Solution Manual - Microelectronics; Circuit Analysis & Desing 3rd Edition ch1

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

GaAs: ni = ( 2.1× 1014 ) ( 300 ) Ge: ni = (1.66 × 1013 ) ( 300 )

3/ 2

3/ 2

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

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

EX1.2 (a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons, n 2 (1.5 × 10 no = i = 1017 po

)

10 2

= 2.25 × 103 cm −3

(b) majority carrier: electrons, no = 5 × 1015 cm −3 minority carrier: holes, n 2 (1.5 × 10 ) = 4.5 × 104 cm −3 po = i = 5 × 1015 no 10 2

EX1.3 For n-type, drift current density J ≅ eμn nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields E = 17.9 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⎠ (a) At x = 0

(1.6 ×10 ) (10 ) (10 ) = 16 A / cm = −19

Jp

16

2

10−3 −3 (b) At x = 10 cm

⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2 ⎝ 10 ⎠

EX1.5 ⎡N N Vbi = VT ln ⎢ a 2 d ⎣ ni

⎡ (1016 )(1017 ) ⎤ ⎤ ⎢ ⎥ or Vbi = 1.23 V = 0.026 ln ( ) ⎥ 6 2 ⎥ ⎢ ⎦ ⎣ (1.8 × 10 ) ⎦

EX1.6

⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ and

−1/ 2

⎡N N ⎤ Vbi = VT ln ⎢ a 2 d ⎥ ⎣ ni ⎦ ⎡ (1017 )(1016 ) ⎤ ⎥ = 0.757 V = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦ 5 ⎞ ⎛ Then 0.8 = C jo ⎜ 1 + ⎟ ⎝ 0.757 ⎠ or C jo = 2.21 pF

−1/ 2

= C jo ( 7.61)

−1/ 2

EX1.7

⎡ ⎛v ⎞ ⎤ iD = I S ⎢exp ⎜ D ⎟ − 1⎥ ⎝ VT ⎠ ⎥⎦ ⎣⎢ ⎡ ⎛ v ⎞ ⎤ so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥ ⎝ 0.026 ⎠ ⎦ ⎣

⎡ 10−3 ⎤ Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥ ⎣10 ⎦ or vD ≅ ( 0.026 ) ln (1010 )

which yields vD = 0.599 V EX1.8 ⎛V ⎞ VPS = I D 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 ≅ 0.864 mA and VD ≅ 0.535 V

EX1.9

(a)

ID =

(b)

ID =

Then R = (c)

VPS − Vγ R VPS − Vγ R

5 − 0.7 ⇒ I D = 1.08 mA 4 VPS − Vγ ⇒R= ID =

8 − 0.7 = 6.79 kΩ 1.075

ID(mA) Diode curve 1.25 1.08

Load lines (b) (a)

0

0.7

2

4 VD(v)

6

8

EX1.10 PSpice analysis EX1.11

Quiescent diode current I DQ =

VPS − Vγ

=

10 − 0.7 = 0.465 mA 20

R Time-varying diode current: V 0.026 We find that rd = T = = 0.0559 kΩ I DQ 0.465 Then id =

vI 0.2sin ω t (V ) = ⋅ or id = 9.97sin ω t ( μ A) rd + R 0.0559 + 20 ( kΩ )

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 ⎟ ⎝ 4 × 10 ⎠ ⎝ IS ⎠ The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V ⎛V ⎞ Now I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ or 1.2 × 10−3 IS = ⇒ I S = 1.07 × 10−10 A 0.4221 ⎛ ⎞ exp ⎜ ⎟ ⎝ 0.026 ⎠

EX1.13 P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA

Also I =

10 − 5.6 = 1.79 ⇒ R = 2.46 kΩ R

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

or ni = 4.76 × 1012 cm −3

3/ 2

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

Ge: ni = (1.66 × 1015 ) ( 400 )

3/ 2

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

or ni = 9.06 × 1014 cm −3 GaAs: ni = ( 2.1× 1014 ) ( 400 )

3/ 2

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

or ni = 2.44 × 109 cm −3 (b) T = 250 K Si: ni = ( 5.23 × 1015 ) ( 250 )

3/ 2

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

or ni = 1.61× 108 cm −3 Ge: ni = (1.66 × 1015 ) ( 250 )

3/ 2

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

or ni = 1.42 × 1012 cm −3 GaAs: ni = ( 2.10 × 1014 ) ( 250 )

3/ 2

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

or ni = 6.02 × 103 cm −3 TYU1.2 (a) n = 5 × 1016 cm −3 , p <<< n, so σ ≅ eμ n n = (1.6 × 10 −19 ) (1350 ) ( 5 × 1016 )

or

σ = 10.8 ( Ω − cm ) (b)

−1

p = 5 × 1016 cm −3 , n <<< p, so σ ≅ eμ p p = (1.6 × 10−19 ) ( 480 ) ( 5 × 1016 )

or

σ = 3.84 ( Ω − cm )

−1

TYU1.3 J = σ E = (10 )(15 ) or J = 150 A / cm2 TYU1.4

(a)

J n = eDn

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

(

)

or J n = 202 A / cm 2 (b)

J p = −eD p

or J p = −24.5 A / cm2 TYU1.5

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

(

)

no = N d = 8 × 1015 cm −3

(a)

10 n 2 (1.5 × 10 ) po = i = = 2.81× 10 4 cm −3 no 8 × 1015 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 + 1014 or p ≅ 1014 cm −3 TYU1.6

(a)

⎡ (1015 )(1017 ) ⎤ ⎡N N ⎤ ⎥ = 0.697 V Vbi = VT ln ⎢ a 2 d ⎥ so Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ni ⎦ ⎣ ⎦

(b)

⎡ (1017 )(1017 ) ⎤ ⎥ = 0.817 V Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦

TYU1.7

⎡ ⎛V I D = I S ⎢exp ⎜ D ⎝ VT ⎣⎢

(a)

⎞ ⎤ ⎟ − 1⎥ ⎠ ⎦⎥

⎛ 0.5 ⎞ I D ≅ 10−14 exp ⎜ ⎟ ⎝ 0.026 ⎠ Then, for VD = 0.5 V, I D = 2.25 μ A VD = 0.6 V, I D = 0.105 mA VD = 0.7 V, I D = 4.93 mA (b) I D ≅ − I S = −10 −14 A for both cases. TYU1.8 ΔT = 100C so ΔVD ≅ 2 × 100 = 200 mV Then VD = 0.650 − 0.20 = 0.450 V TYU1.9 ID(mA) Diode 1.0 艐0.87

Load line

0

1 艐0.54v

2 VD(v)

VD

ID

0.45 0.50 0.55

0.033 0.225 1.54

3

TYU1.10 P = I DVD ⇒ 1.05 = I D ( 0.7 ) so I D = 1.5 mA

4

Now R =

VPS − Vγ ID

=

10 − 0.7 ⇒ R = 6.2 kΩ 1.5

TYU1.11 I 0.8 gd = D = = 30.8 mS VT 0.026 TYU1.12 V 0.026 0.026 rd = T ⇒ 50 = ⇒ ID = ID ID 50 or I D = 0.52 mA TYU1.13 For the pn junction diode, 4 − 0.7 ID = = 0.825 mA 4

For the Schottky diode, I D =

4 − 0.3 = 0.925 mA 4

TYU1.14 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