eng325 lecture 2_2015 (1)

1

ENG325 Semiconductor Devices

Lecture 2

PN Junction in Reverse Bias

Capacitances in PN Junctions

Dr. TOH Eng-Huat

Associate Faculty

School of Science and Technology

SIM University

Email: [email protected]

©2014

Recap… Semiconductor with tailored conductivity

E.-H. Toh ENG325 2014 2

(a) Semiconductor (b) Insulator (c) Conductor

Semiconductor

Bandgap: 0 < Eg < 4

Elements: Si, Ge;

Compounds: GaAs

Tailored conductivity

(impurity, temp., illumination,

and magnetic field)

Conduction Band

Valence Band

Ec

Ev

Eg

Conduction Band

Valence Band

Ec

Ev

Eg

Top of

Conduction Band

Ec

Empty

Filled

Insulator

Bandgap: Eg > 4

r > 108 Ω -cm

Glass: 1011 Ω -cm,

Diamond: 1014 Ω -cm

Conductor

Bandgap: N.A.

r < 10-4 Ω -cm

Ag: 1.6x10-6 Ω -cm,

Cu: 1.7x10-6 Ω -cm,

Al: 2.7x10-6 Ω -cm

Recap… Donor/Acceptor Impurity and Ef

E.-H. Toh ENG325 2014 3

Applets:

http://www.acsu.buffalo.edu/~wie/applet/fermi/fermi.html

Introduce Donor/Acceptor impurities change conductivity and Ef.

Fermi-level (Ef): Energy-level at which probability of finding electron = 1/2

Recap… Ef, Density of states, f(E), and Conc.

E.-H. Toh ENG325 2014 4

Applets:

http://www.acsu.buffalo.edu/~wie/applet/fermi/sfw/show.html

Fermi-level (Ef): Energy-level at which probability of finding electron = 1/2

Recap… p-type and n-type Semiconductor

E.-H. Toh ENG325 2014 5

GP III elements: Accept e, free h

-ve immobile charges (eg. B-)

Efp close to Ev

Majority Carries: holes

Minority Carriers: electrons

GP V elements: free e

+ve immobile charges (eg. As+)

Efn close to Ec

Majority Carries: electrons

Minority Carriers: holes

Φ: Energy to remove

electron from Ef

Φp > Φn

Recap… PN Junction in Equilibrium

E.-H. Toh ENG325 2014 6

Equilibrium Conditions

1. No Light Source

2. Uniform temperature

3. No external applied electric or magnetic fields

Applets:

http://www.acsu.buffalo.edu/~wie/applet/pnformation/pnformation.html


E.-H. Toh ENG325 2014 7

p+ - n: p+ - n+:

Wp = Wn

W large

Vo, Eo low

p - n: p - n+:

Wp > Wn

W small

Vo, Eo high

Wp < Wn

W small

Vo, Eo high

Wp = Wn

W smallest

Vo, Eo highest


E.-H. Toh ENG325 2014 8

nno

xx = 0

pno

ppo

npo

log(n), log(p)

eNa

eNd

M

x

B

h+

p n

M

As+

e

W pW n

Neutral n-regionNeutral p-region

Space charge region

Metallurgical Junction

(a)

(b)

(c)

Wp

Wn(d)

Eo

M

ρ net

ni

ME(x)

(e)x

Eo

Wp Wn0

Vo

V(x)

x

(f)

PE(x)

Electron PE(x)

eVo

x (g)

-eVo

Hole PE(x)

Na.Wp = Nd.Wn

pn = ni2

= 1

=

,ad

dp

NN

NWW

+⋅=

ad

an

NN

NWW

+⋅=

( ) 2

1

2

+=

da

oda

oNNe

VNNW

ε

εεpand

o

WNeWNeE −=−=

=

2ln

i

dao

n

NN

e

kTV

Recap… PN Junction in Equilibrium and Forward Bias

E.-H. Toh ENG325 2014 9

Equilibrium Forward Bias Reverse Bias

Outline

E.-H. Toh ENG325 2014 10

Ideal PN Junction

– PN Junction under Reverse Bias

Capacitances in PN Junction

– Junction Capacitance

– Diffusion Capacitance

N-type SiliconP-type Silicon

-ve +ve

PN Junction in Reverse Bias

E.-H. Toh ENG325 2014 11

N-type P-type REVERSE BIAS:

When n-type is more positive with respect

to p-type. Depletion width increases.

The barrier potential increases to (Vo +

Vr). The electrons and holes thus face a

much larger energy barrier of magnitude

e(Vo + Vr). It is now harder for electrons

and holes to diffuse across the junction.

The diffusion current which is proportional

to exp[-(Vo + Vr)/kT] is now negligible.

There is however a small reverse current

arising from the drift component of the

minority carriers which give rise to – Jo.

This is known as minority carrier

extraction.

Equilibrium Reverse Bias

Depletion Width of RB PN Junction

E.-H. Toh ENG325 2014 12

Large reverse bias implies large barrier, and energy bands bend

more.

The electric field is larger and depletion region is wider.

For any bias V, the barrier is e(Vo-V) and the width of the depletion

region is

For reverse bias, Vo – V = Vo + Vr, which gives

( ) ( ) 2

1

2

−+=

da

oda

NNe

VVNNW

ε

( ) ( ) 2

1

2

++=

da

roda

RBNNe

VVNNW

ε

RB PN Junction: Extraction of Minority Carrier

E.-H. Toh ENG325 2014 13

Under reverse bias, diffusion of

carriers is inhibited.

However drift of carriers remains

unchanged such that

Jp(diff) < Jp(drift); Jn(diff) < Jn(drift)

Minority carrier concentrations in the

neutral p and n regions outside W will

drop below the equilibrium values

(npo and pno) due to the net drift of

carriers.

This is known as minority carrier

extraction.


x

W

HolesElectrons

Diffusion

Drift

ThermallygeneratedEHP

pnonpo

Eo+E

Vr

Minority CarrierConcentration

Wo

RB PN Junction: Thermal Generation

E.-H. Toh ENG325 2014 14

Ideal pn junction

With Thermal Generation

For ideal pn junction, drift current will be

small and limited by the small number of

minority carriers.

As such, the reverse saturation current is

independent of the reverse bias, provided

junction breakdown will not occur.

However, when the depletion width of the

RB junction becomes large, electron-hole

(e-h) pairs generated by thermal generation

could occur in the vicinity of the junction.

These e-h pairs are then separated by the

built-in electric field and would now

contribute to the overall drift current.

RB PN Junction: Thermal Generation

E.-H. Toh ENG325 2014 15

With Thermal Generation

Only those minority carriers that are thermally

generated within one minority carrier diffusion

length away from the edge of W contributes to

the drift current.

Carriers generated outside these two regions will

not contribute when they are more than one

diffusion length away from W, as they recombine

before reaching the depletion region.

These minority carriers will drift across the

junction as soon as it reaches W.

They will, together with the thermally generated

majority carriers, be driven by the weak electric

field in the neutral p and n regions and move

towards both ends of the diode.


x

W

HolesElectrons

Diffusion

Drift


pnonpo

Eo+E

Vr


Wo

RB PN Junction: IV Characteristics

E.-H. Toh ENG325 2014 16

The reverse saturation current would increase

with larger depletion width. It also increases

exponentially with temperature.

Reverse saturation current can also be

increased by optical excitation.

For example shining light near the metallurgical

junction of a reverse biased pn junction will

generate extra electron-hole pairs and increase

the minority carriers supply, i.e. Io.

This is the basis of using a reverse biased pn

junction as a photodetector.

Ideal diode

Space charge layergeneration, surface leakagecurrent, etc.

V

nA

I

mA

(a)


x

W

HolesElectrons

Diffusion

Drift


pnonpo

Eo+E

Vr


Wo

Reverse Bias Diode at Various Temperatures

E.-H. Toh ENG325 2014 17

0.002 0.004 0.006 0.008

1/Temperature (1/K)

10-16

10-14

10-12

10-10

10-8

10-6

10-4

Reverse diode current (A) at V = 5 V

Ge Photodiode323 K

238 K

Slope = 0.63 eV

(b)

Reverse diode current in a Ge pn junction as a function of temperature in a ln(Irev) vs 1/T plot. Above 238

K, Irev is controlled by ni2 and below 238 K it is controlled by ni. The vertical axis is a logarithmic scale

with actual current values. (From D. Scansen and S.O. Kasap, Cnd. J. Physics. 70, 1070-1075, 1992.)

Slope above 238K, Irev is controlled by ni2

and is given by 0.63eV, which is the bandgap

of Ge

Slope below 238K, Irev is controlled by ni and

is given by 0.33eV, which is ~ Eg/2. This is

due to EHP generation in SCL via defects

and impurities through recombination centers.

Slope = 0.33 eV

Trap level

PN Junction in Various Bias Modes

E.-H. Toh ENG325 2014 18

Equilibrium Forward Bias Reverse Bias

Equilibrium (Zero Bias):

At equilibrium, Jdrift = Jdiffusion. Net current = 0.

Reverse Bias (Extraction of minority carrier):

Larger depletion width. Jdrift > Jdiffusion. Net current > 0. Mainly drift current.

At large ξ, avalanche breakdown.

Forward Bias (Injection of minority carrier):

Smaller depletion width. Jdrift < Jdiffusion. Net current > 0. Mainly diffusion current.

PN Junction in FB and RB: Summary

E.-H. Toh ENG325 2014 19

Under FB, Jn(diff) and Jp(diff) increase sharply due to lower potential barrier,

e(Vo – Vf). Jn(drift) and Jp(drift) remain unchanged at their thermal equilibrium

values. Minority carrier injection takes place. Net effect is a large diffusion

current flowing from the p to n-side, which increases sharply with the applied

voltage.

Under RB, Jn(diff) and Jp(diff) decrease sharply due to higher potential barrier,

e(Vo + Vr) and are negligible. Jn(drift) and Jp(drift) remain unchanged at their

thermal equilibrium values. Minority carrier extraction takes place. Net effect is a

constant drift current flowing from n to p side, which is insensitive to the applied

V. This is the generation current (also called the reverse saturation current) of a

pn junction.

PN Junction in FB and RB: Summary

E.-H. Toh ENG325 2014 20

Forward Bias

(V = Vf)

Reverse Bias

(V = -Vr)

Built-in potential Vo - Vf Vo + Vr

Electric Field decreases (external field

opposes built-in field)

increases (external field aids

built-in field)

Depletion width W decreases increases

Conduction

Mechanism

Diffusion

(Injection of minority

carriers)

Drift

(Extraction of minority

carriers)

Ex: EB/E.Field/Carrier Density for a FB pn diode

E.-H. Toh ENG325 2014 21

pn junction in equilibrium

pn junction with FB of 0.6V

p = 1017cm-3; n = 1016cm-3

p = 1017cm-3; n = 1016cm-3

Ex: EB/E.Field/Carrier Density for a FB pn diode

E.-H. Toh ENG325 2014 22


pn junction with FB of 0.6V

p = 1017cm-3; n = 1016cm-3

p = 1017cm-3; n = 1016cm-3

Ex: EB/E.Field/Carrier Density for a RB pn diode

E.-H. Toh ENG325 2014 23


pn junction with RB of 2V

p = 1017cm-3; n = 1016cm-3

p = 1017cm-3; n = 1016cm-3

Ex: EB/E.Field/Carrier Density for a RB pn diode

E.-H. Toh ENG325 2014 24


pn junction with RB of 2V

p = 1017cm-3; n = 1016cm-3

p = 1017cm-3; n = 1016cm-3

Problem 1

E.-H. Toh ENG325 2014 25

A Si p-n junction has donor and acceptor doping concentrations Nd and Na of

1016 cm–3 and 1018 cm–3, respectively. Assume kT = 0.026 eV at 300 K, ni =

1.5x1010 cm–3 and εr = 11.8. Calculate the following parameters at (i) 0 bias,

(ii) forward bias of 0.65 V and (ii) reverse bias of 2 V.

(a) Contact potentials under 0 bias, FB and RB.

(b) The depletion region width

(c) The penetration depth into the n-side and p-side

(d) Peak electric field at the junction and

(e) The space charge in each side of the depletion region, given a cross-

sectional area of 10–4 cm2

(f) Sketch charge density and electric field in the pn junction.

N-type Silicon P-type Silicon

Problem 1: Approach

E.-H. Toh ENG325 2014 26

Parameters Equilibrium Forward Bias Reverse Bias

Contact

Potential

Total

Depletion

Width, W

Depletion

Width,

Wp, Wn

Peak Electric

Field, Eo

Charge, Q

,ad

dp

NN

NWW

+⋅=

ad

an

NN

NWW

+⋅=

aNex ⋅−=)(ρ

dNex ⋅=)(ρ

=

2ln

i

dao

n

NN

e

kTV

( ) 2

1

2

+=

da

oda

oNNe

VNNW

ε

for –Wp < x < 0

for 0 < x < Wn

εεpand

o

WNeWNeE −=−=

N-type Silicon P-type Silicon

W

Wn Wp

fo VV − ro VV +

( ) ( ) 2

1

2

−+=

da

foda

FBNNe

VVNNW

ε ( ) ( ) 2

1

2

++=

da

roda

RBNNe

VVNNW

ε

a p d nQ eN W A Q eN W A− +=− = =

Problem 1: Solution

E.-H. Toh ENG325 2014 27

(a)

Contact potential is the potential difference in the depletion region. This is

nothing but the built-in potential Vo.

The contact potential with zero bias is 0.817 V.

With a FB of 0.65 V, the contact potential changes to

Vo – V = 0.817 – 0.65 = 0.167 V.

With a RB of 2 V, the contact potential changes to

Vo + V = 0.817 + 2 = 2.817 V.

( )

2

18 16

10 2

ln

10 100.026(V)ln 0.817 V

1.5 10

a do

i

o

N NkTV

e n

V

=

× = =

×

Problem 1: Solution

E.-H. Toh ENG325 2014 28

(b)

We can now calculate the zero bias depletion region using

When a Forward Bias is applied the contact potential becomes Vo – V. Thus

the depletion region width will now be

( ) ( ) ( )14 18 161

219 18 16

5

2 8.85 10 11.8 10 10 0.8172 cm

1.6 10 10 10

3.28 10 cm or 0.328μm

a d o

a d

N N VW

eN N

W

ε−

−

−

× × × × + × + = = × × ×

= ×

( )( ) ( ) ( )( )14 18 161

219 18 16

5

2 8.85 10 11.8 10 10 0.817 0.652cm

1.6 10 10 10

1.48 10 cm or 0.148μm

a d oFB

a d

FB

N N V VW

eN N

W

ε−

−

−

× × + − + − = = × × ×

= ×

Problem 1: Solution

E.-H. Toh ENG325 2014 29

When a Reverse Bias is applied the contact potential becomes Vo + V. Thus

the depletion region width will now be

Note that the depletion region width decreases in FB and increases with RB.

( )( ) ( )( )( )14 18 161

219 18 16

5

2 8.85 10 11.8 10 10 0.817 0.652cm

1.6 10 10 10

6.1 10 cm or 0.61μm

a d oRB

a d

RB

N N V VW

eN N

W

ε−

−

−

× × + + + + = = × × ×

= ×

(c) Now we can calculate the extensions of the depletion region into both n

and the p-sides.

With zero bias

5 163

18 16

3.28 10 103.25 10 μm

10 10

dp

a d

W NW

N N

−−× ×

= = ≈ ×+ +

2

Problem 1: Solution

E.-H. Toh ENG325 2014 30

With a FB of 0.65 V

5 184

18 16

3.28 10 103.25 10 cm or 0.325μm

10 10

an

a d

W NW

N N

−−× ×

= = ≈ ×+ +

5 163

18 16

1.48 10 101.47 10 μm

10 10

dp

a d

W NW

N N

−−× ×

= = ≈ ×+ +

5 184

18 16

1.48 10 101.47 10 cm or 0.147 μm

10 10

an

a d

W NW

N N

−−× ×

= = ≈ ×+ +

With a RB of 2 V

5 163

18 16

6.10 10 106.04 10 μm

10 10

dp

a d

W NW

N N

−−× ×

= = ≈ ×+ +

5 184

18 16

6.10 10 106.10 10 cm or 0.610μm

10 10

an

a d

W NW

N N

−−× ×

= = ≈ ×+ +

Obviously, the extensions of the depletion regions also decrease with FB

and increase with RB.

Problem 1: Solution

E.-H. Toh ENG325 2014 31

a pd no

eN WeN WE

ε ε= − = −

19 16 54

14

1.6 10 10 3.25 105 10 V/cm

8.85 10 11.8

d no

eN WE

ε

− −

−× × × ×

= = = ×× ×

19 16 54

14

1.6 10 10 1.47 102.25 10 V/cm

8.85 10 11.8

d no FB

eN WE

ε

− −

−× × × ×

= = = ×× ×

19 16 54

14

1.6 10 10 6.10 109.25 10 V/cm

8.85 10 11.8

d no RB

eN WE

ε

− −

−× × × ×

= = = ×× ×

We can now calculate the peak value of the electric field. With zero bias, we

have

With a FB of 0.65 V, we have

Finally, with a RB of 2 V, we have

The peak value of E-field decreases with FB and increases with RB, consistent

with the contact potential variation.

Problem 1: Solution

E.-H. Toh ENG325 2014 32

a p d nQ eN W A Q eN W A− +=− = =

19 16 5 4

12

1.6 10 10 3.25 10 10 Coul

5.2 10 Coul

d nQ eN W A

Q Q

− − −+

−+ −

= = × × × × ×

= × =

19 16 5 4

12

1.6 10 10 1.47 10 10 Coul

2.35 10 Coul

d nQ eN W A

Q Q

− − −+

−+ −

= = × × × × ×

= × =

19 16 5 4

12

1.6 10 10 6.10 10 10 Coul

9.66 10 Coul

d nQ eN W A

Q Q

− − −+

−+ −

= = × × × × ×

= × =

Use either Q– or Q+ to get the depletion region charge. Thus

With a FB of 0.65 V, we have

With a RB of 2 V, we have

Note once again that the charge in the depletion region decreases with FB and

increases with RB.

(e) The +ve and –ve charges in the depletion region are equal to each other to

maintain charge neutrality.

Problem 1: Solution

E.-H. Toh ENG325 2014 33

Na = 1018 cm–3Nd = 1016 cm–3

Parameters Zero Bias FB

(0.65 V)

RB

(2V)

Contact

Potential (V)

Vo =

0.817

Vo – V =

0.167

Vo + V =

2.817

Wn (um) 0.325 0.147 0.61

Wp (um) 3.25 x10-3 1.47 x10-3 6.04 x10-4

Eo (V/cm) 5 x 104 2.25 x 104 9.25 x 104

p-Sin-Si ρ (x)

x (um)

-1018 q

1016 q

Zero Bias

FB 0.65 V

RB 2Vξ (x)

x (um)

How about voltage and potential energy?

Sketch the energy band diagram with diff bias?

(f) Sketch the charge density and electric field

Problem 1: Solution

E.-H. Toh ENG325 2014 34

Na = 1018 cm–3Nd = 1016 cm–3

Parameters Zero Bias FB

(0.65 V)

RB

(2V)

Contact

Potential (V)

Vo =

0.817

Vo – V =

0.167

Vo + V =

2.817

Wn (um) 0.325 0.147 0.61

Wp (um) 3.25 x10-3 1.47 x10-3 6.04 x10-4

Eo (V/cm) 5 x 104 2.25 x 104 9.25 x 104

p-Sin-Si ρ (x)

x (um)

-1018 q

1016 q

Zero Bias

FB 0.65 V

RB 2Vξ (x)

x (um)

How about voltage and potential energy?

Sketch the energy band diagram with diff bias?

(f) Sketch the charge density and electric field

Ex: Energy Band Diagram for Different Bias

E.-H. Toh ENG325 2014 35


At Equilibrium:

At FB of 0.2V:

At FB of 0.5V:

Na = 1018 cm–3

Nd = 1016 cm–3

Vo = 0.82V

At RB of 0.5V:

At RB of 1.5V:

Applets:

http://www.acsu.buffalo.edu/~wie/

applet/biasedPN/BiasedPN.htm2

Ex: Energy Band Diagram for Different Bias

E.-H. Toh ENG325 2014 36


At Equilibrium:

At FB of 0.2V:

At FB of 0.5V:

Na = 1018 cm–3

Nd = 1016 cm–3

Vo = 0.82V

At RB of 0.5V:

At RB of 1.5V:

Applets:

http://www.acsu.buffalo.edu/~wie/

applet/biasedPN/BiasedPN.htm2

0.2V 0V

0.5V 0V

0.5V

0V

1.5V

0V

W -

Vo & E -

W --

Vo & E --

W +

Vo & E +

W ++

Vo & E ++

Outline

E.-H. Toh ENG325 2014 37

Ideal PN Junction

– PN Junction under Reverse Bias


– Junction Capacitance

– Diffusion Capacitance

CT = Cdep + Cdiff

Capacitor and Capacitance

E.-H. Toh ENG325 2014 38

Capacitance, C = dQ/dV = εA/d

Measured in Farads(F).

Resistance offered by Capacitor is called impedance (Z).

In AC circuits, impedance decreases with increase in frequency and

capacitance. So it can act as short circuit or AC coupling. For low

frequency, it can act as open Circuit to filter out the frequency.

Capacitance reduces Circuit Speed

E.-H. Toh ENG325 2014 39

Capacitances introduced

time delay in switching circuits.


E.-H. Toh ENG325 2014 40

Two kinds of capacitances are associated with pn junctions.

These are junction capacitance (or transition capacitance) and charge

storage capacitance (or diffusion capacitance).

These are modeled by two capacitors, Cdep and Cdiff respectively,

connected parallel to the pn junction.

Cdep

Cdiff

CT = Cdep + Cdiff

Junction Depletion Capacitance

E.-H. Toh ENG325 2014 41

In a pn junction, the width of the depletion region is given by

The depletion region models a parallel plate capacitor of width W,

with the capacitance given by

This is the depletion layer capacitance, also called junction

capacitance.

Under dynamic conditions,

( ) 2

1

)(2

−+=

da

oda

NNe

VVNNW

ε

W

ACdep

ε=

dV

dQCdep =


E.-H. Toh ENG325 2014 42

dQ = Incremental charge

Diode voltage = -Vr

Diode voltage = -(Vr+dVr)

Space charge region

eNd

eNa

Net Space ChargeDensity

MM

W

ACdep

ε=

dV

dQCdep =

The depletion region

behaves like a capacitor.

The charge in the depletion

region depends on the

applied voltage just as in a

capacitor.

The larger the depletion

width, the smaller the

depletion capacitance.


E.-H. Toh ENG325 2014 43

W

ACdep

ε=

dV

dQCdep =

Depletion width decreases with

forward bias and increases with

reverse bias.

The incremental capacitance of

the depletion region increases

with forward bias and decreases

with reverse bias.

Its value is typically in the range

of pF/mm2 of device area.

(10-103) pF/mm2

ForwardReverse

Diode Voltage Vo0

Cdep


E.-H. Toh ENG325 2014 44

The absolute charge and the depletion width in an abrupt junction are:

Substituting Wn or Wp into |Q|:

AWeNAWeNQ pand ==||

pn WWW +=

ad

an

NN

NWW

+⋅=

ad

dp

NN

NWW

+⋅=

( ) 2

1

2

1

)(2)(2||

+−

=

−++

===da

oda

da

oda

da

adpand

NN

VVNNeAA

NeN

VVNN

NN

NeNAWeNAWeNQ

εε

( ) 2

1

)(2

−+=

da

oda

NNe

VVNNW

ε


E.-H. Toh ENG325 2014 45

It is then easy to differentiate and see that:

This expression for Cdep is the same as that for a parallel plate model:

Note that W is a function of applied voltage V.

This capacitance is present under both forward and reverse bias

conditions.

W

A

NN

NNe

VV

A

dV

dQC

da

da

o

dep

εε=

+−==

2

1

)(2||

W

ACdep

ε=


E.-H. Toh ENG325 2014 46

The expression for Cdep is an exact analogy with that of a parallel

plate capacitor, with the depletion width W corresponding to the plate

separation of the capacitor.

Physically as we change the bias V across the pn junction, it takes

time for the majority carriers to respond in order to expose a larger

space charge region under RB, or a smaller space charge region

under FB.

Cdep is associated with this delay time for W to reach its steady state

width.

In other words, W cannot change instantaneously in response to a

sudden change in the bias.

Problem 2

E.-H. Toh ENG325 2014 47

Problem 2: Solution

E.-H. Toh ENG325 2014 48

Problem 2: Solution

E.-H. Toh ENG325 2014 49

Zero Bias

Problem 2: Solution

E.-H. Toh ENG325 2014 50

Reverse Bias

(+6V)

Problem 2: Solution

E.-H. Toh ENG325 2014 51

Forward Bias

(-0.7V)

Diffusion Capacitance

E.-H. Toh ENG325 2014 52

Consider a pn junction in forward bias:

There is a reduction in the barrier height, a

reduction in the depletion region width, and an

injection of minority carriers across the

depletion region into the opposite region, where

they are stored as excess minority carriers.

There is subsequent diffusion of minority

carriers in the two regions.

The density of the excess stored minority

carriers increases with an increase in the FB.

As a result there is stored positive charge (due

to the hole diffusion) in the n-region and stored

negative charge (due to the electron diffusion)

in the p-region.

Diffusion Capacitance: Stored Minority Carriers

E.-H. Toh ENG325 2014 53

x'

pno

pn(0) when V

pn'(0) when V+dV

dQI = Q /τh

V to V+dV

Neutral n-regionSCL

Q

Note that with changing bias V, there will be a

change in the minority carrier concentrations at

sides of the junction, near the depletion edge.

Depletion width become larger upon RB or

smaller upon FB.

The increase in the minority carrier

concentration does not take place

instantaneously with a sudden change in FB.

A delay time is involved in the minority carrier

concentration attaining a steady-state.

This delay is due to capacitive effect as a result

of stored minority carrier charges in the neutral

n and p regions.

This is the origin of diffusion capacitance.

Diffusion Capacitance: Diffusion Current

E.-H. Toh ENG325 2014 54

x'

pno

pn(0) when V

pn'(0) when V+dV

dQI = Q /τh

V to V+dV

Neutral n-regionSCL

Q

The diffusion capacitance is

defined by

Under forward bias, the current is

contributed mainly by diffusion

such that

dV

dQCdiff =

= − 1

Diffusion Capacitance: Dominant in FB

E.-H. Toh ENG325 2014 55

The rate of change of charge Q defines the current I.

If τh is the minority carrier lifetime (for holes in n-region), we can write,

I = Q/τh

This gives

Under reverse bias, even though the same equation for I is applicable,

there is negligible injection and hence no stored charges or diffusion

capacitance.

Diffusion capacitance is predominant only in forward bias.

( )25

mAI

kT

eI

dV

dQC hh

diff

ττ=== at 300K

Diffusion Capacitance: Speed Limitation

E.-H. Toh ENG325 2014 56

It can be seen that Cdiff can be very large under FB due to the large I.

Physically, this can be understood because the storage charge

depends exponentially on the bias voltage V under FB.

This implies that there will be a large change in the amount of storage

charge (dQs) of the minority carriers with changing bias (dV), resulting

in a large Cdiff.

The Cdiff will impose a serious limitation for a FB biased pn junction in

high frequency applications.

Total Capacitance

E.-H. Toh ENG325 2014 57

The total capacitance across the pn junction is

CT = Cdep + Cdiff

During FB, Cdiff is dominant (Cdiff >> Cdep) due to the large change in

the amount of storage charge of the minority carriers, hence CT ≈ Cdiff.

During RB, Cdiff is not critical (Cdiff << Cdep) since there is only very

small change in the amount of storage charge of the minority carriers.

In comparison, the junction capacitance Cdep dominates under RB, i.e.,

CT ≈ Cdep.

The presence of these capacitances will limit the speed of operation of

any circuits using pn junction diodes.

Problem 3

E.-H. Toh ENG325 2014 58

Consider a silicon p+-n junction with Nd = 1016 cm–3. The area of the

diode is 10–4 cm2. Assume kT = 0.026 eV at 300 K, ni = 1.5×1010 cm–3

and εr = 11.8. Holes in the n-region have a lifetime of 0.5 μs with a

diffusion coefficient of 10 cm2/s. Calculate the diffusion capacitance of

the p-n junction at a forward bias of 0.65 V.

Problem 3: Solution

E.-H. Toh ENG325 2014 59

hdiff

e IC

kT

τ=

exp 1soqV

I IkT

= −

2h eso i

h d e a

eD eDI n A

L N L N

= +

The diffusion capacitance is given by

Let us determine the diode current at a FB of 0.65V

where

In this case, we have a pn junction where the p is heavily doped.

Thus, we can write

( )19 2

10 4 15

6 16

1.6 10 101.5 10 10 1.6 10 A

10 0.5 10 10soI

−− −

−

× ×

= × × × = × × × ×

=

Problem 3: Solution

E.-H. Toh ENG325 2014 60

40.65exp 1 1.15 10 A

0.026soI I − = − = ×

6 4

9

0.5 10 1.15 10

0.026

2.23 10 F or 2.23nF

h hdiff

e I IC

kTkT

e

τ τ − −

−

× × ×= = =

= ×

Thus, the diode current is

Finally, we now have the diffusion capacitance as

Compare this value with the FB depletion layer capacitance. It is

obvious that diffusion capacitance is more dominant that depletion layer

capacitance in FB.

e-Learning

E.-H. Toh ENG325 2014 61

Webpage:

http://www.acsu.buffalo.edu/~wie/applet/applet.old

Applets:

http://www.acsu.buffalo.edu/~wie/applet/

Fermi-Level:

http://www.acsu.buffalo.edu/~wie/applet/fermi/fermi.html

Carrier-Concentration:

http://www.acsu.buffalo.edu/~wie/applet/fermi/sfw/show.html

PN Junction and Band Diagram:

http://www.acsu.buffalo.edu/~wie/applet/pnformation/pnformation.html

PN Junction under forward bias

http://www.acsu.buffalo.edu/~wie/applet/students/jiawang/pn.html

PN Junction under bias

http://www.acsu.buffalo.edu/~wie/applet/biasedPN/BiasedPN.htm2

Class Test on Seminar 3

E.-H. Toh ENG325 2014 62

15% of final score

25 MCQ questions (closed book)

Covers materials from seminar 1 & 2

Only a few questions require you to remember

basic equations

Bring your calculator

eng325 lecture 2_2015 (1)

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