Semiconductor Objective Question

2.1

Semiconductor Physics

In the problems assume the parameter given in

eV and material B has a bandgap energy of 1.2 eV.

following follo wing table. Use the temper temperature ature T  = 300 K unless

The rat ratio io of int intrin rinsic sic con concen centra tratio tion n of mat materi erial al A to

otherwise other wise stated.

that of material B is

Property

Si

GaAs

Ge

Bandgap Energy

1.12

1.42

0.66

Dielectric Constant

11.7

13.1

16.0

Effective density of  states in conduction band N c (cm-3)

2.8 ´ 10

Effective density of  states in valence band N v (cm-3)

1.04 04 ´ 1019

Intrinsic carrier concertration ni (cm-3) Mobility Electron Hole

(A) 2016

(B) 47.5

(C) 58.23

(D) 1048

sili lico con n at T  = 300 K the the therma rmal-e l-equi quilib libriu rium m 4. In si concentration of electron is n0 concentrati concen tration on is

19

4.7 ´ 10

17

1.04 04 ´ 10

19

(A) 4.5

7.0 ´ 1018

6.0 ´ 1018

1.5 ´ 10

1.8 ´ 10

2.4 ´ 10

1350 480

8500 400

3900 1900

10

6

18

5. In silicon at T  = 300 K if the Fermi energy is 0.22 eV  (A) 2 ´ 1015 cm -3

(B) 1015 cm -3

(C) 3 ´ 1015 cm -3

(D) 4 ´ 1015 cm -3

6. The thermal-equilibrium concentration of hole p0 in silicon at T  = 300 K is 1015 cm -3. The value of  n0 is

´ 108 cm -3

(B) 4.4 ´ 10 4 cm -3

(C) 2.6 ´ 10 4 cm -3

(D) 4.3 ´ 108 cm -3

(D) 3.6

donorr = 1013 cm -3 and dono

The e th ther erma mall eq equi uili libr briu ium m = 0. Th

concentration p0 is

´ 1014 cm -3

(A) 2.97 97 ´ 10 9 cm -3

(B) 2.68 ´ 1012 cm -3

(C) 2.95 95 ´ 1013 cm -3

(D) 2.4 cm -3

The e ma maxi ximu mum m = 1 ´ 1012 cm -3. Th

temp te mper erat atur ure e al allo lowe wed d fo forr th the e si sili lico con n is (A (Ass ssum ume e

 E g

german manium ium sem semico icondu nducto ctorr at T  = 300 K, th the e 7. In ger concen con centra tratio tion n is N d

(B)) 18.4 ´ 1014 cm -3 (B

2. The intrinsic carrier concentration in silicon is to be no gr grea eate terr th than an ni

´ 1015 m -3

(D) 0.3 ´ 10 -6 m -3

acceptor accept or concen concentratio trations ns is N a

the intrinsic concentration is

´ 1014 cm -3

(B) 4.5

above the valence band energy, the value of  p0 is

germanium ium semico semiconducto nductorr materi material al at T  = 400 K  1. In german

(C) 8.5

´ 1015 cm -3

(C) 0.3 ´ 10 -6 cm -3

(A) 3.8

(A) 26.8 ´ 1014 cm -3

= 5 ´ 10 4 cm -3. The hole

Statement Statem ent for Q.8-9:

In german germanium ium semiconductor semiconductor at T  = 300 K, the

= 1.12 eV)

(A) 300 K

(B) 360 K  

impurity concentration are

(C) 382 K

(D) 364 K  

 N d

= 5 ´ 1015 cm -3 and N a = 0

3. Two semiconductor material have exactly the same

8. The thermal equilibrium electron concentration n0

properties except that material A has a bandgap of 1.0

is

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108

Semiconductor Physics

Chap 2.1

(A) 5 ´ 1015 cm -3

(B) 11 115 . 5 ´ 1011 cm -3

position of Fermi level with respect to the center of the

(C) 11 115 . 5 ´ 10 9 cm -3

(D) 5 ´ 106 cm -3

bandgap is

9. The thermal equilibrium hole concentration p0 is (A) 39 396 . 6 ´ 10

(B) 19 195 . 5 ´ 10

13

(C) 4.36 ´ 1012 cm -3

13

cm

-3

(A)

+0.045 eV

(B)

- 0.046 eV 

(C)

+0.039 eV

(D)

- 0.039 eV 

sili lico con n sam sampl ple e co cont ntai ains ns ac acce cept ptor or ato atoms ms at a 16. A si

(D) 39 396 . 6 ´ 1013 cm -3

concentrati concen tration on of  N a samp mple le of si sili lico con n at T  = 30 300 0 K is do dope ped d wi with th 10. A sa

added

with ´ 1013 cm -3 and with 13 -3 arsenic at a concentration of 1 ´ 10 cm . The

semiconductor such that the Fermi level is 0.215 eV 

material is

donors atoms added are

boro bo ron n at a co conc ncen entr trat atio ion n of 25 .

forming

Dono norr at atoms oms are = 5 ´ 1015 cm -3. Do and type pe comp co mpen ensa sate ted d n - ty

below the conduction band edge. The concentration of 

(A) p - type with p0

= 1.5 ´ 1013 cm -3

(A) 1.2 ´ 1016 cm -3

(B) 4.6 ´ 1016 cm -3

(B) p - type with p0

= 1.5 ´ 10 7 cm -3

(C) 3.9 ´ 1012 cm -3

(D) 2.4 ´ 1012 cm -3

(C) n - type with n0

= 1.5 ´ 1013 cm -3

silico icon n sem semico iconduc nductor tor samp sample le at T  = 300 K is 17. A sil

(D) n - type with n0

= 1.5 ´ 10 cm 7

-3

doped dope d wit with h pho phospho sphorus rus ato atoms ms at a con concen centra tratio tions ns of 

sampl ple e of ga gall lliu ium m ar arse seni nide de at T  = 20 200 0 K, 11.. In a sam 11

= 5 p0 and N a = 0. The value of  n0 is (A) 9.86 ´ 10 9 cm -3 (B) 7 cm -3 n0

(C) 4.86 ´ 10 3 cm -3

(D) 3 cm -3

12. Germanium at T  = 300 K is uniformly doped with an acceptor concentration of  N a concentrati conce ntration on of  N d

-3

= 10 cm and a donor 15

= 0. The position of fermi energy

with respect to intrinsic Fermi level is

1015 cm -3. The position of the Fermi level with respect to the intrinsic Fermi level is (A) 0.3 eV

(B) 0.2 eV  

(C) 0.1 eV

(D) 0.4 eV  

silico icon n cry crystal stal having a cro crossss-sect section ional al are area a of  18. A sil

m m is connected to its ends to a 20 V ba ends batt tter ery y. At T  = 300 K, we want a 0.001 cm 2 and a length of 20 curr cu rren entt of 10 100 0

mA in cr crys ysta tal. l. Th The e co conc ncen entr trat atio ion n of 

donor atoms to be added is

(A) 0.02 eV

(B) 0.04 eV  

(A) 2.4 ´ 1013 cm -3

(B) 4.6 ´ 1013 cm -3

(C) 0.06 eV

(D)0.08 eV  

(C) 7.8 ´ 1014 cm -3

(D) 8.4 ´ 1014 cm -3

13.

In

germanium

at

concentrati conce ntration on are N d

T  = 300

K,

th e

= 1014 cm -3 and

N a

don or

The e = 0. Th

Fermi Fer mi ene energy rgy lev level el wit with h res respec pectt to int intrin rinsic sic Fer Fermi mi level is (A) 0.04 eV

(B) 0.08 eV  

(C) 0.42 eV

(D) 0.86 eV  

19. The cross sectional area of silicon bar is 100

mm 2 .

The Th e le leng ngth th of ba barr is 1 mm mm.. Th The e ba barr is doped doped with with arsenic atoms. The resistance of bar is (A) 2.58 mW

(B) 11.36 kW

(C) 1.36 mW

(D) 24.8 kW

20. A thin film resistor is to be made from a GaAs film 14. A GaAs device is doped with a donor concentration -3

of 3 ´ 10 cm . For the device to operate properly, the 15

intrinsic carrier concentration must remain less than 5%

of

the th e

tota to tall

conc co ncen entr trat atio ion. n.

The Th e

maxi ma ximu mum m

temperature on that the device may operate is

doped n - type. The resistor is to have a value of 2 k W. The resistor length is to be 200

m m and area is to be

10 -6 cm 2 . The doping doping eff effici icienc ency y is known to be 90%. The mobility of electrons is 8000

cm 2 V

- s . Th e

doping needed is

(A) 763 K

(B) 942 K  

(A) 8.7 ´ 1015 cm -3

(B) 8.7 ´ 10 21 cm -3

(C) 486 K

(D) 243 K  

(C) 4.6 ´ 1015 cm -3

(D) 4.6 ´ 10 21 cm -3

Forr a pa part rtic icul ular ar se semi mico cond nduc ucto torr at T  = 300 K   15. Fo

 E g

= 1.5 eV, m = 10 m *  p

* n

and ni

= 1 ´ 10

15

-3

cm . The

21. A silicon sample doped n - type at 1018 cm -3 have a resistance of 10

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W . The sample has an area of  10 -6

Chap 2.1

Semiconductor Physics

109

t sc = 10 -13 s

m m . The doping efficiency of  the sample is (m n = 800 cm 2 V - s )

27. For a sample of GaAs scattering time is

(A) 43.2%

(B) 78.1%

electr ele ctric ic field of 1 kV cm is applied, applied, the dri drift ft velocit velocity y

(C) 96.3%

(D) 54.3%

produced is

cm 2 and a length of 10

22.

Six

v o l ts

is

applied

across

a

2

cm

lo n g  

semiconducto semic onductorr bar bar.. The average drift velocity is 10

-s

(C) 6 ´ 10 4 cm 2 V

(A) 2.6 ´ 106 cm s

(B) 263 cm s

(C) 14 14.8 ´ 106 cm s

(D) 482

28. A gallium arsenide semiconductor at T  = 300 K is (B) 3 ´ 10 4 cm 2 V

-s

. = 0067 mo . I f a n

4

cm s. The electron mobility is (A) 4396 cm 2 V

and ele electr ctron’ on’ss eff effect ective ive mass is m e*

(D) 3333 cm 2 V

-s

doped with impuri impurity ty conce concentrati ntration on N d mobility

-s

= 1016 cm -3. The

m n is 7500 cm 2 V - s. For an applied field of 

10 V cm the drift current density is particular ar semicon semiconductor ductor material following  23. For a particul

(A) 120 A cm 2

(B) 120 A cm 2

parameters paramete rs are observed observed::

(C) 12 ´ 10 4 A cm 2

(D) 12 ´ 10 4 A cm 2

m n = 1000 cm 2 V - s ,

29. In a particular semiconductor the donor impurity

m p = 600 cm V - s , 2

 N c

concentration is N d

= N v = 1019 cm -3

T h e se

parameters,

parameters

temp te mper erat atur ure. e. Th The e

are

meas me asur ured ed

in d e p e n d e n t

m n = 1000 cm 2 V - s,

of  

cond co nduc ucti tivi vity ty of th the e

æ  T  ö  N c = 2 ´ 10 ç ÷ è 300 ø

300 0 K. s = 10 -6 (W - cm) -1 at T  = 30 The conductivity at T  = 500 K is (A) 2 ´ 10 -4 (W - cm) -1 (B) 4 ´ 10 -5 (W - cm) -1

- cm) -1

(D) 6 ´ 10 -3 (W

æ  T  ö  N v = 1 ´ 10 ç ÷ è 300 ø  E g

type e sil silico icon n samp sample le has a res resist istivi ivity ty of 5 24. An n - typ

(C) 11.46 ´ 10

15

cm

-3

(D) 11 . ´ 10

-15

cm

-3

cm -3,

32

19

- cm) -1

300 0 K. Th The e mo mobi bili lity ty is m n = 1350 W - cm at T  = 30 2 cm V - s. The donor impurity concentration is (A) 2.86 ´ 10 -14 cm -3 (B) 9.25 25 ´ 1014 cm -3

32

19

intrin int rinsic sic mat materi erial al is

(C) 2 ´ 10 -5 (W

= 1014 cm -3. Assume the following 

cm -3,

. eV eV.. = 11

  An ele electr ctric ic fie field ld of  E

= 10 V cm is applied. The

electric current density at 300 K is (A) 2.3 A cm 2

(B) 1.6 A cm 2

(C) 9.6 A cm 2

(D) 3.4 A cm 2

Statement Statem ent for Q.30-31 Q.30-31::

In a sil silico icon n samp sample le the ele electr ctron on con concen centra tratio tion n

25.

 A semiconductor has following parameter

drops linearly linearly from 1018 cm -3 to1016 cm -3 over a lengt length h

m n = 7500 cm 2 V - s,

curr rren entt dens densit ity y due to th the e el elec ectr tron on m m. The cu 2 diffusion current is (  Dn = 35 cm s). (A) 9.3 ´ 10 4 A cm 2 (B) 2.8 ´ 10 4 A cm 2 of 2. 2.0 0

(C) 9.3 ´ 10 9A cm 2

m p = 300 cm 2 V - s, ni

(D) 2.8 ´ 10 9 A cm 2

30.

= 3.6 ´ 1012 cm -3

When Wh en

cond co nduc ucti tiv vit ity y

is

mini mi nimu mum, m,

the th e

26. In a GaAs sample the electrons are moving under

concentrati concen tration on is

an

(A) 7.2 ´ 1011 cm -3

(B) 1.8

(C) 144 . ´ 1011 cm -3

(D) 9 ´ 1013 cm -3

electric

field

of

5

kV cm

and an d

the th e

carr ca rriier

concen con centra tratio tion n is uni unifor form m at 10 16 cm -3. Th The e el elec ectr tron on 7

hole ho le

´ 1013 cm -3

velocity is the saturated velocity of  10 cm s. The drift current density is (A) 1.6

´ 10 A cm 4

(C) 1.6 ´ 108 A cm 2

31. The minimum conductivity is 2

2

(A) 0.6 ´ 10 -3 (W

- cm) -1

(B) 1.7 ´ 10 -3 (W

(D) 2.4 ´ 108 A cm 2

(C) 2.4 ´ 10 -3 (W

- cm) -1

(D) 6.8 ´ 10 -3 (W

(B) 2.4 ´ 10 A cm 4

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- cm) -1 - cm) -1

110

32.

Semiconductor Physics

A pa part rtic icul ular ar in intr trin insi sicc

resistivity of 50 (W

semi se mico cond nduc ucto torr

hass ha

Chap 2.1

a Hole concentration p0

- cm) at T  = 300 K and 5 (W - cm) at

æ - x  ö ç ÷ 15 çè  L p  ø÷

cm -3, x

= 10 e

T  = 330 K. If change in mobility with temperature is

= 5 ´ 10

14

neglected, the bandgap energy of the semiconductor is

Electron concentration n0

(A) 1.9 eV

(B) 1.3 eV  

Hole diffusion coefficient D p

(C) 2.6 eV

(D) 0.64 eV  

33.

Three

scattering

mechanism

in

a

Hole diffusion length L p

semi se mico condu nduct ctor or.. If on only ly th the e fi firs rstt me mech chan anis ism m we were re 2

present, the mobility would be 500 cm V

the th e se seco cond nd me mech chan anis ism m we were re pr prese esent nt,, th the e mo mobi bili lity ty

- s . If only third mechanism were present, the mobility would be 1500 cm 2 V - s. The net wou ould ld be 75 750 0 cm2 V mobility is (A) 2750 cm 2 V 2

(C) 818 cm V

-s

(B) 1114 cm 2 V

-s

2

(D) 250 cm V

£0

= 25 cm 2 s

= 5 ´ 10 -4 cm,

Electron diffusion length Ln

- s. If only

cm -3, x

= 10 cm 2 s

Electron Elect ron diffus diffusion ion coeffi coefficients cients Dn

e x is t

æ - x ö ç ÷  L e è  n  ø

³0

= 10 -3 cm

The total current density at x = 0 is (A) 1.2 A cm 2

(B) 5.2 A cm 2

(C) 3.8 A cm 2

(D) 2 A cm 2

germani nium um Hall devic device e is doped doped wi with th 5 ´ 10 15 37. A germa

-s

donor atoms per cm 3 at T  = 300 K. The device has the

-s

concentration varies linearly with distance, as shown

= 5 ´ 10 -3 cm, W  = 2 ´ 10 -2 cm and L = 0.1 cm. The current is I  x = 250 m A, the applied voltage is 100 0 mV mV,, and the magne magneti ticc flux is B z = 5 ´ 10 -2 V  x = 10

in fig. P2.1.34. The diffusion current density is found

tesla. The Hall voltage is

the e el elec ectr tron on di difffu fusi sion on = 0.19 A cm 2 . If th 2 coefficient is Dn = 25 cm s, The elect electron ron concen concentrati tration on

(A) -0.31mV

(B) 0.31 mV  

(C) 3.26 mV

(D) -3.26 mV 

geometry d

34. In a sample of silicon at T  = 300 K, the electron

to be J n at is

Statement Statem ent for Q.38-39 Q.38-39:: 14

  A si sili lico con n Ha Hall ll de devi vice ce at T  = 300 K has the

5´10

     )      3

= 10 -3 cm , W  = 10 -2 cm, L = 10 -1 cm. The follow fol lowing ing par parame ameter terss are mea measur sured: ed: I  x = 0.75 mA, V  x = 15 V, V  H  = +5.8 mV, tesla geometry d

  -

   m    c      (      n

n(0)

0.010

0  x(cm)

majority ty carrier concentrat concentration ion is 38. The majori

Fig. P2.1.34

(A) 8 ´ 1015 cm -3, n - type

(A) 4.86 ´ 108 cm -3

(B) 2.5 ´ 1013 cm -3

(B) 8 ´ 1015 cm -3, p - type

(C) 9.8 ´ 10 26 cm -3

(D) 5.4 ´ 1015 cm -3

(C) 4 ´ 1015 cm -3, n - type

35. The hole concentration in p - type GaAs is given by

æ  è 

 p = 1016 ç 1 -

xö ÷ cm -3 for 0  L ø

(D) 4 ´ 1015 cm -3, p - type

39. The majority carrier mobility is

£  x £ L

where L = 10 m m. The hole diffusion coefficient is

(A) 430 cm 2 V

-s

(B) 215 cm 2 V

-s

(C) 390 cm 2 V

-s

(D) 195 cm 2 V

-s

10 cm 2 s. Th The e ho hole le di diff ffus usio ion n cu curr rren entt de dens nsit ity y at

 x

semicon conduc ductor tor n0 40. In a semi

= 5 mm is

(A) 20 A cm 2

(B) 16 A cm 2

2

2

(C) 24 A cm

(D) 30 A cm

cm -3. The excess-carrier life time is 10 -6 s. The excess holle ho

conc co ncen entr trat atio ion n

is

d p = 4 ´ 1013

cm -3.

electron-hole recombination rate is

partic ticula ularr semi semicon conduct ductor or sam sample ple con consid sider er 36. For a par following parameters:

= 1015 cm -3 and ni = 1010

(A) 4 ´ 1019 cm -3s -1

(B) 4 ´ 1014 cm -3s -1

(C) 4 ´ 10 24 cm -3s -1

(D) 4 ´ 1011 cm -3s -1

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Th e

Chap 2.1

Semiconductor Physics

111

Solutions

semico icondu nducto ctorr in the therma rmall equi equilib libriu rium, m, has a 41. A sem hole concentration concentration of  p0 concentrati conce ntration on of  ni life

ti m e

minori ority ty car carrie rierr = 1010 cm -3. The min

4 ´ 10 -7s.

is

intrin rinsic sic = 1016 cm -3 and an int The Th e

ther th erma mall

equi eq uili libr briu ium m

= Nc Nv

2 i

1. (D) n

recombination rate of electrons is

æ - E g ö - çç ÷  kT  ÷ e è   ø

æ 400 ö = 00345 . = 00259 ç ÷ . è 300 ø

(A) 2.5 ´ 10 22 cm -3 s -1

(B) 5 ´ 1010 cm -3 s -1

V t

´ 1010 cm -3 s -1

(D) 5 ´ 10 22 cm -3 s -1

For Ge at 300 K,

(C) 2.5

 N c

Statement for Q.42-43:

  A nn-ty type pe si sili liccon sa samp mple le co cont ntai ains ns a

dono do norr

minori ority ty car carrie rierr = 106 cm -3. The min hole lifetime is t p p00 = 10 m s. concentrati conce ntration on of  N d

04 ´ 1019, N v = 6.0 ´ 1018 , E g = 0.66 eV  = 1.04 æ  0 .66 ö

3

æ 400 ö ´ e- çè 0 .0345 ø÷ 10 0 ´ 6.0 ´ 1 10 0 ´ç n = 1.04 ´ 1 ÷ è 300 ø 2 i

19

18

Þ ni = 8.5 ´ 1014 cm -3

42. The thermal equilibrium generation rate of hole is

- E g ö - æ  çç ÷  kT   ø÷ è  e

= Nc Nv

2 i

(A) 5 ´ 108 cm -3 s -1

(B) 10 4 cm -3 s -1

2. (C) n

(C) 2.25 ´ 10 9 cm -3 s -1

(D) 10 3 cm -3 s -1

æ  T  ö  e - çè   kT  ÷ø (10 ) = 2.8 ´ 1 10 0 ´ 1.04 ´ 1 10 0 ç ÷ è 300 ø

3

12 2

The e th ther erma mall eq equi uili libr briu ium m ge gene nera rati tion on ra rate te fo forr 43. Th

T e

(A) 112 1125 . 5 ´ 109 cm -3 s -1

(B) 225 . ´ 10 9 cm -3 s -1

(C) 8.9 ´ 10 -10 cm -3 s -1

(D) 4 ´ 10 9 cm -3 s -1

type pe si sili lico con n sa samp mple le co cont ntai ains ns a do dono norr 44. A  n - ty minori ority ty car carrie rierr = 1016 cm -3. The min hole lifetime is t p 0 = 20 m s. The lifetime of the majority carrier is ( ni = 1.5 ´ 1010 cm -3) (A) 8.9 ´ 106 s (B) 8.9 ´ 10 -6 s (C) 4.5 ´ 10

(D) 1.13 13 ´ 10

s

-7

n2 3. (B) iA 2 niB

4. (A) p0

5. (A) p0

s

silico icon n semi semicon conduc ductor tor mat materi erial al the dopi doping  ng  45. In a sil

= 1016 cm -3 and N a = 0. The equilibriu equil ibrium m recomb recombinatio ination n rate is R p p00 = 1011 cm -3s -1 . A 

28 ´ 10 -8 , By trial T  = 382 K  = 9.28

T 

concentrati conce ntration on of  N d

-17

6. (B) p0

=

 e  e

=

-

-

ni2 n0

 E gA  kT   E gB

=

æ  E gA- E - çç  kT   e è 

concentrati conce ntration on of 

The e fa fact ctor or,, by dn = dp = 1014 cm -3. Th

which the total recombination rate increase is (A) 2.3 ´ 1013

(B) 4.4 ´ 1013

(C) 2.3 ´ 10 9

(D) 4.4 ´ 10 9

 ö ÷÷  ø

gB

22575 575 . = 22 Þ

 kT 

(1.5 ´ 1010 ) 2

=

= Nv e

5 ´ 10 4

-

= Nv e

-

04 ´ 10  e = 1.04

( E F - Ev )  kT 

= 47.5

-0 .22 19

 kT 

niA niB

= 4.5 ´ 1015 cm -3

( E F - Ev )

concentrati conce ntration on are N a

unifor uni form m gen genera eratio tion n rat rate e pro produc duces es an exc excessess- car carrie rierr

æ 1 .12 e ö

19

-13´10 3 3

electron is

19

Þ

E F

0 .0259

= 2 ´ 1015 cm -3

æ  N  ö - Ev = kT ln   ç v÷ è   p0  ø

= 1.0 ´ 1019 cm -3 æ 1.04 04 ´ 1019 ö . lnç  E F - Ev = 00259 ÷ = 0.239 eV  15 è  10  ø

  At 300 K, N v

( Ec - EF  )

n0

= Nc e

 kT 

= 2.8 ´ 1019 cm -3 12 - 0.239 = 0.881 eV    Ec - EF  = 1.12 n0 = 4.4 ´ 10 4 cm -3

  At 300 K, N c

***********

7. (C) p0

=

For Ge ni

Na

- N d 2

2

N  ö 2 + æ  ç  N a - d ÷ + ni 2 è   ø

= 2.4 ´ 10 3 2

æ 1013 ö 95 ´ 1013 cm -3  p0 = + ç ÷ + (2.4 ´ 1013) 2 = 2.95 2 è  2  ø 1013

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112

Semiconductor Physics

8. (A) n0

=

5 ´ 1015 2

Nd

=

- N a 2

2

N  ö + æ  ç  N d - a ÷ + ni2 2  ø è 

Chap 2.1

2

æ 1014 ö + ç n0 = ÷ + (2.4 ´ 1013) 2 = 1.055 ´ 1014 cm -3 2 2 è   ø 1014

æ n0 ö æ 1.05 055 ´ 1014 ö 00259 . ln = ÷ ç ÷ 13 è  ni  ø è  2.4 ´ 10  ø

2

æ 5 ´ 1015 ö + ç ÷ + (2.4 ´ 1013) 2 = 5 ´ 1015 cm -3 è  2  ø

E F- E F=i kTln  ç . eV  = 00383

9. (B) p0

=

2 i

n  p0

=

(2.4 ´ 10 )

13 2

2.95 95 ´ 1013

95 ´ 1013 cm -3 = 1.95

10 = 1.5 ´ 10 cm = Na - Nd  = 2.5 ´ 10 - 1 ´ 10 3

p0

3

13

æ 200 ö = 0.01 . 73 eV 11.. (D) kT  = 00259 11 ç ÷ 0173 è 300 ø

ni

-3

 N c

300 . 075 ´ 1015 cm -3 Þ n0 = 30 504 ´ 1014 cm -3 ni = 1.50

 

n

= 4.7 ´ 10 cm ,

N v

= 7.0 ´ 10 , 18

æ 200 ö 1 017 ´ 7.0 ´ 10 1 018 ç = 4.7 ´ 10 ÷ è 300 ø

ni2

= 0.05 n0

2 i

-3

3

E g

= 1.42 eV 

1 .42 ö - æ  ç 0 .0173 ÷ è  ø e

æ - E g ö - çç ÷  kT  ÷ e è   ø

= Nc Nv

For GaAs at T  = 300 K,

 N c

= 4.7 ´ 1017,

n02 5

n0

= 5 ni = 3.3 cm -3 æ 400 ö = 00345 eV ÷ . è 300 ø

= Nc Nv

04 ´ 1019 cm -3 , N v = 6 ´ 1018 cm -3 , E g = 0.66 eV  = 1.04 3

 p0

19

æ  0 .66 ö

18

28 ´ 1014 cm -3 = 8.5 28

=

Na 2

2

N  ö + æ  ç a ÷ + ni2 è  2  ø

æ 1015 ö 10 = + ç ÷ + 7. 274 ´ 10 29 2 è  2  ø

. eV  = 0019

13. (A) n0

=

For Ge at T  = 300 K, ni

300 0ö æ 1 .42 ´30

-3

 Ec - EF  ö - æ  ç ÷ è  e  kT   ø

19

0 .215 ö - æ  ç 0 .0259 ÷ è   ø e

19 ´ 1016 cm -3 = 1.19

F

 L  RA

=

20 V  =  I  100m 2 ´ 10 -3

(200) (000 0001 . 1)

= 200 W = 0.01( W - cm) -1

1350.. s »  em n n0 , For Si, m n = 1350 01 = ( 1.6 ´ 10 10 -19)(1350) n0 Þ 0.01 n0 = 4.6 ´ 1013 cm -3, n0 >> ni Þ n0 = N d

19. (B) N d

= 2.4 ´ 10 cm

4

æ m p*  ö ÷ . eV  ç m p* ÷ = 00446 è   ø

kT ln  ç

æ  N  ö   ç d÷ - E F=i kT ln è  N i  ø æ  1015 ö 00259 . ln . eV  = ç ÷ = 0287 10 1 . 5 1 0 ´ è   ø 17. (A) E

2

N  ö 2 + æ  ç d ÷ + ni è  2  ø 13

3

= 5 ´ 10 + 2.8 ´ 10 15

 N d

s=

489 ´ 1015 cm -3 Þ  p0 = 1.48 æ  p0 ö æ 1.4 89 89 ´ 1015 ö ln 00345 . ln = = E FE k T   ç ÷ ç ÷ i F   28 ´ 1014  ø è  n0  ø è 8.5 28

Nd 2

= 1.42 eV 

= 2.8 ´ 1019 cm -3

18. (D) R =

2

15

midga= p

= Nd - Na = Nc

For Si, N c

æ 400 ö e- çè 0 .0345 ø÷ = 7. 274 ´ 10 29 10 0 ´ 6´1 10 0 ç n = 1.04 ´ 1 ÷ è 300 ø ni

Fi E

16. (A) n0

- E g ö - æ  çç ÷  kT  ÷ e è   ø

2 i

E g

18

 

For Ge at 300 K,

 N c

17

15. (A) E

. 12. (A) kT  = 00259 ç

ni2

= 7 ´ 1018 ,

æ  T  ö  e- çè  0 .0259T  ø÷ (1.504) = 4.7 ´ 10 10 ´ 7 ´ 1 10 0 ç ÷ è 300 ø By trial and error T  = 763 K 

Þ ni = 1.48 cm -3 = n0 p0 = 5 p02 =

N v

3

2

ni2

2

N  ö 2 + æ  ç d ÷ + ni è  2  ø

Þ n0 = 1.5 ´ 1015 + (1.5 ´ 1015) 2 + (0.05 n0) 2

For GaAs at 300 K, 17

=

14. (A) n0

> N d , thus material is p-type ,

10. (A) Since N a

Nd 2

 R =

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L s A

=

>> ni Þ

 L , em n Nd A

n0

= N d , s »  e m n n0 ,

Chap 2.1

Semiconductor Physics

0.1

=

(1.6 ´ 10

-19

)(1100)(5 ´ 10 )(100 ´ 10 )

L , s A

20. (A) R =

Þ n0 = n0

-8

16

s »  e m n n0 ,

R=

1 0 -19)(10 7)(1016 ) = 1.6 A cm 2 = evn = (1.6 ´ 10

26. (A) J

L em n n0 A

J

L , n s A 0

= 8.7 ´ 1015 cm -3

10 ´ 10 -4

=

(1. 6 ´ 10 1 0 -19)( 800)(10 -6 )(10)

Efficiency

=

n0  N d

´ 100 =

V  6 22. (D) E = =  L 2

mn

v = d  E

23. (D)

=

10 4 3

L

=

em n AR

= 7.81 ´ 1017 cm -3

7.8 ´ 1017

´ 100 = 78.1 %

1018

= 3 V/cm, vd = m n E,

= Nc Nv

2 i

= 3333 cm V - s

s1 = eni (m n + m p )

10 -19)(1000 + 600) ni = (1.6 ´ 10  At T  = 300 K, ni = 3.91 91 ´ 10 9 cm -3 æ  E g ö - çç ÷  kT  ÷ e è   ø

æ  N N  ö Þ E g = kTln  ç c 2 v ÷ è  ni  ø æ  1019 ö . ) ln ç . eV  Þ  E g = 2(00259 ÷ = 1122 9 91 ´ 10  ø è 3.91 æ 500 ö = 0.04  At T  = 500 K , kT  = 00259 eV,, . 32 eV ç ÷ 0432 è 300 ø n

ni2

= Nc Nv

= (1019) 2 e

1 .122 ö - æ  ç 0 .0432 ÷ è   ø

30. (A)

-3

29 ´ 10 cm Þ ni = 2.29 -19 29 ´ 1013)(1000 + 600) = (1.6 ´ 10 )(2.29 = 5.86 ´ 10 -3(W - cm) -1

 N d

=

r= 1

r em n

25. (B) J n

1

=

=0 =

1 5(1.6 ´ 10 -19)(1350)

= eDn

ni2  p0

25 ´ 1014 cm -3 = 9.25

dn dx

æ 1018 - 1016 ö 10 -19)( 35)ç = (1.6 ´ 10 = 2.8 ´ 10 4 A cm 2 -4 ÷ è  2 ´ 10  ø

ni2  p0

+  em p p0 ,

( -1) em n ni2

 p02

+  em p 1

= 7. 2 ´ 1011 cm -3 31. (B)

2s i

m pm n = 2  en i m pm n m p + m n

smin =

1 0 -19( 3.6 ´ 10 1 012 ) (7500)( 300) = 2 ´ 1.6 ´ 10

= 1.7 ´ 10 -3(W - cm) -1 32. (B)

s=

1

r

1

r1

-

=

ni1 ni 2

s  em n N d =

18 ´ 1019 = 7.18

æ m  ö 2 7500 ö 2 Þ  p0 = ni çç n ÷÷ = 3.6 ´ 1012 æ  ç ÷ è  300  ø è m p  ø

r2

1

1 .1 ö - æ  ç 0 .0259 ÷ è   ø )  e

19

1

1

24. (B)

- E g ö - æ  çç ÷  kT   ø÷ è  e

s = em n n0 + em p p0 and n0 =

cm -3,

13

= N d

Þ ni = 8.47 ´ 10 9 cm -3 Nd >> ni Þ Nd = n0 J = s E = em n n0 E = (1.6 ´ 10 -19)(1000)(1014 )(100) = 1.6   A cm2

ds dp0

10 -6

2 i

10 = (2 ´ 10 )(1 ´ 10 19

Þ s =  em n

2

n0

= em n n0 E = (1.6 ´ 10 -19)(7500)(1016 )(10) = 120   A cm2

29. (D) n

s »  e m n n0 , R =

(0.06 0 67)(9.1 ´ 10 -31)

>> ni Þ

28. (A) N d

= 0.9 N d

21. (B)

=

(1.6 ´ 10 -19)(10 -13)(10 5)

= 26.2 ´ 10 3 m s = 2.6 ´ 106 cm s

em n AR

(0.9)(1. 6 ´ 10 -19)( 8000)(10 -6)(2 ´ 10 3)

et sc E m e*

=

27. (A) vd

L

20 ´ 10 -4

=

. kW = 1136

113

0.1 =  e

-

=

 e -

 e

 E g æ  1

ç

2 kè  T1

-

=  emni ,  E g 2 kT 1  E g 2 kT 2

1 ö ÷ T2 ø 

æ  330 - 300 ö ç ÷ = ln 10 2 k è  330 ´ 300 ø 300) ln 10 = 1.31 eV   E g = 22( k30  E g

33. (D)

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1

m

=

1

m1

+

1

m2

+

1

m3

114

1

m

Semiconductor Physics

1

=

500

+

1 750

1

+

ni2 41. (C) n0 =  p0

Þ m = 250 cm 2 V - s

1500

æ 5 ´ 1014 - n(0) ö 10 -19)(25)ç = (1.6 ´ 10 ÷, 0010 . è   ø 1 013 cm -3 n(0) = 2.5 ´ 10

0.19

16 dp d æ  16 æ   x ö ö e10 D p 35. (B) J = -eD p = -  eDp ç 10 ç 1 - ÷ ÷ = dx dx è   L è   L ø ø

(1.6 ´ 10 10 -19)(1016 )(10) 10 ´ 10 -4

= e Dn

J = J p

+ Jn   =

=

=

1015 eD p

 L p

5 ´ 1014 eDn

1015 eD p

 L p

+

5 ´ 1014 eDn

 Ln

= N d = 106 cm -3

25 ´ 10 4 cm -3 = 2.25 25 ´ 10 9 cm -3s -1 = 2.25

10 ´ 10 -6

Þ

G

25 ´ 10 9 cm -3s -1 = Rn 0 = Rp 0 = 2.25

tp 0

,

ni2 n0

= N d , p0 = n0  p0

tn 0 =

n02 ni2

tp 0 =

t p 0

= -0.313 mV  45. (D) N d  p0

I B  p = x z   eV eV H  d

=

 R p 0

(1.6 ´ 10 -19)(5.8 ´ 10 -3)(10 -5)

 R p

= 8.08 ´ 10 21 m -3 = 8.08 ´ 1015 cm -3 I x L 39. (C) u p =  epV x Wd

ni2 n0

=

=

 R p  R p 0

=

p0

t p 0

>> ni

Þ

(1.5 ´ 1010 ) 2 16

10

Þ

n0

= Nd  

= 2.25 ´ 10 4 cm -3

t p 0 =

p0  R p 0

=

2. 25 ´ 10 4 1011

(1.6 ´ 10 -19)( 8.08 ´ 10 21 )(15)(10 -4 ) (10 -5)

=

4.44 ´ 10 20 1011

= 4.44 ´ 10 9

***********

m p = 3.9 ´ 10 m V - s = 390 cm V - s 2

40. (A) n-type semiconductor

dp 4 ´ 1013 = = 4 ´ 1019 cm -3s -1 -6 10 t p 0

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= 2.25 ´ 10 7 s

1014 dp = = 4.44 ´ 10 20 cm -3s -1 25 ´ 10 -7 t p 0 2.25

(0.75 75 ´ 10 -3)(10 -3)

 R =

p0

2

ned

2

tn 0

=

>> ni

 N d

(0.75 75 ´ 10 -3)(10 -1)

-2

n0

æ  1016 ö =ç ÷ ´ 20 ´ 10 -6 = 8.9 ´ 106 s 10 è 1.5 ´ 10  ø

38. (B) V  H  is positive p-type

=

ni2 , n0 n0

Recombination rate.

n0

- I x Bz

(5 ´ 10 10 21 )(1.6 ´ 10 10 -19)(5 ´ 10 10 -5)

 p =

=

, p0

t p 0

2.25 25 ´ 10 4

14

(250 ´ 10 -6 )(5 ´ 10 -2 )

V  H  =

= 2.5 ´ 1010 cm -3s -1

44. (A) Recombination rates are equal

æ  10 (10) 5 ´ 10 (25) ö = 1.6 ´ 10 -19ç + ÷ = 5. 2 A cm 2 -4 -3 1 5 1 0 10 ´ è   ø

I xB z  epd

p0

=

-7

= 10 4 cm -3

 Ln

15

=

4 ´ 10

1016

=

 Rn 0

10

10 4

(1.5 ´ 1010 ) 2

=

16

carr ca rrie ierr ar are e eq equa uall . Th The e ge gene nera rati tion on ra rate te is eq equa uall to

dn0 dx  x = 0

37. (A) V  H  =

tn 0

=

=

(1010 ) 2

43. (B) Recombination rate for minority and majority

2

dp 36. (B) J p = -eDp 0 dx  x = 0 Jn

n0

42. (C) Rn 0  p0

 J  = 16 A cm

=

 Rn 0

dn 34. (B) J n = eDn dx

=

Chap 2.1