Title | Diodos Exercicios resolvidos |
---|---|
Course | Eletronica |
Institution | Centro Universitário de Lins |
Pages | 7 |
File Size | 728.9 KB |
File Type | |
Total Downloads | 54 |
Total Views | 183 |
Download Diodos Exercicios resolvidos PDF
R1 , R2 , R3 D1 Ω
E
Ω
R1 = 220Ω R2 = R3 = 470Ω
E D1 ID1 E R3
E R1
R3
I1 −E + R1 ∗ I1 + VR3 = 0 E = R1 ∗ I1 + VR3 R3 I1 = IR3 + ID1 =
ID1 = IR2 =
VR 2 R2
VR3 + ID1 R3
VR3 = VR2 + VD1 = VR2 + 0, 7
=
VR3 = VR2 + 0, 7 = ID1 R2 + 0, 7 =
∗ 470Ω + 0, 7
I1 = 19, 289 I1 VR3
R1
E = R1 ∗ I1 + VR3 = 9, 1267 D1
=
ID1
VR − VD1 VR3 − E VR3 + + 3 =0 R1 R1 R2
VR3 =
E/R1 + VD1 /R2 = 4, 883 1/R1 + 1/R2 + 1/R3
ID1 ID1 = IR2 =
VR2 VR3 − VD1 = 8, 9 = R2 R2
R = 150Ω RL = 3, 3 Ω VZ
PZmax
Vi
Vi
RL
Vi
Vimin
VL = VZ VL
VL
≥ VZ VL = Vi RL /(RL + R)
< VZ
Vimin = Vimin =
(RL + R)VZ RL
(3, 3 Ω + 150Ω)6, 2 3, 3 Ω
= 6, 4818
Vi IZmax = PZmax /VZ
IZ = IR − IRL
IRL = VL /RL = VZ /RL
IRmax = IZmax + IRL Vimax = IRmax R + VZ Vimax = (IZmax + IRL )R + VZ Vimax = (PZmax /VZ + VL /RL )R + VZ Vimax = (PZmax /VZ + VZ /RL )R + VZ Vimax = (500 RL
/6, 2 + 6, 2 /3, 3 Ω)150Ω + 6, 2
= 18, 579
Vi = 15
RL
RLmin = RVZ /(Vi − VZ ) = 150Ω ∗ 6, 2/(15 − 6, 2) = 105, 68Ω
V1 = 3 vi (t) = 10
t
vo (t)
• •
vi (t)
•
D2 D3
D4
D1 Vi < V1 + 0, 7
V1 Vi > V1 + 0, 7 VD1 < 0, 7 D1 =
VD1 > 0, 7
D1
Vi < V1 + 0, 7 Vi > V1 + 0, 7
Vo Vo =
Vi < V1 + 0, 7 Vi > V1 + 0, 7
Vi V1 + 0, 7
Vo =
Vi < 3, 7 Vi > 3, 7
Vi 3, 7
vi (t)
• D1
V1 D2 D3
−0, 7 ∗ 3
Vi > −2, 1
D4
Vi > Vi < −2, 1
D3 =
D1 D2
Vi > −2, 1
Vi < −2, 1
Vo Vo =
Vi −(VD1 + VD2 + VD3 )
Vo =
Vi −2, 1
Vi > −2, 1
Vi < −2, 1
Vi > −2, 1
Vi < −2, 1 Vo V1 + 0, 7
vo (t)
vi (t)
vi (t)
vo (t) vo (t)
V i(t)
V o(t)
Vo
D1
D2
Vo Vo = 2 Ω ∗ IT IT
−20 + VD1 + VD2 + 2 ΩIT + 2 ΩIT = 0 20 − VD1 − VD2 4 Ω
IT =
IT =
20 − 0, 7 − 0, 3 4 Ω IT = 4, 75
Vo = 2 Ω ∗ IT = 9, 5 Vo
Vo Vo = 4, 7 Ω ∗ IT − 2 IT −10 + 1, 2 Ω ∗ IT + VD + 4, 7 Ω ∗ IT − 2 = 0
IT =
IT =
10 − VD + 2 1, 2 Ω + 4, 7 Ω
10 − 0, 7 + 2 1, 2 Ω + 4, 7 Ω
IT = 1, 9153 Vo = 7, 0017 Vo
• Vin = 8 • Vin = 3, 3 • Vin = 1, 5
Vo Vo = 4, 7 Ω ∗ ID + VD = 4, 7 Ω ∗ ID + 0, 7 IT −Vin + 1, 2ID + 4, 7ID + VD = 0
−Vin + 1, 2ID + 4, 7ID + 0, 7 = 0
ID =
Vin − 0, 7 5, 9
Vo = 4, 7 ∗ ID + 0, 7 = 4, 7
Vo = 4, 7 •
Vin − 0, 7 + 0, 7 5, 9
Vin − 0, 7 + 0, 7 5, 9
Vin Vo = 4, 7
8 − 0, 7 + 0, 7 5, 9
Vo = 6, 5153...