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PSPICE 计算机仿真 Simulation Program with Integrated Circuit Emphasis
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CH12 FREQUENCY RESPONSE 频率响应
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12.1 specifying frequency variation and number
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Sinusoidal Linear, Logarithmic Decade, octave
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12.2 frequency response output Rectangular, polar, decibel
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Example 18 illustrates how to analyze the frequency response of a parallel RLC circuit with PSpice.
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Fig. 104 ddb
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Example 18 a) The current source in the circuit shown in Fig. 104 is 50cosωt mA. Use Probe to plot Vo versus f from 1000 to 2000 Hz in increments of 10 Hz on a linear frequency scale. b) From the Probe plot, estimate the resonant frequency, the bandwidth, and the quality factor of the circuit. c) Compare the results obtained in b) with an analytic solution for f0, β, and Q.
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Solution a & b
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Fig. 105 sch
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Fig. 106 setting
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Using the Probe Cursor, we note that the peak amplitude( 峰值振幅 ) of about 400 V occurs at a frequency of 1590 Hz in Fig. 107a. Thus we estimate the resonant frequency( 衰减频率 ) at 1590 Hz.
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Fig. 107a probe
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To estimate the bandwidth, we use both cursors to find the frequencies where Vo=399.697/1.414=282.67 V. The closest values are 281.992 at 1552 Hz and 1632.1 Hz. (Fig. 107b) Thus we estimate the bandwidth to be 1632.1- 1552, or about 80 Hz.
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Fig. 107b
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We calculate the quality factor from the relationship,
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Solution c) A direct analysis of the circuit yields,
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Comparison QuantityAnalysisPSpice f01591.551590 f11551.761552 f21631.341632.1 β79.5880 Q2019.88
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Example 19
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Modify the PSpice schematic for Example 18 to step the capacitor values from 0.15 μF through 0.35 μF. Then use Probe to display the frequency response characteristics for all values of capacitance. Comment on the effect of the changing capacitance.
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Solution
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Fig. 108 sch
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Fig. 108a param property
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Fig. 108b setting
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Fig. 108c Param_setting
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Fig. 109 Probe
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Result The smallest value of capacitance produced the plot farthest to the right. We expect this result because the equation for resonant frequency for an RLC circuit is:
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Furthermore, as the capacitance increases, the resonant peak becomes sharper. This result, too, comes as no surprise because the equation for Q in a parallel RLC circuit is:
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12.3 Bode plots with probe
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Fig. 110 ddb
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Example 20 compares the exact dB voltage magnitude versus log frequency and phase angle versus log frequency plots to the Bode straight-line approximation using Probe.
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Example 20 Construct a PSpice schematic and associated analysis to generate the frequency response of the circuit shown in Fig. 110 for three different values of resistance: 5 Ω, 50 Ω, and 500 Ω. Then use Probe to plot the output voltage magnitude in dB and output voltage phase angle versus log frequency. Finally, use the Label tool in Probe to overlay a straight-line Bode approximation plot and comment.
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Fig. 111 sch
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Because this is a series RLC circuit, we know that the centre frequency and the bandwidth are given by:
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For the circuit shown in Fig. 110, the centre frequency is:
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The bandwidth ranges:
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Fig. 111a param_property
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Fig. 111b vac_property
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Fig. 111c setting1
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Fig. 111d setting2
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Fig. 112a Plot/Add plot to window
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选中上面窗口 (SEL>> 表示选中 !) Trace/Add trace 先选择右侧 DB() 再选择左侧 V(out) 底部出现 : DB(V(out))
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Fig. 112b select DB(V(out))
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选中下面窗口 (SEL>> 表示选中 !) Trace/Add trace 先选择右侧 P() 再选择左侧 V(out) 底部出现 : P(V(out))
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Fig. 112c select P(V(out))
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Fig. 112 probe
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12.4 Filter design
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Example 21 demonstrates the use of PSpice and Probe in verifying the behavior of a high-Q bandpass filter.
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Fig. 113 ddb
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Example 21 The circuit in Fig. 113 is an active high-Q bandpass filter. Using 1 nF capacitors and an ideal op amp, design values for the three resistors to yield a centre frequency of 10 KHz, a quality factor of 10, and a passband gain of 3. Use Probe to verify that the resistor values you compute produce a filter that satisfies the three frequency response specifications.
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Solution The resistor design equations are given by:
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The scaling factors are
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After scaling,
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Fig. 114 Sch
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Fig. 114a setting
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Fig. 115 probe
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