半導(dǎo)體物理與器件第四版課后習(xí)題答案7參考

文檔來源為:文檔來源為:從網(wǎng)絡(luò)收集整理.word版本可編輯.歡迎下載支持 .PAGE7PAGE7文檔來源為 :從網(wǎng)絡(luò)收集整理.word版本可編輯 .7.1

Chapter7

Ge:

Vbi

0.432V(a)

GaAs:

Vbi

1.28V2 2

0.0259ln

102101.510

7.3Silicon(

300K)0.611V2

1015

2

For

Na Nd

14 310 cm

Vbi

0.4561VVbi

0.0259ln

1.5

10210

1510 ;15

V0.671V2 2

1016

; V(iii)

Vbi

0.0259ln

102101.510

1017

; V0.731V

GaAs(

300K)(b)

210171.5

2 10210

Vbi

For Na

143Nd 10 cm ;30.731V2

2

1015

; V(ii)

Vbi

0.0259ln

10210171017

1016

; V0.790V

10

1.282V2 2

Silicon(400

kT (iii)

Vbi

0.0259ln

102101.510

n

10

cm30.850V

iFor Na

N 1014cm3;d7.2d

15Si: ni15

1.5

10310 cm3

Vbi

0.2582VGe: ni

2.4

1013cm3

10 ;

VGaAs:

6310cm3

1610 ;16

VVbiVt

Vtln

NaNdn2ni

and

1017

9;GaAs(400K),9

VN

1014cm3, N

1017cm3'

ni 3.29

10 cm

14 3aThen a

Vbi

V

For

Na N

10 cm ;Ge:

Vbi

Vbi

0.7129V1515

GaAs:16

Vbi3

1.10V

10 ;

VNa 5

d 5 10316310 cm

cmSi:

,Vbi

16V16

10 ;

1.031VGe:

Vbi

1017

; VN

GaAs:1710 cm

Vbia3, Na

1.25V1017cm3

7.4(a) n-sideSi:

V orEF Ep-sideor

eV

7.7

0.695V3(b)or

EFi EF

eV

200K;300K;1.81061.810ini

kT ; kT ;cm3cm

1.38cmV(c)

400K; 9

0.034533;3or ni

3.28 10 cmVbi(d)or

0.7363V4

For200For300

VVxn4Nowor4

0.426 10

m

For400K;

1.023VxpWehaveor

0.0213

10

m

7.8

So Nd 3Na2 10

0.710max

3.29

104V/cm

or 3Na 1.5

exp

0.0259

n-side

whichyields Na16 3

7.766

15 310 cmorEF E

eV

Nd 2.33 10xn 9.93

cm610 cmp-sideor

or

m5EFi EF

eV

10 cm(b)or

0.7306V

or xpNow

0.2979

m

410V/cm4(c)or

Frompart(a),wecanwritewhichyields(d)or

xn

0.7305V104104cm0.154m104cm0.154m

Na 8.127Nd 2.438

15 310 cm1016cm3510 cm5Bysymmetry5

or x mxp 0.154

n3.973

10 cm7.6

Nowormax

4.75

104V/cm

7.9

or xp 0.39734.45

m410V/cm4N

ni

EF EFikT

or(b)

Vbi

0.635Vor Nd

1.98

cm3 oror(c)

Na 5.12

15 310cm10

xNowor

410

0.8644 mxor

0.08644

10

0.08644 m

xor

2.43

10 cmmax

1.34

104V/cm

max

3.75

102V/cm7.10

17(a)17

Vbi

0.0259ln

2 10

16410162

7.14

Assumesilicon,soor0.80813V

NN

8 10

cm3, LD16 3D

mVbidecreasesAtT

increases astemperature300K,wecanwrite

dLD Nd

2.2 10 cm ,m17 38 10 cm ,At

287K,

0.024778eV

LD 0.004577 mSoThen

2

1019

NowVbi

ni0.7427Vni7.11

0.82494VWefindUsingtheprocedurefromProblem7.10,VbiVbiAlsoThen

VVwecanwrite,for

T 300K,

xnxn

m0.2178 mAtFor

TVbi

300K,0.68886VV,

300K

xnNowL

mAt

380

kT eV

(a)xn

0.1320AlsoThen7.12For Ndor

V16 310 cm ,

V

LD(b)xnLD(c)xn7.15

0.1267EF E

eV

WefindFor Nor

1015cm3

(a)(i)For17 14EF EThen

eV

Na 10(ii)15

,N

10 ;

0.6350Vor7.13or

Vbi

V

10

(iii)(iv)

VVV(b)or

0.456V

(i)ForNa

1710 4Nd4

1014;xn(c)

2.43

107cm

max

(ii)

10V/cm

1510 ;1543or 43

104V/cm(iii)

10 ;

10 cm1616

104V/cm17

or

m464(iv)

10 ;

5.97 10 cm

104V/cm

or x

0.0597 m(b)For14 14

or

104cm0.3584 mNa 10 ,Nd 10 ;Vbi(ii)

0.4561V

Also W

xn xp

0.3584 m1510 15

(iii)

0.5157V

(c)

max

2VRW

20.8081 41041610 ;16

0.5754V

105V/cm12

0.6350V

or

10 F5.78pF(i)ForNa14

1410 ,4

7.18

WefindNd 10

max

0.265

V/cm

Nd 2.684

1015

cm3(ii)

10 ;

Na

1017cm3164164

104V/cm

(iii)

10 ;

10 cm176176

104V/cm

or

m4(iv)4

10 ;

2.83

10 cm

10 V/cm

or x

0.0283 m(b) max increasesasthedopingincreases,andtheelectricfieldextendsfurtherintothelow-dopedsideofthepnjunction.

(c)

max

2VbiW

VR

104V/cm1/27.16

V

C

e2Vbi

sNaNdVR Na Nd

VR 0,

105cm

C

109F/cm2or WFor VR

m45V,4

7.19

V1/22.738

10 cm

C 3NSo

3N

3 1.732or

m

C Na NaFor VR 0,

Foralargerdoping,thespacechargewidthnarrowswhichresultsinalargermax

20.67674

410V/cm4

capacitance.(ii)For

10VR 5V,

7.20orVbi

0.766V2

410V/cm

Now7.17

max

2.738

1040.8081V4

ororsothatVbi VR

73.53V15(b)15

whichyieldsVRor

72.8V0.826V

1.162VCVR1SoCVR2

VbiVbi

VR2VR1WehavesothatVbi VR

8.008V

7.24

whichyields

VR2 Vwhichyields

0.6889VVRor

7.18V0.886V

VR 0,C pFWehavesothatVbi VRwhichyields

1.456V

For VRCFor VR

5V,2.633pF1.157V0,7.21

VR(a)or

0.570V

For7.25

C 6.178pFVR 5V,C pFWefind

V18 15

C

1012F3VbiA

ln

10

1010102

V

(b)

10 H 3.306mHFor VR

1V, C pF51018

1016

10Hz MHzVbiB

0.0259ln

102

0.8139V

For VR

5V, C pFWefindor(b)or

1.510

7.26

10Hz MHz6Let V6(c)

Nd 1.88

10

cm37.22

orWehave

7.27

Nd 3.01

1015cm3or Bytrialanderror,Foror

VRVbi

10V,wefind1.137V

NN

1.5046.0161.10

1015cm3,15 310 cm ,(b)ThenNowso

Frompart(a),BytrialandNd 2.976

101516

cm3,3Wecanthenwrite Na 1.19 10 cm ,whichyields

Vbi

1.135VNa 1.23and

1016cm3

7.287.23

Nd 3.07

15 310 cm

or(b)

Vbi

0.5574Vorxp 5.32

10 cm

orWehave

2 0 at x xn6Also6

sothatfor x

x xn,wehaveorxn 2.66

104cm

Wealsohave Then

1at x xO(c)For xn 30 m,wewhichbecomesWefind

whichgivesThenfor

x

wehave7.29

VR 70.4V

7.34

(a)

d2 x2dx

x d xs dxAn n

junctionwith N

14 310 cm ,

For 2

1 m,

eNd(a) Aone-sidedjunctionandassumeVR Vbi. Thenor

SoAt x 2 So

xO, 0whichyields ThenVR(b)so

193V

At x 0, 0 or40

1,so7.7264

10 V/cm10

m

(c)Magnitudeofpotentialdifferenceis(c)

Let 0at

xO,thenor Thenwecanwrite7.30

max

7.65

104V/cm

At x 1 or0.7305V14

1 3.863VPotentialdifferenceacrosstheintrinsicFor For VRFor VR

13V,5V,

C C C

10 F1410 F141410 F14

regionor

2 15.45V7.31

Nd 5.36

15 310 cm

Bysymmetry,thepotentialdifferenceacrossthep-regionspace-chargeregionisalso3.863V. Thetotalreverse-biasvoltageA

105cm2

then7.32

Plot

VR V

VR7.35or

23.863

15.45

V7.33

ThenOr N

NB NaN

1.29416

10163

cm3(c)p-regionorWehave

7.36

B a 2.59 10 cmThenfor

0 at x xp xp x 0

eNaOxps

7.37For

15 310 cm16 3d 10 cm ,fromFigure7.15,n-region, 0 x xOorn-region, x

VB 75VdFor N 1015cm3,d7.38

VB 450V

each

voltageisapproximately300V. So,incase,breakdownisreachedfirst.(a) FromEquation(7.36),Set

max

crit

and

VR VB

7.42

18Impuritygradien18Then

V

a 2

1022cm4Vbi VB

51.77V

2 104So VB(b)

V

7.43

FromFigure7.15, VB 15VThenSo

V206V

ForthelinearlygradedjunctionThenNow7.39

At xSo

and x

xO, 0Forasilicon p15Nd 5 10

n junctionwith3andVB 100V,then,3

Then(b)Set 0 at

xO,thenor

Vbi

wehave4

Then7.40

xnminWefind

5.09 10 cm

5.09

7.44WehavethatThen20whichyields2010181018

a 10 cm4Vbi

ln

1021010

V 7.45Nowsowhichyields

Let N

15 3N5 10 cm << N3 1017

51015xn 6.47

610 cm

Then

Vbi

0.0259ln

2101.51010NowThenwhichyields

NowA

0.7648V5 105 VR Vor

For VR V, Cj 0.598pFVR V

For VR 0

Cj 1.24pF7.41

Assumesilicon: Foran

p junctionAssume

Vbi VRFor xp 75 mwhichyieldsVR 4.35

310VFor xp 150 mwhichyieldsVR 1.74

410 V文檔來源為:文檔來源為:從網(wǎng)絡(luò)收集整理.word版本可編輯.歡迎下載支持 .PAGE11PAGE11文檔來源為 :從網(wǎng)絡(luò)收集整理.word版本可編輯 .Note: FromFigure7.15,thebreakdown衛(wèi)生管理制度1 總則1.1 為了加強(qiáng)公司的環(huán)境衛(wèi)生管理，創(chuàng)造一個整潔、文明、溫馨的購物、辦公環(huán)境，根據(jù)《公共場所衛(wèi)生管理條例》的要求，特制定本制度。1.2 集團(tuán)公司的衛(wèi)生管理部門設(shè)在企管部，并負(fù)責(zé)將集團(tuán)公司的衛(wèi)生區(qū)域詳細(xì)劃分到各部室，各分公司所轄區(qū)域衛(wèi)生由分公司客服部負(fù)責(zé)劃分，確保無遺漏。2 衛(wèi)生標(biāo)準(zhǔn)2.1 室內(nèi)衛(wèi)生標(biāo)準(zhǔn)2.1.1 地面、墻面：無灰塵、無紙屑、無痰跡、無泡泡糖等粘合物、無積水，墻角無灰吊、無蜘蛛網(wǎng)。2.1.2 門、窗、玻璃、鏡子、柱子、電梯、樓梯、燈具等，做到明亮、無灰塵、無污跡、無粘合物，特別是玻璃，要求兩面明亮。2.1.3 柜臺、貨架：清潔干凈，貨架、柜臺底層及周圍無亂堆亂放現(xiàn)象、無灰塵、無粘合物，貨架頂部、背部和底部干凈，不存放雜物和私人物品。2.1.4 購物車（筐）、直接接觸食品的售貨工具（包括刀、叉等）：做到內(nèi)外潔凈，無污垢和粘合物等。購物車（筐）要求每天營業(yè)前簡單清理，周五全面清理消毒；售貨工具要求每天消毒，并做好記錄。2.1.5 商品及包裝：商品及外包裝清潔無灰塵（外包裝破損的或破舊的不得陳列）。2.1.6 收款臺、服務(wù)臺、辦公櫥、存包柜：保持清潔、無灰塵，臺面和側(cè)面無灰塵、無灰吊和蜘蛛網(wǎng)。桌面上不得亂貼、亂畫、亂堆放物品，用具擺放有序且干凈，除當(dāng)班的購物小票收款聯(lián)外，其它單據(jù)不得存放在桌面上。2.1.7 垃圾桶：桶內(nèi)外干凈，要求營業(yè)時間隨時清理，不得溢出，每天下班前徹底清理，不得留有垃圾過夜。2.1.8 窗簾：定期進(jìn)行清理，要求干凈、無污漬。2.1.9 吊飾：屋頂?shù)牡躏椧鬅o灰塵、無蜘蛛網(wǎng)，短期內(nèi)不適用的吊飾及時清理徹底。2.1.10 內(nèi)、外倉庫：半年徹底清理一次，無垃圾、無積塵、無蜘蛛網(wǎng)等。2.1.11 室內(nèi)其他附屬物及工作用具均以整潔為準(zhǔn)，要求無灰塵、無粘合物等污垢。2.2 室外衛(wèi)生標(biāo)準(zhǔn)2.2.1 門前衛(wèi)生：地面每天班前清理，平時每一小時清理一次，每周四營業(yè)結(jié)束后有條件的用水沖洗地面（冬季可根據(jù)情況適當(dāng)清理），墻面干凈且無亂貼亂畫。2.2.2 院落衛(wèi)生：院內(nèi)地面衛(wèi)生全天保潔，果皮箱、消防器械、護(hù)欄及配電箱等設(shè)施每周清理干凈。垃圾池周邊衛(wèi)生清理徹底，不得有垃圾溢出。2.2.3 綠化區(qū)衛(wèi)生：做到無雜物、無紙屑、無塑料袋等垃圾。3 清理程序3.1 室內(nèi)和門前院落等區(qū)域衛(wèi)生：每天營業(yè)前提前10分鐘把所管轄區(qū)域內(nèi)衛(wèi)生清理完畢，營業(yè)期間隨時保潔。下班后5-10分鐘清理桌面及衛(wèi)生區(qū)域。3.2 綠化區(qū)衛(wèi)生：每周徹底清理一遍，隨時保持清潔無垃圾。4 管理考核4.1 實行百分制考核，每月一次（四個分公司由客服部分別考核、集團(tuán)職能部室由企管部統(tǒng)一考核）。不符合衛(wèi)生標(biāo)準(zhǔn)的，超市內(nèi)每處扣0.5分，超市外每處扣1分。4.2 集團(tuán)堅持定期檢查和不定期抽查的方式監(jiān)督各分公司、部門的衛(wèi)生工作。每周五為衛(wèi)生檢查日，集團(tuán)檢查結(jié)果考核至各分公司，各分公司客服部的檢查結(jié)果考核至各部門。4.3 集團(tuán)公司每年不定期組織衛(wèi)生大檢查活動，活動期間的考核以通知為準(zhǔn)。7.1

Chapter7

Ge:

Vbi

0.432V(a)

GaAs:

Vbi

1.28V2 2

0.0259ln

102101.510

7.3Silicon(

300K)0.611V2

1015

2

For

Na Nd

14 310 cm

Vbi

0.4561VVbi

0.0259ln

1.5

10210

1510 ;15

V0.671V2 2

1016

; V(iii)

Vbi

0.0259ln

102101.510

1017

; V0.731V

GaAs(

300K)(b)

210171.5

2 10210

Vbi

For Na

143Nd 10 cm ;30.731V2

2

1015

; V(ii)

Vbi

0.0259ln

10210171017

1016

; V0.790V

10

1.282V2 2

Silicon(400

kT (iii)

Vbi

0.0259ln

102101.510

n

10

cm30.850V

iFor Na

N 1014cm3;d7.2d

15Si: ni15

1.5

10310 cm3

Vbi

0.2582VGe: ni

2.4

1013cm3

10 ;

VGaAs:

6310cm3

1610 ;16

VVbiVt

Vtln

NaNdn2ni

and

1017

9;GaAs(400K),9

VN

1014cm3, N

1017cm3'

ni 3.29

10 cm

14 3aThen a

Vbi

V

For

Na N

10 cm ;Ge:

Vbi

Vbi

0.7129V1515

GaAs:16

Vbi3

1.10V

10 ;

VNa 5

d 5 10316310 cm

cmSi:

,Vbi

16V16

10 ;

1.031VGe:

Vbi

1017

; VN

GaAs:1710 cm

Vbia3, Na

1.25V1017cm3

7.4(a) n-sideSi:

V orEF Ep-sideor

eV

7.7

0.695V3(b)or

EFi EF

eV

200K;300K;1.81061.810ini

kT ; kT ;cm3cm

1.38cmV(c)

400K; 9

0.034533;3or ni

3.28 10 cmVbi(d)or

0.7363V4

For200For300

VVxn4Nowor4

0.426 10

m

For400K;

1.023VxpWehaveor

0.0213

10

m

7.8

So Nd 3Na2 10

0.710max

3.29

104V/cm

or 3Na 1.5

exp

0.0259

n-side

whichyields Na16 3

7.766

15 310 cmorEF E

eV

Nd 2.33 10xn 9.93

cm610 cmp-sideor

or

m5EFi EF

eV

10 cm(b)or

0.7306V

or xpNow

0.2979

m

410V/cm4(c)or

Frompart(a),wecanwritewhichyields(d)or

xn

0.7305V104104cm0.154m104cm0.154m

Na 8.127Nd 2.438

15 310 cm1016cm3510 cm5Bysymmetry5

or x mxp 0.154

n3.973

10 cm7.6

Nowormax

4.75

104V/cm

7.9

or xp 0.39734.45

m410V/cm4N

ni

EF EFikT

or(b)

Vbi

0.635Vor Nd

1.98

cm3 oror(c)

Na 5.12

15 310cm10

xNowor

410

0.8644 mxor

0.08644

10

0.08644 m

xor

2.43

10 cmmax

1.34

104V/cm

max

3.75

102V/cm7.10

17(a)17

Vbi

0.0259ln

2 10

16410162

7.14

Assumesilicon,soor0.80813V

NN

8 10

cm3, LD16 3D

mVbidecreasesAtT

increases astemperature300K,wecanwrite

dLD Nd

2.2 10 cm ,m17 38 10 cm ,At

287K,

0.024778eV

LD 0.004577 mSoThen

2

1019

NowVbi

ni0.7427Vni7.11

0.82494VWefindUsingtheprocedurefromProblem7.10,VbiVbiAlsoThen

VVwecanwrite,for

T 300K,

xnxn

m0.2178 mAtFor

TVbi

300K,0.68886VV,

300K

xnNowL

mAt

380

kT eV

(a)xn

0.1320AlsoThen7.12For Ndor

V16 310 cm ,

V

LD(b)xnLD(c)xn7.15

0.1267EF E

eV

WefindFor Nor

1015cm3

(a)(i)For17 14EF EThen

eV

Na 10(ii)15

,N

10 ;

0.6350Vor7.13or

Vbi

V

10

(iii)(iv)

VVV(b)or

0.456V

(i)ForNa

1710 4Nd4

1014;xn(c)

2.43

107cm

max

(ii)

10V/cm

1510 ;1543or 43

104V/cm(iii)

10 ;

10 cm1616

104V/cm17

or

m464(iv)

10 ;

5.97 10 cm

104V/cm

or x

0.0597 m(b)For14 14

or

104cm0.3584 mNa 10 ,Nd 10 ;Vbi(ii)

0.4561V

Also W

xn xp

0.3584 m1510 15

(iii)

0.5157V

(c)

max

2VRW

20.8081 41041610 ;16

0.5754V

105V/cm12

0.6350V

or

10 F5.78pF(i)ForNa14

1410 ,4

7.18

WefindNd 10

max

0.265

V/cm

Nd 2.684

1015

cm3(ii)

10 ;

Na

1017cm3164164

104V/cm

(iii)

10 ;

10 cm176176

104V/cm

or

m4(iv)4

10 ;

2.83

10 cm

10 V/cm

or x

0.0283 m(b) max increasesasthedopingincreases,andtheelectricfieldextendsfurtherintothelow-dopedsideofthepnjunction.

(c)

max

2VbiW

VR

104V/cm1/27.16

V

C

e2Vbi

sNaNdVR Na Nd

VR 0,

105cm

C

109F/cm2or WFor VR

m45V,4

7.19

V1/22.738

10 cm

C 3NSo

3N

3 1.732or

m

C Na NaFor VR 0,

Foralargerdoping,thespacechargewidthnarrowswhichresultsinalargermax

20.67674

410V/cm4

capacitance.(ii)For

10VR 5V,

7.20orVbi

0.766V2

410V/cm

Now7.17

max

2.738

1040.8081V4

ororsothatVbi VR

73.53V15(b)15

whichyieldsVRor

72.8V0.826V

1.162VCVR1SoCVR2

VbiVbi

VR2VR1WehavesothatVbi VR

8.008V

7.24

whichyields

VR2 Vwhichyields

0.6889VVRor

7.18V0.886V

VR 0,C pFWehavesothatVbi VRwhichyields

1.456V

For VRCFor VR

5V,2.633pF1.157V0,7.21

VR(a)or

0.570V

For7.25

C 6.178pFVR 5V,C pFWefind

V18 15

C

1012F3VbiA

ln

10

1010102

V

(b)

10 H 3.306mHFor VR

1V, C pF51018

1016

10Hz MHzVbiB

0.0259ln

102

0.8139V

For VR

5V, C pFWefindor(b)or

1.510

7.26

10Hz MHz6Let V6(c)

Nd 1.88

10

cm37.22

orWehave

7.27

Nd 3.01

1015cm3or Bytrialanderror,Foror

VRVbi

10V,wefind1.137V

NN

1.5046.0161.10

1015cm3,15 310 cm ,(b)ThenNowso

Frompart(a),BytrialandNd 2.976

101516

cm3,3Wecanthenwrite Na 1.19 10 cm ,whichyields

Vbi

1.135VNa 1.23and

1016cm3

7.287.23

Nd 3.07

15 310 cm

or(b)

Vbi

0.5574Vorxp 5.32

10 cm

orWehave

2 0 at x xn6Also6

sothatfor x

x xn,wehaveorxn 2.66

104cm

Wealsohave Then

1at x xO(c)For xn 30 m,wewhichbecomesWefind

whichgivesThenfor

x

wehave7.29

VR 70.4V

7.34

(a)

d2 x2dx

x d xs dxAn n

junctionwith N

14 310 cm ,

For 2

1 m,

eNd(a) Aone-sidedjunctionandassumeVR Vbi. Thenor

SoAt x 2 So

xO, 0whichyields ThenVR(b)so

193V

At x 0, 0 or40

1,so7.7264

10 V/cm10

m

(c)Magnitudeofpotentialdifferenceis(c)

Let 0at

xO,thenor Thenwecanwrite7.30

max

7.65

104V/cm

At x 1 or0.7305V14

1 3.863VPotentialdifferenceacrosstheintrinsicFor For VRFor VR

13V,5V,

C C C

10 F1410 F141410 F14

regionor

2 15.45V7.31

Nd 5.36

15 310 cm

Bysymmetry,thepotentialdifferenceacrossthep-regionspace-chargeregionisalso3.863V. Thetotalreverse-biasvoltageA

105cm2

then7.32

Plot

VR V

VR7.35or

23.863

15.45

V7.33

ThenOr N

NB NaN

1.29416

10163

cm3(c)p-regionorWehave

7.36

B a 2.59 10 cmThenfor

0 at x xp xp x 0

eNaOxps

7.37For

15 310 cm16 3d 10 cm ,fromFigure7.15,n-region, 0 x xOorn-region, x

VB 75VdFor N 1015cm3,d7.38

VB 450V

each

voltageisapproximately300V. So,incase,breakdownisreachedfirst.(a) FromEquation(7.36),Set

max

crit

and

VR VB

7.42

18Impuritygradien18Then

V

a 2

1022cm4Vbi VB

51.77V

2 104So VB(b)

V

7.43

FromFigure7.15, VB 15VThenSo

V206V

ForthelinearlygradedjunctionThenNow7.39

At xSo

and x

xO, 0Forasilicon p15Nd 5 10

n junctionwith3andVB 100V,then,3

Then(b)Set 0 at

xO,thenor

Vbi

wehave4

Then7.40

xnminWefind

5.09 10 cm

5.09

7.44WehavethatThen20whichyields2010181018

a 10 cm4Vbi

ln

1021010

V 7.45Nowsowhichyields

Let N

15 3N5 10 cm << N3 1017

51015xn 6.47

610 cm

Then

Vbi

0.0259ln