Difference between revisions of "Analog Tutorial Lesson 8: Designing Active Higher-Order Cascaded Op-Amp Filters"
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Revision as of 15:23, 21 September 2015
What You Will Learn
In this tutorial you will cascade two second-order lowpass filters to design fourth-order Butterworth and Chebyshev lowpass filters.
Designing a Cascaded Fourth-Order Butterworth Lowpass Filter
The following is a list of parts needed for this part of the tutorial lesson:
Part Name | Part Type | Part Value |
---|---|---|
VCC | DC Voltage Source | 15V |
VEE | DC Voltage Source | -15V |
V1 | AC Voltage Source | 1V |
R1 - R4 | Resistor | 10k |
C1 | Capacitor | 17n |
C2 | Capacitor | 15n |
C3 | Capacitor | 42n |
C4 | Capacitor | 6n |
X1 - X2 | LM741 Op-Amp | Defaults |
You can cascade two second-order lowpass filters to design a fourth-order lowpass filter as shown in the opposite figure. The transfer function of the cascaded systems is given by:
[math]H(s) = H_1(s) . H_2(s) = \frac {\omega_{01} ^2 \omega_{02} ^2 } { \left( s - s_1 \right) \left( s - s_2 \right) \left( s - s_3 \right) \left( s - s_4 \right) } [/math]
The four poles of the fourth-order filter are nothing but the two pairs of poles of the constituent second-order lowpass stages. These poles can be determined according to the type of the filter response specification. For example, for the Butterworth filter, the poles are given by:
N = 4 | Butterworth Poles |
---|---|
Pole 1 | -0.924 + j 0.383 |
Pole 2 | -0.924 - j 0.383 |
Pole 3 | -0.383 + j 0.924 |
Pole 4 | -0.383 - j 0.924 |
Based on the Butterworth poles, and assuming R1 = R2 = R3 = R4 = R = 10k, we can calculate the values of the four capacitors:
C1 = 17nF
C2 = 15nF
C3 = 42nF
C4 = 6nF
Place and connect the parts with the specified values as shown in the above figure. Run an AC frequency sweep of your active filter with the stop frequency of the sweep set to 50kHz. You should get a frequency response like the figure shown below with a cutoff frequency of 1kHz. You can see from the graph that frequency response of the fourth-order filter drops after the cutoff much faster and the value of gain at f = 5kHz is below -55dB.
Start Frequency | 1Hz |
---|---|
Stop Frequency | 50kHz |
Steps/Interval | 50 |
Interval Type | Decade |
Preset Graph Plots | Custom: Gain = VDB(OUT) - VDB(IN) |
Designing a Cascaded Fourth-Order Chebyshev Lowpass Filter
The following is a list of parts needed for this part of the tutorial lesson:
Part Name | Part Type | Part Value |
---|---|---|
VCC | DC Voltage Source | 15V |
VEE | DC Voltage Source | -15V |
V1 | AC Voltage Source | 1V |
R1 - R4 | Resistor | 10k |
C1 | Capacitor | 187n |
C2 | Capacitor | 1.5n |
C3 | Capacitor | 77.2n |
C4 | Capacitor | 16.7n |
X1 - X2 | LM741 Op-Amp | Defaults |
Next, we consider the Chebyshev filter design. With a ripple factor ε = 1, the maximum peak-to-peak ripples would be 3dB. For a fourth-order design, the poles are given by:
N = 4 | Chebyshev Poles |
---|---|
Pole 1 | -0.085 + j 0.947 |
Pole 2 | -0.085 - j 0.947 |
Pole 3 | -0.206 + j 0.392 |
Pole 4 | -0.206 - j 0.392 |
Based on the 3-dB Chebyshev poles, and assuming R1 = R2 = R3 = R4 = R = 10k, we can calculate the values of the four capacitors:
C1 = 187nF
C2 = 1.5nF
C3 = 77.2nF
C4 = 16.7nF
Place and connect the parts with the specified values as shown in the above figure. Run an AC frequency sweep of your active filter with the stop frequency of the sweep set to 50kHz. You will get a frequency response like the figure shown below with a cutoff frequency of 1kHz. You can see from the graph that frequency response of this fourth-order filter drops even faster than the previous case and the value of gain at f = 5kHz is -70dB. However, now you have introduced a ripple in the frequency response with a maximum peak-to-peak amplitude of 3dB!
Start Frequency | 1Hz |
---|---|
Stop Frequency | 50kHz |
Steps/Interval | 50 |
Interval Type | Decade |
Preset Graph Plots | Custom: Gain = VDB(OUT) - VDB(IN) |