ece606: solid state devices lecture 16 p-n diode ac...

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

ECE606: Solid State DevicesLecture 16

p-n diode AC Response

Gerhard [email protected]


Topic Map

2

Equilibrium DC Small signal

Large Signal

Circuits

Diode

Schottky

BJT/HBT

MOSFET

Diode in Non-Equilibrium(External DC+AC voltage applied)


Why should we study AC Response?

3

Series Resistance

Conductance

Diffusion Capacitance

Junction Capacitance

www.sci-toy.com

Motivation

Radio


Outline

4

1) Conductance and series resistance

2) Majority carrier junction capacitance

3) Minority carrier diffusion capacitance

4) Conclusion

Ref. SDF, Chapter 7


Forward Bias Conductance

5

( )( )/ 1A Sq V R I moI I e −= −β

0

1

( )FBS

m

q IgR

Iβ+

+=

0

ln ( )oA S

I Iq V R I

I m

β+ = −

( )A

So

m dVR

q I I dI= −

+β

RS

G

CJ

V

2

45

1

3

6,7

ln(I)

Cdiff

m = RG (2), diff (1), Ambipolar (2)

Forward Bias Conductance


Reverse Bias Conductance

6

( )( )/

0

1A Sq V R I moI I e

I

−= −

≈ −

β

02i

bi A

qnB V V

τ− −

01

2i

RB bi A

qn B

g V Vτ=

−

02i

bi A

qnB V V

τ− −

RS

G

CJ

Cdiff

V

2

45

1

3

6,7

ln(I)

Reverse Bias Conductance


Outline

7




4) Conclusion



8

VDC

Series Resistance

Conductance



Depletion width modulation

Charge modulation

Majority carrier effect

Forward biased diode + AC signal

Fn

Fp

VA> 0

VA< 0


Majority Carrier Junction Capacitance

9

0 0

0 02 2( )

s sJ

n p s sbi A

D A

K A K AC

W W K KV V

qN qN

= =+

+ −

ε εε ε

Cj

Va

Measure

0

x

∆ρ

x

∆ρ

VA < 0

VA > 0

Series Resistance

Conductance



Majority Carrier Junction Capacitance


Measurement of Built-in Potential

10

2 20

1 2( )

( ) AJ D s

biV VC qN x K Aε

≈ −

plot CJ-2

VA

measureCJ

VA

(Assume single sided p+-n junction)

Vbi

Derive Vbi from CJmeasurements


.. And Variable Doping

11

2 20

2

( )(

1)bi

sDA

J

V VNq KC Ax ε

≈ −

plot CJ-2

VA

measureCJ

VA

Charge

VA<0VA=0

VA>0

220

2 1

(1 )( )

s AJD qK A d C

NV

xdε

=

Measure doping concentration as a function of position


Dielectric Relaxation Time (majority side)

12

1 nn n

dn dJR G

dt q dx= − +n N NJ qn E qD n= + ∇µ

VDC

( ) ( )1 ND N

d n d qn dN

dt q dx dx

µµ

∆= =

E E

( )00

D AS

d qp n n N N

dx k ε= − − ∆ + −E

( )0 0

0D

S S

NqNd n nn

dt k kεµ

εσ∆ ∆= − ∆ = −

0

00 0( ) S d

t t

kn t n e n eε τσ− −

∆ = =

0 0.1 pssd

K ετ = ≈σVery fast

How long does it take for the signal to cross the junction?

Majority side Neglect

N-side


Outline

13




4) Conclusion


Diffusion Capacitance for Minority Carriers

14

1 nN N

n dJr g

t q dx

∂ = − +∂

N N N

dnqn qD

dxµ= +J E

( ) ( )20 0

2

j t j t j tdc ac dc ac dc ac

N

n

n n n e d n n n e n n eD

t dx

∂ + ∆ + ∆ + ∆ + ∆ ∆ + ∆= −∂

ω ω ω

τ

VDC

np

x

Minority Carrier side

DC AC

ACAC


Diffusion Capacitance for Minority Carriers

15

( ) ( )20 0

2

dc dj t j t

acc dcN

j tac

n

acn e nn d nD

t d

n n

x

e n en ωω ω

τ∂ + + + + += −

∂∆∆ ∆∆∆ ∆

2 2

2 2ac dc ac

ac Ndcj t j t j

n

t

n

d dn n n nj n D

d dxe

xe eω ω ωω

τ τ∆ ∆ ∆ ∆∆ = + − −

2

2C 0D : n n

x x

dc dcN

n

L Ldcn Ae B

d n nD

dxe

τ− +∆ ∆

⇒ ∆ +− ==

( )2

2* * *AC : 0 1 n n n

x x x

L L Lac acN an

nc

d n nD j

dn Ce De C

xeωτ

τ− + −

⇒∆ ∆ ∆ = +− + →=

( ) ( )* */ 1 / 1n n n n n n nL D j jτ ωτ τ τ ωτ= + = +


AC Boundary Conditions

16

( )

2

2

( 0) 1

1

j tac

c

d

d

c V e

qVi kT

dcA

qkTj t

ac j tA ac

Vi

dc n

nn x e

N

eenp e

n

N

ω

ωω

+

∆ = = −

∆ + =

+ ∆ −∆

( )2

1dc

j tacqV q

i kT kTdc a

V e

cA

j t nn n ee e

N

ω

ω

∆ + ∆ ≈ −

2

( 0)dcqV

ac kTac

i

A

qVn x e C

kT

n

N∆ = = =

2

1 1dcqV

ackit

T

A

jqVe

n e

N kT

ω ≈ + −

xn

n

Taylor expansion


AC Current and Impedance

17

2

*0

dcqVac n i kT

ac nx n A

acVd n qD q nJ qD e

dx L kT N=

∆= − =

2

( 0)dcqV

ac i kTac

A

qV nn x e

kT NC∆ = = =

2 2

0*1

dcq

nn

Vac n i kT

acac A

J q D nY e G

V kTj

NLωτ+= = ≡

* * *( ) n n n

x x x

L L Lacn x e De eC C

− + −∆ = + →Finally…

AC Impedance

AC Current


Diffusion Conductance and Capacitance

18

1/ 22 20

1/ 22 20

1 12

1 12

D n

D n

GG

GC

τ

ω ω

ω

τ

= + +

= + −

0 1ac D D nY G j C G j= + ≡ +ω ωτ

Separate in real & imaginary parts …

DG ω∝

1/DC ω∝Product of GD and CDfrequency-independent


Small Signal pn-diodeConclusion

19

1) Small signal response relevant for many analog

applications.

2) Small signal parameters always refer to the DC operating

conditions, as such the parameter changes with bias

condition.

3) Important to distinguish between majority and minority

carrier capacitance. Their relative importance depends on

specific applications.


ECE606: Solid State Devicesp-n diode

Large Signal Response

Gerhard [email protected]


Topic Map

21

Equilibrium DC Small signal

Large Signal

Circuits

Diode

Schottky

BJT/HBT

MOSFET

Digital Signals:switch on and off


Outline

22

1) Large signal response and charge control model

2) Turn-off characteristics

3) Turn-on characteristics

4) Other applications

5) Conclusion

Ref. SDF, Chapter 8


Digital, Large Signal Applications

23

V

ln(I)

‘1’ to ‘0’

‘0’ to ‘1’

1-2

---:If transition is very fast

AV‘1’:1V‘0’:-2V

:If transition is slow, every point is in quasi-equlibrium�treat them like DC


A Closer Look to Fast Transition

24

V

ln(I)

‘1’ to ‘0’

AV‘1’:1V‘0’:-2V

-0.1IR

-IR

IF

tstrr

tr

i(t)

t

VA(t)

1-2

Before transition occur

constant voltage, current change from IF to IR

constant current, voltage change form 1V to 0V


Definitions

25

ts … Charge storage time

tr … Recovery time

trr .. Reverse recovery time-0.1IR

-IR

IF

tstrr

tr

i(t)

t

VA(t)


Continuity Equations

26

1P P P

pr g

t q

∂ = ∇ • − +∂

J

( )D AD q p n N N+ −∇ • = − + −

P P Pqp qD pµ= − ∇J E

1N N N

nr g

t q

∂ = ∇ • − +∂

J

N N Nqn qD nµ= + ∇J E

Full analytical solution impossible for large signal….

,n n

n diffn

Q Qi

t

∂ = −∂ τ

,p p

p diff

n

Q Qi

t

∂= −

∂ τ

Charge control equations:Approximation when you havelarge transient response


forward bias, minority carrier has been injected

How Does Current Flip Without Voltage Flipping?

27

VDC

n(x,t)

x

n(x,t)

x

Recombination�n(x,t) exponentially dying outWhere did the charge go?

1. Back to the left-hand side2. Recombine with the trap and the

majority carrier

n(x,t)

x

When voltage start to change, n(x,t) suddenly bended downwards, dn/dxchange sign and current flip from p�n to n�p

0<dx

dn

0>dx

dn


Large Signal Charge Control Model

28

1 nN N

n dJr g

t q dx

∂ = − +∂ N N N

dnqn qD

dxµ= +J E

( ) ( )2

2Nn

n d n nD

t dx τ∂ ∆ ∆ ∆= −

∂

minority carrier


Large Signal Charge Control Model

29

( )0

pW

Q qA n dx≡ ∆∫

( ) ( )2

2Nn

n d n nD

t dx τ∂ ∆ ∆ ∆= −

∂

( ) ( )0p

N Nnx W x

d qA n d qA nQ QD D

t dx dx= =

∆ ∆∂ = − −∂ τ

diffn

Q Qi

t

∂ = −∂ τ

Area under thecurve @ given time

n(x,t)

xx(qA), integrate

Current going out Current coming in

Recombination

( ) ( )0 0 0

p n nW W W

Nn

dx dx dxd

ddD

t

qAqA n n n

dx x

qA

τ∆∂

= −∂

∆∆∫ ∫ ∫

pW pW

(Total charge that is building in)=(net electrons flowing in) – (recombination)


Outline

30





5) Conclusion


Turn-off Characteristics: Determine (ts)

31

diffn

Q Qi

t

∂ = −∂ τ

(0 )0 F

n

Q Qt I

t

−∂< = −∂ τ

AV‘1’‘0’

P

N

S R

VRVF

IF

IR

Vj

+

-

-0.1IR

-IR

IF

tstrr

tr

i(t)

Why does the current remain constanteven with t>0?

=0 (Time Independent)


Boundary Condition

32

diffn

Q Qi

t

∂ = −∂ τ

(0 )0 F

n

Q Qt I

t

−∂< = −∂ τ

(0 ) (0 )F nQ I Q− += =τ

Note. For a capacitor, voltage can not change instantly. So charge can not change instantly ….

AV‘1’‘0’

P

N

S R

VRVF

IF

IR

Vj

+

-

=0


Turn-off Current Transient

33

P

N

S R

VRVF

IF

IR

Vj

+

-

Since Vj can’t be larger than the band gap, which is smaller than VR, the diode will be forced to supply the negative current IR


Turn-off Current Transient

34

diff

n

Q Qi

t

∂ = −∂ τ

0 Rn

Q Qt I

t

∂> = − −∂ τ

( )

(0 ) 0

s sQ t t

QR

n

dQdt

QI+

=

− +

∫ ∫τ

(0 ) /ln

( ) /R n

s n

R s n

I Qt

I Q t

++=+

τττ

P

N

S R

VRVF

IF

IR

Vj

+

-

n(x,t)

x

n(x,t)

x

Approximately constant!

See how the slope change


Storage Time

35

(0 ) /ln ln

( ) /R n R F

s n n

R s n R

I Q I It

I Q t I

++ += =+

ττ ττ

V

ln(I)

‘1’ to ‘0’

‘0’ to ‘1’

P

N

S R

VRVF

IF

IR

Vj

+

-

n(x,t)

x

Shorten it for fast response!

A

i

N

n2

Q(ts)


Turn-off Voltage Transient

36

(0, )( ) ln p

Apo

n tkTt

q nυ =

( )/( ) ln ( ) pt

n p R R FQ t I I I e−= − + + ττ

Allows easy calculation of np(0,t).

t

Va

(t)

ts

P

N

S R

VRVF

IF

IR

Vj

+

-

n(x,t)

x


Recovery Time

37

1 0.1

r

p

t

r F

p Rr

p

t e Ierf

It

−

+ = +τ

τ πτ

V

ln(I)

‘1’ to ‘0’

‘0’ to ‘1’

tr0.51.27 0.15

1 0.15( ) 1 ex

xxerf x

+−+

= −

Useful formula …


Recovery Time

38

V

ln(I)

‘1’ to ‘0’

‘0’ to ‘1’

tr

22

2

( )2

( )2

p Rp n

n F

rr r f

p Rp n

F

W IW L

D It t t

IW L

I

−

−

= + ≈

≪

≫

τ

Ref. Sze/Ng, p. 117

ts


Outline

39





5) Conclusion


Turn-on Characteristics: Boundary Condition

40

diffn

Q Qi

t

∂ = −∂ τ

( )F

n

Q Q tt I

t

∂ = ∞→ ∞ = −∂ τ

( ) F nQ t I= ∞ = τ

AV‘1’‘0’

P

N

S R

VRVF

IF

IR

Vj

+

-


Turn-on Characteristics

41

diffn

Q Qi

t

∂ = −∂ τ

0 Fn

Q Qt I

t

∂> = −∂ τ

( ) ( ) 1 1n n

t t

F nQ t Q t e I e− −

= → ∞ − = −

τ ττ

np(x,t)time

x

P

N

S R

VRVF

IF

IR

Vj

+

-

Check:Q(t=0)=0Q(t �∞)=IF τn


Outline

42

1) large signal response and charge control model

2) turn-off characteristics

3) turn-on characteristics

4) other applications

5) conclusion


Diffusion Time

43

diff

n

Q Qi

t

∂ = −∂ τ

(0 ) ( )

dififf

fd

Q Q ti

τ

+ − = ∞ =

2

2p

n

W

D=

np(x,t)

x

Q(0+)

Wp(0)

2(0 )(0)di

p

d

p

pn

ff

p

iff

nq W

Qn

qDW

iτ

+

∆ = = ∆

Diffusion velocity

No recombination( , ) 'n x t C Dx∆ = +

Electrons get scattered to the other side, the average time is…

)/(2

1~

pn

p

WD

W×

Only half of themgoes to the right

recombination


Two line derivation of Steady State Diode Current

44

diffn

Q Qi

t τ∂ = −∂

ddiff

iff

Qi

τ=

( )( )

2

2

2

11

2

12

A

AqVn idif

qV

f

ip

diff p p A

A

n

nq e W

D nQi q e

W

D

N

NWβ

β

τ

× −=

×= = −

np(x,t)

x

Q

Exact Expression!

n diffτ τ→


Large Signal pn-diodeConclusion

45

Large signal response of devices of great importance for digital applications.

Analytical solution of partial differential equation often difficult (if not impossible), therefore approximate methods like Charge-control approximation often help simplify the solution and still provide a great deal of insight into the dynamics of switching operation.

Be careful in using the boundary condition which is often dictated by external circuit conditions.

ece606: solid state devices lecture 16 p-n diode ac...

Documents