半导体物理与器件第四版课后习题答案8

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Semiconductor Physics and Devices: Basic Principles, 4th edition

Chapter 8

By D. A. Neamen

Problem Solutions

______________________________________________________________________________________

1

Chapter 8

8.1

In forward bias

kTeVIISfexp

Then

212121expexpexpVVkTekTeVIkTeVIIISSff

or

2121lnffIIekTVV

(a)

For 1021ffII, then

10ln0259.021VV

or

6.5921VVmV60mV

(b)

For 10021ffII, then

100ln0259.021VV

or

3.11921VVmV120mV

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8.2

4152102108125.2108105.1aipoNnncm3

515210210125.1102105.1dinoNnpcm3

tanonnVVpxpexp

tapoppVVnxnexp

(a) 45.0aVV, 0259.045.0exp10125.15nnxp

121095.3cm3

0259.045.0exp108125.24ppxn

or

111088.9ppxncm3

(b) 55.0aVV,

0259.055.0exp10125.15nnxp

141088.1cm3

0259.055.0exp108125.24ppxn

131069.4cm3

(c) 55.0aVV

0259.055.0exp10125.15nnxp

0

0259.055.0exp108125.24ppxn

0

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8.3

516262101.8104108.1aipoNnncm3

4162621024.310108.1dinoNnpcm3

(a) 90.0aVV,

0259.090.0exp1024.34nnxp

11100.4cm3

0259.090.0exp101.85ppxn Semiconductor Physics and Devices: Basic Principles, 4th edition

Chapter 8

By D. A. Neamen

Problem Solutions

______________________________________________________________________________________

2

10100.10cm3

(b) 10.1aVV

0259.010.1exp1024.34nnxp

141003.9cm3

0259.010.1exp101.85ppxn

141026.2cm3

(c) 0nnxp

0ppxn

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8.4

(a) 162102105105.1aipoNnn

3105.4cm3

152102105105.1dinoNnp

4105.4cm3

(i)tanonnVVpxpexp

or nonntapxpVVln

415105.41051.0ln0259.0

599.0V

(ii) n-region - lower doped side

(b) 152102107105.1aipoNnn

410214.3cm3

162102103105.1dinoNnp

3105.7cm3 (i) poatanNVV1.0ln

41510214.31071.0ln0259.0

6165.0V

(ii) p-region - lower doped side

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8.5

(a) tanponpnVVLneDxJexp

tanonaiVVDNenexp2

8162619105205105108.1106.1

0259.010.1exp

849.1A/cm2

849.1103pnnxAJIA

or 85.1nImA

(b) tapnopnpVVLpeDxJexp

tappdiVVDNenexp02

81626191080.910108.1106.1

0259.010.1exp

521.4A/cm2

521.4103nppxAJIA

or 52.4pImA

(c) 37.652.485.1pnIIImA

_______________________________________

8.6

For an pn silicon diode Semiconductor Physics and Devices: Basic Principles, 4th edition

Chapter 8

By D. A. Neamen

Problem Solutions

______________________________________________________________________________________

3

nOnaiSDNAenI12

616210194102510105.1106.110

or

15108.1SIA

(a) For 5.0aVV,

taSDVVIIexp

0259.05.0exp108.115

or

71036.4DIA

(b) For 5.0aVV,

10259.05.0exp108.115DI

or

15108.1SDIIA

_______________________________________

8.7

0211ppdnonaisDNDNenJ

21319104.2106.1

617615102481021102901041

410568.1sJA/cm2

(a) tasVVAJIexp

0259.025.0exp10568.11044 41044.2A

or 244.0ImA

(b) 4410568.110ssAJII

810568.1A

_______________________________________

8.8

(a) 0211ppdnonaisDNDNenJ

21019105.1106.1

81571710810108110251051

1110145.5sJA/cm2

11410145.5102ssAJI

1410029.1A