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Catégorie :Category: mViewer GX Creator Lua TI-Nspire
Auteur Author: kennet67
Type : Classeur 3.6
Page(s) : 22
Taille Size: 1.14 Mo MB
Mis en ligne Uploaded: 12/09/2019 - 01:56:08
Uploadeur Uploader: kennet67 (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a2316865

Description 

Chapter 5
1. (a) If the dc power supply is set to zero volts, the amplification will be zero.

(b) Too low a dc level will result in a clipped output waveform.

(c) Po = I2R = (5 mA)22.2 kΩ = 55 mW
Pi = VCCI = (18 V)(3.8 mA) = 68.4 mW
P (ac) 55 mW
η= o = = 0.804 ⇒ 80.4%
Pi (dc) 68.4 mW

2. −

1 1
3. xC = = = 15.92 Ω
2π fC 2π (1 kHz)(10 μ F)
f = 100 kHz: xC = 0.159 Ω
Yes, better at 100 kHz

4. −

Vi 10 mV
5. (a) Zi = =
I i 0.5 mA
= 20 Ω (=re)

(b) Vo = IcRL
= αIcRL
= (0.98)(0.5 mA)(1.2 kΩ)
= 0.588 V

Vo 0.588 V
(c) Av = =
Vi 10 mV
= 58.8

(d) Zo = ∞ Ω

Io α Ie
(e) Ai = = = α = 0.98
Ii Ie

(f) Ib = Ie − Ic
= 0.5 mA − 0.49 mA
= 10 μA




52
Vi 48 mV
6. (a) re = = = 15 Ω
I i 3.2 mA

(b) Zi = re = 15 Ω

(c) IC = αIe = (0.99)(3.2 mA) = 3.168 mA

(d) Vo = ICRL = (3.168 mA)(2.2 kΩ) = 6.97 V

Vo 6.97 V
(e) Av = = = 145.21
Vi 48 mV

(f) Ib = (1 − α)Ie = (1 − 0.99)Ie = (0.01)(3.2 mA)
= 32 μA

26 mV 26 mV
7. (a) re = = = 13 Ω
I E (dc) 2 mA
Zi = βre = (80)(13 Ω)
= 1.04 kΩ

IC α Ie β Ie I
(b) Ib = = = ⋅ = e
β β β +1 β β +1
2 mA
= = 24.69 μA
81

Io I L
(c) Ai = =
Ii Ib
r (β Ib )
IL = o
ro + RL
ro
⋅ β Ib
ro + RL ro
Ai = = ⋅β
Ib ro + RL
40 kΩ
= (80)
40 kΩ + 1.2 kΩ
= 77.67

RL ro 1.2 kΩ 40 kΩ
(d) Av = − =−
re 13 Ω
1.165 kΩ
= −
13 Ω
= −89.6




53
8. (a) Zi = βre = (140)re = 1200
1200
re = = 8.571 Ω
140
V 30 mV
(b) Ib = i = = 25 μA
Z i 1.2 kΩ

(c) Ic = βIb = (140)(25 μA) = 3.5 mA

ro I c (50 kΩ)(3.5 mA)
(d) IL = = = 3.321 mA
ro + RL 50 kΩ + 2.7 kΩ
I 3.321 mA
Ai = L = = 132.84
Ii 25 μ A

Vo − Ai RL (2.7 kΩ)
(e) Av = = = −(132.84)
Vi Zi 1.2 kΩ
= −298.89

VCC − VBE 12 V − 0.7 V
9. (a) re: IB = = = 51.36 μA
RB 220 kΩ
IE = (β + 1)IB = (60 + 1)(51.36 μA)
= 3.13 mA
26 mV 26 mV
re = = = 8.31 Ω
IE 3.13 mA
Zi = RB || βre = 220 kΩ || (60)(8.31 Ω) = 220 kΩ || 498.6 Ω
= 497.47 Ω
ro ≥ 10RC ∴ Zo = RC = 2.2 kΩ

RC −2.2 kΩ
(b) Av = − = = −264.74
re 8.31 Ω

(c) Zi = 497.47 Ω (the same)
Zo = ro || RC = 20 kΩ || 2.2 kΩ
= 1.98 kΩ

− RC ro −1.98 kΩ
(d) Av = = = −238.27
re 8.31 Ω
Ai = −AvZi/RC
= −(−238.27)(497.47 Ω)/2.2 kΩ
= 53.88




54
RC R 4.7 kΩ
10. Av = − ⇒ re = − C = − = 23.5 Ω
re Av (−200)
26 mV 26 mV 26 mV
re = ⇒ IE = = = 1.106 mA
IE re 23.5 Ω
I 1.106 mA
IB = E = = 12.15 μA
β +1 91
V − VBE
IB = CC ⇒ VCC = IBRB + VBE
RB
= (12.15 μA)(1 MΩ) + 0.7 V
= 12.15 V + 0.7 V
= 12.85 V

VCC − VBE 10 V − 0.7 V
11. (a) IB = = = 23.85 μA
RB 390 kΩ
IE = (β + 1)IB = (101)(23.85 μA) = 2.41 mA
26 mV 26 mV
re = = = 10.79 Ω
IE 2.41 mA
IC = βIB = (100)(23.85 μA) = 2.38 mA


(b) Zi = RB || βre = 390 kΩ || (100)(10.79 Ω) = 390 kΩ || 1.08 kΩ
= 1.08 kΩ
ro ≥ 10RC ∴Zo = RC = 4.3 kΩ

RC −4.3 kΩ
(c) Av = − = = −398.52
re 10.79 Ω

RC ro (4.3 kΩ) (30 kΩ) 3.76 kΩ
(d) Av = − =− =− = −348.47
re 10.79 Ω 10.79 Ω

12. (a) Test βRE ≥ 10R2
?
(100)(1.2 kΩ) ≥ 10(4.7 kΩ)
120 kΩ > 47 kΩ (satisfied)

Use approximate approach:
RV 4.7 kΩ(16 V)
VB = 2 CC = = 1.721 V
R1 + R2 39 kΩ + 4.7 kΩ
VE = VB − VBE = 1.721 V − 0.7 V = 1.021 V
V 1.021 V
IE = E = = 0.8507 mA
RE 1.2 kΩ
26 mV 26 mV
re = = = 30.56 Ω
IE 0.8507 mA




55
(b) Zi = R1 || R2 || β re
= 4.7 kΩ || 39 kΩ || (100)(30.56 Ω)
= 1.768 kΩ
ro ≥ 10RC ∴ Zo ≅ RC = 3.9 kΩ

RC 3.9 kΩ
(c) Av = − =− = −127.6
re 30.56 Ω

(d) ro = 25 kΩ

(b) Zi(unchanged) = 1.768 kΩ
Zo = RC || ro = 3.9 kΩ || 25 kΩ = 3.37 kΩ

( RC ro ) (3.9 kΩ) (25 kΩ) 3.37 kΩ
(c) Av = − =− =−
re 30.56 Ω 30.56 Ω
= −110.28 (vs. −127.6)

?
13. βRE ≥ 10R2
(100)(1 kΩ) ≥ 10(5.6 kΩ)
100 kΩ > 56 kΩ (checks!) & ro ≥ 10RC
Use approximate approach:
R R 3.3 kΩ
Av = − C ⇒ re = − C = − = 20.625 Ω
re Av −160
26 mV 26 mV 26 mV
re = ⇒ IE = = = 1.261 mA
IE re 20.625 Ω
V
IE = E ⇒ VE = IERE = (1.261 mA)(1 kΩ) = 1.261 V
RE
VB = VBE + VE = 0.7 V + 1.261 V = 1.961 V
5.6 kΩ VCC
VB = = 1.961 V
5.6 kΩ + 82 kΩ
5.6 kΩ VCC = (1.961 V)(87.6 kΩ)
VCC = 30.68 V

14. Test βRE ≥ 10R2
?
(180)(2.2 kΩ) ≥ 10(56 kΩ)
396 kΩ < 560 kΩ (not satisfied)

Use exact analysis:

(a) RTh = 56 kΩ || 220 kΩ = 44.64 kΩ
56 kΩ(20 V)
ETh = = 4.058 V
220 kΩ + 56 kΩ
ETh − VBE 4.058 V − 0.7 V
IB = =
RTh + ( β + 1) RE 44.64 kΩ + (181)(2.2 kΩ)



56
= 7.58 μA
IE = (β + 1)IB = (181)(7.58 μA)
= 1.372 mA
26 mV 26 mV
re = = = 18.95 Ω
IE 1.372 mA

(b) VE = IERE = (1.372 mA)(2.2 kΩ) = 3.02 V
VB = VE + VBE = 3.02 V + 0.7 V
= 3.72 V
VC = VCC − ICRC
= 20 V − βIBRC = 20 V − (180)(7.58 μA)(6.8 kΩ)
= 10.72 V

(c) Zi = R1 || R2 || βre
= 56 kΩ || 220 kΩ || (180)(18.95 kΩ)
= 44.64 kΩ || 3.41 kΩ
= 3.17 kΩ

RC ro
ro < 10RC ∴ Av = −
re
(6.8 kΩ) (50 kΩ)
= −
18.95 Ω
= −315.88

VCC − VBE 20 V − 0.7 V
15. (a) IB = =
RB + ( β + 1) RE 390 kΩ + (141)(1.2 kΩ)
19.3 V
= = 34.51 μA
559.2 kΩ
IE = (β + 1)IB = (140 + 1)(34.51 μA) = 4.866 mA
26 mV 26 mV
re = = = 5.34 Ω
IE 4.866 mA

(b) Zb = βre + (β + 1)RE
= (140)(5.34 kΩ) + (140 + 1)(1.2 kΩ) = 747.6 Ω + 169.9 kΩ
...

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603.38 Ko KB Ti_NspireTS/01-11.tns
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