Analisis y diseño
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Publicado: 25 de marzo de 2011
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
⎛ ⎞ −1.4 ⎟ or ni = 1.8 × 106 cm −3 exp ⎜ ⎜ 2 (86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠
3/ 2
⎛ ⎞ −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,
n2 (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 no10 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
−19 16
10−3 −3 (b) At x = 10 cm
Jp
(1.6 ×10 ) (10 ) (10 ) = 16 A / cm =
2
⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89A / cm 2 ⎝ 10 ⎠
EX1.5
⎡N N Vbi = VT ln ⎢ a 2 d ⎣ ni ⎡ (1016 )(1017 ) ⎤ ⎤ ⎥ or Vbi = 1.23 V = ( 0.026 ) ln ⎢ ⎥ ⎢ (1.8 × 106 )2 ⎥ ⎦ ⎣ ⎦
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) (b)
ID = ID =
VPS − Vγ R VPS − Vγ R
5 − 0.7 ⇒ I D = 1.08 mA 4 VPS − Vγ ⇒R= ID =...
⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠
GaAs: ni = ( 2.1× 1014 ) ( 300 ) Ge: ni = (1.66 × 1013 ) ( 300 )
3/ 2
⎛ ⎞ −1.4 ⎟ or ni = 1.8 × 106 cm −3 exp ⎜ ⎜ 2 (86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠
3/ 2
⎛ ⎞ −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,
n2 (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 no10 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
−19 16
10−3 −3 (b) At x = 10 cm
Jp
(1.6 ×10 ) (10 ) (10 ) = 16 A / cm =
2
⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89A / cm 2 ⎝ 10 ⎠
EX1.5
⎡N N Vbi = VT ln ⎢ a 2 d ⎣ ni ⎡ (1016 )(1017 ) ⎤ ⎤ ⎥ or Vbi = 1.23 V = ( 0.026 ) ln ⎢ ⎥ ⎢ (1.8 × 106 )2 ⎥ ⎦ ⎣ ⎦
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) (b)
ID = ID =
VPS − Vγ R VPS − Vγ R
5 − 0.7 ⇒ I D = 1.08 mA 4 VPS − Vγ ⇒R= ID =...
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