EE130/230A Discussion 7

Peng Zheng

Minority-Carrier Charge Storage• Under forward bias (VA > 0), excess minority carriers are

stored in the quasi-neutral regions of a pn junction:

Pnn

x nP

LxpqA

dxxpqAQn

)(

)(

Npp

x pN

LxnqA

dxxnqAQp

)(

)(

EE130/230A Fall 2013 Lecture 13, Slide 2

Charge Control Model Summary• Under forward bias, minority-carrier charge is stored in the

quasi-neutral regions of a pn diode.

– Long base:

– Narrow base:

PkTqV

D

iP Le

N

nqAQ A 1/

2

NkTqV

A

iN Le

N

nqAQ A 1/

2

PkTqV

A

iN We

N

nqAQ A 1

2

1 /2

NkTqV

D

iP We

N

nqAQ A 1

2

1 /2

Lecture 13, Slide 3EE130/230A Fall 2013

• The steady-state diode current can be viewed as the charge supply required to compensate for charge loss via recombination (for long base) or collection at the contacts (for narrow base).

– Long base (both sides):

– Narrow base (both sides):

where and

p

P

n

N QQI

ττ

ptr

P

ntr

N QQI

,, ττ

P

P

p

P

N

N

n

N

L

DL

L

DL

τ and

τNote that

N

Pntr D

W

2τ

2

,

P

Nptr D

W

2τ

2

,

Lecture 13, Slide 4EE130/230A Fall 2013

Sample Problem-Charge Control Model

Sample Problem-Charge Control ModelSince the minority carrier concentrations (np and pn) are enhanced within the quasi-neutral regions, the diode is forward biased. The majority carrier concentrations (pp and nn) are not significantly enhanced, however, so low-level injection conditions prevail. a) Since low-level injection conditions prevail, the “Law of the Junction” holds: within the depletion region and at

the edges of the depletion region, np=ni2 exp{qVA/kT}.

np and pn each are enhanced by a factor 1010 at the edges of the depletion region, so 1010 = exp{qVA/kT} VA = (kT/q) ln(1010) = 10 × (kT/q) ln(10) = 10 × (60 mV) = 0.6 V.

b) pp = NA = 1016 cm-3 and nn = ND = 1018 cm-3 c) np(-xp) = np(-xp) – np0(-xp) = 1014 – 104 1014 cm-3. pn(xn) = pn(xn) – pn0(xn) = 1012 – 102 1012 cm-3

The majority carrier concentrations (pp and nn) are not significantly enhanced within the quasi-neutral regions, so low-level injection conditions prevail.

d) From Lecture 4, Slide 17 the electron mobility for NA =1016 cm-3 is n =1200 cm2/Vs and the hole mobility for ND =1018 cm-3 is p =150 cm2/Vs. The electron diffusion constant Dn= n (kT/q)=1200×0.026=31.2 cm2/s. The hole diffusion constant, Dp= p (kT/q)=150×0.026=3.9 cm2/s.

The electron minority carrier diffusion length

Small-Signal Model Summary

qkT

IG DC

/

qkT

IC

/

τ DCD

W

AC s

J

)1( /0

A kTqVDC eII

DJ CCC

Depletion capacitance

Diffusion capacitance

Conductance

Lecture 13, Slide 7EE130/230A Fall 2013R. F. Pierret, Semiconductor Device Fundamentals, p. 302

Sample Problem-Small Signal ModelFrom the previous problem, IDC= 8.9×10-8 A. The small-signal resistance R = 1/G =(kT/q)/ IDC=0.026/8.9×10-8

=2.9×105 . Since the n-type side is degenerately doped (ND = 1018 cm-3), we should find the reduction in band gap energy on the n side:

8 1 3 3003.5 10GE N

T =35meV

The built-in potential is then

ln( )2

G G Abi

i

E E kT NV

q q n

= (1.12-0.035)/2 + 0.026×ln(1016/1010)=0.902V

The depletion width 12

19 16

2 ( ) 2 10 (0.902 0.6)0.194

1.6 10 10s bi A

A

V VW m

qN

Depletion capacitance 12 8

145

10 100 105.15 10

1.94 10s

j

AC F

W

Diffusion capacitance

In your HW, it is not degenerately doped.

Transient Response of pn Diode

Lecture 13, Slide 9

• Suppose a pn-diode is forward biased, then suddenly turned off at time t = 0. Because of CD, the voltage across the pn junction depletion region cannot be changed instantaneously.

The time delay in switching between

the FORWARD-bias and REVERSE-bias states is due to the time required to change the amount of excess minority carriers stored in the quasi-neutral regions.

EE130/230A Fall 2013R. F. Pierret, Semiconductor Device Fundamentals, Fig. 8.2

Decay of Stored ChargeConsider a p+n diode (Qp >> Qn):

t

i(t)

t

vA(t)

ts

ts0 pxx

n

qAD

i

dx

dp

n

For t > 0:

pn(x)

Lecture 13, Slide 10EE130/230A Fall 2013

R. F. Pierret, Semiconductor Device Fundamentals, Fig. 8.3

Qualitative Examples

t

i(t)

ts

Increase IF

t

i(t)

ts

Increase IR

t

i(t)

ts

Decrease tp

EE130/230A Fall 2013 Lecture 13, Slide 21

Illustrate how the turn-off transient response would change: