Respostas do capitulo 3 razavi
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3.1 (a)IX =
VX
R1
0
VX < 0
VX > 0
IX
VX (V)
Slope = 1/R1
3.2
IX =
VX
R1
0
VX < 0
VX > 0
Plotting IX (t), we have
0
0
−V0 /R1
−π/ω
0 t π/ω
−V0
VX (t) (Dotted)
IX (t) for VB = 1 V (Solid)
V0
3.3
IX =
0
VX −VB
R1
VX < VB
VX > VB
Plotting IX vs. VX for VB = −1 V and VB = 1 V, we get:
IX
VB = −1 V
VB = 1 V
Slope = 1/R1
−1
Slope = 1/R1
1
VX (V)
3.4
IX =
0
VX −VB
R1
VX < VB
VX > VB
Let’s assume V0 > 1 V. Plotting IX (t) for VB = −1 V, we get
(V0 − VB )/R1
0
0
VB
−π/ω
Plotting IX (t) for VB = 1 V, we get
0 t π/ω
−V0
VX (t) (Dotted)
IX (t) for VB = −1 V (Solid)
V0
IX (t) for VB = 1 V (Solid)
(V0 − VB )/R1
0
0
−π/ω
0
t π/ω −V0
VX (t) (Dotted)
V0
VB
3.5
IX =
VX −VB
R1
∞
VX < 0
VX > 0
Plotting IX vs. VX for VB = −1 V and VB = 1 V, we get:
IX
IX for VB = −1 V
IX for VB = 1 V
1/R1
Slope = 1/R1
−1
VX (V)
−1/R1
Slope = 1/R1
3.6 First, note that ID1 = 0 always, since D1 is reverse biased by VB (due to the assumption that VB > 0).
We can write IX as
IX = (VX − VB )/R1
Plotting this, we get:
IX
VB
VX (V)
Slope = 1/R1
3.7
VX −VB
R1
VX −VB
R1 R2
IX =
IR1 =
VX < VB
VX > VB
VX − VB
R1
Plotting IX and IR1 for VB = −1 V, we get:
IX
IX for VB = −1 V
IR1 for VB = −1 V
Slope = 1/R1 + 1/R2
−1
Slope = 1/R1
Plotting IX and IR1 for VB = 1 V, we get:
VX (V)
IX
IX for VB = 1 V
IR1 for VB = 1 V
Slope = 1/R1 + 1/R2
1
VX (V)
Slope = 1/R1
3.8
IX =
IR1 =
0
VX
R1
+
VB
R1 +R2
VX
R1
VX −VB
R2
VX <
VX >
VX <
VX >
VB
R1 +R2 R1
VB
R1 +R2 R1
VB
R1 +R2 R1
VB
R1 +R2 R1
Plotting IX and IR1 for VB = −1 V, we get:
IX for VB = −1 V
IR1 for VB = −1 V
Slope = 1/R1 + 1/R2
−VB /R2
Slope = 1/R1
VB
R
R1 +R2 1
VB
R1 +R2
Plotting IX and IR1 for VB = 1 V, we get:
VX (V)