Microelectronic circuits analysis and design 3rd edition rashid solutions manual 1

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Solution Manual for Microelectronic Circuits Analysis and Design 3rd Edition Rashid 1305635167 9781305635166 Download full solution manual at: https://testbankpack.com/p/solution-manual-for-microelectronic-circuitsanalysis-and-design-3rd-edition-rashid-1305635167-9781305635166/

Chapter 5 5.1

= 0.62V , V = 25.8 mV

I = 2.2 x10−14 A, V s

BE

T

(a) Using Eq. (5.6)

I C = I S eVBE /VT = 2.2 x10−14 e0.62/ 0.0258 = 6.011 x10 −4 A −4

I C 6.011 x10 = = 6.011 x10 −6 A F 100

(b)

IB =

(c)

I = I + I = 6.011 x10 −6 + 6.011x10 −4 = 6.17 x10 −4 A E

B

C −4

(d)

6.011 x10 I  = C = 6.17 x10−4 = 0.974 F I E

5.2

I = 2.2 x10 −14 A,  = 100, V S

F

= 0.65V

EB

(a) Using Eq. (5.6)

I C = I S eVEB /VT = 2.2 x10 −14 e0.65/0.0258 = 1.9228 x10−3 A −3

(b)

I = B

(c)

I = I + I = 1.9228 x10−5 +1.9228 x10 −3 = 1.94 x10−3 A E

(d)

I C = 1.9228x10 = 1.9228 x10 −5 A 100 

B

C

1.9228 x10 −3 I = 0.991 F = C = 1.94x10 −3 IE

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