Lecture 10: PN Junction & MOS Capacitorsinst.cs.berkeley.edu/~ee105/fa03/handouts/lectures/Lecture10.pdf · EECS 105Fall 2003, Lecture 10 Prof. A. Niknejad PN Junction in Thermal

Department of EECS University of California, Berkeley

EECS 105 Fall 2003, Lecture 10

Lecture 10: PN Junction & MOS Capacitors

Prof. Niknejad


EECS 105 Fall 2003, Lecture 10 Prof. A. Niknejad

Lecture Outline

� Review: PN Junctions Thermal Equilibrium

� PN Junctions with Reverse Bias (3.3-3.6)

� MOS Capacitors (3.7-3.9):– Accumulation, Depletion, Inversion

– Threshold Voltage

– CV Curve



Results of MT #1

� Good Job!

� This is only 17% of your grade

13STD DEV

99MAX

34MIN

74AVG

Homework 15%Laboratory 20%Midterm #1 17%Midterm #2 18%Final 30%



PN Junction in Thermal Equilibrium

� Contact potential develops between P and N region

� Diffusion current balanced by drift current

� Depletion region is a “space-charge” region where the concentration of free carriers is low

� The depletion region is charged due to the immobile background ions (donors and acceptors)

� Used the “Depletion Approximation” to estimate the charge density � calculate the electric fields and potential variation using electrostatics in 1D



Have we invented a battery?

� Can we harness the PN junction and turn it into a battery?

� Numerical example:

2lnlnln

i

ADth

i

A

i

Dthpnbi n

NNV

n

N

n

NV =

+=−≡ φφφ

mV60010

1010logmV60lnmV26

20

1515

2=×==

i

ADbi n

NNφ

?



Contact Potential

� The contact between a PN junction creates a potential difference

� Likewise, the contact between two dissimilar metals creates a potential difference (proportional to the difference between the work functions)

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

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

mnpmbi φφφ ++=0

pnmnφ

pmφ

+

−biφ )( mnpmbi φφφ +−=



PN Junction Capacitor

� Under thermal equilibrium, the PN junction does not draw any current

� But notice that a PN junction stores charge in the space charge region (transition region)

� Since the device is storing charge, it’s acting like a capacitor

� Positive charge is stored in the n-region, and negative charge is in the p-region:

nodpoa xqNxqN =



Reverse Biased PN Junction

� What happens if we “reverse-bias” the PN junction?

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

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

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

+

−Dbi V+−φ

DV 0<DV



Current Under Reverse Bias

� Under thermal equilibrium current is zero � If we apply a reverse bias, we are increasing the barrier

against diffusion current� Drift current is low since the field only moves minority

carriers across junction� In fact, current is not zero but very small since the minority

carrier concentration is low. Minority carriers within one diffusion length of junction can contribute to a reverse bias current. This is more or less independent of the applied bias

pn

+− DV−pφ

nφ Dn V+φ

pφ0dX )( Dd VX

0E 0E



Voltage Dependence of Depletion Width

� 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:

+−=+=

da

DbisDnDpDd NNq

VVxVxVX

11)(2)()()(

φε

bi

Dn

da

a

d

DbisDn

Vx

NN

N

qN

VVx

φφε −=

+−= 1

)(2)( 0

bi

Dp

da

d

a

DbisDp

Vx

NN

N

qN

VVx

φφε −=

+−= 1

)(2)( 0

bi

DdDd

VXVX

φ−= 1)( 0



Charge Versus Bias

� As we increase the reverse bias, the depletion region grows to accommodate more charge

� Charge is not a linear function of voltage

� This is a non-linear capacitor

� We can define a small signal capacitance for small signals by breaking up the charge into two terms

bi

DaDpaDJ

VqNVxqNVQ

φ−−=−= 1)()(

)()()( DDJDDJ vqVQvVQ +=+



Derivation of Small Signal Capacitance

� From last lecture we found

� Notice that

�++=+ DV

DDJDDJ v

dV

dQVQvVQ

D

)()(

RD VVbipa

VV

jDjj

VxqN

dV

d

dV

dQVCC

==

−−===

φ1)( 0

bi

D

j

bi

Dbi

paj

V

C

V

xqNC

φφφ −

=−

=112

00

da

da

bi

s

da

d

a

bis

bi

a

bi

paj NN

NNq

NN

N

qN

qNxqNC

+=

+

==

φεφε

φφ 2

2

220

0



Physical Interpretation of Depletion Cap

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

� This looks like a parallel plate capacitor!

da

da

bi

sj NN

NNqC

+=

φε

20

0

1

0

11

2 d

s

dabissj XNN

qC

εφε

ε =

+=

−

)()(

Dd

sDj VX

VCε=



A Variable Capacitor (Varactor)

� Capacitance varies versus bias:

� Application: Radio Tuner

0j

j

C

C



P-type Si Substrate

N-type Diffusion RegionOxide

“Diffusion” Resistor

� Resistor is capacitively isolation from substrate – Must Reverse Bias PN Junction!

– PN Junction creates a distributed capacitance with substrate (RC transmission line)



MOS Capacitor

� MOS = Metal Oxide Silicon� Sandwich of conductors separated by an insulator � “Metal” is more commonly a heavily doped polysilicon

layer n+ or p+ layer� NMOS � p-type substrate, PMOS � n-type substrate

Oxide (SiO2)

Body (p-type substrate)

Gate (n+ poly)

011.7sε ε=

03.9oxε ε=

Very Thin!

~ 1nmoxt

x

0



P-I-N Junction

� Under thermal equilibrium, the n-type poly gate is at a higher potential than the p-type substrate

� No current can flow because of the insulator but this potential difference is accompanied with an electric field

� Fields terminate on charge!

ln ap

i

NkT

q nφ = − 550mV

nφ + ≈


Gate (n+ poly)



Fields and Charge at Equilibrium

� At equilibrium there is an electric field from the gate to the body. The charges on the gate are positive. The negative charges in the body come from a depletion region


++++++++++++++++++− − − − − − − − −− − − − − − − −

− − − − − − − − − 0dX

+

−oxV +

−BV oxE



Good Place to Sleep: Flat Band

� If we apply a bias, we can compensate for this built-in potential

� In this case the charge on the gate goes to zero and the depletion region disappears

� In solid-state physics lingo, the energy bands are “flat” under this condition

( )FB pnV φ φ+= − −

( ) 0G GB FBQ V V= =


+−0FBV <



Accumulation

� If we further decrease the potential beyond the “flat-band” condition, we essentially have a parallel plate capacitor

� Plenty of holes and electrons are available to charge up the plates

� Negative bias attracts holes under gate

( )G ox GB FBQ C V V= −


−+GB FBV V<++++++++++++++++++−−−−−−−−−−−−−−−−−−

B GQ Q=



Depletion

� Similar to equilibrium, the potential in the gate is higher than the body

� Body charge is made up of the depletion region ions

� Potential drop across the body and depletion region


+−GB FBV V> + + + + + + + + + +

( )B a d GBQ qN X V= −− − − − − − − − −− − − − − − − −

( )G GB BQ V Q= −




+ + + + + + + + + +

Inversion

� As we further increase the gate voltage, eventually the surface potential increases to a point where the electron density at the surface equals the background ion density

� At this point, the depletion region stops growing and the extra charge is provided by the inversion charge at surface

+−GB TV V=− − − − − − − − −− − − − − − − −

sq

kTs i an n e N

φ

= = s pφ φ= −

sφ



Threshold Voltage

� The threshold voltage is defined as the gate-body voltage that causes the surface to change from p-type to n-type

� For this condition, the surface potential has to equal the negative of the p-type potential

� We’ll derive that this voltage is equal to:

12 2 ( 2 )Tn FB p s a p

ox

V V q NC

φ ε φ= − + −



Inversion Stops Depletion

� A simple approximation is to assume that once inversion happens, the depletion region stops growing

� This is a good assumption since the inversion charge is an exponential function of the surface potential

� Under this condition:

,max( )G Tn BQ V Q≈ −

,max( ) ( )G GB ox GB Tn BQ V C V V Q= − −



Q-V Curve for MOS Capacitor

� In accumulation, the charge is simply proportional to the applies gate-body bias

� In inversion, the same is true

� In depletion, the charge grows slower since the voltage is applied over a depletion region

GQ

( )GBV VTnVFBV

inver

sion

accu

mula

tion

depletion

,maxBQ−

( )N GBQ V−



Numerical Example

� MOS Capacitor with p-type substrate:

� Calculate flat-band:

� Calculate threshold voltage:

20nmoxt = 16 35 10 cmaN −= ×

( ) (550 ( 402)) 0.95VFB pnV φ φ+= − − = − − − = −

12 2 ( 2 )Tn FB p s a p

ox

V V q NC

φ ε φ= − + −

13

-6

3.45 10 F/cm

2 10 cmox

oxox

Ct

ε −×= =×

19 12 162 1.6 10 1.04 10 5 10 2 0.4.95 2( 0.4) 0.52VTn

ox

VC

− −× × × × × × × ×= − − − + =



Num Example: Electric Field in Oxide

� Apply a gate-to-body voltage:

� Device is in accumulation

� The entire voltage drop is across the oxide:

� The charge in the substrate (body) consist of holes:

2.5GB FBV V= − <

56

2.5 0.55 ( 0.4) V8 10

2 10 cmGB pox n

oxox ox

VVE

t t

φ φ+

−

+ − − + − −= = = = − ××

7 2( ) 2.67 10 C/cmB ox GB FBQ C V V −= − − = ×



Numerical Example: Depletion Region

� In inversion, what’s the depletion region width and charge?

,max 2 0.8VB s p p p pV φ φ φ φ φ= − = − − = − =

2,max ,max

1

2a

B ds

qNV X

ε

=

,max,max

2144nms B

da

VX

qN

ε= =

7 2,max ,max 1.15 10 C/cmB a dQ qN X −= − = − ×

Documents

Lecture 10: PN Junction & MOS Capacitorsinst.cs.berkeley.edu/~ee105/fa03/handouts/lectures/Lecture10.pdf · EECS 105Fall 2003, Lecture 10 Prof. A. Niknejad PN Junction in Thermal