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Microelectronics: Circuit Analysis and Design, 4 edition Chapter 1By D. A. Neamen Exercise Solutions Chapter 1####### Exercise SolutionsEX1.3/ 2 exp 2g iE nBT kT⎛⎞− = ⎜⎟ ⎝⎠GaAs: ()()()()14 3/ 2 61. 2 10 300 exp 2 86 10 300ni −⎛⎞ − =× ⎜⎟ ⎜⎟× ⎝⎠or63 nci 1 10 m− =×Ge: ()()()()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛×− = × − 286 10...

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Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

Chapter 1 Exercise Solutions EX1.1 ⎛ −E ⎞ ni = BT 3 / 2 exp ⎜ g ⎟ ⎝ 2kT ⎠

GaAs: ni = ( 2.1 ×1014 ) ( 300 )

⎛ ⎞ − 1.4 ⎟ or ni = 1.8 ×106 cm−3 exp ⎜ −6 ⎜ 2 (86 × 10 ) (300 ) ⎟ ⎝ ⎠

3/2

⎛ − 0.66 ⎞ ⎟ or ni = 2.40 × 1013 cm−3 exp⎜⎜ ⎟ −6 × ( ) ⎝ 2 86 10 300 ⎠ ______________________________________________________________________________________

(

)

Ge: n i = 1.66 × 10 15 (300 )

3/ 2

(

)

EX1.2 (a) (i)

no = N d = 2 ×1016 cm 2

po =

(

2

= 1.125 ×10 4 cm −3

−3

(

)

2

ni 1.5 × 1010 = po 10 15

no =

no = N d = 2 × 1016 cm 2

po =

(ii)

2

ni 1.5 × 1010 = no 2 × 1016

(ii) p o = N a = 10 15 cm

(b) (i)

)

(

ni 1.8× 106 = no 2 × 1016

p o = N a = 1015 cm 2

(

−3

)

= 2.25 × 105 cm −3 −3

2

=1.62 ×10 −4 cm −3

−3

)

2

ni 1.8 × 10 6 = = 3.24 × 10− 3cm −3 po 10 15 ______________________________________________________________________________________ no =

EX1.3 (a) For n-type;

ρ = (b) J =

1

ρ

1 1 = = 0.046 ohm-cm −19 eμ n N d 1.6 × 10 (6800 ) 2 × 1016

(

)

(

)

⋅ Ε ⇒ Ε = ρJ = (0.046)(175) = 8.04 V/cm

--------------------------------------------------------------------------------------------------------------------------------EX1.4 Diffusion current density due to holes: dp J p = −eD p dx ⎛ − 1⎞ ⎛ −x ⎞ = − eD p ( 1016 ) ⎜ ⎟ exp ⎜ ⎟ ⎜L ⎟ ⎜L ⎟ ⎝ p⎠ ⎝ p⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (a) At x = 0 ( 1.6× 10−19 ) (10 )(1016 ) 16 A / cm 2 Jp = = 10 −3 (b) At x = 10−3 cm ⎛ − 10−3 ⎞ J p = 16exp ⎜ − 3 ⎟ = 5.89 A / cm 2 ⎝ 10 ⎠ ______________________________________________________________________________________ EX1.5

( )( ) ( ) ( )( ) ( )

⎡ 1016 1017 ⎤ (a) Vbi = (0.026) ln ⎢ ⎥ = 1.23 V ⎢⎣ 1.8× 106 2 ⎥⎦ ⎡ 10 16 10 17 ⎤ (b) Vbi = (0.026) ln⎢ ⎥ = 0.374 V ⎢⎣ 2.4 ×1013 2 ⎥⎦ ______________________________________________________________________________________

EX1.6 −1/ 2

⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ and ⎡N N ⎤ Vbi = VT ln ⎢ a 2 d ⎥ ⎣ ni ⎦

⎡( 1017 )( 1016) ⎤ ⎥ = 0.757 V = ( 0.026) ln ⎢ ⎢ (1.5 ×1010 )2 ⎥ ⎣ ⎦ − 1/ 2

5 ⎞ ⎛ Then 0.8 = C jo ⎜ 1 + ⎟ ⎝ 0.757 ⎠ or C jo = 2.21 pF

= C jo ( 7.61)

−1/ 2

______________________________________________________________________________________ EX1.7 ⎛I (a) VD = VT ln⎜⎜ D ⎝I S

⎞ ⎟ ⎟ ⎠

⎛ 50 × 10− 6 ⎞ ⎟ = 0.563 V (i) V D = ( 0.026) ln⎜⎜ −14 ⎟ ⎝ 2× 10 ⎠ ⎛ 10 −3 ⎞ ⎟ = 0.641 V (ii) V D = (0.026) ln⎜⎜ −14 ⎟ ⎝ 2× 10 ⎠ ⎛ 50 × 10 −6 ⎞ ⎟ = 0.443 V (b) (i) V D = (0.026) ln⎜⎜ −12 ⎟ ⎝ 2× 10 ⎠ ⎛ 10 −3 ⎞ ⎟ = 0.521 V (ii) V D = (0.026) ln⎜⎜ −12 ⎟ ⎝ 2× 10 ⎠ ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ EX1.8 ⎛V ⎞ VPS = ID R + VD and I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ ( 4 − VD ) so 4 = I D ( 4 × 103 ) + VD ⇒ I D = 4 ×103 and ⎛ V ⎞ I D = (10− 12 ) exp ⎜ D ⎟ ⎝ 0.026 ⎠ By trial and error, we find I D ≅ 0.866 mA and VD ≅ 0.535 V. ______________________________________________________________________________________

EX1.9

V PS − Vγ

8 − 0.7 = 6.08 k Ω ⇒R = R 1.20 4 − 0 .7 = 0.9429 mA (b) ID = 3 .5 PD = I DV D = (0.9429 )(0.7 ) = 0.66 mW ______________________________________________________________________________________ ID =

(a)

EX1.10 PSpice Analysis ______________________________________________________________________________________ EX1.11 8 − 0 .7 = 0.365 mA 20 V 0.026 rd = T = ⇒ 71.2 Ω I D 0.365

(a) I D =

0.25 sin ωt ⇒ 12.5 sin ω t ( μ A) 20 + 0.0712 8 − 0 .7 = 0.73 mA (b) I D = 10 0.026 rd = ⇒ 35.6 Ω 0.73 0.25 sin ω t ⇒ 24.9 sin ω t ( μ A) id = 10 + 0.0356 ______________________________________________________________________________________ id =

EX1.12 ⎛I ⎞ ⎛ 1.2 ×10 −3 ⎞ or VD = 0.6871 V For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026) ln ⎜ −15 ⎟ ⎝ IS ⎠ ⎝ 4 × 10 ⎠ The Schottky diode voltage will be smaller, so V D = 0.6871 − 0.265 = 0.4221 V ⎛V ⎞ Now ID ≅ IS exp ⎜ D ⎟ ⎝ VT ⎠

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ or

1.2× 10− 3 ⇒ I S = 1.07× 10− 10 A ⎛ 0.4221 ⎞ exp ⎜ ⎟ ⎝ 0.026 ⎠ ______________________________________________________________________________________ IS =

EX1.13 P = I ⋅VZ ⇒ 10 = I ( 5.6) ⇒ I = 1.79 mA 10 − 5.6 = 1.79 ⇒ R = 2.46 kΩ R ______________________________________________________________________________________

Also I =

Test Your Understanding Solutions TYU1.1 (a) T = 400K ⎛ −E g ⎞ Si: ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2 kT ⎠ ni = (5.23 ×1015

) (400 )

3/2

⎡ ⎤ − 1.1 ⎥ exp ⎢ ⎢ 2 ( 86 ×10− 6 ) ( 400 ) ⎥ ⎣ ⎦

or ni = 4.76 × 1012 c m− 3

Ge: ni = (1.66 ×1015 )( 400 )

3/2

⎡ ⎤ − 0.66 ⎥ exp ⎢ −6 ⎢⎣ 2 (86 ×10 ) (400 ) ⎥⎦

or ni = 9.06 ×10 14 cm GaAs:

−3

ni = ( 2.1 ×10 14 ) ( 400 )

⎡ ⎤ − 1.4 ⎥ exp ⎢ −6 ⎢ 2 ( 86 ×10 ) (400 )⎥ ⎣ ⎦

3/2

or ni = 2.44 × 109 c m−3 (b) T = 250 K

Si: ni = ( 5.23× 10 15) ( 250)

3/2

⎡ ⎤ − 1.1 exp ⎢ ⎥ ⎢⎣ 2 ( 86 × 10 −6 ) ( 250 ) ⎥⎦

or ni = 1.61× 108 cm −3

Ge: n i = ( 1.66 × 10 15) ( 250 ) or ni = 1.42 × 1012 c m −3

3/2

⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 250) ⎥ ⎣ ⎦

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ GaAs: ni = ( 2.10 ×1014 )( 250)

3/2

⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 ×10 ) (250 )⎥⎦

or −

ni = 6.02 × 103 c m 3 ______________________________________________________________________________________

TYU1.2

(

)

(

)

(a) σ = eμ p N a = 1.6 ×10 −19 ( 480) 2 ×1015 = 0.154 (ohm-cm)

ρ=

1

σ

=

1 = 6.51 Ω -cm 0.1536

(

)

(

)

(b) σ =e μn N d = 1.6 ×10 −19 (1350) 2 ×10 17 = 43.2 (ohm-cm)

ρ=

1

σ

=

−1

−1

1 = 0.0231Ω -cm 43.2

________________________________________________________________________ TYU1.3 J = σ Ε = ( 0.154)( 4) = 0.616 A/cm

(a)

2

(b) J = σ Ε = (43.2)(4) = 172.8 A/cm 2 ______________________________________________________________________________________ TYU1.4

(a)

J n = eDn

⎛ 1015 − 1016 ⎞ dn Δn so J n = 1.6 × 10− 19 (35 ) ⎜ = eDn −4 ⎟ Δx dx ⎝ 0 − 2.5 ×10 ⎠

(

)

or J n = 202 A / cm 2

(b)

J p = − eD p

⎛ 1014 − 5 × 1015 ⎞ dp Δp 19 = − eD p so J p = − 1.6 ×10 − (12.5 ) ⎜ −4 ⎟ dx Δx ⎝ 0 − 4 ×10 ⎠

(

)

or

J p = −24.5 A / cm 2 ______________________________________________________________________________________ TYU1.5 (a) no = Nd = 8× 1015 cm −3 ni2 ( 1.5 × 10 ) = = 2.81× 104 cm −3 no 8 × 1015 10

po =

2

(b) n = no + δ n = 8 ×1015 + 0.1× 1015 or n = 8.1 ×1015 cm −3 p = po + δ p = 2.81× 10 4 + 10 14 or − p ≅1014 cm 3 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU1.6 ⎛N N (a) Vbi = VT ln⎜⎜ a 2 d ⎝ ni

(

)(

⎡ 1015 5× 1016 ⎞ ⎟ = (0.026) ln⎢ 2 ⎟ ⎢⎣ 1.5 ×10 10 ⎠

( )( ( ( )( (

(

)

)⎤⎥ = 0.679 V ⎥⎦

)

⎡ 1015 5 ×1016 ⎤ (b) Vbi = (0.026) ln⎢ ⎥ = 1.15 V ⎢⎣ 1.8 × 10 6 2 ⎥⎦ ⎡ 1015 5 ×1016 ⎤ (c) Vbi = (0.026 )ln⎢ ⎥ = 0.296 V ⎢⎣ 2.4 × 1013 2 ⎥⎦ ______________________________________________________________________________________

)

)

)

TYU1.7 ⎛V (a) (i) I D = I S exp⎜⎜ D ⎝ VT

(

)

(

)

⎞ ⎛ 0.55 ⎞ ⎟ = 10 −16 exp ⎜ ⎟ ⇒ 0.154 μ A ⎟ ⎝ 0.026 ⎠ ⎠

(

)

⎛ 0.65 ⎞ ⎟ ⇒ 7.20 μ A (ii) I D = 10 −16 exp⎜ ⎝ 0.026 ⎠ ⎛ 0.75 ⎞ (ii) ID = 10 −16 exp⎜ ⎟ ⇒ 0.337 mA ⎝ 0.026 ⎠

(b) (i) ID = − 10 −16 A (ii) I D = − 10 − 16 A ______________________________________________________________________________________ TYU1.8 ΔT = 100C so ΔVD ≅ 2 ×100 = 200 mV Then VD = 0 .650 − 0 .20 = 0.450 V ______________________________________________________________________________________ TYU1.9

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU1.10 (a) ID = 0 2 − 0.7 = 0.325 mA 4 5 − 0.7 = 1.075 mA (c) ID = 4 (d) ID = 0 (e) ID = 0 ______________________________________________________________________________________

(b)

ID =

TYU1.11 P = I DVD ⇒ 1.05 = I D (0.7 ) so I D = 1 .5 mA VPS − Vγ

10 − 0.7 = ⇒ R = 6.2 kΩ ID 1.5 ______________________________________________________________________________________

Now R =

TYU1.12 ID 0.8 = = 30.8 mS VT 0.026 ______________________________________________________________________________________ gd =

TYU1.13 V T 0.026 = = 2.6 k Ω I D 0.010 0.026 rd = ⇒ 260 Ω 0.10 0.026 rd = ⇒ 26 Ω 1 ----------------------------------------------------------------------------------------------------------------------------rd =

TYU1.14 rd =

VT ID

⇒ 50 =

0.026 0.026 ⇒ ID= 50 ID

or I D = 0.52 mA ______________________________________________________________________________________

TYU1.15 For the pn junction diode, 4 − 0.7 ID = = 0.825 mA 4 4 − 0.3 = 0.925 mA 4 ______________________________________________________________________________________

For the Schottky diode, I D =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

TYU1.16 Vz = Vzo + I z rz ⇒ Vzo = Vz − I z rz so Vzo = 5.20 − (10−3 )( 20) = 5.18 V

Then Vz = 5.18+ (10× 10−3 ) ( 20) ⇒ Vz = 5.38 V ______________________________________________________________________________________ TYU1.17 P 6.5 = = 1.81 mA V Z 3.6 V PS = I Z R + V Z = (1.81)( 4) + 3.6 = 10.8 V ______________________________________________________________________________________ P = I ZVZ ⇒ I Z =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

Chapter 2 Exercise Solutions EX2.1

V S −V B −Vγ

12 − 4.5− 0.6 = 27.6 mA = 0.25 R υ R ( max) = VS + V B = 12 + 4.5 = 16.5 V Conduction cycle: υ I = 12 sin ω t1 = 4.5 + 0.6 = 5.1 V or ⎛ 5.1⎞ ω t1 = sin− 1⎜ ⎟ = 25.15 ° ⎝ 12 ⎠ ωt 2 = 180 − 25.15 = 154.85° 154. 85 − 25.15 Percent time = × 100% = 36.0% 360 ______________________________________________________________________________________ iD (peak ) =

EX2.2 (a)

v O = 12sin θ1 − 1.4 = 0

1.4 = 0.1166 12 which yields θ1 = 6.7° By symmetry, θ 2 = 180 − 6.7 = 173.3° Then 173.3 − 6.7 % time = × 100% = 46.3% 360 1.4 = 0 .35 sin θ1 = 4 (b) which yields θ1 = 20 .5 ° By symmetry, θ 2 = 180 − 20.5 = 159.5° 159.5 − 20.5 × 100% = 38.6% % time = 360 Then

or sin θ 1 =

______________________________________________________________________________________ EX2.3 (a) C =

12 VM ⇒ 125 μ F = 2 fRV r 2( 60) 2× 103 (0.4)

(

)

VM 12 ⇒ 250 μ F = fRVr (60 ) 2 ×10 3 (0.4 ) ______________________________________________________________________________________

(b) C =

(

)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ EX2.4 Vr =

VM f RC

⇒ R=

VM

R=

f CVr

75

( 60) (50 ×10 −6 ) (4 )

or R = 6.25 k Ω Then ______________________________________________________________________________________ EX2.5 10 ≤ V PS ≤ 14 V , V Z = 5.6 V , 20 ≤ RL ≤ 100 Ω 5.6 = 0.28 A , 20 5.6 I L ( min ) = = 0.056 A 100 ⎡VPS (max ) − VZ ⎤⎦ ⋅ I L (max ) ⎡⎣V PS (min ) − V Z ⎤⎦ ⋅ I L (min ) − I Z (max) = ⎣ V PS (min )− 0.9V Z − 0.1V PS ( max ) V PS ( min ) − 0.9V Z − 0.1V PS ( max )

I L ( max) =

I L ( max) = or

or

(14 − 5.6 )(280 ) − (10 −5.6 )(56 ) 10 − (0.9 )(5.6 ) − (0.1 )(14 )

I L ( max ) = 591.5 mA

Power(min) = I Z ( max) ⋅VZ = ( 0.5915)( 5.6) So Power(min) = 3.31 W VPS ( max ) − VZ 14 − 5.6 Ri = = IZ (max ) + IL (min ) 0.5915 +0.056 or Ri ≅ 13 Ω Now ______________________________________________________________________________________ EX2.6 13.6 − 9 = 0.2383 A 15.3 + 4 = 9 + ( 4)( 0.2383) = 9.9532 V

For v PS = 13.6 V , I Z = vL,max

11 − 9 = 0.1036 A 15.3 + 4 = 9 + (4 )(0.1036 ) = 9.4144 V

For v PS = 11 V , I Z = v L ,min

Δ vL 9.9532 − 9.4144 × 100% = × 100% 13.6 − 11 Δv PS or Source Reg = 20.7% 13.6 − 9 For I L = 0, I Z = = 0.2383 A 15.3 + 4 v L, noload = 9 + ( 4 )( 0.2383) = 9.9532 V Source Reg =

For I L = 100 mA, which yields

IZ =

13.6 − ⎣⎡ 9 + I Z ( 4)⎦⎤ 15.3

− 0.10

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ I Z = 0.1591 A vL, full load = 9 + (4 )(0.1591 A ) = 9.6363 V vL ,noload − vL , full load × 100% Load Reg = vL , full load 9.9532 − 9.6363 × 100% 9.6363 or Load Reg = 3.29% ______________________________________________________________________________________ =

EX2.7

5 , on ⇒ VO = −5 V For vI < V D2 Then, V2 = 4.3 V. D1 turns on when v1 = 2.5 V, Then, V1 = 1.8 V. Δv R2 1 1 vI > 2.5 V, O = ⇒ = 3 3 v R R Δ I 1+ 2 For

So that R1 = 2R2 ______________________________________________________________________________________ EX2.8 For υO = +2 V, D is on. Δυ I = 10 V, so Δ υ O = 10 V. Output = Square wave between +2 and − 8 V. ______________________________________________________________________________________ EX2.9 10 − 4.4 = 0.5895 mA 9.5 v I = 4.4 − 0.6 − ( 0.5895) ( 0.5) = 3.505V

v O = 4.4 V , I =

Set I = ID1, then Summary: For 0 ≤ vI ≤ 3 .5 V , vO = 4.4 V

V 3.5 V, D2 For vI > turns on and when v I ≥ 9.4 V , vO = 10

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

______________________________________________________________________________________ EX2.10 (a) If D1 is on, υO = υI − Vγ − VB = 5 − 0.7 − 1 = 3.3 V. 3.3 − 0.7 = 0.65 mA Then I D 2 = 4 5 −3.3 = 1.0 mA, but I D 2 < I R1 is impossible. Now I R1 = 1.7 D 1 is cutoff and I D 1 = 0 5 − 0.7 = 0.754 mA Then I R 1 = I D 2 = 1.7 + 4 υO = 0.7 + (0.754 )(4 ) = 3.72 V (b) υI = 10 V, Both D1 and D2 are on. υ O = 10 − 0.7 − 1 ⇒ υ O = 8.3 V 8.3 − 0.7 I D2 = = 1.9 mA 4 1.7 I R1 = = 1.0 mA 1.7 I D1 = 1.9 −1.0 = 0.9 mA ______________________________________________________________________________________ EX2.11 D 2 cutoff, I D 2 = 0

− 0.7 − (− 5 ) = 2.15 mA 2 5 − 0.7 − ( − 10) 14.3 I D1 = = 1.19 mA = 8+ 4 R1 + R 2 V A = 5 − (1.19)( 8) = −4.53 V V A < VB so that D 2 is cutoff. ______________________________________________________________________________________ VB = −0.7 V, I D3 =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

EX2.12

(a)

I ph = η eΦ

⎡ 6.4 × 10− 2 ⎤ − ⎥ (0.5 ) I = ( 0.8) (1.6 ×10 19 ) ⎢ ⎢ ( 2 ) (1.6 ×10− 19 ) ⎥ ⎣ ⎦ so I = 12 .8 mA or ph v = (12.8)(1) = 12.8 V . (b) We have O 12.8 V The diode must be reverse biased so that VPS > ______________________________________________________________________________________

EX2.13 The equivalent circuit is

I =

So

5 −1.7 −0.2 =15 mA rf + R

15 − 1.7 − 0.2 3.1 = = 0.207 k Ω 15 15 Or Then R = 207 − 15 ⇒ R = 192 Ω ______________________________________________________________________________________ rf + R =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

Test Your Understanding Solutions TYU2.1 (a) iD ( peak) for V B = 4 V. 15 − 0.7 − 4 i D ( peak) = 18 = ⇒ R = 572 Ω R 15 − 0.7 − 8 (b) iD = = 11.0 mA 0.572 Then 11.0 ≤ iD ( peak ) ≤ 18 mA For VB = 4 V,

⎛ 4.7 ⎞ 15 sin ωt1 = 4.7 ⇒ ωt1 = sin −1 ⎜ ⎟ = 18.26° ⎝ 15 ⎠ ω t2 = 180 − 18.26 = 161.74° 161.74 − 18.26 duty cycle = × 100 % = 39.9% 360 For VB = 8 V, ⎛ 8.7 ⎞ 15 sin ωt 1 = 8.7 ⇒ ωt1 = sin− 1 ⎜ ⎟ = 35.45 ° ⎝ 15 ⎠ ωt 2 = 180 − 35.45 = 144.55° 144.55 − 35.45 × 100 % = 30.3% duty cycle = 360 Then 30.3 ≤ duty cycle ≤ 39.9% ______________________________________________________________________________________

TYU2.2 vI = 120sin ( 2π 60t ) , Vγ = 0.7 V ,

and R = 2.5 kΩ Full-wave rectifier: Turns ratio 1:2 so that vS = vI VM = 120 − 0.7 = 119.3 V V r = 119.3 −100 = 19.3 V C =

119.3 VM = 2 f RV r 2 ( 60 ) (2.5x10 3 ) (19.3 )

or C = 20.6 μ F So _____________________________________________________________________________ TYU2.3 vI = 50 sin ( 2π 60t ) , Vγ = 0.7 V , C =

( 50 − 1.4)

and R = 10 k Ω. Full-wave rectifier

VM = 2 f RV r 2 ( 60 ) (10 ×103 )( 2)

or C = 20.3 μ F ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

TYU2.4 Using Equation (2.16)