1. Diffusion: the flow of charge due to density gradients.

a. Diffusion Current

The diffusion currents are expressed in terms of the electron or hole diffusion

coefficient Dn or Dp, as shown below:

Jn,diff = ( q)Dn

dn

dx= qDn

dn

dx

Jp,diff = (+q)Dp

dn

dx= qDp

dp

dx

b. Einstein Relation:

Dn =kT

qμn

Dp =kT

qμp

2. Total Current: J = Jdiff + Jdrift

EXAMPLE1: We have a piece of uniformly doped silicon with a acceptor concentration

of Na=2E16cm-3

. By shedding light on one end (x=0) of the semiconductor, we creat an

electron concentration of n(x)=1E13*exp(-x/2um). Calculate the electron diffusion

current as a function of x.

SOLUTION1:

Na = 2E16 μn = 1050cm2 /Vs

Dn =kT

qμn = 0.026 1050 = 27.3cm2 / s

Jn,diff = qDndn

dx= 1.6 10 19 27.3

d

dx1013e

x / 2 10 4( )( )= 0.22e x /2μmA / cm2

3. Integrated Passives

a. Resistors

-Diffusion Resistor: ion implantation/diffusion

100ohm/sq(unsilicided),10ohm/sq(silicided)

-Poly Film Resistor: deposit a thin film of heavily doped poly-Si

10-100ohm/sq(unsilicided),1ohm/sq(silicided)

b. Capacitors

-IC MIM Capacitor: by forming a Metal plate-Insulator(thin oxide)-Metal plate structure.

-Parallel Plate Capacitor:

C =A

d

-Nonlinear capacitor: Q-V curve is not a straight line

Small signal capacitance: differential capacitance

Q = Q0 + q f (Vs) +df (V )

dV V =Vs

vs

Cdf (V )dV V =Vs

4. PN Junction

a. Structure

b. Distribution of charge, electric field, and potential

-Depletion Approximation: The transition region is completely depleted of free carriers,

the only charges exist are immobile dopants. This region is called “Depletion Region”

0(x)qNa ( xpo < x < 0)

+qNd (0 < x < xno)

-Electric Field in Depletion Region:

E0(x) =

qNa

s

x + xp0( )( xpo < x < 0)

qNd

s

xn0 x( )(0 < x < xno)

-Potential in Depletion Region

p (x) = p +qNa

2 s

x + xp0( )2

n (x) = n +qNd

2 s

x xn0( )2

-Boundary Conditions:

Potential continuous:

p (0) = p +qNa

2 s

x p0( )2

= n (0) = n +qNd

2 s

xn0( )2

Electrical Field continuous: qNaxp0 = qNd xn0

-Depletion Width & Built-in Potential

Xd 0 = xn0 + xp0 =2 s bi

q

1

Na

+1

Nd

xn0 =2 s bi

qNd

Na

Na + Nd

xp0 =2 s bi

qNa

Nd

Na + Nd

bi n p =Vth lnNd

ni

+ ln

Na

ni

=Vth lnNaNd

ni> 0

c. PN Junction Capacitor

-PN Junction under Bias

xn (VD ) =2 s bi VD( )

qNd

Na

Na + Nd

= xn0 1

VD

bi

xp (VD ) =2 s bi VD( )

qNa

Nd

Na + Nd

= xp0 1

VD

bi

C j =C j 0

1VD

bi

=s

Xd 0

C j0 =q s

2 bi

NaNd

Na + Nd

=s

Xd

C j (VD ) =s

Xd (VD )

-EXAMPLE2: Consider the following silicon P-N junction:

(a) If Na = 4E16cm-3 in the P-region and Nd = 1E17cm-3 in the N-region, under increasing reverse bias, which region (N or P) will become completely depleted first? (b) What is the reverse bias at this condition? (c) What is the small-signal capacitance (F/cm2) at the bias condition? SOLUTION2:

(a) Under reverse bias, the depletion region will expand. The side which has fewer dopants will fully deplete first.

Na*Wp=4E16*1.2>Nd*Wn=1E17*0.4 So N side will deplete first

(b) bi =Vth lnNaNd

ni= 0.81V

xn0 =2 s bi

qNd

Na

Na + Nd

= 0.055um

0.4 = xn (VD ) = xn0 1VD

bi

So VD=-42V

(c) The small signal capacitance is determined by the depletion region width. Xn=Wn=0.4um, Xp=(Nd/Na)*Xn=(1E17/4E16)*0.4=1um

Xd=Xn+Xp=0.4+1=1.4um Cd= s/Xd=8.85E-14*12/1.4E-4=7.6E-9F/cm2

5. PN Junction Diode

-Thermal Equilibrium:

small diffusion current: high barrier

small drift current: few minority carriers and wide depletion region

net current: none

-Reverse Bias:

drift current: remains small

diffusion current: reduced exponentially with bias

net current: small reverse current

-Forward Bias:

drift current: remains small

diffusion current: increase exponentially with bias

net current: large forward current

-IV curve:

-Minority Carrier Concentration

with recombination, long base diode

without recombination or short base diode:

-Diode Current Densities

Jdiff = Jndiff

+ Jpdiff

= qni2 Dp

NdWn

+Dn

NaWp

eqVAkT 1

-Small Signal Model

ID + iD = IS eq Vd +vd( )

kT 1

ISe

qVdkT e

qvdkT

iDqvdkT

= gdvd

-Charge Storage

Cd =1

2

qIdkT

-EXAMPLE3 Consider an ideal, long-base, silicon PN junction diode with uniform cross section and constant doping on either side of the junction. P side is heavily doped and N side has an doping concentration of 1E16cm-3 .

For the n-side of the junction: (a) Calculate the density of the minority carriers as a function of x (distance from the junction) when the applied voltage is 0.589V (which is 23kT/q). (b) Find the minority carrier currents as functions of x (distance from the junction)

SOLUTION3

(a) ( ) 3/14//2

101)( == cmeeeN

nxp ppa

LxLxkTqV

d

i .

(b) Minority current:

( ) 2///2

/5.01)(

)( cmAeeeDLN

nq

dx

xdpqDxJ ppa

LxLxkTqV

p

pd

ipp === .