14.10 Frequency Response Using PSpice

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14.10 Frequency Response Using PSpice

PSpice is a useful tool in the hands of the modern circuit designer for obtaining the frequency response of circuits. The frequency response is obtained using the AC Sweep as discussed in Section D.5 (Appendix D). This requires that we specify in the AC Sweep dialog box Total Pts, Start Freq, End Freq, and the sweep type. Total Pts is the number of points in the frequency sweep, and Start Freq and End Freq are, respectively, the starting and final frequencies, in hertz. In order to know what frequencies to select for Start Freq and End Freq, one must have an idea of the frequency range of interest by making a rough sketch of the frequency response. In a complex circuit where this may not be possible, one may use a trial-and-error approach.

There are three types of sweeps:

• Linear: The frequency is varied linearly from Start Freq to End Freq with Total Pts equally spaced points (or responses).
• Octave: The frequency is swept logarithmically by octaves from Start Freq to End Freq with Total Pts per octave. An octave is a factor of 2 (e.g., 2 to 4, 4 to 8, 8 to 16).
• Decade: The frequency is varied logarithmically by decades from Start Freq to End Freq with Total Pts per decade. A decade is a factor of 10 (e.g., from 2 to 20 Hz, 20 to 200 Hz, 200 Hz to 2 kHz).

It is best to use a linear sweep when displaying a narrow frequency range of interest, as a linear sweep displays the frequency range well in a nar row range. Conversely, it is best to use a logarithmic (octave or decade) sweep for displaying a wide frequency range of interest—if a linear sweep is used for a wide range, all the data will be crowded at the highor low-frequency end and insufficient data at the other end.

With the abo ve specifications, PSpice performs a steady-state sinusoidal analysis of the circuit as the frequency of all the independent sources is varied (or swept) from Start Freq to End Freq.

The PSpice A/D program produces a graphical output. The output data type may be specified in the Trace Command Box by adding one of the following suffixes to V or I:

• M Amplitude of the sinusoid.
• P Phase of the sinusoid.
• dB Amplitude of the sinusoid in decibels, that is, 20 log ¹⁰ (amplitude).

Example 14.15 Determine the frequency response of the circuit shown in Fig. 14.50.

Solution:

We let the input voltage vs be a sinusoid of amplitude 1 V and phase 0°. Figure 14.51 is the schematic for the circuit. The capacitor is rotated 270° counterclockwise to ensure that pin 1 (the positive terminal) is on top. The voltage marker is inserted to the output voltage across the capacitor. To perform a linear sweep for 1 < f < 1,000 Hz with 50 points, we select Analysis/Setup/AC Sweep, DCLICK Linear, type 50 in the Total Pts box, type 1 in the Start Freq box, and type 1000 in the End Freq box. After saving the file, we select Analysis/Simulate to simulate the circuit. If there are no errors, the PSpice A/D window will

Figure 14.50 For Example 14.15.

The schematic for the circuit in Fig. 14.50.

display the plot of V(C1:1), which is the same as Vo or H(ω) = Vo∕1, as shown in Fig. 14.52(a). This is the magnitude plot, since V(C1:1) is the same as VM(C1:1). To obtain the phase plot, select Trace/Add in the PSpice A/D menu and type VP(C1:1) in the Trace Command box. Figure 14.52(b) shows the result. By hand, the transfer function is

$$H(\omega) = \frac{V_o}{V_s} = \frac{1,000}{9,000 + j\omega 8}$$

or

$$H(\omega) = \frac{1}{9 + j16\pi \times 10^{-3}}$$

showing that the circuit is a low-pass filter as demonstrated in Fig. 14.52. Notice that the plots in Fig. 14.52 are similar to those in Fig. 14.3 (note that the horizontal axis in Fig. 14.52 is logrithic while the horizontal axis in Fig. 14.3 is linear.)

Figure 14.53 For Practice Prob. 14.15.

For Practice Problem 14.15: (a) magnitude plot, (b) phase plot of the frequency response.

For Example 14.16.

Example 14.16 Use PSpice to generate the gain and phase Bode plots of V in the circuit of Fig. 14.55.

Solution:

The circuit treated in Example 14.15 is first-order while the one in this example is second-order. Since we are interested in Bode plots, we use decade frequency sweep for 300 < f < 3,000 Hz with 50 points per de cade. We select this range because we know that the resonant frequency of the circuit is within the range. Recall that

$$\omega_0 = \frac{1}{\sqrt{LC}} = 5 \text{ krad/s} \qquad \text{or} \qquad f_0 = \frac{\omega}{2\pi} = 795.8 \text{ Hz}$$

After drawing the circuit as in Fig. 14.55, we select Analysis/Setup/AC Sweep, DCLICK Decade, enter 50 in the Total Pts box, 300 as the Start Freq, and 3,000 in the End Freq box. Upon saving the file, we simulate it by selecting Analysis/Simulate. This will automatically bring up the PSpice A/D window and display V(C1:1) if there are no errors. Since we are interested in the Bode plot, we select Trace/Add in the PSpice A/D menu and type dB(V(C1:1)) in the Trace Command box. The result is the Bode magnitude plot in Fig. 14.56(a). For the phase plot, we select

For Example 14.16: (a) Bode plot, (b) phase plot of the response.

Trace/Add in the PSpice A/D menu and type VP(C1:1) in the Trace Command box. The result is the Bode phase plot of Fig. 14.56(b). Notice that the plots confirm the resonant frequency of 795.8 Hz.

Consider the network in Fig. 14.57. Use PSpice to obtain the Bode plots Practice Problem 14.16 for Vo over a frequency from 1 to 100 kHz using 20 points per decade.

For Practice Prob. 14.16.

Answer: See Fig. 14.58.