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Passive RC Filter Circuits
A passive low-pass filters using RC is shown below. The reactance of the capacitor is
dependent on frequency, X c = 1 2π f c . As the frequency goes up, the reactance, X c of the
capacitor goes down. The resistance of a resistor is constant with changes in frequency. By
examining the low-pass filter, it is a voltage divider made up of R and X c . The output is
taken across X c . At low frequencies, X c is much larger than R , and most of the input
voltage is across the output. At high frequencies, X c becomes smaller than R , and most of
the input voltage is dropped across R , causing the output to be small.
1 1
V jωC jωC 1 1 1
H (ω ) = out = = = ; H (ω ) = =
Vin R + 1 jω RC + 1 1 + jω RC 1 + jω RC 1 + ω 2 ( RC )
2
jωC jωC
⎛V ⎞
ω = 2π f ; RC = time constant ; dB = 20 × log10 ⎜ out ⎟
⎝ Vin ⎠
ω H (ω ) Approximation for H (ω )
1 1
10RC 1 + j ( 0.1)
1 or 0 dB
1 1
RC 1+ j
0.707 or -3 dB
10 1
RC 1 + j (10 )
0.1 or -20 dB
100 1
RC 1 + j (100 )
0.01 or -40 dB
1
Note: = 2π f c = cutoff frequency
RC
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For the high-pass filter, the components have been reversed. The high frequencies will be
passed and the low frequencies will be attenuated.
Vout R 1 1 1 1
H (ω ) = = == ; H (ω ) = = =
Vin R + 1 1 1 j 1
1+ 1+ 1− 1+
jωC jω RC jω RC ω RC ω ( RC )
2 2
⎛V ⎞
ω = 2π f ; RC = time constant ; dB = 20 × log10 ⎜ out ⎟
⎝ Vin ⎠
ω H (ω ) Approximation for H (ω )
1 1
100RC 1 − j100 0.01 or -40 dB
1 1
10RC 1 − j10 0.1 or -20 dB
1 1
RC 1− j 0.707 or -3 dB
10 1
RC 1 − j ( 0.1) 1 or 0 dB
No
Fig. 1 shows a low-pass filter and Fig. 2 shows a high-pass active filter. The circuits consist
of a passive RC filter connected to an op-amp follower stage. The voltage at the output of the
follower is the same as the voltage at the non-inverting input.
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First-order Active Filters
Fig. 1 shows a low-pass filter and Fig. 2 shows a high-pass active filter. The circuits
consist of a passive RC filter connected to an op-amp follower stage. The voltage at the
output of the follower is the same as the voltage at the non-inverting input.
_ +V
R1
+
−V RL
VS C1
Fig. 1: Low-pass Filter
_ +V
C1
+
−V RL
VS R1
Fig. 2: High-pass Filter
These circuits have the same output as single-stage passive filters. Since the RC filter is
connected to the high impedance input of the non-inverting amplifier, the RC filter will not
be loaded by the next stage. It is possible to replace the op-amp follower circuit with an op-
amp non-inverting amplifier and obtain voltage gain.
Filters can be classified by the steepness of their roll-off. The roll-off is a measurement of
how a reference output signal in the pass-band to the output signal at a given frequency
beyond cutoff. Fig. 3 shows the Bode plot for three different orders of filters. The passive
filters are limited to a roll-off of 20 dB/decade or 6 dB/octave and are called first-order
filters. A roll-off of 20 dB/decade means that as the frequency is increased by ten times, the
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amplitude of the output signal is reduced to one-tenth. First-order filters use only one RC
network and are also called single-pole filters.
0 db 1V Cutoff frequency
Roll-off 20 dB/decade
-20 dB 100 mV
Roll-off 40 dB/decade
- 40 dB 10 mV
10 KHz
Roll-off 60 dB/decade
- 60 dB 1 mV
- 80 dB 0.1 mV
100 KHz
1 KHz
10 KHz
1 MHz
Fig. 3: Bode plot of different order filters
Higher-order Active Filters:
Filters can be made closer to the ideal by increasing the number of RC networks or poles. If a
filter has a roll-off of −40 dB/decade, it has two RC networks and is referred to as a second-
order, or double-pole, filter. If a filter has a roll-off of −60 db/decade, it has three RC
networks and it is referred to as a third order or three-pole filter. Adding additional RC
networks will increase the steepness of the roll-off.
The roll-off of a filter can be expressed in dB per octave. For example, a first order filter has
a roll-off of 20 dB/decade, which is equal to 6 dB/octave. Fig. 3 shows a roll-off of 20
dB/decade equal to 6 dB/octave, a roll-off of 40 dB/decade equal to 12 dB/octave, and a roll-
off of 60 dB/decade equal to 18 dB/octave.
By putting two first-order active filters in series, a second-order filter can be formed.
However, this requires two op-amps. Figs 4 & 5 show a method of obtaining a second-order
filter using only one op-amp. Fig. 4 is a low-pass filter, and Fig. 5 is a high-pass filter. The
gain of the circuit is unity, and the filtering components must have the relationship shown in
Figs 4 & 5 for proper operation. The cutoff frequency of both circuits can be found using the
1
following formula: f cutoff =
2π R1 R 2C1C2
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C2
_ +V AV = 1
R1 = R 2
R1 R2
2 C 1 = C2
+
−V RL Rf = 0Ω
VS C1 R in = ∞ Ω
Fig. 4: Second-order Low-pass filter with unity gain
R2
_ +V AV=1
C1 C2 R1 = 2 R 2
C 1 = C2
+
−V RL Rf =0Ω
VS R1 R in = ∞ Ω
Fig. 5: Second-order High-pass filter with unity gain
Figs 4 & 5 show a second-order filter that uses equal value filtering capacitors and resistors.
In order for this circuit to work properly, the gain must be set to 1.586. The gain is calculated
using the normal voltage gain formula for a non-inverting amplifier, Av = 1 + ( R f R in ) .
When equal value filtering components are used, the cutoff frequency for the total circuit is
calculated by calculating the cutoff frequency for one pole using the
formula: f cutoff = 1 ( 2π RC ) .
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Higher-order filters can be made by cascading stages of first- and second-order filters. For
example, Fig. 6a shows a third-order low-pass filter constructed by using a second-order low-
pass filter followed by a first-order low-pass filter. The roll-off of the first stage is −40
db/decade, and the second stage has a roll-off of −20 db/decade. The total roll-off is equal to
60 db/decade.
Fig. 6b shows a high-pass active filter with a total roll-off of 60 db/decade. If steeper roll-offs
are needed, they can be obtained by adding more stages. If all the filtering capacitors and
resistors are equal, the formula f cutoff = 1 ( 2π RC ) can be used to calculate the cutoff
frequency of the filter circuit. In order to use equal value filtering components, the voltage
gain of each stage must be set to a specific value using resistors Rf and Ri in each stage.
The calculations for these gains are not discussed here. Table 1 shows the voltage gains
needed to design up to six-order filter circuits using equal value components.
C2 R f2
R f1
R i2 +V
_
R i1 +V
_
R1 +
+ R3 −V
RL
R2 −V C3
C1
VS
Fig. 6a: Third-order low-pass filter (60 dB/decade roll-off)
R2 R f2
R f1
R i2 +V
_
R i1 +V
_
C1 +
+ C3 −V
RL
C2 −V R3
VS R1
Fig. 6b: Third-order high-pass filter (60 db/decade roll-off)
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Table 1
Order Roll-off 1st stage 2nd stage 3rd stage Total
DB/decade Poles Av Poles Av Poles Av Av
1 20 dB 1 Optional Optional
2 40 dB 2 1.586 1.586
3 60 dB 2 2.000 1 2.000 4.000
4 80 dB 2 1.152 2 2.235 2.575
5 100 dB 2 2.000 2 1.382 1 2.382 6.583
6 120 dB 2 1.068 2 1.586 2 2.382 4.204
Note: The voltage gain of the total circuit is equal to the product of each stage.
Bandpass Filters
Bandpass filters are designed to pass a band of frequencies between a low cutoff frequency,
f1, and a high cutoff frequency, f2. The frequency in the center of the band is called the center
frequency, f0. Both response curves in Figure 7 have an output signal of 1 V at the center
frequency, f0, of 100 KHz. Since they are bandpass filters, each response curve has a low
cutoff frequency, f1 and a high cutoff frequency, f2. The magnitude of the output signal at the
cutoff frequency is equal to 70.7% of the magnitude at the center frequency, or in this case
0.707 V. The bandwidth (BW) is defined as the difference between the high and low cutoff
frequencies (BW = f2 - f1 ). The response curve in Figure 7a shows a bandwidth of 20 KHz,
whereas the response curve in Figure 7b shows a bandwidth of 2 KHz.
V f 1 = 0.707 V V f 1 = 0.707 V
V f 0 = 1V V f 0 = 1V
V f 2 = 0.707 V V f 2 = 0.707 V
f1 f0 f2 f1 f 0 f 2
f 1 = 90 KHz BW = 20 KHz f 1 = 99 KHz BW = 2 KHz
f 0 = 100 KHz Q = 100 20 = 5 f 0 = 100 KHz Q = 100 2 = 50
f 2 = 110 KHz Roll -off = 20 dB / decade f 2 = 101 KHz Roll -off = 40 dB / decade
(a) (b)
Fig. 7: Bandwidth comparison
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Q is equal to the center frequency divided by the bandwidth (Q = f0 /BW). The Q value is
useful when comparing filter circuits. For example, both response curves in Fig. 7 have a
center frequency of 100 KHz, but the Q in Fig. 7a is 5, and the Q in Fig. 7b is 50. The higher
the Q, the more selective the filter will be. Filter circuits with a Q below 10 are called low-Q
filters, and filter circuits with a Q above 10 are called high-Q filters. Low-Q filters are wide
bandpass circuits, and high-Q filters are narrow bandpass circuits.
Fig. 8a shows the frequency-response curve for a bandpass filter circuit, and Fig. 8b shows
the Bode plot for the same circuit. The Bode plot shows the output signal as a constant 2 V
across the bandwidth, and then a constant rate of roll-off beyond cutoff. The filter has a low-
end roll-off and high-end roll-off. Normally both ends roll off at the same rate (as shown in
Fig. 8), but it is possible to have a band-pass filter circuit with different roll-off rates on the
low and high ends. To calculate the high-end roll-off in dB/decade, the center frequency
output signal is divided into the output signal at a frequency of one decade above cutoff. To
calculate the low-end roll-off, the center frequency output signal is divided into the output
signal at a frequency of one decade below cutoff. Example 1 shows how to evaluate a band-
pass filter by examining its frequency-response curve.
Example 1: Calculate the center frequency, bandwidth, Q, and roll-off for the filter
response curve shown in Fig. 8b.
2V 2V
0 dB
-10 dB 1.414 V 1.414 V
-20 dB
-30 dB
-40 dB
0.2 V 0.2 V
-50 dB
0.9 9 10 11 110 KHz 0.9 9 10 11 110 KHz
(a) (b)
Fig. 8: Band response and Bode plot curves
Step 1: From the Bode plot, determine the low cutoff frequency. f 1 = 9 KHz
Step 2: From the Bode plot, determine the high cutoff frequency. f 2 = 11 KHz
Step 3: Calculate the center frequency. For a circuit with a Q of two or higher, the center
frequency can be calculated by dividing the sum between f2 and f1 by two.
f 0 = ( f 1 + f 2 ) 2 = ( 9 KHz + 11 KHz ) 2 =10 KHz
Step 4: Calculate the bandwidth of the filter: BW = 11 KHz − 9 KHz = 2 KHz
Step 5: Calculate the Q of the filter: Q 0 = f 0 BW = 10 KHz 2 KHz = 5
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Step 6: Calculate the low-end roll-off in dB/decade. The low cutoff frequency is 9 KHz.
Therefore, read the magnitude of the signal at one-tenth of the cutoff (900 Hz)
from Fig. 8b. The output at 900 Hz is 0.02 V.
⎛ V0.1 f 1 ⎞ ⎛ 0.02 V ⎞
Low-end roll-off = 20 log ⎜ ⎟ = 20 log ⎜ ⎟ = −40 dB / decade
⎜ Vf ⎟ ⎝ 2V ⎠
⎝ 0 ⎠
Step 7: Calculate the high-end roll-off in dB/decade. The high cutoff frequency is 11
KHz. Therefore, read the magnitude of the signal at ten times of the cutoff (110
KHz) from Fig. 8b. The output at 110 KHz is 0.02 V.
⎛ V10 f 1 ⎞ ⎛ 0.02 V ⎞
High-end roll-off = 20 log ⎜ ⎟ = 20 log ⎜ ⎟ = −40 dB / decade
⎜ Vf ⎟ ⎝ 2V ⎠
⎝ 0 ⎠
Fig. 9 shows a bandpass filter using a second-order low-pass filter followed by a second-
order high-pass filter. The upper cutoff frequency (f2 ) is determined by calculating the cutoff
frequency of the low-pass filter, and the lower cutoff frequency (f1 ) is determined by
calculating the cutoff frequency of the high-pass filter. This type of filter is used for wide
bandpass filters. The Q of this filter will be below two; consequently, the center frequency
should be calculated using the formula: f 0 = f 1 f 2 . This formula is valid for calculating
the center frequency of all bandpass filters; however, for bandpass filters with Qs higher than
two, the formula f 0 = ( f 1 + f 2 ) 2 gives approximately the same results.
Example 2: Calculate the low and high cutoff frequencies, the bandwidth, the center
frequency, and the Q of the filter circuit as shown in Fig. 9. Draw a Bode plot of the
frequency response curve.
C 2 0.001 μ F R 3 6.8 K
R f 1 5.86 K R f 2 5.86 K
R i1 10 K +V
_ R i 2 10 K
_ +V
R1 R2 C3 C4
+
+
2.2 K 2.2 K −V
0.001 μ F 0.001 μ F RL
C1 R4 −V
VS
0.001 μ F 6.8 K
Fig. 9: Bandpass filter
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Step 1: Calculate the low cutoff frequency. The low cutoff frequency is a function of the
high-pass filter stage.
1 1
f1 = = = 23 KHz
2π R 3C 3 2π × 6.8K × 0.001μ F
Step 2: Calculate the high cutoff frequency. The high cutoff frequency is a function of the
low-pass filter stage.
1 1
f2 = = = 72 KHz
2π R1C 2 2π × 2.2 K × 0.001μ F
Step 3: Calculate the center frequency:
f0 = f 1 × f 2 = 23 KHz × 72 KHz = 40.7 KHz
Step 4: Calculate the bandwidth: BW = f 2 − f 1 = 72 KHz − 23 KHz = 49 KHz
Step 5: Calculate the Q: Q = f 0 BW = 40.7 KHz 49 KHz = 0.83
Step 6: The Bode plot:
40 dB/decade
23 KHz 72 KHz
Fig. 10: Bandpass Filter Frequency response
Multiple-feedback Bandpass Filter:
C f 0.005μ F
R f 20 K
1
+V f0 =
( R // R ) R C C
Ri Ci
_ 2π i 1 f i f
1K
0.01 μ F
+
Q = π f 0 R f C iC f
R1 −V
Vin
1K V out
Fig. 11: Multiple-feedback bandpass filter
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Fig. 11 is a multiple-feedback bandpass filter. The circuit is essentially an inverting
amplifier with two feedback paths. At frequencies in the bandpass, capacitors Ci and C f
are at moderate impedance values, permitting the circuit to function as an inverting
amplifier. In the bandpass area, the main factors controlling voltage gain are resistors R f
and R i ( Av = R f R i ). At frequencies below the bandpass, the main factors controlling
the voltage gain are the resistance of R f and the impedance of capacitor Ci
( Av ≈ R f C i ) .Since Ci has high impedance at low frequencies, the gain is low, resulting
in a small output signal. At frequencies above the bandpass, the main factors controlling
the voltage gain are the impedance of capacitor C f and the resistance of R i
( Av ≈ C f R i ). Since C f has low impedance at high frequencies, the gain is low,
resulting in a small output signal.
Once the Q and center frequency, f0, are known, the bandwidth can be calculated by using
the formula, BW = f 0 / Q . Since approximately half of the bandwidth is above the center
frequency and half below, the upper and lower cutoff frequencies can be approximated.
Example 3: Calculate the center frequency, the Q, the bandpass, the low cutoff
frequency, and the high cutoff frequency for the circuit in Fig. 11.
Step 1: Calculate the center frequency:
1 1
f0 = = = 7.1 KHz
2π ( Ri // R1 ) R f C i C f 2π (1K //1K ) 20 K × 0.01μ F × 0.005μ F
Step 2: Calculate the Q:
Q = π f 0 R f C i C f = π ( 7.1 KHz )( 20 K Ω ) 0.01 μ F × 0.005 μ F = 3
Step 3: Calculate the Bandwidth: BW = f 0 Q = 7.1 KHz 3 = 2.36 KHz
Step 4: Calculate the low cutoff frequency (f1 ):
f 1 = f 0 − ( BW 2 ) = 7.1 KHz − ( 2.36 KHz 2 ) = 5.92 KHz
Step 5: Calculate the high cutoff frequency (f2 ):
f 1 = f 0 + ( BW 2 ) = 7.1 KHz + ( 2.36 KHz 2 ) = 8.28 KHz
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