6.012 - electronic devices and circuits lecture 7 - p-n junctions: i...

6.012 - Electronic Devices and Circuits Lecture 7 - p-n Junctions: I-V Relationship - Outline

• Announcements Handout - Lecture Outline and Summary First Hour Exam - In 2 wks, Oct. 8, 7:30-9:30 pm; thru this week, PS #4

• Review Depletion approximation for an abrupt p-n junction

Cq

Depletion charge storage and depletion capacitance (recitation topic) DP(vAB) = – AqNApxp = – A[2eq(fb-vAB){NApNDn/(NAp+NDn)}]1/2

dp(VAB) ≡ ∂qDP/∂vAB|VAB = A[eq{NApNDn/(NAp+NDn)}/2(fb-VAB)]1/2

• I-V relationship for an abrupt p-n junction Assume: 1. Low level injection

2. All applied voltage appears across junction:3. Majority carriers in quasi-equilibrium with barrier4. Negligible SCL generation and recombination

Relate minority populations at QNR edges, -xp and xn, to vAB Use excesses at -xp, xn to find hole and electron currents in QNRs Connect currents across SCL to get total junction current, iD

• Features and limitations of modelEngineering the minority carrier injection across a junction Deviations at low and high current levels, and large reverse bias

Clif Fonstad, 9/03 Lecture 7 - Slide 1

Abrupt p-n junctions: The Depletion Approximation: an informed first estimate of r(x)

Assume full depletion for -xp < x < xn, where xp and x are two n unknowns yet to be determined. This leads to:

r(x)Ï 0 for x < -xÔ

p qNDn

Ô-qNAp for -x < x < 0pr(x) = Ì -xp

Ô qNDn for 0 < x < xn xn x -qNApÔÓ 0 for xn < x

Integrating the charge once gives the electric field Ï 0 for x < -xÔ

p

Ô-qNAp (x + xp ) for - x < x < 0pÔ

E(x) = Ì eSi Ô qNDn (x - xn ) for 0 < x < xnÔ eSiÔÓ 0 for xn < x


The Depletion Approximation, cont.: Insisting E(x) is continuous at x = 0 yields our first equation relating xn and xp, our unknowns:

NApxp = NDnxn

Integration again gives the electrostatic potential: Ï f for x < -xpÔ

p

Ô f +qNAp (x + xp )

2 for - x < x < 0p pÔ

f(x) = Ì 2eSi 2Ô f -

qNDn (x - xn ) for 0 < x < xn nÔ 2eSiÔÓ fn for x < xn

Requiring that the potential be continuous at x = 0 gives us our second relationship between xn and xp:

2 qNDn 2f +qNAp x = f - xp p n2eSi 2eSi

n


The Depletion Approximation, cont.: Combining our two equations and solving for xp and xn gives:

xp = 2eSi fb q

NDn

NAp NAp + NDn( ), xn =

2eSi fb q

NAp

NDn NAp + NDn( ) where we have introduced the built-in potential, fb:

ˆkT Ê kT NAp ˜ =

kT NDnNApfb ≡ f - fp = ln NDn - - ln lnn

q ni ËÁ

q ni ¯ q ni 2

We also care about the total width of the depletion region, w:

2eSi fb (NAp + NDn )w = x + x = p n

q NApNDn

And we want to know the peak electric field, | Epk |: qNApxp qNDnxn 2q fb NApNDn = = =Epk eSi eSi eSi (NAp + NDn )


The Depletion Approximation, cont.: An important case: asymmetrically doped junctions

A p+-n junction (NAp >> NDn): 2eSi fb 2q fb NDnx >> xp , w ª x ª , ª n n qN

Epk

Dn eSi

An n+-p junction (NDn >> NAp): 2q fb NAp2eSi fbxp >> xn , w ª x ª

qNAp

, ª p eEpk

Si

Note that in both cases the depletion region exists predominately on the lightly doped side, and it is the doping level of the more lightly doped junction that matters (i.e., dominates).

Note also that as the doping level increases the depletion width decreases and the peak E-field increases.


How does the depletion approximation compare with a full

: An abrupt p-n junction with

NAp= 1017 cm-3, NDn= 5 x 1016 cm-3

solution?

Charge

E-field

Potential

nie±qf(x)/kTpo o(x)

An example

(x), n


Images courtesy of Jesus Del Alamo. Used with permission.

cc_rchakr

Text Box

(Courtesy of Professor Jesus Del Alamo.)

So…just how good is the depletion approximation?

The plots look good, but equally important is that 1. It gives an excellent model for making hand calculations

and gives us good values for quantities we care about: • Depletion region width • Peak electric field • Potential step

2. It gives us the proper dependences of these quantities on the doping levels (relative and absolute) and the bias voltage.

Applying bias; what happens? Two things happen 1. The depletion width changes

• fb is replaced by (fb - vAB) in the Depletion Approximation Eqs. • This was discussed in recitation yesterday

2. Currents flow • This is the main topic of today’s lecture


Biased p-n junctions: The Depletion Approximation with bias applied: Assume all the applied potential appears across the junction:

fb Æ(fb - vAB )

The depletion region width and peak electric field become:

w(vAB ) = 2eSi fb - vAB( )

q

NAp + NDn( ) NApNDn

2q (fb - vAB ) NApNDn = e

Epk

Si (NAp + NDn ) Note that with increasing reverse bias, both w and |Epk|

increase. The Depletion Approximation continues to give us a good fit

to the more accurate solutions, as the figures on the next two foils illustrate.


Comparing the depletion approximation and a full solution, cont.:

Applied reverse bias, - 3Volts :

Charge

E-field

Potential

nie±qf(x)/kT

Same sample, rev. bias

p (x) n (x)



cc_rchakr

Text Box


Comparing the depletion approximation and a full solution, cont.:

Applied forward bias, 0.75 Volts :

Charge

E-field

Potential

nie±qf(x)/kT

Same sample, fwd. bias

p (x) n (x)



cc_rchakr

Text Box


Biased p-n junctions: depletion region changes, cont. Depletion charge stores and depletion capacitance: The depletion regions represent charge stores, like those on

either side of a parallel plate capacitor, except that they are non-linear functions of the applied voltage:

qNApNDn

DP (vAB ) = -AqNApxp (vAB ) = -A 2qeSi (fb - vAB ) (NAp + NDn ) For small variations, vab, in the applied voltage, vAB, about an

DC operating (bias) point, VAB, the charge varies linearly with the small variation in the applied voltage:

qDP (vAB ) =QDP (VAB ) + qdp (vab ) ªQDP (VAB ) + Cdp (VAB)vab

The proportionality factor is called the “depletion capacitance”,and it is:

C = AqeSi NApNDn

dp (VAB) ≡∂qDP ∂vAB 2(fb -VAB ) (NAp + NDn )vAB =VAB

Note: “A” is the cross-sectional Clif Fonstad, 9/03 area of the junction. Lecture 7 - Slide 11

Biased p-n junctions: depletion region changes, cont. Comparing depletion charge stores with a parallel plate

capacitor: r(x)

d/2

-d/2

qA

qB A)( = -q

x

Parallel plate capacitor

qA,PP = Ae vAB

d

Cpp (VAB) ≡∂qA,PP

∂vAB vAB =VAB Ae

= d

Many similarities; important differences.

qN

r(x)

Dn

-xp

qA

qB A)( = -q

x xn

-qNAp

Depletion region charge store

qA,DP (vAB ) = -AqNApxp (vAB )

= -A 2qeSi [fb - vAB ] NApNDn

[NAp + NDn ]

Cdp (VAB) ≡∂qA,DP

∂vAB vAB =VAB

= AqeSi NApNDn A eSi=

2[fb -VAB ] [NAp + NDn ] w(VAB )


Biased p-n junctions: Current flow: Our goal is to determine the relationship between iD and vAB. We are now ready to find the boundary conditions set by the

applied voltage across the space charge region. Ohmic

p n

contact iD

+ -vAB

Uniform p-type Uniform n-type Ohmiccontact

A B

x -w -x 0 x wp p n n

Quasineutral Space charge Quasineutral region I region region II

Flow problem, QN region I - Excitation:n’(-wp ( p) AB

gL = 0p’(wn n) AB

) = 0, n’ -x = function of v

) = 0, p’(x = function of v

gL = 0 B.C.’s:

Flow problem, QN region II - Excitation:B.C.’s:

Clif Fonstad, 9/03 Getting these is our objective now. Lecture 7 - Slide 13

Biased p-n junctions: current flow, cont. Boundary condtions at the edges of the space charge layer:

What are n’(-xp) and p’(xn)? Begin by looking at the situation in thermal equilibrium, where we

have: p(-xp ) = NAp and p(xn ) = ni

2 NDn

qfb

qf If the population of holes at the top ofthe potential “hill” is related to thepopulation at the bottom by aBoltzman factor, then we should alsofind that:

p(xn ) = p(-xp )e -qfb / kT -x 0 xp n

Do we? 2

-qfb / kTfb = kT

ln NApNDn fi

ni = NApe2q ni NDn 2

-qfb / kT

NThus : p(xn ) =

ni = NApe -qfb / kT = p(-xp )e

Dn

YES, we do, and the Boltzman relationship holds. Clif Fonstad, 9/03 Lecture 7 - Slide 14

x

Biased p-n junctions: current flow, cont.

What are n’(-xp) and p’(xn) with vAB applied?

We propose that the populations are still related by the Boltzmanfactor, which is now: exp[-q(fb-vAB)/kT]

Thus:p(xn ) = p(-xp )e -q[fb -vAB ] / kT

Under low level injection conditions, the majority carrier population is unchanged, so p(-xp) remains NAp, so:

2

p(xn ) = NAp e -q[fb -vAB ] / kT ni eqvAB / kT =

NDn

And the excess population we seek is: 2

eqvAB / kT -1)N

p'(xn ) = p(xn ) - pon = ni (Dn

Similarly at -xp: 2

eqvAB / kT -1)N

n'(-xp ) = ni (Ap



What is the current, iD?

Knowing p’(xn) and n’(-xp), we know: Jh (x) for x < x < wn n

and Je(x) for - w p < x < -xp

But we still don’t know the total current because we don’t know both currents at the same position, x:

= A Jh (x) + Je ( x)] Have to be at same “x” iD = AJTOT [ To proceed we make the assumption that there is negligible recom

bination of holes and electrons in the depletion region, so whatgoes in comes out and:

Jh (xn ) = Jh (-xp ) and Je(xn ) = Je(-xp )

With this assumption, we can write:

[iD = AJTOT = A Jh (xn ) + Je (-xp )] Values at edges of SCL



What is the current, iD, cont,?

Both Jh(xn) and Je(-xp), are proportional to p’(xn) and n’(-xp),respectively, which in turn are both proportional to (eqv/kT -1):

eqv AB / kT -1] eqv AB / kT -1]Jh (xn ) µp'(xn ) µ[ and Je(-xp) µn'(xp) µ[ Thus the diode current is also proportional to (eqv/kT -1):

[ µ eqvAB / kT eqvAB / kT -1]iD = A Jh (xn ) + Je (-xp )] [ -1] fi iD = Is [

(IS is called the reverse saturation current of the diode.)

** Notice that non-linearity, i.e., the exponential dependence of thediode current on voltage, arises because of the exponentialdependence of the minority carrier populations the edges ofthe space charge layer (depletion region). The flow problemsthemselves are linear.



The saturation current of several diodes:

Short-base diode, wn << Lh, wp << L :e

p’(x), n’(x)

p’(xn)

n’(-xp) x

2 Dh kT -1 n]) = q ni

NDn

-w -x x w/[¸ÔÔ˝

Jh (xn eqv AB p p n

È ˘( ) Dh De w x-

NAp w(kT -1]

p’ ’(x)

x n’(-xp)

p’(xn)

(x), n

Í2iD = Aqni /[n n ˙

˙eqv AB+2ni De )NDn( )ÎÍÔ

Ô

w x x-eqv AB / kT -1] -[J (-xp) = q n n p pe NAp (wp - )xp

Long-base diode, wn >> Lh, wp >> L :e

= q 2ni

NDn

Dh eqv AB[ -w -x x wnkT -1¸ÔÔ˝

]/Jh (xn ) p p n

È ˘Lh N

Dh

DnLh NApL

De ]/ kT -12iD = Aqni [˚ ˙ eqv ABÍ

Î+2ni D

N= q e eqv AB

Ap Le [ kT -1 Ô

Ô]/Je(-xp) e

ÈGeneral diode: ˘Dh De ]/ kT -12iD = Aqni [˚˙ eqv ABÍ

Î+

NDnwn,eff NApwp,eff

Hole injection into n-side Electron injection into p-side Clif Fonstad, 9/03 Lecture 7 - Slide 18

Biased p-n junctions: current flow, cont. Limitations of the model - times/places it isn’t perfect (but are any of us?)

• Large forward bias: High level injection (c)

Series voltage drop (d)

• Large reverse bias: See pg. 105 from Sze, S.M. Physics of Semiconductor Devices. ISBN 471-842-90-7.

Reverse breakdown

• Very low bias levels: SCL generation and

recombination (a, e)

Ref: Figure 18 in S. M. Sze, “Physics of Semiconductor Devices” 1st. Ed (Wiley, 1969)


6.012 - Electronic Devices and Circuits

Lecture 7 - p-n Junctions: I-V Relationship - Summary • I-V relationship for an abrupt p-n junction

Focus is on minority carrier diffusion on either side of SCL Voltage across SCL sets excess populations -xp and xn:

n'(-xp) = nnoe-q[fb – vAB]/kT-n = npo(eqvAB/kT-1) = (ni2/NAp)(eqvAB/kT -1)po

p'(xn) = ppoe-q[fb – vAB]/kT-pno = pno(eqvAB/kT-1) = (ni2/NDn)(eqvAB/kT -1)

Flow problems in QNR regions give minority currents: Je(-wp<x<-xp) = q(De/Le)[cosh(wp-x)/sinh(wp-xp)](ni

2/NAp)(eqvAB/kT -1)Jh(xn<x<wn) = q(Dh/Lh)[cosh(wn-x)/sinh(wn-xn)](ni

2/NDn)(eqvAB/kT -1)Total current is found from continuity across SCL:

iD(vAB) = A [Je(-xp) + Jh(xn)] = IS (eqvAB/kT -1), with* *IS ≡A q ni

2 [(Dh/NDn wn ) + (De/NAp wp )](hole component) (electron component) *Note: wp

* and wn are the effective widths of the p- and n-sidesIf Le >> wp, then wp* ≈ (wp - xp), and if Le << wp, then wp* ≈ Le If Lh >> wn, then wn* ≈ (wn - xn), and if Lh << wn, then wn* ≈ Lh

• Features and limitations of modelExponential dependence enters via boundary conditions Injection is predominantly into more lightly doped side Saturation current, IS, goes down as doping levels go up Limits: 1. SCL g-r may dominate at low current levels

2. Series resistance may reduce junction voltage at high currents3. Junction may breakdown (conduct) at

Lecture 7 - Slide 20 Clif Fonstad, 9/03 large reverse bias

6.012 - electronic devices and circuits lecture 7 - p-n junctions: i...

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