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Active Filters
Topic 13
Active Filters Active filters are another important application of operational amplifiers. When a signal of mixed frequencies is passed through a filter circuit, some of the frequencies are reduced in amplitude. Some may be removed altogether. The simplest filter consists of a resistor and a capacitor (p. 44). As it is built from two passive components it is known as a passive filter.
FIRST-ORDER ACTIVE FILTER The diagram below shows one of the simplest active filters. It consists of a resistor-capacitor passive filter followed by a non-inverting amplifier to restore or possibly increase the amplitude of the signal. R
VIN
TL081C
Passive Component
vOUT
IC1
1k6
RF 56k
C
One that does not use an external power source. Examples are resistors, capacitors, inductors, and diodes.
100n
RA 100k
0V
Other passive filters may consist of a network of several resistors and capacitors, or may be combinations of resistors and inductors, or capacitors and inductors. A passive filter must have at least one reactive component (a capacitor or an inductor) in order to make its action frequency-dependent. Passive filters are often used, but a large number of individual filters must be joined together to effectively cut off unwanted frequencies. Such filters take up a lot of space on the circuit board and can be expensive to build. When filtering low frequencies the inductors are necessarily large, heavy and expensive. Another serious disadvantage of passive filters is that passive components can only reduce the amplitude of a signal. A signal that has been passed through a multi-stage passive filter is usually very weak. For these reasons, we prefer active filters, based on active components, especially op amps. When these are used we can obtain much sharper cutoff with relatively few filtering stages, there is no need to use inductors, and the filtered signal may even show amplitude gain. Active Component One that uses an external power source. Examples include transistors, and operational amplifiers.
Electronics: Circuits and Systems. r 2011 Owen Bishop. Published by Elsevier Ltd. All rights reserved.
passive filter
non-inverting amplifier
FIGURE 13.1 In a first-order active filter, the signal from the RC filter network is amplified by a non-inverting op amp amplifier.
With the values shown, the voltage gain of the amplifier is 1.56. The graph below displays the signals in this filter when it is supplied with a 1 V sinusoid at 1 kHz. 1.0 v/V 0.5 0.0 –0.5 –1.0 0
1
2
t / ms
3
FIGURE 13.2 The action of a first-order active filter on a 1 kHz sinusoid (dark grey). The RC filter network reduces its amplitude and delays its phase (light grey). The amplifier then increases the amplitude (black).
The RC network filters the input sinusoid (dark grey line). At the output of the network (light grey)
111
112
PART | 1
the filtered signal has an amplitude of only 0.7 V. This is not the only change. The output reaches its peaks 125 μs behind the input. This is a delay of 1/8 of the period of the signal. In terms of angle, this is 360/8 5 45 . Because it is a delay, it has negative value. The output is out of phase with, or lags the input, by 245 . The black curve represents the output of the op amp. With the values shown for RF and RA, the gain of the amplifier is 1.56. The output from the RC network has amplitude 0.7 V, so the overall gain of the active filter is 0.7 3 1.56 5 1.1. The amplitude of the output is 1.1 V.
Circuits
Example With the values of R and C shown in in Fig 13.1, the cutoff point is: fC 5
1 2πRC
fC 5 1=ð2π 3 1:6 3 100Þ 5 1 kHz
½1=ðkΩ 3 nFÞ
As frequency is increased beyond 1 kHz, the amplitude rolls off even faster and eventually the curve plunges downward as a straight line. If we read off frequency and amplitude at two points on this line we find that for every doubling of frequency, the amplitude falls by 6 dB. We say that the roll-off is 26 dB per octave.
Gain Gain of non-inverting amplifier 5 (RA 1 RF)/RA
Octave
The amplifier does not change the phase any further, so the output lags the input by 45 .
FREQUENCY RESPONSE The frequency response (see Topic 5) is plotted below. The plot (black) shows that amplitude is the full 1.56 V up to about 300 Hz. From then on amplitude begins to fall off slightly, reaching 0.7 V (equivalent to 23 dB, or half power) at 1 kHz. The bandwidth of the filter is 1 kHz. The frequency of fC, the 23 dB point or cut-off point, depends on the values of the resistor and capacitor.
An interval over which frequency doubles or halves.
Another way of expressing this is to measure how much the amplitude falls for a ten-times increase in frequency, or decade. In the plot, the roll-off is 20 dB per decade. Note that these roll-off rates are characteristic of single-stage passive or active filters and do not depend on the values of the capacitor and resistor. The plot also shows how the phase change of the output varies with frequency. The scale shows phase angle on the right margin of the figure. We can see that the phase change at 1 kHz is 245 , confirming what we found in the output plot in Fig 13.3.
0 0 dB = 1.56 V
v / V(dB)
Phase
–3dB
180 160 140 120
–10
100 80
–20
60
–6
40
dB ve cta
/o
–30 –45°
–40
20 0 –20 –40 –60 –80 –100
–50
–120 –140 –160
–60 1
3
10
30
100 300
1k
3k
10k 30k
–180
100k 300k 1M
f / Hz FIGURE 13.3 The Bode plot of the frequency response of the first-order lowpass filter shows that it falls to its half-power level (amplitude 0.7 V) and has phase lag of 45 at the cut-off.
Topic | 13
113
Active Filters
Self Test A first-order lowpass filter has a 47 kΩ resistor and a 220 pF capacitor. What is its cut-off frequency?
A phase lag of 245 at the cut-off point is another characteristic of single-stage passive or active filters. The phase lag is less at lower frequencies. At 0 Hz (DC) the phase change is zero. It increases at frequencies above the cut-off point.
HIGHPASS FIRST-ORDER ACTIVE FILTER The lowpass filter is converted to a highpass filter by transposing the resistor and capacitor. C
TL081C
VIN
VOUT
IC1
100n
RF 56k
R 1k6
RA 100k
The amplitude curve rises up from 0 Hz at 16 dB per octave, or 120 dB per decade. The equation for calculating the cut-off point is the same for both lowpass and highpass filters, so this is 1 kHz as before. Beyond the cut-off point, amplitude rises to 0 dB above about 4 kHz. The phase response curve shows that the filter has a 45 lead at its cut-off frequency. This is the opposite of the phase change found in the lowpass filter. Looking more closely at the frequency response, we note that the amplitude starts to fall again from about 400 kHz upward. This is not the effect of the RC filter but is due to the fall in gain of the op amp at high frequencies. The transition frequency of the TL081C is 3 MHz. If we were to plot the response at higher frequencies, we should see a steep roll-off. In practice, if we extend the operation of this filter into higher frequencies it acts as a bandpass filter.
DESIGNING FIRST-ORDER FILTERS 1 Decide on a capacitor value. 2 Apply the equation: R 5 1/2πfCC.
Example
0V passive filter
To design a lowpass filter with cut-off point at 40 kHz. 1 For stability, try the calculation for a 220 pF capacitor, so that a ceramic plate capacitor with zero tempco can be used. 2 Applying the equation, R 5 18086 Ω. Use an 18 kΩ, 1% tolerance, metal film resistor. For a highpass filter, exchange the resistor and capacitor.
non-inverting amplifier
FIGURE 13.4 A first order highpass active filter consists of a highpass passive filter followed by a non-inverting amplifier.
When this is done, the frequency response has the appearance shown below:
0 dB = 1.56 V
0
–3 dB
v / V(dB)
Phase
180 160 140 120
–10
100 80
–20
60
+45°
40 20 0
–30
–20 –40
–40
–60 –80 –100
–50
–120 –140 –160
–60 1
3
10
30
100 300
1k
3k
10k 30k f / Hz
FIGURE 13.5
–180
100k 300k 1M
114
PART | 1
SECOND-ORDER ACTIVE FILTER Adding an extra stage to the first-order lowpass filter allows feedback to be introduced. The circuit below has two RC filters, one of which is connected to 0 V, as in the first-order lowpass filter. The other is part of a feedback loop from the filter output.
R1
R2
1k6
1k6
TL081C
VIN
VOUT
IC1
C1
C2
100n
100n
RF 56k RA feedback
100k
0V FIGURE 13.6 The second-order active filter has positive feedback from the output to the capacitor of the first stage.
The feedback action makes the filter resonate at frequencies around the cut-off frequency. Resonance is not strong, but there is enough extra feedback around this frequency to push up the ‘knee’ of the frequency response curve. 0 dB = 1.56 V
0 v/V(dB)
–5 –10
Circuits
This sharpening of the ‘knee’ is accomplished in a passive filter by using an inductor-capacitor network (p. 57) to produce the resonance. Using an op amp instead provides the same effect but without the disadvantages of inductors. A highpass version of the second order filter is built by substituting resistors for capacitors and capacitors for resistors. Higher-order filters with even steeper roll-off often consist of two or more secondorder filters and possibly a first-order filter cascaded together.
BANDPASS FILTER A bandpass filter may be constructed by cascading a lowpass filter and a highpass filter (see in Fig 13.8). The cut-off frequency of the lowpass filter is made higher than that of the highpass filter so that there is a region of overlap in which both filters pass frequencies at full strength. This area of overlap can be made narrow or broad by choosing suitable cut-off frequencies. Both lower and upper cut-off frequencies are calculated using the equation f 5 1/2πRC. In the example opposite (Fig 13.7), the cut-off frequency of the lowpass filter is 4.8 kHz, while that of the highpass filter is 1 kHz. The frequency response shows a broad passband from 645 Hz to 7440 Hz, with a central frequency of 2.2 kHz. The bandwidth of the filter is 6795 Hz. Another approach to bandpass filtering is the multiple feedback filter (see Fig 13.11). This has two loops, each providing negative feedback.
–15 –20 –25
VIN
–30
C2
R1
TL081C 100n
–35
C1
–50
100n
–45 1
3
10
30
100
300
1k
3k
10k f / Hz
FIGURE 13.7 The Bode plot of the response of the second-order filter (black) also shows the amplitude of the feedback signal (grey), which pushes up output at the knee of the curve. The doubled filter network produces a steep roll-off at high frequencies.
The grey curve is a plot of the voltage across C1, and it can be seen that the strongest feedback occurs around 1 kHz. The response of the filter turns sharply down above 1 kHz. From then on, there is a steady and more rapid fall in response, the rate depending on the number of stages. With two stages in the filter, the roll-off is 212 dB per octave or 240 dB per decade.
VOUT
IC1
330R
R2
RF 56k
1k6
RA 100k
0V FIGURE 13.8 A broad-band bandpass filter can be made by cascading a lowpass filter and a highpass filter. This is followed by a non-inverting amplifier to restore the amplitude to a reasonable level.
The result (Fig 13.10) is a passband of narrow bandwidth. The centre frequency is 1.8 kHz. The lower cutoff frequency is 1 kHz and the upper cutoff frequency is 3 kHz. This gives a bandwidth of 2 kHz. The centre frequency (but not the bandwidth or the amplitude) may be tuned by varying the value of R3.
Topic | 13
115
Active Filters
The centre frequency of the filter is calculated from the equation: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 R1 1R3 f0 5 2πC R1 R2 R3
V(dB) 0 –3
–10
100
1k
10k
Hz
FIGURE 13.9 The bandpass filter in Figure 13.8 produces a broad passband, with response falling off at 6 dB per octave on either side.
The response curve in Fig 13.10 shows an upturn in the higher frequency range. The is the result of the transition frequency of the op amp. The gain of the op amp is reduced at higher frequencies and therefore the higher frequencies in the signal are not completely cancelled out by the negative feedback.
BANDSTOP FILTERS 0
V(dB)
–10
–20 100
10k
1k
100k Hz
FIGURE 13.10 The bandpass filter in Figure 13.11 produces a narrow passband, with response falling off at 6 dB per octave on either side.
G
G
R1 and C1 form a highpass filter; high frequency signals passing through this are fed back negatively, partly cancelling out the high frequencies in the original signal. R2 and C2 form a lowpass filter; low frequency signals passing through this are fed back negatively, partly cancelling out the low frequencies in the original signal.
Only signals in the medium range of frequency are able to pass through the filter at full strength. C1 100n
R2
VIN
C2
R1
100n
3k3
R3
1k6
IC1
VOUT
TL081 C
560R
0V FIGURE 13.11 A multiple feedback filter cancels out both high and low frequencies by feeding them back negatively, leaving the middle frequencies unaffected.
Bandstop filters, often known as notch filters, are used to remove a limited range of frequencies from a signal. For example, such a filter may be used to remove a 50 Hz mains ‘hum’ that has been picked up by electromagnetic interference. This may be necessary when measuring small voltage signals from the human body. There are two approaches to bandstop filtering, both of which show interesting applications of electronic principles. The first circuit (below) depends on one of the basic applications of op amps.
Bandpass filter
Intermediate frequencies, inverted RF
RA2
VIN
RA1 All frequencies
Adder
VOUT Low and high frequencies
0V
FIGURE 13.12 This bandstop filter uses two op amps, one as part of an active filter, the other as an adder.
The signal is sent to an op amp adder, both directly and indirectly through a bandpass filter. The filter could be a multiple feedback filter (see p. 114). This not only filters the signal, removing low and high frequencies, but also inverts it. The second op amp mixes the unfiltered signal with the inverted filtered signal. The result is to subtract the medium frequencies from the original signal, creating a notch in the frequency response curve. The depth of the notch depends on the relative values of RA1 and RA2. Another type of bandstop filter depends on an LC resonant network Fig 13.13. This is combined with a resistor and an op amp wired as an inverting amplifier.
116
PART | 1
G
C
VIN
TL081C
220u
VOUT
IC1
G G
100u
RF 56k
R 22R
RA 100k 0V FIGURE 13.13 This bandstop filter uses an LC resonant network to produce a very narrow notch in its frequency response (below).
v(dB) 0
–4
–8 500
1k
Hertz
2k
FIGURE 13.14 The LC bandstop filter has a steep-sided notch at 1.07 kHz.
At the resonant frequency, the capacitor and inductor have equal impedance. Their impedance in parallel is a maximum. This reduces the amplitude of signals of that frequency, producing the narrow notch in the frequency response. To calculate the notch frequency we use the equation on p. 57. The depth of the notch is determined by the value of R. The smaller the value of R, the deeper the notch.
ACTIVITIES — ACTIVE FILTERS It is easier to work this activity on a circuit simulator, and this has the advantage that an analysis of frequency response can be done very quickly. But you should build and test at least two filter circuits using a breadboard. The active filter circuits that you can investigate include: G G G G
First-order (single stage) lowpass filter. First-order (single stage) highpass filter. Second order lowpass filter. Second order highpass filter.
Circuits
Bandpass filter consisting of a lowpass filter cascaded with a highpass filter (cutoff point for the lowpass filter must be higher than that for the highpass filter). Multiple feedback bandpass filter. Bandstop filter based on resonant LC network.
For each filter connect a signal generator to its input and a dual-trace oscilloscope to its input and output. Observe and explain what happens to the signal as it passes through the filter. Observe the effect of the filter on the amplitude of the signal and on the shape of the waveform, trying this with both sinusoidal and square waves, and at different frequencies. For example, when testing a lowpass filter, use a signal with a frequency well below the cut-off point, try another signal at the cut-off point and a try a third signal well above the cut-off point. Summarise your results in a table or by sketching frequency response curves. Remember that the transition frequency of the op amp itself may be about 1 MHz, so do not use frequencies higher than 100 kHz in these investigations.
QUESTIONS ON ACTIVE FILTERS 1 List and discuss the advantages of op amp active filters, compared with passive filters. 2 Design a first-order lowpass filter with a cut-off frequency of 15 kHz. 3 Design a first-order highpass filter with a cut-off frequency of 20 Hz. 4 Explain the action of a second-order op amp lowpass filter and state why its performance is superior to that of a first-order filter. 5 Design a bandpass filter consisting of a lowpass and a highpass first-order filter cascaded. The specification is: low cut-off point 5 2 kHz; upper cut-off point 5 5 kHz; gain at the centre frequency 5 1.5. 6 Explain the action of a multiple feedback bandpass filter. In what way is its frequency response different from that of cascaded lowpass and highpass filters? 7 Outline the structure of three different types of active filter. Describe two examples of how each type can be used.
MULTIPLE CHOICE QUESTIONS 1 A A B C D
passive filter can be built from: a capacitor, a resistor and an op amp a capacitor and an inductor resistors only a resistor, and an op amp.
Topic | 13
Active Filters
2 An op amp used in a highpass filter should have: A a high gain-bandwidth product B low slew rate C low input bias current D high output resistance. 3 At the cut-off point of a lowpass filter the phase change is: A 190 B 0 C 245 D 290 . 4 The cut-off frequency of a first-order active high pass filter is 5 kHz. It has a 1 nF capacitor. The nearest E12 value for the resistor is: A 200 kΩ B 31.8 Ω C 33 kΩ D 220 kΩ. 5 An active filter must have: A a resistor B a diode C no more than two passive components D a least one reactive component. 6 The roll-off of a 2nd-order highpass filter is: A 16 dB per octave B 140 dB per decade C 212 dB per decade D 220 dB per octave.
117
7 An active bandpass filter is built from a lowpass filter cascaded with a highpass filter. It has a lower cutoff frequency of 200 Hz, and a bandwidth of 800 Hz. The cutoff points of its two filters are: A lowpass 5 200 Hz, highpass 5 1 kHz. B lowpass 5 1 kHz, highpass 5 200 Hz. C lowpass 5 200 Hz, highpass 5 800 Hz. D lowpass 5 800 Hz, highpass 5 1 kHz. 8 In a multiple feedback bandpass filter, there is: A negative feedback of low frequencies B negative feedback of middle frequencies C positive feedback of middle frequencies D negative feedback of low and high frequencies. 9 A bandstop active filter produces a notch in the frequency response because: A of the inverting action of the op amp B the reactances of the capacitor and inductor are equal at the notch frequency C the reactance of the capacitor and inductor in parallel is a maximum at the notch frequency D the LC network resonates strongly at the centre frequency. 10 A bandstop active filter has a resonant network in which C 5 10 nF, and L 5 27 μH. The centre frequency of the notch is at: A 589 kHz B 306 kHz C 962 Hz D 1.85 MHz.
Active Filters Active filters are another important application of operational amplifiers. When a signal of mixed frequencies is passed through a filter circuit, some of the frequencies are reduced in amplitude. Some may be removed altogether. The simplest filter consists of a resistor and a capacitor (p. 44). As it is built from two passive components it is known as a passive filter.
FIRST-ORDER ACTIVE FILTER The diagram below shows one of the simplest active filters. It consists of a resistor-capacitor passive filter followed by a non-inverting amplifier to restore or possibly increase the amplitude of the signal. R
VIN
TL081C
Passive Component
vOUT
IC1
1k6
RF 56k
C
One that does not use an external power source. Examples are resistors, capacitors, inductors, and diodes.
100n
RA 100k
0V
Other passive filters may consist of a network of several resistors and capacitors, or may be combinations of resistors and inductors, or capacitors and inductors. A passive filter must have at least one reactive component (a capacitor or an inductor) in order to make its action frequency-dependent. Passive filters are often used, but a large number of individual filters must be joined together to effectively cut off unwanted frequencies. Such filters take up a lot of space on the circuit board and can be expensive to build. When filtering low frequencies the inductors are necessarily large, heavy and expensive. Another serious disadvantage of passive filters is that passive components can only reduce the amplitude of a signal. A signal that has been passed through a multi-stage passive filter is usually very weak. For these reasons, we prefer active filters, based on active components, especially op amps. When these are used we can obtain much sharper cutoff with relatively few filtering stages, there is no need to use inductors, and the filtered signal may even show amplitude gain. Active Component One that uses an external power source. Examples include transistors, and operational amplifiers.
Electronics: Circuits and Systems. r 2011 Owen Bishop. Published by Elsevier Ltd. All rights reserved.
passive filter
non-inverting amplifier
FIGURE 13.1 In a first-order active filter, the signal from the RC filter network is amplified by a non-inverting op amp amplifier.
With the values shown, the voltage gain of the amplifier is 1.56. The graph below displays the signals in this filter when it is supplied with a 1 V sinusoid at 1 kHz. 1.0 v/V 0.5 0.0 –0.5 –1.0 0
1
2
t / ms
3
FIGURE 13.2 The action of a first-order active filter on a 1 kHz sinusoid (dark grey). The RC filter network reduces its amplitude and delays its phase (light grey). The amplifier then increases the amplitude (black).
The RC network filters the input sinusoid (dark grey line). At the output of the network (light grey)
111
112
PART | 1
the filtered signal has an amplitude of only 0.7 V. This is not the only change. The output reaches its peaks 125 μs behind the input. This is a delay of 1/8 of the period of the signal. In terms of angle, this is 360/8 5 45 . Because it is a delay, it has negative value. The output is out of phase with, or lags the input, by 245 . The black curve represents the output of the op amp. With the values shown for RF and RA, the gain of the amplifier is 1.56. The output from the RC network has amplitude 0.7 V, so the overall gain of the active filter is 0.7 3 1.56 5 1.1. The amplitude of the output is 1.1 V.
Circuits
Example With the values of R and C shown in in Fig 13.1, the cutoff point is: fC 5
1 2πRC
fC 5 1=ð2π 3 1:6 3 100Þ 5 1 kHz
½1=ðkΩ 3 nFÞ
As frequency is increased beyond 1 kHz, the amplitude rolls off even faster and eventually the curve plunges downward as a straight line. If we read off frequency and amplitude at two points on this line we find that for every doubling of frequency, the amplitude falls by 6 dB. We say that the roll-off is 26 dB per octave.
Gain Gain of non-inverting amplifier 5 (RA 1 RF)/RA
Octave
The amplifier does not change the phase any further, so the output lags the input by 45 .
FREQUENCY RESPONSE The frequency response (see Topic 5) is plotted below. The plot (black) shows that amplitude is the full 1.56 V up to about 300 Hz. From then on amplitude begins to fall off slightly, reaching 0.7 V (equivalent to 23 dB, or half power) at 1 kHz. The bandwidth of the filter is 1 kHz. The frequency of fC, the 23 dB point or cut-off point, depends on the values of the resistor and capacitor.
An interval over which frequency doubles or halves.
Another way of expressing this is to measure how much the amplitude falls for a ten-times increase in frequency, or decade. In the plot, the roll-off is 20 dB per decade. Note that these roll-off rates are characteristic of single-stage passive or active filters and do not depend on the values of the capacitor and resistor. The plot also shows how the phase change of the output varies with frequency. The scale shows phase angle on the right margin of the figure. We can see that the phase change at 1 kHz is 245 , confirming what we found in the output plot in Fig 13.3.
0 0 dB = 1.56 V
v / V(dB)
Phase
–3dB
180 160 140 120
–10
100 80
–20
60
–6
40
dB ve cta
/o
–30 –45°
–40
20 0 –20 –40 –60 –80 –100
–50
–120 –140 –160
–60 1
3
10
30
100 300
1k
3k
10k 30k
–180
100k 300k 1M
f / Hz FIGURE 13.3 The Bode plot of the frequency response of the first-order lowpass filter shows that it falls to its half-power level (amplitude 0.7 V) and has phase lag of 45 at the cut-off.
Topic | 13
113
Active Filters
Self Test A first-order lowpass filter has a 47 kΩ resistor and a 220 pF capacitor. What is its cut-off frequency?
A phase lag of 245 at the cut-off point is another characteristic of single-stage passive or active filters. The phase lag is less at lower frequencies. At 0 Hz (DC) the phase change is zero. It increases at frequencies above the cut-off point.
HIGHPASS FIRST-ORDER ACTIVE FILTER The lowpass filter is converted to a highpass filter by transposing the resistor and capacitor. C
TL081C
VIN
VOUT
IC1
100n
RF 56k
R 1k6
RA 100k
The amplitude curve rises up from 0 Hz at 16 dB per octave, or 120 dB per decade. The equation for calculating the cut-off point is the same for both lowpass and highpass filters, so this is 1 kHz as before. Beyond the cut-off point, amplitude rises to 0 dB above about 4 kHz. The phase response curve shows that the filter has a 45 lead at its cut-off frequency. This is the opposite of the phase change found in the lowpass filter. Looking more closely at the frequency response, we note that the amplitude starts to fall again from about 400 kHz upward. This is not the effect of the RC filter but is due to the fall in gain of the op amp at high frequencies. The transition frequency of the TL081C is 3 MHz. If we were to plot the response at higher frequencies, we should see a steep roll-off. In practice, if we extend the operation of this filter into higher frequencies it acts as a bandpass filter.
DESIGNING FIRST-ORDER FILTERS 1 Decide on a capacitor value. 2 Apply the equation: R 5 1/2πfCC.
Example
0V passive filter
To design a lowpass filter with cut-off point at 40 kHz. 1 For stability, try the calculation for a 220 pF capacitor, so that a ceramic plate capacitor with zero tempco can be used. 2 Applying the equation, R 5 18086 Ω. Use an 18 kΩ, 1% tolerance, metal film resistor. For a highpass filter, exchange the resistor and capacitor.
non-inverting amplifier
FIGURE 13.4 A first order highpass active filter consists of a highpass passive filter followed by a non-inverting amplifier.
When this is done, the frequency response has the appearance shown below:
0 dB = 1.56 V
0
–3 dB
v / V(dB)
Phase
180 160 140 120
–10
100 80
–20
60
+45°
40 20 0
–30
–20 –40
–40
–60 –80 –100
–50
–120 –140 –160
–60 1
3
10
30
100 300
1k
3k
10k 30k f / Hz
FIGURE 13.5
–180
100k 300k 1M
114
PART | 1
SECOND-ORDER ACTIVE FILTER Adding an extra stage to the first-order lowpass filter allows feedback to be introduced. The circuit below has two RC filters, one of which is connected to 0 V, as in the first-order lowpass filter. The other is part of a feedback loop from the filter output.
R1
R2
1k6
1k6
TL081C
VIN
VOUT
IC1
C1
C2
100n
100n
RF 56k RA feedback
100k
0V FIGURE 13.6 The second-order active filter has positive feedback from the output to the capacitor of the first stage.
The feedback action makes the filter resonate at frequencies around the cut-off frequency. Resonance is not strong, but there is enough extra feedback around this frequency to push up the ‘knee’ of the frequency response curve. 0 dB = 1.56 V
0 v/V(dB)
–5 –10
Circuits
This sharpening of the ‘knee’ is accomplished in a passive filter by using an inductor-capacitor network (p. 57) to produce the resonance. Using an op amp instead provides the same effect but without the disadvantages of inductors. A highpass version of the second order filter is built by substituting resistors for capacitors and capacitors for resistors. Higher-order filters with even steeper roll-off often consist of two or more secondorder filters and possibly a first-order filter cascaded together.
BANDPASS FILTER A bandpass filter may be constructed by cascading a lowpass filter and a highpass filter (see in Fig 13.8). The cut-off frequency of the lowpass filter is made higher than that of the highpass filter so that there is a region of overlap in which both filters pass frequencies at full strength. This area of overlap can be made narrow or broad by choosing suitable cut-off frequencies. Both lower and upper cut-off frequencies are calculated using the equation f 5 1/2πRC. In the example opposite (Fig 13.7), the cut-off frequency of the lowpass filter is 4.8 kHz, while that of the highpass filter is 1 kHz. The frequency response shows a broad passband from 645 Hz to 7440 Hz, with a central frequency of 2.2 kHz. The bandwidth of the filter is 6795 Hz. Another approach to bandpass filtering is the multiple feedback filter (see Fig 13.11). This has two loops, each providing negative feedback.
–15 –20 –25
VIN
–30
C2
R1
TL081C 100n
–35
C1
–50
100n
–45 1
3
10
30
100
300
1k
3k
10k f / Hz
FIGURE 13.7 The Bode plot of the response of the second-order filter (black) also shows the amplitude of the feedback signal (grey), which pushes up output at the knee of the curve. The doubled filter network produces a steep roll-off at high frequencies.
The grey curve is a plot of the voltage across C1, and it can be seen that the strongest feedback occurs around 1 kHz. The response of the filter turns sharply down above 1 kHz. From then on, there is a steady and more rapid fall in response, the rate depending on the number of stages. With two stages in the filter, the roll-off is 212 dB per octave or 240 dB per decade.
VOUT
IC1
330R
R2
RF 56k
1k6
RA 100k
0V FIGURE 13.8 A broad-band bandpass filter can be made by cascading a lowpass filter and a highpass filter. This is followed by a non-inverting amplifier to restore the amplitude to a reasonable level.
The result (Fig 13.10) is a passband of narrow bandwidth. The centre frequency is 1.8 kHz. The lower cutoff frequency is 1 kHz and the upper cutoff frequency is 3 kHz. This gives a bandwidth of 2 kHz. The centre frequency (but not the bandwidth or the amplitude) may be tuned by varying the value of R3.
Topic | 13
115
Active Filters
The centre frequency of the filter is calculated from the equation: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 R1 1R3 f0 5 2πC R1 R2 R3
V(dB) 0 –3
–10
100
1k
10k
Hz
FIGURE 13.9 The bandpass filter in Figure 13.8 produces a broad passband, with response falling off at 6 dB per octave on either side.
The response curve in Fig 13.10 shows an upturn in the higher frequency range. The is the result of the transition frequency of the op amp. The gain of the op amp is reduced at higher frequencies and therefore the higher frequencies in the signal are not completely cancelled out by the negative feedback.
BANDSTOP FILTERS 0
V(dB)
–10
–20 100
10k
1k
100k Hz
FIGURE 13.10 The bandpass filter in Figure 13.11 produces a narrow passband, with response falling off at 6 dB per octave on either side.
G
G
R1 and C1 form a highpass filter; high frequency signals passing through this are fed back negatively, partly cancelling out the high frequencies in the original signal. R2 and C2 form a lowpass filter; low frequency signals passing through this are fed back negatively, partly cancelling out the low frequencies in the original signal.
Only signals in the medium range of frequency are able to pass through the filter at full strength. C1 100n
R2
VIN
C2
R1
100n
3k3
R3
1k6
IC1
VOUT
TL081 C
560R
0V FIGURE 13.11 A multiple feedback filter cancels out both high and low frequencies by feeding them back negatively, leaving the middle frequencies unaffected.
Bandstop filters, often known as notch filters, are used to remove a limited range of frequencies from a signal. For example, such a filter may be used to remove a 50 Hz mains ‘hum’ that has been picked up by electromagnetic interference. This may be necessary when measuring small voltage signals from the human body. There are two approaches to bandstop filtering, both of which show interesting applications of electronic principles. The first circuit (below) depends on one of the basic applications of op amps.
Bandpass filter
Intermediate frequencies, inverted RF
RA2
VIN
RA1 All frequencies
Adder
VOUT Low and high frequencies
0V
FIGURE 13.12 This bandstop filter uses two op amps, one as part of an active filter, the other as an adder.
The signal is sent to an op amp adder, both directly and indirectly through a bandpass filter. The filter could be a multiple feedback filter (see p. 114). This not only filters the signal, removing low and high frequencies, but also inverts it. The second op amp mixes the unfiltered signal with the inverted filtered signal. The result is to subtract the medium frequencies from the original signal, creating a notch in the frequency response curve. The depth of the notch depends on the relative values of RA1 and RA2. Another type of bandstop filter depends on an LC resonant network Fig 13.13. This is combined with a resistor and an op amp wired as an inverting amplifier.
116
PART | 1
G
C
VIN
TL081C
220u
VOUT
IC1
G G
100u
RF 56k
R 22R
RA 100k 0V FIGURE 13.13 This bandstop filter uses an LC resonant network to produce a very narrow notch in its frequency response (below).
v(dB) 0
–4
–8 500
1k
Hertz
2k
FIGURE 13.14 The LC bandstop filter has a steep-sided notch at 1.07 kHz.
At the resonant frequency, the capacitor and inductor have equal impedance. Their impedance in parallel is a maximum. This reduces the amplitude of signals of that frequency, producing the narrow notch in the frequency response. To calculate the notch frequency we use the equation on p. 57. The depth of the notch is determined by the value of R. The smaller the value of R, the deeper the notch.
ACTIVITIES — ACTIVE FILTERS It is easier to work this activity on a circuit simulator, and this has the advantage that an analysis of frequency response can be done very quickly. But you should build and test at least two filter circuits using a breadboard. The active filter circuits that you can investigate include: G G G G
First-order (single stage) lowpass filter. First-order (single stage) highpass filter. Second order lowpass filter. Second order highpass filter.
Circuits
Bandpass filter consisting of a lowpass filter cascaded with a highpass filter (cutoff point for the lowpass filter must be higher than that for the highpass filter). Multiple feedback bandpass filter. Bandstop filter based on resonant LC network.
For each filter connect a signal generator to its input and a dual-trace oscilloscope to its input and output. Observe and explain what happens to the signal as it passes through the filter. Observe the effect of the filter on the amplitude of the signal and on the shape of the waveform, trying this with both sinusoidal and square waves, and at different frequencies. For example, when testing a lowpass filter, use a signal with a frequency well below the cut-off point, try another signal at the cut-off point and a try a third signal well above the cut-off point. Summarise your results in a table or by sketching frequency response curves. Remember that the transition frequency of the op amp itself may be about 1 MHz, so do not use frequencies higher than 100 kHz in these investigations.
QUESTIONS ON ACTIVE FILTERS 1 List and discuss the advantages of op amp active filters, compared with passive filters. 2 Design a first-order lowpass filter with a cut-off frequency of 15 kHz. 3 Design a first-order highpass filter with a cut-off frequency of 20 Hz. 4 Explain the action of a second-order op amp lowpass filter and state why its performance is superior to that of a first-order filter. 5 Design a bandpass filter consisting of a lowpass and a highpass first-order filter cascaded. The specification is: low cut-off point 5 2 kHz; upper cut-off point 5 5 kHz; gain at the centre frequency 5 1.5. 6 Explain the action of a multiple feedback bandpass filter. In what way is its frequency response different from that of cascaded lowpass and highpass filters? 7 Outline the structure of three different types of active filter. Describe two examples of how each type can be used.
MULTIPLE CHOICE QUESTIONS 1 A A B C D
passive filter can be built from: a capacitor, a resistor and an op amp a capacitor and an inductor resistors only a resistor, and an op amp.
Topic | 13
Active Filters
2 An op amp used in a highpass filter should have: A a high gain-bandwidth product B low slew rate C low input bias current D high output resistance. 3 At the cut-off point of a lowpass filter the phase change is: A 190 B 0 C 245 D 290 . 4 The cut-off frequency of a first-order active high pass filter is 5 kHz. It has a 1 nF capacitor. The nearest E12 value for the resistor is: A 200 kΩ B 31.8 Ω C 33 kΩ D 220 kΩ. 5 An active filter must have: A a resistor B a diode C no more than two passive components D a least one reactive component. 6 The roll-off of a 2nd-order highpass filter is: A 16 dB per octave B 140 dB per decade C 212 dB per decade D 220 dB per octave.
117
7 An active bandpass filter is built from a lowpass filter cascaded with a highpass filter. It has a lower cutoff frequency of 200 Hz, and a bandwidth of 800 Hz. The cutoff points of its two filters are: A lowpass 5 200 Hz, highpass 5 1 kHz. B lowpass 5 1 kHz, highpass 5 200 Hz. C lowpass 5 200 Hz, highpass 5 800 Hz. D lowpass 5 800 Hz, highpass 5 1 kHz. 8 In a multiple feedback bandpass filter, there is: A negative feedback of low frequencies B negative feedback of middle frequencies C positive feedback of middle frequencies D negative feedback of low and high frequencies. 9 A bandstop active filter produces a notch in the frequency response because: A of the inverting action of the op amp B the reactances of the capacitor and inductor are equal at the notch frequency C the reactance of the capacitor and inductor in parallel is a maximum at the notch frequency D the LC network resonates strongly at the centre frequency. 10 A bandstop active filter has a resonant network in which C 5 10 nF, and L 5 27 μH. The centre frequency of the notch is at: A 589 kHz B 306 kHz C 962 Hz D 1.85 MHz.