Semiconductor Devices A brief review

Semiconductor DevicesA brief review

Dr. K. Fobelets

Purpose of the course

• Study bipolar devices in more detail– Diodes and BJTs– Closer to reality: recombination – What causes the delays in these devices when

switching?

The most frequently used sentence in this course will be:

Excess minority carrier concentration

Structure

• 1. Lectures : 10 hrs– Basic principles based on Q&A session– Recombination and how does it impact the

characteristics– LONG pn diode – correct and approximated

solutions– LONG BJT– Switching of pn diodes and BJTs

• 2. Classes: solving past exam papers

Review

• Electrons and holes• Minority and majority carriers• Energy band diagram

Intrinsic Si

Si Si Si Si

Si Si Si Si

Si Si Si Si

Movement: kT

Si Si Si Si

Si Si Si Si

Si Si Si Si

Thermal energy: kT

Si Si Si Si

Si Si Si Si

Si Si Si Si

Si

Covalent bond

Free charged carriers in Si

Extrinsic Si

Si B Si Si

Si Si Si Si

Si Si Si Si

NA

Extrinsic Si

Si As Si Si

Si Si Si Si

Si Si Si Si

ND

Obtained by dopingB

As

Extrinsic Si

p-type n-type

In semiconductors two types of free charged carriers exist: electrons and holes.

Q1: What are holes?

a) Spherical voids in a semiconductorb) A positively charged Si atom that has lost its electronc) A positively charged particle that is the result of quantum mechanics

SiSi

SiSiSi

+ SiSi

Si

Si

SiSi

SiSiSi

SiSi

Si

Si

CThe two charged particles describe together the conduction in semiconductors.

Electron e- with charge q=-e and mass mn = m0 m*n

Hole h+ with charge q=+e and mass mp = m0 m*p

Intrinsic silicon (Si) has a small number of both free electrons and holes such that n i=pi.In order to increase the free carrier concentration, the semiconductor can be doped. With donors ND more electrons are created, with acceptors NA more holes are generated.

Q2: When intrinsic Si is doped with donor atoms, which of the following statements is correct?

a) n = p = ni = pi

b) n > ni & p < ni

c) n > p > ni

d) p > n > ni

n: electron concentrationp: hole concentrationni: intrinsic electron concentrationpi: intrinsic hole concentration

Bn > ni & p < ni in an n-type semiconductor.

n-type semiconductorn = ND p = ni

2/ND

p-type semiconductorn = ni

2/NA p = NA By heart

The concept of majority carrier and minority carrier is important in semiconductor devices. Majority carrier is the carrier type in a doped semiconductor with the highest concentration. Minority carrier is the carrier type with the lowest concentration.

Q3: True or False? The holes are the majority carriers in a p-type semiconductor (doped with acceptor atoms NA).

TRUEp-type semiconductor

pp

holeconcentration

p-typesemiconductor

np

electronconcentration

p-typesemiconductor

>

n-type semiconductor

nn

electronconcentration

n-typesemiconductor

np

holeconcentration

n-typesemiconductor

>

MAJORITY CARRIERS MINORITY CARRIERS

Drift and diffusion

• Two types of carrier movement– As a result of an electric field → DRIFT– As a result of a carrier gradient → DIFFUSION

Drift of carriers under influence of an electric field: E

E+ -

E+ -

EqJqJ

carriers ofnumber

v carriers ofnumber

Diffusion of carriers due to a carrier gradient

carriers ofnumber D

gradiention concentratconstant diffusion

dxdqJ

qJ

x

The purpose of semiconducting devices is to generate a current/voltage in response to an applied voltage/current. Two different types of current can exist in a semiconductor: drift and diffusion current. The expression of the total current that can flow in a semiconductor is given by the drift-diffusion equation:

Q4: Which statement is true?

a) Term (1) is drift current and (2) diffusion currentb) Term (2) is drift current and (1) diffusion currentc) Only term (1) can exist in a semiconductord) Only term (2) can exist in a semiconductor

dxxdpeDxExpexJ

dxxdneDxExnexJ

ppp

nnn

)()()()(

)()()()(

(1) (2)

ADrift current is proportional to the carrier concentration and the electric fieldDiffusion current is proportional to the carrier gradient.

E(x) Jndrift

Jpdrift

  n(x) Jn

diff

  p(x) Jpdiff

Motion of free charged carriers in a semiconductor.

Q5: If a p-type semiconductor at room temperature is conducting carriers due to drift, which of the following motion paths would be followed by the holes?

a)

(b)

c)

(d)

E+ - E+ -

E+ - E+ -

BWhen carriers move in a semiconductor they are scattered along the way. This means that they will be accelerated by the electric field (in this case) and then interact with atoms, impurities, other carriers that makes them lose some of their kinetic energy = scattering. Therefore the carriers will travel with an average velocity in amplitude and direction.

me

Ev

Q6: Solve diffusion processes

p+ n p

1. Draw arrows indicating the direction of diffusion of carriers.2. Identify the type of carriers that is diffusing.

Solution

p+ n p

Holes

Electrons

p+ n p

1. Because hole diffusion and electron diffusion cancel each other.2. Because an internal electric field is built up across each junction

causing drift of holes/electrons that cancel the diffusion of .holes/electrons.

3. Because holes and electrons diffuse automatically back to where they came from.

Q7: Why is there no net current while diffusion is happening?

p+ n p

Holes

Electrons

2. Because an internal electric field is built up across each junction causing drift of holes/electrons that cancel the diffusion of .holes/electrons.

Holes

Electronsdiffusion drift

+- E + -E

p-Si

Si B Si Si

Si Si Si Si

Si Si Si Si

NA n-Si

Si As Si Si

Si Si Si Si

As Si Si Si

ND

Depletion

Si

B

As

Si

Si

B

Cap

aciti

ve e

ffec

t

E+ -

--

B- : boron atom ionised

Si

Si

Si

Cap

aciti

ve e

ffec

t

E- +

As+ : arsenic atom ionised

+

+

Q8: True - False

The position of the Fermi level EF determines the type of the semiconductor.

Ec

Ev

EF

Q9: Multiple choice

1. This is the energy band diagram of an n-type semiconductor.2. This is the energy band diagram of a p-type semiconductor.3. This is the energy band diagram of an intrinsic semiconductor.

Ec

Ev

EF

Bottom of conduction bandEc

Top of valence bandEv

EiIntrinsic “level”. Is the position of the Fermi level EF when the semiconductor is intrinsic.

EG Bandgap. No energy levels in this energy region.

Position of Fermi level is determined by the doping type and densityFor n-type Si:

D

CFc

D

CCFc

FcC

NN

kTEE

NN

nN

kTEE

kTEE

Nn

ln

exp

exp

EF

Devices• A combination of n and p type

semiconductors plus ohmic contacts to apply the external voltages/currents makes devices

• When combining a-similar materials diffusion will occur and as a result an internal electric field will be built up to an amount that opposes diffusion current.

Energy band diagram

e.g.p-Si – n-Sip-Si – n-Si – p-Si

It is possible to start from the knowledge on workfunctions, and the energy reference: the vacuum level, Evac. The workfunction is dependent on the doping concentration!

Evac

n-Sie×n-Si

EF

p-Si

e×p-Si

EF

Evac

p-Si

e×p-Si

EF

Evac

n-Sie×n-Si

EF

Evac

p-Si

EF EF

Depleted region on both sides

Ec

Ev

Ec

Ev

e×p-Si

Evac

n-Sie×n-Si

Evac

SinSipeVe 0

Diffusion and drift can occur at the same time.

E

Both also always occur across junctions

A charge packet

A look at the short pn-diode

PN diode I

V

p n

p n

p n

E

Short PN diodeI

V

p n

p n

p n

E

DIFFUSION

Short PN diodeI

V

p n

p n

p n

E

DIFFUSION

Short PN diodeI

V

p n

p n

p n

E

Linear variation of minority carrier concentration

How do we find the current?

DIFFUSION

distanceMin

ority

car

rier c

once

ntra

tion

Apply diffusion current formula to the minority carrier variation

Short PN diodeI

V

p n

p n

Ep n

Only few carriers can contribute to the current

Contents of course this year

• Long pn diode– Introducing the concept of recombination of carriers.– Switching of the pn diode, where does the delay come

from?

• Bipolar junction transistor– Internal functioning– Switching delays

p n

Long

But what happens in a long pn diode?

p n

Ln Lp

Minority carrier diffusion length

Short

In long semiconductors recombination of the minority carriers will occur whilst

diffusing

Loss of both carrier type, but felt most in excess minority carriers. Remember: the amount of majority carriers is much larger than the excess.

Excess holes, in an n-type semiconductor will recombine with the large amount of available electrons.

p

In long semiconductors recombination of the minority carriers will occur whilst

diffusing

• Diffusing minority carriers (e.g. holes) recombine with majority carriers (electrons) within a diffusion length LpIn

ject

ion

of c

arrie

rs

x

Loss of both carrier type, but felt most in excess minority carriers. Remember: the amount of majority carriers is much larger than the excess.

Lp

Excess holes, in an n-type semiconductor will recombine with the large amount of available electrons.

p

Generation-recombination

• Generation of carriers and recombination is continuously happening at the same time such that the equilibrium carrier concentrations are maintained.

Charge neutral

R=G

Recombination - generation

• In case there is an excess carrier concentration then the recombination rate R of the excess, will be larger than its generation rate, G: R>G

When there is a shortage, then G > R

Recombination - generation

• Simple model: Recombination/generation rate is proportional to excess carrier concentration.

• Thus no net recombination/generation takes place if the carrier density equals the thermal equilibrium value.

Recombination of e- in p-type semiconductor

p

n

p

nnppp

n

p

n

ppnnn

pppGRU

nnnGRU

0

0

Recombination of h+ in n-type semiconductor

Diffusion, drift and recombination of carriers

What is the consequence of this recombination on the characteristics of the pn diode with neutral regions

larger than the diffusion lengths of the minority carriers?

In the pn diode the carrier gradient determines the current thus we have

to find the function p(x) of the minority carrier concentration.

• Note, reasoning done for p(x). For n(x) analogous approach.

Mathematical description of diffusion and recombination

xx x+x

Jp(x) Jp (x+x)A

p

pp

xxx

px

xxJxJqt

p

)()(1

Rate of hole variation

Variation of hole concentration in x x A/s

Recombination rate= +

Mathematical description of diffusion and recombination

p

p px

Jqt

p

1

= bulk defined + excess concentration

Jp : total current = drift + diffusion

Neglect drift current (no electric field applied)

p

pp

xxx

px

xxJxJqt

p

)()(1

p

p px

Jqt

txpx

1),(:0

ppp 0

D

in N

npp2

0 0

with

Mathematical description of diffusion and recombination

pp

pp

p

p

px

pDtp

ppp

pxpDp

xJ

qtp

2

2

0

2

21

= bulk defined+ excess concentration

dxxdpeDxJ pp)()(

D

in N

npp2

0 0

with

Solve equation in steady state

22

2

0

ppp Lp

Dp

xp

tp

Diffusion length

Boundary conditions:ppx

pXx n

00

General solution of 2nd order differential equation:

21 sinh)( C

LxCxpp

x

p

Xn0

pcontact

p

n

p

nL

Xx

LX

pxp sinh

sinh

)(

Too complicated

• Short approximation • Long approximation

Xn << Lp

p

n

p

nL

Xx

LX

pxp sinh

sinh

)(

Xn >> Lp

LINEAR EXPONENTIAL

Short semiconductor• Xn ≤ Lp carriers do not have time to recombine (=∞) !• Taking linear approximation.

pn(x)

x0

pn0

p

pn(x

)

Xn

NO recombination : variation of the excess carrier concentration linear

pn(x)= pn0+ p (1–x/Xn)

pn(x)

Contact imposes pn(Xn)=0

p’n

Diffusion and recombination• Xn >> Lp carriers do have time to recombine (t<∞) !• Taking exponential approximations

When recombination occurs and Xn >> Lp variation of the excess carrier concentration is exponential

pn(x)

x0

pn0

p

pn(x

)

pn(x)=pn0+p’n

LpContact imposes pn(Xn)=0

Xn

p

n

p

p

nLX

Lx

LX

p expexp

exp1

pn(x)

p

n

p

p

nLX

Lx

LX

p expexp

exp1pn(x)=

pn still too complex for quick calculations

• Take really extreme case• Xn >>> Lp or Xn → ∞

pLxp exp

Note: I and Q of both expressions of for the same

I for same as for linear approximation when Xn=Lp

pn(x) Xn → ∞

pn(x)=

pLxp exp

Diffusion and recombination

When recombination occurs and Xn → ∞ variation of the excess carrier concentration is exponential

pn(x)

x0

pn0

p

pn(x

)

pn(x)=pn0+p e-x/Lpp’n

Lp

• Xn >>> Lp carriers do have time to recombine (t<∞) !• Taking exponential approximations

Imposes pn(Xn)=0∞

SHORT ↔ LONGapproximation

Short

Boundary of short

LongIntermediate

Correct solutionExponential solutionLinear solution

pn(x)

pn(x)

pn(x)

pn(x)Lp=200 nm, Xn=400nm

Lp=200 nm, Xn=20nm Lp=Xn=200nm

Lp=200 nm, Xn=1000nmx

x

x

x

• Calculation of currents in pn diode with neutral regions larger than the diffusion length, using the long semiconductor approximation

→• Exponential variation of the excess minority

carrier concentration.

Carrier injections: forward bias

• Carrier injection across junction

-wp wn0

p ne-

diff

h+diff

• Creates minority carrier concentration gradients

np(-x)n’p

pn(x)

p’n

np0=ni2/NA & pp=NA

pn0= ni2/ND & nn=ND

pn0

x

np0

-x

Tnn

Tpp

VVpp

VVnn

exp'

exp'

0

0

Carrier injections: reverse bias

• Minority carriers are swept across junction V<0

-wp wn0

p ne-

drift

h+drift

• Small amount of minority carriers → small current

pn0

x

np0

-x

np(-x)

n’’p

pn(x)

p’’n

Tnn

Tpp

VV

pp

VV

nn

exp''

exp''

0

0

Thus

pn = pn0 (eeV/kT -1)

-wp wn0

p ne-

diff

h+diff

np(-x)n’p

pn(x)

p’n

pn0

x

np0

-x

np = np0 (eeV/kT -1)

nppn

nLx

pp enxn)(

)(

pLx

nn epxp)(

)(

Two methods to calculate current

x-wp wn0 I

nppn

x-x

np pn

Slope

1. Gradient excess carrier concentration2. Re-supply of recombined excess charge

0 0

Qn

Qp

1. Excess carrier concentration gradient

-wp wn

np pn

x-x

np pn

Slope

e-

In = e A Dn dnp/dx = max @ x=0

h+

Ip = -e A Dp dpn/dx = max @ x=0

Maximum diffusion currents at the edges of the transition region

0 0

1. Excess carrier concentration gradient

e- h+

Fill in expression for excess carrier concentration

1exp

1exp

1exp)(

0

0

0

max

0

)(

max

)(

kTeV

LDeAn

I

dx

ekTeVdn

eADI

ekTeVnxn

n

npdiffn

x

Lx

p

ndiffn

Lx

pp

n

n

1exp

1exp

1exp)(

0

0

0

max

0

)(

max

)(

kTeV

LDeAp

I

dx

ekTeVdp

eADI

ekTeVpxp

p

pndiffp

x

Lx

n

pdiffp

Lx

nn

p

p

In Ip

Changing gradient!→

Changing diffusion current density

p n

ItotIp In

Itot=In + Ip

x

diffntotxdriftp

Lx

n

npxdiffn

III

ekTeV

LDeAn

I n

)(

1exp0

InIp

x

diffptotxdriftn

Lx

p

pnxdiffp

III

ekTeV

LDeAp

I p

)(

1exp0

x-wp wn0 I

np pn

x

np0

-x

np pn

pn0

In

Ip

np = np e-(-x)/Ln

pn = pn e-(x)/LpQn

Qp

0 0

2. Re-supply of recombined excess carriers

Excess carrier charge Q recombines every seconds (carrier life time).For steady state Q has to be re-supplied every seconds → current

-wp wn0

np pn

x

np0

-x

np pn

pn0

In

Qn = -e A ∫-∞0np dx

In = Qn/n = e A Ln np /n

Ip

Qp = e A ∫0∞pn dx

Ip = Qp/p = e A Lp pn /p

Charge – minority carrier life time ratio

np = np e-(-x)/Ln

pn = pn e-(x)/LpQn

Qp

0 0

2. Re-supply of recombined excess carriers

Charge = area under excess carrier concentration: integrate-∞ and + ∞ are the contacts: excess charge = 0!

Total current

• I = Ip(0) + In(0) = e A (Dp pn0 /Lp + Dn np0/Ln )(eeV/kT -1)

• I = I0 (eeV/kT -1)

• With I0 = e A (Dp pn0/Lp + Dn np0/Ln)

Reverse bias current

Same equation as short diode with length exactly equal to the minority carrier diffusion lengths

SHORT ↔ LONGapproximation error on current calculation:

ratio of currents

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5

Xn/Lp

Ireal/Ia

ppro

x

Ireal/IexpIreal/Ilin

Error on linear and exponential approximation

same when Xn=Lp

• Non-idealities in the pn diodes

Log(I)

V

a)b)

c)

idealreal

(a) Low voltage: low injection of carriers

V

Log(I)

V

a)

idealreal

1nkTeV

stot eII

(c) High voltage: high injection of carriers

n’p ≈ pp

p’n ≈ nn

Log(I)

V

c)

idealreal

a) n=2b) n=1c) n=2

(d) Higher currents

Log(I)

d)

idealreal

V

Current determined by resistance

Switching of p-n diodes• When a p-n diode is forward biased, excess carrier

concentrations exists at both sides of the depletion region edge.

• To switch the diode from forward to off or reverse bias, this excess carrier concentration needs to be removed.

• The transients resulting from the time it takes to remove the excess carriers will lead to the equivalent capacitance.

-wp wn0

p nnp pn

Switching off

on

off

i

t0-wp wn0

p nnp pn

e-

h+

-wp wn0

p n

Steady state snap shots

How do we go from this:p

x

pn

pno

pn

To this?

Off: NO current flows!!!

Excess carrier concentration+pno

Variation of the excess carrier concentration as a function of time.

p(x,t)

p

pp

p

p

ppcontactp

p

contact

p

contactp

contact

tQI

dttdQ

tQJ

eeAJ

eeA

dttdQ

dxpeAdxx

Je

eAdxt

txpeA

)()(

)()(

),(

0

000

Relationship for charge Qp

p

p px

Jqt

p

1

Transient during switching off

i(t)= I + dQ/dt = Q/ + dQ/dt

Excess charge due to charge injection at any instance of timeAverage lifetime of minority carriers

Recombination termCharge depletion term (or buildup)

Since no current in “off”, charge has to disappear byrecombination!

For switch from on to off:

At t<0 → Ion=Ion (Von)At t≥0 → Ioff = 0 (Voff = 0)And at t=-0 Q(0)=Ion At t→∞ Q(∞)=0 Q(t)=Ion e-t/

t > 00 = Q/ + dQ/dt

Transient during switching offvariation of the excess carrier concentration as a function of time

t=0

gradient→ i≠0

p

x

Variation in timepn

i=0→gradient=0

A voltage, vd will exists across the diode as long as charge remains

Qp(t)=eA∫p(x,t)dx=Ippe-t/p

p(x,t)=p(vd(t)) e-x/Lp

Revision

• When a pn diode switches, the excess minority carrier concentration needs to change. The removal of the excess minority carrier concentration causes the delay in the pn diode.

• The variation of the excess carrier concentration as a function of time given by:

dttdQtQ

ti p

p

pp

)()()(

ON-OFF (open circuit)take: p+n → Itot ≈ Ip

dttdQtQ

ti p

p

pp

)()()(

p+ nIp

t=0

ppONp

p

p

p

p

p

pONp

tItQ

dttdQtQ

it

QRVIit

exp)(

)()(0

0)0(;0@

)0()0(;0@

vd

R

V

OFF (open circuit) → ONtake: p+n → Itot ≈ Ip

dttdQtQ

ti p

p

pp

)()()(

p+ n

Ip

t=0

vd

pONp

pONpONpp

pONp

ONpp

pONpONpp

p

tONpp

pONpp

p

p

pONpp

p

p

pON

ONp

pp

tItIItQ

tI

ItQ

tIItQ

tItQ

dtItQ

tdQ

tQIdt

tdQ

dttdQtQ

I

RVIit

Qit

exp1exp)(

)(ln

ln)(ln

)(ln

)()(

)()(

)()(

)0(;0@

0)0(;0)0(;0@

0

V

R integrate

Reverse recovery transientSwitch the diode from forward to reverse bias

on

off

i

t0-wp wn0

p nnp pn

e-

h+

Steady state snap shots

How do we go from this:

Reverse bias current flows!!!

Excess carrier concentration 0-wp wn

e-

h+

x

p

pn

0pn

To this?

Transients when switching to reverse biase(t)

tE

-Ep n

e(t)i(t) R If≈E/R

Ir≈-E/R

I

V

If

-Ir

x

pIf → gradient≠0

Ir → gradient≠0

t

v(t)

ti(t)

t

-E

Storage delay time: tsd

i(t)If

t

-Ir

v(t)

Time required for the stored charge to disappear

tsd = minority carrier ln(1 + If/Ir)

tsd

Calculate storage delay time: tsd

dttdQtQ

ti p

p

pp

)()()(

i(t)IF

t

-IR

v(t)

tsd

0)(;@

)0()0(;0@

)0()0(;0@

sdsd

RpR

FpF

tQtt

IQIit

IQIit

dttdQtQ

I

Ititt

p

p

pR

Rpsd

)()(

)(0

X !

Calculated storage delay time: tsd

i(t)IF

t

-IR

v(t)

tsd

pFpRpRpp

FpRp

pRp

p

pRppRpp

tpRp

p

pRp

p

p

p

p

pRp

p

p

pR

tIIItQ

IItQIt

QItQIt

tQIt

tQItdQdt

dttdQtQI

dttdQtQ

I

exp)(

)(exp

)0(ln)(ln

)(ln

)()(

)()(

)()(

0

integrate

Calculated storage delay time: tsd

i(t)IF

t

-IR

v(t)

tsd

R

FRp

FR

Rpsd

p

sdFpRpRp

sd

pFpRpRpp

III

IIIt

tIII

tt

tIIItQ

lnln

exp0

exp)(

i(t)IF

t

-IR

v(t)

tsd

After: tsd

0

0)(

d

sdp

v

tQ

Evd

Build-up of depletion region

deplbu RCt

Small signal equivalent circuit

• Junction capacitance • Diffusion capacitance

p n

w

• Cj = A/w

• w function of bias→ C voltage variable capacitance

• Important in reverse bias

• Due to charge storage effects

-wp wn0

p nnp pn

• Due to depletion region

• Cd = dQ/dV = d (I )/dV

= e/kT I

• Important in forward bias

• Diffusion capacitance

Equivalent conductances

• Diffusion conductance

• gd = dI/dV = e/kT I0 eeV/kT

≈ e/kT I

• Slope of the current voltage characteristic in forward bias

• Series resistance rs

• Due to n and p region + contact resistance

• Vd = Vappl – rs I

rd

rs

Cj

CdOnly linear circuit elements present

Large signal equivalent circuit

C

Rs

Reverse bias: depletion capacitanceForward bias: diffusion capacitance

Non-linear circuit elements present

Conclusions

• The characteristics in a pn diode are based upon excess minority carrier diffusion.– Excess carrier concentrations are being formed

by injection of carriers across the junction.– The gradient of the excess minority carrier

concentration at the junction determines the magnitude of the current.

– Delay times are due to the storage of excess minority charge in the layers.

Revision

• When recombination is taken into account, the excess minority carrier concentration reduces while diffusing through the neutral regions of the diode.

• The variation of the excess carrier concentration is then given by:

pp

px

pDtp

2

2

Lifetime of minority carrier holes

Revision

• The steady state solution for the excess minority carrier concentration is then:

• This is considered too complex for quick calculations and approximations are used in the case of a short or long neutral region.

p

n

p

nL

Xx

LX

pxp sinh

sinh

)(

Revision

• Short: Xn ≤ Lp

pn(x)

x0

pn0

p

pn(x

)

Xn

linear

pn(x)= pn0+ p (1–x/Xn)

pn(x)

Contact imposes pn(Xn)=0

p’n

Revision

• Long: Xn >>> Lp exponential

pn(x)

x0

pn0

p

pn(x

) pn(x)=pn0+p e-x/Lpp’n

Lp Imposes pn(Xn)=0∞

pn(x)=pn0+

p

n

p

p

nLX

Lx

LX

p expexp

exp1

Revision

• These approximation make some errors in the calculation of the current and the charge stored in the neutral regions.

• However we will see that:

1. I and Q for simplified and non-simplified exponential variation of pn(x) for Xn → ∞ is the same

2. I for is same as for linear approximation when Xn=Lp

pn(x) =

pLxp exp

Errors on current

0200400600800

100012001400

1 2 3 4

Xn (nm)

Cur

rent

(a.u

.)

Series1Series2Series3

Lp=20 nm

1020 40 200

CorrectExponentialLinear

Short = good approximation up to Xn = Lp

Long = good approximation up to Xn > 5 ×Lp