Transcript PSPICE计算机仿真

PSPICE计算机仿真
Simulation Program with Integrated Circuit Emphasis
CH6
INDUCTORS, CAPACITORS, AND
NATURAL RESPONSE
电感、电容和零输入响应
You should know,

Resistor
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Inductor (initial current)
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The value of resistance (In ohms)
The value of inductance (In henries)
Capacitor (initial voltage)
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The value of capacitance (In farads)
6.1 Transient Analysis
(暂态分析)
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You use transient analysis to examine
the response of a circuit as a function of
time.
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You may examine the output of a
transient analysis in two different ways.
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First, you may direct PSpice to write the
results of transient analysis to the output
file.
Second, you may use Probe to generate
plots of voltage and current versus time.
6.2 Natural Response
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To find the natural response of a circuit,
you first find the initial inductor currents
and capacitor voltages.
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Example 10 illustrates how to use
PSpice to find the natural response of a
series RLC circuit. We incorporated a
preliminary analytic solution as part of
the example to permit checking the
validity of the PSpice solution.
Fig. 65
Example 10
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a) analyze the natural response of the
circuit shown in Fig. 65 with respect to
the type of damping, the peak value of
Vc, and the frequency of oscillation.
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b) Use PSpice to analyze the circuit’s
natural response. Then use Proble to
generate a plot of the voltage Vc versus
t, and use the Proble cursor to identify
the peak value of the voltage.
Solution a)
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For the series circuit shown in Fig. 65,
R

 1000
2L
2
6
  10
1
2
6
 
 50  10
LC
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Hence the response is underdamped(欠阻尼),
and the roots of the characteristic equation
are:
s1  1000  j 7000
s2  1000  j 7000
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Therefore, the form of the solution for Vc is,
vc  B1 cos 7000t  B2 sin 7000t e
1000t
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From the initial conditions,
vc 0  10
dvc 0
 45,000
dt
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Solving for B1 and B2 yields,
vc   10 cos 7000t  5 sin 7000t e
1000t
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Based on the solution for Vc, the peak
value of Vc is 7.70 V, and it occurs at
t=362.69 µs. The damped frequency(衰
件频率) is 7000 rad/s, and thus the
damped period(衰减周期) is 897.60 µs.
Solution b)
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To get the signs right, you must pay attention
to the orientation of these parts. (Fig. 66)
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The initial voltage is positive at the left-most
terminal of the capacitor, or bottom terminal if the
part is rotated once.
The initial current enters the left-most terminal of
the inductor, or terminal of the inductor, or bottom
terminal if the part is rotated once.
Fig. 66 schematics
Set the initial values(设定初值)
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The initial capacitor voltage is 10 V.
The initial inductor current is -90 mA.
These values are set in the Property
Editor, in the column labeled IC. (Fig.
66a)
Fig. 66a set the initial values
Simulation
Fig. 67 setting
Fig. 68 probe
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Trace/Cursor/Display
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显示坐标
Trace/Cursor/Peak
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显示峰值
Result
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We see from the cursor output window
that the peak output voltage is 7.7347 V
(书上：7.7255 V), which occurs at
t=371.222 us (书上：372.181 us）,
which is in good agreement with our
analytic results in (a).