Electronics Chapter02

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CHAPTER 2

2.1 Based upon Table 2.1, a resistivity of 2.6 μΩ-cm < 1 mΩ-cm, and aluminum is a conductor.

2.2

Based upon Table 2.1, a resistivity of 1015

Ω-cm > 105 Ω-cm, and silicon dioxide is an insulator.

2.3

I max = 107 A

cm2

⎛

⎝ ⎜

⎠⎟ 5μ m( )1μ m( ) 10−8cm2

μ m2

⎝ ⎜

⎠⎟ = 500 mA

2.4

⎟ ⎠

⎞⎜⎝

⎛ −= − T x

E BT n G

i 5

3

1062.8exp

For silicon, B = 1.08 x 1031

and EG = 1.12 eV:

ni = 2.01 x10-10

/cm3

6.73 x109 /cm

38.36 x 10

13 /cm

3.

For germanium, B = 2.31 x 1030

and EG = 0.66 eV:

ni = 35.9/cm3

2.27 x1013 /cm

38.04 x 10

15 /cm

3.

2.5 Define an M-File:

function f=temp(T)ni=1E14;f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T));

ni = 1014 /cm

3for T = 506 K ni = 10

16 /cm

3for T = 739 K

2.6

ni = BT 3 exp − E G8.62 x10−5

T

⎛

⎝ ⎜ ⎞

⎠⎟with B = 1.27x1029 K −3cm−6

T = 300 K and EG

= 1.42 eV: ni= 2.21 x10

6 /cm

3

T = 100 K: ni = 6.03 x 10-19

/cm3

T = 500 K: ni = 2.79 x1011 /cm

3

20



2.7

vn = −μ n E = −700cm2

V − s

⎝ ⎜

⎠⎟ 2500

V

cm

⎛

⎝ ⎜

⎠⎟ = −1.75 x106 cm

s

v p = +μ p E = +250

cm2

V − s

⎛

⎝ ⎜

⎞

⎠⎟ 2500

V

cm

⎛

⎝ ⎜

⎞

⎠⎟ = +6.25 x105 cm

s

jn = −qnvn = −1.60 x10−19C ( )1017 1

cm3

⎛

⎝ ⎜

⎞

⎠⎟ −1.75 x106 cm

s

⎛

⎝ ⎜

⎞

⎠⎟ = 2.80 x104 A

cm2

j p = qnv

p = 1.60 x10−19C ( )103 1

cm3

⎛

⎝ ⎜

⎞

⎠⎟ 6.25 x105 cm

s

⎛

⎝ ⎜

⎞

⎠⎟ =1.00 x10−10 A

cm2

2.8

ni

2 = BT 3 exp −

E G

kT

⎝ ⎜

⎠⎟ B = 1.08 x1031

1010( )2

=1.08 x1031T 3 exp −1.12

8.62 x10−5T

⎛

⎝ ⎜

⎞

⎠⎟

Using a spreadsheet, solver, or MATLAB yields T = 305.22K

Define an M-File:function f=temp(T)f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T));

Then: fzero('temp',300) | ans = 305.226 K

2.9

scm

cmC cm A

Q jv 5

2

2

10 / 01.0 / 1000 −=−==

2.10

22

67

34104

sec104.0

cm

MA

cm

A x

cm

cm

C Qv j ==⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ ==

21



2.11

vn = −μ n E = −1000cm

2

V − s

⎝ ⎜

⎠⎟ −2000

V

cm

⎛

⎝ ⎜

⎞

⎠⎟ = +2.00 x106 cm

s

v p = +μ p E = +400cm

2

V − s

⎛

⎝ ⎜

⎞

⎠⎟ −2000

V

cm

⎛

⎝ ⎜

⎞

⎠⎟ = −8.00 x105 cm

s

jn = −qnvn = −1.60 x10−19C ( )103 1

cm3

⎛

⎝ ⎜

⎞

⎠⎟ +2.00 x106 cm

s

⎛

⎝ ⎜

⎞

⎠⎟ = −3.20 x10−10 A

cm2

j p = qnv p = 1.60 x10−19C ( )1017 1

cm3

⎛

⎝ ⎜

⎞

⎠⎟ −8.00 x105 cm

s

⎛

⎝ ⎜

⎞

⎠⎟ = −1.28 x104 A

cm2

2.12

( ) ( ) ( ) V100101010 b 50001010

5 45

4=⎟

⎠ ⎞

⎜⎝ ⎛ === −

− cm xcm

V V

cm

V

cm x

V E a

2.13

j p = qpv p = 1.60 x10−19C ( )1019

cm3

⎛

⎝ ⎜

⎞

⎠⎟ 107 cm

s

⎛

⎝ ⎜

⎞

⎠⎟ =1.60 x107 A

cm2

i p = j p A = 1.60 x107 A

cm2

⎛

⎝ ⎜

⎞

⎠⎟ 1 x10−4

cm( )25 x10−4cm( )= 4.00 A

2.14

For intrinsic silicon, σ = q μ nni + μ pni( )= qni μ n + μ p( )

σ ≥1000 Ω − cm( )−1

for a conductor

ni ≥σ

q μ n + μ p( )=

1000 Ω − cm( )−1

1.602 x10−19C 100 + 50( ) cm2

v − sec

=4.16 x1019

cm3

n i

2 =1.73 x10

39

cm6= BT 3 exp −

E G

kT

⎛

⎝ ⎜

⎞

⎠⎟ with

B = 1.08 x1031 K −3cm−6 , k = 8.62x10-5eV/K and EG =1.12eV

This is a transcendental equation and must be solved numerically by iteration. Using the HP

solver routine or a spread sheet yields T = 2701 K. Note that this temperature is far above themelting temperature of silicon.

22



2.15

For intrinsic silicon, σ = q μ nni + μ pni( )= qni μ n + μ p( )

σ ≤10−5 Ω − cm( )−1

for an insulator

ni ≥ σ q μ n + μ p( )=

10−5 Ω − cm

( )

−1

1.602 x10−19C ( )2000 + 750( ) cm2

v − sec

⎛

⎝ ⎜

⎞

⎠⎟

= 2.270 x10

10

cm3

n i

2 =5.152 x1020

cm6= BT 3 exp −

E G

kT

⎛

⎝ ⎜

⎞

⎠⎟ with

B =1.08 x1031 K −3cm−6, k = 8.62x10-5eV/K and EG =1.12eV

Using MATLAB as in Problem 2.5 yields T = 316.6 K.

2.16

Si Si Si

SiB

Si Si Si

P

Donor electron

fills acceptorvacancy

No free electrons or holes (except those corresponding to ni).

2.17

(a) Gallium is from column 3 and silicon is from column 4. Thus silicon has an extra electron

and will act as a donor impurity.(b) Arsenic is from column 5 and silicon is from column 4. Thus silicon is deficient in oneelectron and will act as an acceptor impurity.

2.18 Since Ge is from column IV, acceptors come from column III and donors come from columnV. (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi

23



2.19 (a) Germanium is from column IV and indium is from column III. Thus germanium has oneextra electron and will act as a donor impurity.(b) Germanium is from column IV and phosphorus is from column V. Thus germanium hasone less electron and will act as an acceptor impurity.

2.20

( ) field.electricsmalla,20002.0100002 cm

V cm

cm

A j

j E =−Ω⎟

⎠

⎞⎜⎝

⎛ === ρ σ

2.21

jn

drift = qnμ n E = qnvn = 1.602 x10−19( )1016( ) C

cm3

⎝ ⎜

⎠⎟ 107 cm

s

⎝ ⎜

⎠⎟ =16000

A

cm2

2.22

N = 1015

atomscm3

⎛

⎝ ⎜

⎞

⎠⎟ 1μ m( )10μ m( )0.5μ m( ) 10

−4

cmμ m

⎛

⎝ ⎜

⎞

⎠⎟

3

= 5,000 atoms

2.23

N A > N D: N A − N D =1015 −1014 = 9 x1014 /cm3

If we assume N A − N D >> 2ni =1014 / cm3 :

p = N A − N D = 9 x1014 /cm3 | n =ni

2

p=

251026

9 x1014= 2.78 x1012 /cm3

If we use Eq. 2.12: p =

9 x1014 ± 9 x1014( )2

+ 4 5 x1013( )2

2 = 9.03 x1014

and n = 2.77 x1012 /cm

3. The answers are essentially the same.

2.24

35

16

222314

3113161616

10502104

10104

102210410105

/cm x. x p

n| n /cm x N N p

/cm xn /cm x x N : N N N

i D A

i D A D A

====−=

=>>=−=−>

2.25

N D > N A: N D − N A = 3 x1017

− 2 x1017

=1 x1017

/cm3

2ni = 2 x1017 /cm3; Need to use Eq. (2.11)

n =1017 ± 1017( )

2

+ 4 1017( )2

2=1.62 x1017

/cm3

p =ni

2

n=

1034

1.62 x1017= 6.18 x1016

/cm3

24



2.26

N D − N A = −2.5 x1018 / cm3

U sing Eq. 2.11: n =−2.5 x1018 ± −2.5 x1018( )

2

+ 4 1010( )2

2

Evaluating this with a calculator yields n = 0, and n =ni

2

p= ∞.

No, the result is incorrect because of loss of significant digits

within the calculator. It does not have enough digits.

2.27

(a) Since boron is an acceptor, NA = 6 x 1018 /cm3. Assume ND = 0, since it is not specified.The material is p-type.

3

318

6202318

i

318310

7.16106

10 and 106So

2n>> / 106 and10re,temperaturoomAt

/cm /cm x

/cm

p

nn /cm x p

cm x N N /cmn

i

D Ai

=====−=

(b)

At 200K, ni

2 =1.08 x1031

200( )3

exp −1.12

8.62 x10−5 200( )⎛

⎝ ⎜⎜

⎞

⎠⎟⎟ = 5.28 x10

9 /cm6

ni = 7.27 x104 /cm3 N A − N D >> 2ni , so p = 6 x1018 /cm3 and n =5.28 x109

6 x1018= 8.80 x10−10 /cm3

2.28 (a) Since arsenic is a donor, ND = 3 x 1017 /cm3. Assume NA = 0, since it is not specified. Thematerial is n-type.

3

317

6202317

i

317310

i

333103

10 and 103So

2n>> / 103 and / 10nre,temperaturoomAt

/cm /cm x

/cm

n

n p /cm xn

cm x N N cm

i

A D

====

=−=

(b) At 250K, ni

2 =1.08 x1031 250( )3

exp −1.12

8.62 x10−5 250( )⎛

⎝ ⎜⎜

⎠⎟⎟ = 4.53 x1015 /cm6

ni = 6.73 x10

7

/cm

3

N D − N A >> 2ni , so n = 3 x10

17

/ cm

3

and n =

4.53x1015

3x1017 = 0.0151/ cm

3

2.29

(a) Arsenic is a donor, and boron is an acceptor. ND = 2 x 1018 /cm3, and NA = 8 x 1018 /cm3.

Since NA > ND, the material is p-type.

25



3

318

6202318

i

318310

i

716106

10 and 106So

2n>> / 106 and / 10nre,temperaturoomAt(b)

/cm. /cm x

/cm

p

nn /cm x p

cm x N N cm

i

D A

====

=−=

2.30

(a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 1017 /cm3, and NA = 5 x 1017 /cm3.

Since NA > ND, the material is p-type.

3

317

6202317

i

317310

333103

10 and 103So

2n>> / 103 and10re,temperaturoomAt(b)

/cm /cm x

/cm

p

nn /cm x p

cm x N N /cmn

i

D Ai

====

=−=

2.31

ND = 4 x 1016 /cm3. Assume NA = 0, since it is not specified.

N D > N A : material is n - type | N D − N A = 4 x1016 / cm3 >> 2ni = 2 x1010 / cm

3

n = 4 x1016 / cm3 | p =

n i

2

n=

1020

4 x1016= 2.5 x103 / cm

3

N D + N A = 4 x1016 / cm3 | Using Fig. 2.13, μ n =1030

cm2

V − sand μ p = 310

cm2

V − s

ρ =1

qμ nn=

1

1.602 x10−19C ( )1030cm

2

V − s

⎛

⎝ ⎜

⎞

⎠⎟

4 x1016

cm3

⎛

⎝ ⎜

⎞

⎠⎟

= 0.152 Ω − cm

26



2.32

NA = 1018 /cm

3. Assume ND = 0, since it is not specified.

N A > N D : material is p - type | N A − N D =1018 / cm3 >> 2ni = 2 x1010 / cm3

p = 1 018 / cm3 | n =n i

2

p

=10

20

1018

=100/ cm3

N D + N

A =1018 / cm3 | Using Fig. 2.13, μ n = 375cm2

V − sand μ p =100

cm2

V − s

ρ =1

qμ p p=

1

1.602 x10−19C 100

cm2

V − s

⎛

⎝ ⎜

⎞

⎠⎟

1018

cm3

⎛

⎝ ⎜

⎞

⎠⎟

= 0.0624 Ω − cm

2.33

Indium is from column 3 and is an acceptor. NA = 7 x 1019 /cm

3. Assume ND = 0, since it is not

specified.

N A > N

D : material is p - type | N A − N

D = 7 x1019 /cm3 >> 2ni = 2 x1010 /cm3

p = 7 x1019 /cm3 | n=

ni

2

p=

1020

7 x1019

=1.43 /cm3

N D + N A = 7 x1019 / cm3 | Using Fig. 2.13, μ n =120cm2

V − sand μ p = 60

cm2

V − s

ρ =1

qμ p p=

1

1.602 x10−19C 60cm2

V − s

⎛

⎝ ⎜

⎞

⎠⎟

7 x1019

cm3

⎛

⎝ ⎜

⎞

⎠⎟

= 1.49 mΩ − cm

2.34

Phosphorus is a donor : N D = 5.5 x1016 / cm3 | Boron is an acceptor : N A = 4.5 x1016 / cm3

N D > N A : material is n - type | N D − N A =1016 / cm3 >> 2ni = 2 x1010 / cm3

n =1016 /cm3 | p =ni

2

p=

1020

1016

=104 /cm3

N D + N A =1017 / cm3 | Using Fig. 2.13, μ n = 800cm2

V − sand μ p = 230

cm2

V − s

ρ =1

qμ nn=

1

1.602 x10−19C 800cm2

V − s⎛ ⎝ ⎜ ⎞

⎠⎟

1016

cm3

⎛ ⎝ ⎜ ⎞

⎠⎟

= 0.781 Ω − cm

27



2.35

ρ =1

qμ p p| μ p p =

1

1.602 x10−19C ( )0.054Ω − cm( )=

1.16 x1020

V − cm − s

An iterative solution is required. Using the equations in Fig. 2.8:

NA μp μp p

1018 96.7 9.67 x 1020

1.1 x1018 93.7 1.03 x 1020

1.2 x 1017 91.0 1.09 x 1020

1.3 x 1019 88.7 1.15 x 1020

2.36

ρ =1

qμ p p

| μ p p =1

1.602 x10−19

C ( )0.75Ω − cm( )

=8.32 x1018

V − cm − s


NA μp μp p

1016 406 4.06 x

1018

2 x 1016 363 7.26 x

1018

3 x 1016 333 1.00 x

1019

2.4 x 1016 350 8.40 x 1018

2.37

Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsicbecause it contains impurities. Addition of the impurities has increased the resistivity.

2.38

ρ =1

qμ nn| μ nn ≈ μ n N D =

1

1.602 x10−19C

( )2Ω − cm

( )

=3.12 x1018

V − cm − s


ND μn μnn

1015 1350 1.35 x 1018

2 x 1015 1330 2.67 x 1018

2.5 x 1015 1330 3.32 x 1018

28



2.3 x 1015 1330 3.06 x 1018

29



2.39 (a)

ρ =1

qμ nn| μ nn ≈ μ n N D =

1

1.602 x10−19C ( )0.001Ω − cm( )=

6.24 x1021

V − cm − s


ND μn μnn

1019 116 1.16 x 1021

7 x 1019 96.1 6.73 x 1021

6.5 x 1019 96.4 6.3 x 1021

(b)

ρ =1

qμ p p| μ p p ≈ μ p N A =

1

1.602 x10−19C ( )0.001Ω − cm( )

=6.24 x10

21

V − cm − s

An iterative solution is required using the equations in Fig. 2.8:

NA μp μp p

1.3 x 1020 49.3 6.4 x 1021

2.40

Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, butthe hole and electron concentrations remain unchanged. See Problem 2.37 for example.However, it is physically impossible to add exactly equal amounts of the two impurities.

2.41

(a) For the 1 ohm-cm starting material:

ρ =1


1

1.602 x10−19C ( )1Ω − cm( )

=6.25 x1018

V − cm − s


NA μp μp p

1016 406 4.1 x 1018

1.5 x 1016

383 5.7 x 1018

1.7 x 1016 374 6.4 x 1019

30



To change the resistivity to 0.25 ohm-cm:

ρ =1


1

1.602 x10−19C ( )0.25Ω − cm( )

=2.5 x1019

V − cm − s

NA μp μp p

6 x 1016 276 1.7 x 1019

8 x 1016 233 2.3 x 1019

1.1 x 1017 225 2.5 x 1019

Additional acceptor concentration = 1.1x1017

- 1.7x1016

= 9.3 x 1016 /cm

3

(b) If donors are added:

ND ND + NA μn ND - NA μnn

2 x 1016 3.7 x 1016 1060 3 x 1015 3.2 x 1018

1 x 1017 1.2 x 1017 757 8.3 x 1016 6.3 x 1019

8 x 1016 9.7 x 1016 811 6.3 x 1016 5.1 x 1019

4.1 x 1016 5.8 x 1016 950 2.4 x 1016 2.3 x 1019

So ND = 4.1 x 1016 /cm

3must be added to change achieve a resistivity of 0.25 ohm-cm. The

silicon is converted to n-type material.

2.42

Phosphorus is a donor: ND = 1016 /cm3 and μn = 1250 cm2 /V-s from Fig. 2.8.

σ = qμ nn ≈ qμ n N D = 1.602 x10−19C ( )1250( )1016

( )= 2.00Ω − cm

Now we add acceptors until σ = 5.0 (Ω-cm) -1:

σ = qμ p p | μ p p ≈ μ p N A − N D( )=5 Ω − cm( )

−1

1.602 x10−19C =

3.12 x1019

V − cm − s

NA ND + NA μp NA - ND μp p

1 x 1017 1.1 x 1017 250 9 x 1016 2.3 x 1019

2 x 1017 2.1 x 1017 176 1.9 x 1017 3.3 x 1019

1.8 x 1017

1.9 x 1017

183 1.7 x 1016

3.1 x 1019

31



2.43

Boron is an acceptor: NA = 1016 /cm3 and μp = 405 cm2 /V-s from Fig. 2.8.

σ = qμ p p ≈ qμ p N A = 1.602 x10−19C ( )405( )1016( )=

0.649

Ω − cm

Now we add donors until σ = 5.5 (Ω-cm) -1:

σ = qμ nn | μ nn ≈ μ n N D − N A( )=5.5 Ω − cm( )

−1

1.602 x10−19C =

3.43 x1019

V − cm − s

ND ND + NA μn ND - NA μp p

8 x 1016 9 x 1016 832 7 x 1016 5.8 x 1019

6 x 1016 7 x 1016 901 5 x 1016 4.5 x 1019

4.5 x 1016 5.5 x 1016 964 3.5 x 1016 3.4 x 1019

2.44

V T =kT

q=

1.38 x10−23T

1.602 x10−19= 8.62 x10−5T

T (K) 50 75 100 150 200 250 300 350 400

VT (mV) 4.31 6.46 8.61 12.9 17.2 21.5 25.8 30.1 34.5

2.45

j = −qDn −dn

dx

⎛

⎝ ⎜

⎞

⎠⎟ = qV T μ n

dn

dx

j = 1.602 x10−19C ( )0.025V ( ) 350

cm2

V − s

⎛

⎝ ⎜

⎞

⎠⎟

1018 − 0

0 −10−4

⎛

⎝ ⎜

⎞

⎠⎟

1

cm4

= −14.0kA

cm2

2.46

j = −qD p

dp

dx= −1.602 x10−19C ( )15

cm2

s

⎝ ⎜

⎠⎟ −

1019 / cm3

2 x10−4cm

⎝ ⎜

⎠⎟exp −

x

2 x10−4cm

⎛

⎝ ⎜ ⎟

j =1.20 x105 exp −5000 x

cm

⎛

⎝ ⎜

⎞

⎠⎟

A

cm2

I 0( )= j 0( ) A = 1.20 x105 A

cm2

⎛

⎝ ⎜

⎞

⎠⎟ 10μ m2( )10−8cm2

μ m2

⎛

⎝ ⎜

⎞

⎠⎟ =12.0 mA

32



2.47

j p = qμ p pE − qD p

dp

dx= qμ p p E −V T

1

p

dp

dx

⎝ ⎜

⎠⎟ = 0 → E = V T

1

p

dp

dx

E ≈ V T

1

N A

dN A

dx

= 0.025−1022 exp −104 x( )

1014 +1018 exp −104 x( ) E 0( )= −0.025

1022

1014 +1018= −250

V

cm

E 5 x10−4cm( )= −0.025

1022 exp −5( )1014 +1018 exp −5( )

= −246V

cm

2.48

jn

drift = qμ nnE = 1.60 x10−19C ( )350cm2

V − s

⎝ ⎜

⎠⎟

1016

cm3

⎝ ⎜

⎠⎟ −20

V

cm

⎛

⎝ ⎜

⎠⎟ = −11.2

A

cm2

j pdrift = qμ p pE = 1.60 x10−19C ( )150

cm2

V − s

⎛

⎝ ⎜

⎞

⎠⎟

1.01 x1018

cm3

⎛

⎝ ⎜

⎞

⎠⎟ −20

V

cm

⎛

⎝ ⎜

⎞

⎠⎟ = −484

A

cm2

jn

diff = qDn

dn

dx= 1.60 x10−19C ( )350 ⋅ 0.025

cm2

s

⎛

⎝ ⎜

⎞

⎠⎟

104 −1016

2 x10−4cm4

⎛

⎝ ⎜

⎞

⎠⎟ = −70.0

A

cm2

j pdiff = −qD p

dp

dx= −1.60 x10−19

C ( )150 ⋅ 0.025cm

2

s

⎛

⎝ ⎜

⎞

⎠⎟

1018 −1.01 x1018

2 x10−4cm

4

⎛

⎝ ⎜

⎞

⎠⎟ = 30.0

A

cm2

jT = −11.2 − 484 − 70.0 + 30.0 = −535 A

cm2

2.49 NA = 2ND

EC

EA

NA

Holes

EV

NA

ND

NA

ND

NA

ND

ED

2.50

λ =hc

E =

6.626 x10−34 J − s( )3 x108 m / s( )1.12eV ( )1.602 x10−19

J / eV ( )=1.108 μ m

33



2.51

p-type silicon

n-type silicon

Si02

Al - Anode Al - Cathode

2.52 An n-type ion implantation step could be used to form the n+ region following step (f) in Fig.2.17. A mask would be used to cover up the opening over the p-type region and leave theopening over the n-type silicon. The masking layer for the implantation could just bephotoresist.

n-type silicon

p-type silicon

Si02

Photoresist

Structure after exposure anddevelopment of photoresist layer

Mask

n-type silicon

p-type silicon

Structure following ionimplantation of n-type impurity

n+

Ion implantation

Side view

Top View

Mask for ion implantation

2.53

(a) N = 81

8

⎝ ⎜

⎠⎟ + 6

1

2

⎝ ⎜

⎠⎟ + 4 1()= 8 atoms

(b) V = l3 = 0.543 x10−9

m( )3

= 0.543 x10−7cm( )

3

=1.60 x10−22cm

3

(c) D =

8 atoms

1.60 x1022cm3 = 5.00 x10

22 atoms

cm3

(d ) m = 2.33g

cm3

⎛

⎝ ⎜

⎞

⎠⎟1.60 x1022

cm3 = 3.73 x10−22

g

(e) From Table 2.2, silicon has a mass of 28.086 protons.

mp =3.73 x10−22

g

28.082 8( ) protons=1.66 x10−24 g

proton

Yes, near the actual proton rest mass.

34

Electronics Chapter02

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Transcript of Electronics Chapter02