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Passive Band Pass Filter

Passive Band Pass Filter

Passive Band Pass Filters can be made by connecting together a low pass filter with a high pass filter

The Passive Band Pass Filter can be used to isolate or filter out certain frequencies that lie within a particular band or range of frequencies. The cut-off frequency or ƒc point in a simple RC passive filter can be accurately controlled using just a single resistor in series with a non-polarized capacitor, and depending upon which way around they are connected, we have seen that either a Low Pass or a High Pass filter is obtained.

One simple use for these types of passive filters is in audio amplifier applications or circuits such as in loudspeaker crossover filters or pre-amplifier tone controls. Sometimes it is necessary to only pass a certain range of frequencies that do not begin at 0Hz, (DC) or end at some upper high frequency point but are within a certain range or band of frequencies, either narrow or wide.

By connecting or “cascading” together a single Low Pass Filter circuit with a High Pass Filter circuit, we can produce another type of passive RC filter that passes a selected range or “band” of frequencies that can be either narrow or wide while attenuating all those outside of this range. This new type of passive filter arrangement produces a frequency selective filter known commonly as a Band Pass Filter or BPF for short.

Typical Band Pass Filter Circuit

passive rc band pass filter circuit

 

Unlike the low pass filter which only pass signals of a low frequency range or the high pass filter which pass signals of a higher frequency range, a Band Pass Filters passes signals within a certain “band” or “spread” of frequencies without distorting the input signal or introducing extra noise. This band of frequencies can be any width and is commonly known as the filters Bandwidth.

Bandwidth is commonly defined as the frequency range that exists between two specified frequency cut-off points ( ƒc ), that are 3dB below the maximum centre or resonant peak while attenuating or weakening the others outside of these two points.

Then for widely spread frequencies, we can simply define the term “bandwidth”, BW as being the difference between the lower cut-off frequency ( ƒcLOWER ) and the higher cut-off frequency ( ƒcHIGHER ) points. In other words, BW = ƒH – ƒL. Clearly for a pass band filter to function correctly, the cut-off frequency of the low pass filter must be higher than the cut-off frequency for the high pass filter.

The “ideal” Band Pass Filter can also be used to isolate or filter out certain frequencies that lie within a particular band of frequencies, for example, noise cancellation. Band pass filters are known generally as second-order filters, (two-pole) because they have “two” reactive component, the capacitors, within their circuit design. One capacitor in the low pass circuit and another capacitor in the high pass circuit.

Frequency Response of a 2nd Order Band Pass Filter

band pass filter bode plot

 

The Bode Plot or frequency response curve above shows the characteristics of the band pass filter. Here the signal is attenuated at low frequencies with the output increasing at a slope of +20dB/Decade (6dB/Octave) until the frequency reaches the “lower cut-off” point ƒL. At this frequency the output voltage is again 1/√2 = 70.7% of the input signal value or -3dB (20*log(VOUT/VIN)) of the input.

The output continues at maximum gain until it reaches the “upper cut-off” point ƒH where the output decreases at a rate of -20dB/Decade (6dB/Octave) attenuating any high frequency signals. The point of maximum output gain is generally the geometric mean of the two -3dB value between the lower and upper cut-off points and is called the “Centre Frequency” or “Resonant Peak” value ƒr. This geometric mean value is calculated as being ƒr 2 = ƒ(UPPER) x ƒ(LOWER).

A band pass filter is regarded as a second-order (two-pole) type filter because it has “two” reactive components within its circuit structure, then the phase angle will be twice that of the previously seen first-order filters, ie, 180o. The phase angle of the output signal LEADS that of the input by +90o up to the centre or resonant frequency, ƒr point were it becomes “zero” degrees (0o) or “in-phase” and then changes to LAG the input by -90o as the output frequency increases.

The upper and lower cut-off frequency points for a band pass filter can be found using the same formula as that for both the low and high pass filters, For example.

cut-off frequency equation

 

Then clearly, the width of the pass band of the filter can be controlled by the positioning of the two cut-off frequency points of the two filters.

Band Pass Filter Example No1.

A second-order band pass filter is to be constructed using RC components that will only allow a range of frequencies to pass above 1kHz (1,000Hz) and below 30kHz (30,000Hz). Assuming that both the resistors have values of 10kΩ, calculate the values of the two capacitors required.

The High Pass Filter Stage

The value of the capacitor C1 required to give a cut-off frequency ƒL of 1kHz with a resistor value of 10kΩ is calculated as:

high pass filter cut-off frequency

 

Then, the values of R1 and C1 required for the high pass stage to give a cut-off frequency of 1.0kHz are: R1 = 10kΩ and to the nearest preferred value, C1 = 15nF.

The Low Pass Filter Stage

The value of the capacitor C2 required to give a cut-off frequency ƒH of 30kHz with a resistor value of 10kΩ is calculated as:

low pass filter cut-off frequency

 

Then, the values of R2 and C2 required for the low pass stage to give a cut-off frequency of 30kHz are, R = 10kΩ and C = 530pF. However, the nearest preferred value of the calculated capacitor value of 530pF is 560pF, so this is used instead.

With the values of both the resistances R1 and R2 given as 10kΩ, and the two values of the capacitors C1 and C2 found for both the high pass and low pass filters as 15nF and 560pF respectively, then the circuit for our simple passive Band Pass Filter is given as.

Completed Band Pass Filter Circuit

second order band pass filter

 

The Filters Resonant Frequency

We can also calculate the “Resonant” or “Centre Frequency” (ƒr) point of the band pass filter were the output gain is at its maximum or peak value. This peak value is not the arithmetic average of the upper and lower -3dB cut-off points as you might expect but is in fact the “geometric” or mean value. This geometric mean value is calculated as being ƒr 2 = ƒc(UPPER) x ƒc(LOWER) for example:

Centre Frequency Equation

resonant frequency center point

  • Where, ƒr is the resonant or centre frequency
  • ƒL is the lower -3dB cut-off frequency point
  • ƒH is the upper -3db cut-off frequency point

and in our simple example above, the calculated cut-off frequencies were found to be ƒL = 1,060 Hz and ƒH = 28,420 Hz using the filter values.

Then by substituting these values into the above equation gives a central resonant frequency of:

resonant frequency

 

Band Pass Summary

A simple passive Band Pass Filter can be made by cascading together a single Low Pass Filter with a High Pass Filter. The frequency range, in Hertz, between the lower and upper -3dB cut-off points of the RC combination is know as the filters “Bandwidth”.

The width or frequency range of the filters bandwidth can be very small and selective, or very wide and non-selective depending upon the values of R and C used.

The centre or resonant frequency point is the geometric mean of the lower and upper cut-off points. At this centre frequency the output signal is at its maximum and the phase shift of the output signal is the same as the input signal.

The amplitude of the output signal from a band pass filter or any passive RC filter for that matter, will always be less than that of the input signal. In other words a passive filter is also an attenuator giving a voltage gain of less than 1 (Unity). To provide an output signal with a voltage gain greater than unity, some form of amplification is required within the design of the circuit.

A Passive Band Pass Filter is classed as a second-order type filter because it has two reactive components within its design, the capacitors. It is made up from two single RC filter circuits that are each first-order filters themselves.

If more filters are cascaded together the resulting circuit will be known as an “nth-order” filter where the “n” stands for the number of individual reactive components and therefore poles within the filter circuit. For example, filters can be a 2nd-order, 4th-order, 10th-order, etc.

The higher the filters order the steeper will be the slope at n times -20dB/decade. However, a single capacitor value made by combining together two or more individual capacitors is still one capacitor.

Our example above shows the output frequency response curve for an “ideal” band pass filter with constant gain in the pass band and zero gain in the stop bands. In practice the frequency response of this Band Pass Filter circuit would not be the same as the input reactance of the high pass circuit would affect the frequency response of the low pass circuit (components connected in series or parallel) and vice versa. One way of overcoming this would be to provide some form of electrical isolation between the two filter circuits as shown below.

Buffering Individual Filter Stages

buffering filter stages

One way of combining amplification and filtering into the same circuit would be to use an Operational Amplifier or Op-amp, and examples of these are given in the Operational Amplifier section. In the next tutorial we will look at filter circuits which use an operational amplifier within their design to not only to introduce gain but provide isolation between stages. These types of filter arrangements are generally known as Active Filters.

95 Comments

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  • Morgan Peter

    very educative

  • Yan A.

    I am looking to create a band pass filter with cutoff frequencies from 2kHz to 6kHz. I have e12 resistors and a few capacitors at hand. i am having a hard time selecting values for the capacitors and resistors to have the desired bandwidth. Can someone help me out please?
    thanks in advance

  • Ankit Bendwar

    I want a circuit for the 60 watt horn mid drivers, for band bass frequency so which components and how many value of thats circuits may be used so please tell me ideas for these band pass circuit

  • Naser

    In real life we use active filters, right?

  • Harry Gray

    perhaps some comment is needed on the choice of the resistors. When i simulate in circuit lab or with TI-inspire I find that with C1 = 15.9u, R1 = 1k (fL = 10Hz) and R2 = 1ohm, C2 = 15.9u (fR = 10Hz) i get Vo/Vi = 0.5 in the pass band (large attenuation). Scaling up R2 to 1k and C2 down to 15.9n (same fR as before) i get Vo/Vi = 1.0 (no attenuation), which is what i would hope regardless of selection of R’s.

    • Wayne Storr

      The selection of either R and C components is determined by the required bandwidth of the particular application. As explained in the tutorial, passive circuits (even a single resistance) will add attenuation to any signal, so Vo/Vi = 1.0 is not possible in a real world application

      • Harry Gray

        THanks for the response! I also discovered setting R2 > 10R1 for lpf ->hpf and R2 = R1 for hpf->lpf does the trick then Vo/Vi is almost 1 in magnitude (about 0.98 as mentioned for said reasons, but a whole lot better than 0.5). Harry

        • jon e

          Indeed, Harry G, you discovered a ‘good’ passive filter design. In general, one should follow the ’10x’ rule: increase the resistor value by 10x anytime you add on another stage. This makes the input impedance of the newly added stage much larger than the output impedance of the previous stage, as desired. The basic idea is to leave the earlier stages’ behavior intact by bleeding off only minimal current into the next stage. The second stage input essentially measures the first stage output. This is sort of like making a good measurement of bike tire pressure by only bleeding off a little bit of air pressure so we don’t deflate the tire we are trying to measure! (Electronics-tutorial.ws is a fabulous and well written site in general, but this section should really be updated with a good and proper design.)

  • Jim

    Why are all the commercial 7 MHz bandpass filters LC instead of RC? Thanks!

    • Jim

      Taking a swag here, but, I am guessing the resistors are pure resistance, and an inductor or capacitor would be reactance and they wouldn’t cause as much signal loss. So an LC bandpass filter would be better for weak signals, and an RC bandpass filter would only be good if the signal you are wanting to keep is pretty strong to begin with. I don’t know, but that’s my guess. I pretty electronic ignorant, which is why I am here in the first place. Thanks for any insight on the subject.

  • Ahmed morsy

    Very nice information

  • gracia

    muucha gracia

  • Tariq Baig

    Nice

  • SANJIT MANDAL

    kindly sent me circuit diagram for LN B range up to 6 .8 G HZ

  • Surender Dalal

    this fillter is effect on little on bass

  • Mohammed Elgaily

    There are two conditions in band pass filter (narrow- and wide-band), and we all know the center frequency has two equation, the first one is for wide-band pass filter which equal to (square root for (f1*f2)), and the second one is for narrow-band pass filter which equal to ( (f1+f2)/2 ).

    My question is when we use the first equation and the second equation for find the center frequency of band pass filter?

    Thanks.

    • Wayne Storr

      The band or range of frequencies between two corner frequency points of a Bandpass Filter (BPF) defines the filters Bandwidth (BW). Q is a measure of the filters selectivity. The higher the value of Q the narrower is the bandwidth, and the more selective is the filter. While a low value of Q the wider is the bandwidth and less selective is the filter. A BPF with a Q of less than 10 is defined as wide, while a BPF with a Q greater than 10 is defined as narrow.
      Q = fc / Bandwidth = fc / (fh – fl) where: fc = sqr-root(fh*fl)

    • Suman Kumar

      Geometric mean, due to logarithmic scale and averaging the critical frequencies and so named center frequency. In narrow band pass filter the critical frequencies are so close that we can take the average of frequencies to get center frequency rather to take logarithmic of frequencies.

  • Suman Kumar

    Can we connect first low pass filter and then high pass filter to get band pass filter, in that working is not changed of band pass filter?

    • Deon

      In that case it would be a Band Stop filter.

    • Wayne Storr

      Passive High and Low pass filters can be cascaded together to create a selective filter with a Bandpass characteristic. However, electrical isolation between passive stages is required if the centre frequency and corner frequencies are to be as calculated.

      • Suman Kumar

        Yes I know but my question was different that:

        Generally we connect high pass filter first then cascade it with low pass filter.

        But can we do reverse that taking output from high pass filter and going input into low pass filter.

        • Wayne Storr

          It’s the same. A High pass section will attenuate signals below the cut-off frequency point and the Low pass section will attenuate signals above the cut-off frequency point no matter which way around they are connected

  • zoya

    hi

  • Deekshitha

    What is the magnitude for the above examples

  • Francis Jansz

    Hello . I am Looking at having a Simple High Pass First order Filter At 7000 kHz at ,4 Ohms impedience. What value Capacitor and Resisters will I need and how can I wire them please I am a novice please send me The Parts Zlist and it’s Valued plus a simple Wiring Diagram .
    Kind Rgds
    Francis Jansz
    Passive High Pass Filter. ,7000.<,kHz at 4 Ohms

  • Steve Rohe

    I am making a radio bandpass filter at 3.975khz I am using 3kohm R1 and 29nf C1 for HP and 15nf C2 and 2kohm R2…what wattage value should I use? Would 1/4 watt resistors be ok?

  • Vartul Sharma

    Hi, I wad trying to derive the pole zero equation of the BPF but was unable to get the required answers. Can you post the derivation too? How we come up to fc1 = 1/2πR1C1 and fc2 = 1/2πR2C2?

  • Pyae Phyo

    Hello

  • Zubair ali

    Hi

    Hallo