Motivation - Konkukhome.konkuk.ac.kr/~jintkim/ckt1_17/data/lecturenote_chap2.pdf · - IC plays a central role in modern electronic devices, and ... - To understand how integrated

Fundamentals of Microelectronics Jintae Kim

1

Basic Semiconductor Physics Motivation

ttv

. Intel Itanium Processor die photograph (21.5mm x 32.5mm)

- Modern electronic devices are built upon “Integrated Circuit (IC)” - IC is a small piece of “semiconductor” where billions of “transistors” are connected to perform sophisticated digital computation and analog signal processing - IC plays a central role in modern electronic devices, and both analog and

digital circuits are absolutely necessary to realize modern high-performance electronic devices.

- To understand how integrated circuit works, we will start learning underlying physics behind IC technology What is the “Semiconductor”?


2

Charge Carriers in Semiconductor - Let’s start with a simple theory of current conduction

I = C∙dV/dt = dQ/dt current is a “flow of charge”

- First Question to answer: what is the “semiconductor” and who carries the charge in a semiconductor?

- Semiconductor is a solid-state material in a periodic table that belongs to group III ~ group V with 3~5 valence electrons - The electrical conductivity of semiconductor falls between that of a conductor

and an insulator. But we can somehow “control” the conductivity of semiconductor by mixing different semiconductors.

- Group IV semiconductor is called “intrinsic semiconductor”: e.g., Si and Ge

- Si has 4 valence electrons a crystal formed by covalent bond by sharing 4 valence electrons with its neighbors


3

- As temperature T ↑, thermal energy dislodges e- in covalent-bond free electron the free electrons can move around with negative charge! electron is a charge carrier with negative charge

- Question: what happens to the empty state in the covalent bond where the electron has left? “holes” are generated

- Hole is an empty state of covalent bond, and has positive charge (remember:

Si is electrically neutral, so lack of electron means positive charge)

- Holes can be filled by recombining with nearby free electrons In other words, holes can move around within a silicon with positive charge hole is a charge carrier with positive charge

- Electron and hole are created at the same time in semiconductor (called

electron-hole pair) # of electrons = # of holes in intrinsic semiconductor

- Electron and holes can also be recombined anytime and both disappear

Carrier Density in Intrinsic Semiconductor - Question: how many e/h pairs are generated in silicon? - We use the unit “carrier density” # of carrier per cubic centimeter cm3

- The carrier density in pure silicon is given as follows

3150

3100

32/315

/1054.1)600(

/1008.1)300(

/2

exp102.5

cmelectronsKTn

cmelectronsKTn

cmelectronskT

ETn

i

i

gi


4

- Notation: n electron, p hole: subscript i means “intrinsic”, Eg is “bandgap energy”

- Note that electron density is a function Eg and T

- Eg is the minimum energy to break an electron off from its covalent bond (Eg of Si = 1.12eV whereas Eg of insulator 5~6eV) Ex) T = 300k ,ni = 1.08*1010 electrons/cm3

- Is this large enough? # of atoms in Si crystal = 5*1022

- One out of 1012 gives out an electron to increase the conductivity, we need “doping”

Carrier Density in Extrinsic Semiconductor - Doping = Adding “impurities (or dopants)” to Si Extrinsic Semiconductor

- N-type : Use group V element P as dopant has 5 electrons in valence band

- Every dopant donates 1 excess electron (hence called “donor”)

- We use donor density =ND >> ni free electron density n≈ND ( n>> ni)

- What happens to the holes? With a lot more electrons nearby, hole has higher chance of recombining with electrons hole density p↓


5

- It can be proved that

n∙p=ni2

(called mass-action law. no proof given here)

- P-type : Use group III element (B) as dopant has 3 electrons in valence band

- Every dopant accepts 1 electron from nearby Si, creating 1 hole (called “acceptor”)

- We use acceptor density =NA >> ni hole density p≈NA ( p>> ni)

- n∙p=ni2 still holds

- Summary of charge density in doped silicon

Majority Carrier

Minority Carrier

Electron density (n)

Hole Density (p)

comment

N-type electron hole ND ni2/ND n∙p=ni

2 holds

P-type hole electron ni2/NA NA


6

Charge Transportation and Current Density

Q) How does the charged carrier move in semiconductor? By “Drift” and “Diffusion”

1) “Drift” - Drift: charged carrier moves when E-field is applied

∙ E-field is created when voltage is applied to a semiconductor device

∙ Velocity of carrier is proportional to mobility and the strength of electric field

∙ n =1350 cm2/V·s p = 480 cm2/V·s Electron is ~3x faster than hole

- Calculation of drift current by electron

The volume = W·h·v = A·v

Number of electrons = A·v·n

Assumption:

1) There are n electrons in 1 cm3 in this device

2) The electrons moves at the speed of v cm/s

v

After 1 second

v

∙ I = dQ/dt = the amount of charge that passes through a surface A in 1 second. Assume 1) The electron density is n/cm3

2) the electron moves with velocity v [cm/s]

Ev

Ev

ne

ph


7

∙ Each electron carries the net charge of -q [columbs] (q=1.6x10-19)

∙ The total amount of charge that passes through area A=W·h in 1 second is n·q·W·h·v

∙ E-Field direction : Left Right Electron movement: Right Left Direction of current : Left Right

∴ Idrift,n = −q ⋅ n ⋅ �� ⋅ w ⋅ h

∙ In terms of current density (current per unit area)

∴ 𝐽𝑑𝑟𝑖𝑓𝑡,𝑛 =

I

A= −q ⋅ n ⋅ �� = +q ⋅ n ⋅ μn��

- The drift current by hole can similarly derived as

∴ 𝐽𝑑𝑟𝑖𝑓𝑡,𝑝 = +q ⋅ p ⋅ μp��

- There always are holes and electrons in semiconductor The total current density is

𝐽𝑑𝑟𝑖𝑓𝑡,𝑡𝑜𝑡 = 𝐽𝑑𝑟𝑖𝑓𝑡,𝑛 + 𝐽𝑑𝑟𝑖𝑓𝑡,𝑝

= +q ⋅ (n ⋅ μn�� + p ⋅ μp��)

2) “Diffusion”: movement of charged carriers from a region of high density to low density by random motion

n(x)

Electron

Density

Ex) Electron Diffusion

x

dx

dn<0

electron

Current I < 0


8

- Diffusion Current Calculation

∙ The current depends on the gradient (=slope) of the charge density

∙ The diffusion current due to electron density gradient is

Idiff,n = +Dn ⋅ A ⋅dn(x)

𝑑𝑥⋅ q

∙ Similarly, the diffusion current due to hole density gradient is

Idiff,p = −Dp ⋅ A ⋅dp(x)

𝑑𝑥⋅ q

Note: Dn = 34 𝑐𝑚2/s, Dp = 12 𝑐𝑚2/s

∙ Total diffusion current density is given as

Jdiff,tot = Jdiff,n + Jdiff,p = q ⋅ (Dn ⋅dn

𝑑𝑥− Dp ⋅

dp

𝑑𝑥)

Ex) find diffusion current density at x

n(x)

Electron

Density

x

n(x)=N·exp(-x/L)

External

Injection

L

Jn(x)=

Note: Diffusion constant and mobility are related “Einstein Relation”

Dn

𝜇𝑛=

Dp

𝜇𝑝=

𝑘𝑇

𝑞

k = 1.38 × 10−23 [Joule

K] , q = 1.602 × 10−19 [columb],

𝑘𝑇

𝑞= 25.86𝑚𝑉 𝑎𝑡 𝑇 = 300 K


9

P-N Junction

- What happens if P and N type semiconductors are joined together “PN-

Junction”

- Depletion Region

Step 1)

P-Type N-Type

h+

e-

h+

: hole: electron

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+ e

-

e-

e-

e-

e-

e-

e-

e-

e- e

-

e-

e-

e-

e- e

-

e-

e-

e-

h+

h+

h+

e-

e-

e-

e-

e-

at t= 0P/N interface

Electron density

Hole density

np = ni2/NA nn = ND

pp = NA pn =ni2/ND

∙ Near the P/N interface, there exists large carrier gradient “diffusion” process starts Electrons in N-type moves toward P-type Holes in P-type moves toward N-type

Step 2)

P-Type N-Type

h+

e-

h+

: hole: electron

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

e-

e-

e-

e-

e-

e- e

-

e-

e-

e- e

-

e-

e-

e-h

+ e-

e-

e-

e-

e-

at t= t1

e-

e-

e-

Positive donor ion

Negative acceptor ion

∙ Electrons that moved from N to P leave fixed positive donor ions

∙ Holes that moved from P to N leave fixed negative acceptor ions


10

Step 3)

P-Type N-Type

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

e-

e-

e-

e-

e-

e- e

-

e-

e-

e- e

-

e-

e-

e- e

-

e-

e-

e-

e-

at t= t2

e-

e-

e-

Electric

Field

Depletion

Region

Total

Current

I=0

∙ The fixed ions create non-zero net charge near the interface called “Depletion Region” Electric field builds up in depletion region electrons and holes are affected by this E-filed due to “drift”

∙ The direction of the carrier movement by drift and diffusion are opposite

Diffusion by Carrier Gradient Drift by Electric Field Electron Right Left Left Right

Hole Left Right Right Left

∙ This P/N junction carriers no current because the terminals are open In other words the net carrier movement must be zero At equilibrium, the E-field over the depletion region opposes the movement of carrier by diffusion such that |I,diff| = |I,drift|, or

𝐽𝑡𝑜𝑡𝑎𝑙 = 𝐽𝑑𝑖𝑓𝑓 + 𝐽𝑑𝑟𝑖𝑓𝑡 = 0 (holds for both electron and hole)

- The concept of “Built-In Potential” and “Potential Barrier”

∙ We know that the electric filed is the negative gradient of the potential

�� = −𝑑𝑉

𝑑𝑥

This means that for non-zero E-field, there must be potential difference between two locations. We define the potential difference across entire depletion region as “potential-barrier”. Let’s find out how large this built-in potential is.


11

x

P-Type N-Type

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

e-

e-

e-

e-

e-

e- e

-

e-

e-

e- e

-

e-

e-

e- e

-

e-

e-

e-

e-

e-

e-

e-

Electric Field

Depletion

Region

Total

Current

I=0

x2x1

V(x)Electron density

Hole density

np = ni2/NA nn = ND

pp = NA pn =ni2/ND

Using 𝐽𝑡𝑜𝑡𝑎𝑙 = 𝐽𝑑𝑖𝑓𝑓 + 𝐽𝑑𝑟𝑖𝑓𝑡 = 0 for holes

−Dp ⋅dp(x)

𝑑𝑥⋅ q + q ⋅ p ⋅ μp�� = 0

Using �� = −𝑑𝑉

𝑑𝑥

Dp ⋅dp(x)

𝑑𝑥⋅= −p ⋅ μp

𝑑𝑉

𝑑𝑥

Dp ⋅1

p⋅ dp =⋅ −μp𝑑𝑉

Using the boundary condition

at x = x1, pp = NA and let′s assume V(x1) = V1

at x = x2, pn = ni2/ND and let′s assume V(x2) = V2

By performing the integration from x1 to x2,

Dp ⋅ ∫1

p⋅ dp

𝑥2

𝑥1

=⋅ −μp ∫ d𝑉2

𝑉1

𝑉

Dp ⋅ [ln(𝑝𝑛) − ln(pp)] =⋅ −μp(𝑉2 − 𝑉1)

∴ 𝑉2 − 𝑉1 = 𝑉𝑏𝑖 =𝐷𝑝

𝜇𝑝⋅ ln (

pp

pn) =

𝑘𝑇

𝑞⋅ ln (

NA ⋅ ND

𝑛𝑖2 ) = 𝑉𝑇 ⋅ ln (

NA ⋅ ND

𝑛𝑖2 )


12

EX) T=300K, NA=2x1016, ND=4x1014 When equilibrium, barrier height = built-in potential Vbi

∙ Note: barrier height is a function of doping density NA and ND

𝑉𝑏𝑖 = 𝑉𝑇 ⋅ ln (pp

pn)

→ pn = pp ⋅ exp (−Vbi

𝑉𝑇)

Can show that for electron,

→ np = nn ⋅ exp (−Vbi

𝑉𝑇)

In other words, minority carrier density in one side can be calculated from built-in potential and majority carrier density in other side. (We will use these equations later when we find the current in the forward-biased PN junction)


13

PN Junction with forward bias (non-equilibrium)

P-Type N-Type

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

e-

e-

e-

e-

e-

e- e

-

e-

e-

e- e

-

e-

e-

e-

e-

e-

e-

e-

e-

e-

Depletion

Region

nn = NDpp = NA

h+

h+

h+

Itot > 0

External Voltage Source

Electric Field by Built-in Potential

Electric Field when Vext>0

Vext

∙ By adding external voltage source, the loop is closed and non-zero Itot can flow

in the circuit

∙ Assume we apply forward-bias (Vext>0)

E-field by Vext reduces the net E-field strength in depletion region

The potential barrier across the depletion region is lowered

|Idiffusion| > |Idrift|

Net current flows from P to N!

∙ Calculation of the total current in forward-biased PN junction

- With external VF > 0, the potential barrier height = 𝑉𝑏𝑖 − 𝑉𝐹

𝐼𝑡𝑜𝑡 ∝ number of extra charge carriers that diffuse over the potential barrier

- With VF>0, hole density at the N-side is given by


14

= pp exp(−𝑉0−𝑉𝐹

𝑉𝑇) − pp exp(−

𝑉0

𝑉𝑇)

= pp exp(−𝑉0

𝑉𝑇) [exp (

𝑉𝐹

𝑉𝑇) − 1]

- When considering both hole and electron, the total current can be expressed as

Itot = 𝐼𝑆 [exp (𝑉𝐹

𝑉𝑇) − 1]

∙ PN-junction in forward-biased region is a “nonlinear” device! (Compare it with linear device such as resistor)


15

PN Junction with reverse bias

P-Type N-Type

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

h+

e-

e-

e-

e- e

-

e-

e-e

-e

-

e-

e-

e-

e-

e-

e-

e-

Depletion region

: very few mobile

charge carriers

nn = NDpp = NA

h+

h+

h+

External Voltage Source

Electric Field by Built-in Potential

Electric Field when VR>0

VR

Itot

Depletion Region Width

∙ With reverse biased VR>0, electric field pushes even more number of holes and electrons away from the PN junction interface Increases the depletion region width

- Note that the drift current due to VR is negligible because there are very few mobile charge carriers in the depletion region (not many holes end electrons in depletion region), so the current in reverse-bias is negligible

∙ Reverse biased PN Junction carrier almost no current, and can be used as a “capacitor” (Also called Junction capacitor)

VRN

P

VR

0

0

0

0

1

2

1

VNN

NNqC

V

V

CC

DA

DAsij

R

j

j

Documents

Motivation - Konkukhome.konkuk.ac.kr/~jintkim/ckt1_17/data/lecturenote_chap2.pdf · - IC plays a central role in modern electronic devices, and ... - To understand how integrated