potential on p-side (parabolic)

l Do integral on n-side

l Potential must be continuous at interface (field finite at interface)

20 )(

2)( p

s

ap

po xxqNx ++=

eff

20 )(

2)( n

s

dnn xxqNx --=

eff

)0(22

)0( 20

20 pp

s

apn

s

dnn xqNxqN f

ef

eff =+=-=


21

Solve for Depletion Lengthsl We have two equations and two unknowns. We are

finally in a position to solve for the depletion depths

20

20 22 p

s

apn

s

dn xqNxqN

ef

ef +=-

nodpoa xqNxqN =

(1)

(2)

÷÷ø

öççè

æ+

=da

a

d

bisno NN

NqN

x fe2÷÷ø

öççè

æ+

=ad

d

a

bispo NN

NqN

x fe2

0>-º pnbi fff


22

Sanity Checkl Does the above equation make sense?l Let’s say we dope one side very highly. Then

physically we expect the depletion region width for the heavily doped side to approach zero:

l Entire depletion width dropped across p-region

02lim0 =+

=¥®

ad

d

d

bis

Nn NNN

qNx

d

fe þ

a

bis

ad

d

a

bisNp qNNN

NqN

xd

fefe 22lim0 =÷÷ø

öççè

æ+

=¥®


23

Total Depletion Widthl The sum of the depletion widths is the “space

charge region”

l This region is essentially depleted of all mobile charge

l Due to high electric field, carriers move across region at velocity saturated speed

÷÷ø

öççè

æ+=+=

da

bisnpd NNqxxX 112000

fe

µ11012150 »÷øö

çèæ=

qX bisd

fecmV10

µ1V1 4=»pnE


24

Have we invented a battery?l Can we harness the PN junction and turn it into a

battery?

l Numerical example:

2lnlnlni

ADth

i

A

i

Dthpnbi n

NNVnN

nNV =÷÷

ø

öççè

æ+=-º fff

mV600101010logmV60lnmV26 20

1515

2 =´==i

ADbi n

NNf

?


25

Contact Potentiall The contact between a PN junction creates a

potential differencel Likewise, the contact between two dissimilar

metals creates a potential difference (proportional to the difference between the work functions)

l When a metal semiconductor junction is formed, a contact potential forms as well

l If we short a PN junction, the sum of the voltages around the loop must be zero:

mnpmbi fff ++=0

pnmnf

pmf

+

−bif )( mnpmbi fff +-=


26

PN Junction Capacitorl Under thermal equilibrium, the PN junction does

not draw any (much) currentl But notice that a PN junction stores charge in the

space charge region (transition region)l Since the device is storing charge, it’s acting like a

capacitorl Positive charge is stored in the n-region, and

negative charge is in the p-region:

nodpoa xqNxqN =


27

Reverse Biased PN Junctionl What happens if we “reverse-bias” the PN

junction?

l Since no current is flowing, the entire reverse biased potential is dropped across the transition region

l To accommodate the extra potential, the charge in these regions must increase

l If no current is flowing, the only way for the charge to increase is to grow (shrink) the depletion regions

+

−Dbi V+-f

DV 0<DV


28

Voltage Dependence of Depletion Width

l Can redo the math but in the end we realize that the equations are the same except we replace the built-in potential with the effective reverse bias:

÷÷ø

öççè

æ+

-=+=

da

DbisDnDpDd NNq

VVxVxVX 11)(2)()()( fe

bi

Dn

da

a

d

DbisDn

VxNN

NqN

VVxf

fe-=÷÷

ø

öççè

æ+

-= 1)(2)( 0

bi

Dp

da

d

a

DbisDp

VxNN

NqN

VVxf

fe-=÷÷

ø

öççè

æ+

-= 1)(2)( 0

bi

DdDd

VXVXf

-= 1)( 0


29

Charge Versus Biasl As we increase the reverse bias, the depletion

region grows to accommodate more charge

l Charge is not a linear function of voltagel This is a non-linear capacitorl We can define a small signal capacitance for small

signals by breaking up the charge into two terms

bi

DaDpaDJ

VqNVxqNVQf

--=-= 1)()(

)()()( DDJDDJ vqVQvVQ +=+


30

Derivation of Small Signal Capacitance

l Do a Taylor Series expansion:

l Notice that

!++=+ DV

DDJDDJ v

dVdQVQvVQ

D

)()(

RD VVbipa

VV

jDjj

VxqNdVd

dVdQ

VCC== ú

úû

ù÷÷ø

öççè

æ--===f

1)( 0

bi

D

j

bi

Dbi

paj V

CV

xqNC

fff -

=-

=112

00

da

da

bi

s

da

d

a

bis

bi

a

bi

paj NN

NNqNN

NqN

qNxqNC

+=÷÷

ø

öççè

æ+÷÷

ø

öççè

æ==

fefe

ff 22

220

0


31

Physical Interpretation of Depletion Cap

l Notice that the expression on the right-hand-side is just the depletion width in thermal equilibrium

l This looks like a parallel plate capacitor!

da

da

bi

sj NN

NNqC+

=fe20

0

1

011

2 d

s

dabissj XNN

qC efe

e =÷÷ø

öççè

æ+=

-

)()(

Dd

sDj VXVC e

=


32

A Variable Capacitor (Varactor)l Capacitance varies versus bias:

l Application: Radio Tuner

0j

j

CC



Currents in PN Junctions


34

Diode under Thermal Equilibrium

l Diffusion small since few carriers have enough energy to penetrate barrierl Drift current is small since minority carriers are few and far between: Only

minority carriers generated within a diffusion length can contribute currentl Important Point: Minority drift current independent of barrier!l Diffusion current strong (exponential) function of barrier

p-type n-type

DN AN

---

-

---

------

+++++++++++++

0Ebiqf

,p diffJ

,p driftJ

,n diffJ

,n driftJ

−

−

+

+

−

−

ThermalGeneration

Recombination Carrier with energybelow barrier height

Minority Carrier Close to Junction


35

Reverse Bias

l Reverse Bias causes an increases barrier to diffusion

l Diffusion current is reduced exponentially

l Drift current does not change l Net result: Small reverse current

p-type n-type

DN AN

----

---

+++++++

( )bi Rq Vf +

+−


36

Forward Biasl Forward bias causes an exponential increase in

the number of carriers with sufficient energy to penetrate barrier

l Diffusion current increases exponentially

l Drift current does not change l Net result: Large forward current

p-type n-type

DN AN

----

---

+++++++

( )bi Rq Vf +

+−


37

Diode I-V Curve

l Diode IV relation is an exponential function

l This exponential is due to the Boltzmann distribution of carriers versus energy

l For reverse bias the current saturations to the drift current due to minority carriers

1dqV

kTd SI I e

æ ö= -ç ÷

è ø

dqVkT

d

s

II

1-

( )d d SI V I®-¥ = -


38

Minority Carriers at Junction Edges

Minority carrier concentration at boundaries of depletion region increase as barrier lowers … the function is

=-==

)()(

pp

nnxxpxxp (minority) hole conc. on n-side of barrier

(majority) hole conc. on p-side of barrier

kTEnergyBarriere /)(-=

(Boltzmann’s Law)

kTVq DBe /)( --= f

A

nnNxxp )( =


39

“Law of the Junction”

Minority carrier concentrations at the edges of thedepletion region are given by:

kTVqAnn

DBeNxxp /)()( --== f

kTVqDpp

DBeNxxn /)()( --=-= f

Note 1: NA and ND are the majority carrier concentrations onthe other side of the junction

Note 2: we can reduce these equations further by substitutingVD = 0 V (thermal equilibrium)

Note 3: assumption that pn << ND and np << NA


40

Minority Carrier Concentration

The minority carrier concentration in the bulk region for forward bias is a decaying exponential due to recombination

p side n side

-Wp Wn xn -xp

0

AqVkT

np e

0 0( ) 1A

p

xqVLkT

n n np x p p e e-æ ö

= + -ç ÷è ø

0np0pn

0

AqVkT

pn e

Minority CarrierDiffusion Length


41

Steady-State Concentrations

Assume that none of the diffusing holes and electrons recombine à get straight lines …

p side n side

-Wp Wn xn -xp

0

AqVkT

np e

0np0pn

0

AqVkT

pn e

This also happens if the minority carrier diffusion lengths are much larger than Wn,p , ,n p n pL W>>


42

Diode Current Densities

0 1A

p

qVpdiff n kT

n n ppx x

dn DJ qD q n edx W

=-

æ ö= » -ç ÷

è ø

0 0( )( )

AqVkT

p p p

p p

dn n e nx

dx x W-

»- - -

p side n side

-Wp Wn xn -xp

0

AqVkT

np e

0np

0pn

0

AqVkT

pn e

0 1A

n

qVpdiff n kT

p p nx x n

DdpJ qD q p edx W=

æ ö= - » - -ç ÷

è ø

2 1AqV

pdiff n kTi

d n a p

D DJ qn eN W N W

æ öæ ö= + -ç ÷ç ÷ç ÷è øè ø

2

0i

pa

nnN

=


43

Fabrication of IC Diodes

l Start with p-type substratel Create n-well to house diodel p and n+ diffusion regions are the cathode and annodel N-well must be reverse biased from substratel Parasitic resistance due to well resistance

p-type

p+

n-well

p-type

n+

annodecathode

p


44

Diode Small Signal Modell The I-V relation of a diode can be linearized

( )

1d d d dq V v qV qvkT kT kT

D D S SI i I e I e e+æ ö

+ = - »ç ÷è ø

( )1 d dD D D

q V vI i IkT+æ ö+ » + +ç ÷

è øL

2 3

12! 3!

x x xe x= + + + +L

dD d d

qvi g vkT

» =


45

Diode Capacitance l We have already seen that a reverse biased diode

acts like a capacitor since the depletion region grows and shrinks in response to the applied field. The capacitance in forward bias is given by

l But another charge storage mechanism comes into play in forward bias

l Minority carriers injected into p and n regions “stay” in each region for a while

l On average additional charge is stored in diode

01.4Sj j

dep

C A CXe

= »


46

Charge Storage

l Increasing forward bias increases minority charge density l By charge neutrality, the source voltage must supply equal

and opposite chargel A detailed analysis yields:

p side n side

-Wp Wn xn -xp

( )

0

d dq V vkT

np e+

0np0pn

( )

0

d dq V vkT

pn e+

12

dd

qICkT

t=Time to cross junction(or minority carrier lifetime)


47

Diode Circuitsl Rectifier (AC to DC conversion)l Average value circuitl Peak detector (AM demodulator)l DC restorerl Voltage doubler / quadrupler /…

Documents

Module 2.3: PN Junctions Prof. Ali M. Niknejad Prof. …rfic.eecs.berkeley.edu/105/pdf/module2-3_pn.pdfEE 105Fall 2016 Prof. A. M. Niknejad 9 PN Junctions: Overview l The most important