Designing & Implementing

Active BP filter

 

Abstract:

This report shows the procedure done to design and implement an active band-pass filter to some given specifications. The report explains and discuses the theory in the introduction, and then shows the hand-calculations needed to meet the specifications. A conclusion follows the results, and discuses them.

 

Specifications:

 

Introduction:

            Electrical filters were in use long before IC operational amplifiers (Op Amps) were available. Typically they were constructed with passive components such as resistors, capacitors, and inductors. And that is why they are called passive filters. With low-cost and reliable amplifying active devices available, such as Op Amps, filter designs are now mainly active types. For most applications, active filters are easy to design and they eliminate the need for high-cost inductors.

            In this report I will describe the way a band-pass filter is designed for the given specifications. To begin with, figure 1 shows the ideal low-pass, high-pass, and band-pass filters. It is easy to realize that a band-pass filter consists of a low-pass and high-pass filters, and this is the way I am going to design the band-pass filter. The result filter, however, will not have a frequency response like the ideal one, but will be like that shown in figure 2.

            The simple passive filters cannot provide consistent filtering if the output must be applied across a low or varying load resistance. An Op Amp can serve to buffer or isolate the load from the filter. Generally, active filters use components or devices that are able to amplify, such as transistors or Op Amps. Passive filters, on the other hand, consist of passive components only. The simplest first-order filter has only one resistor and one capacitor. A second-order filter has two resistors and two capacitors. A third-order has three and three, and so on.

            The band-pass filter I will design will consists of the low-pass and high-pass filters shown in figures 3 and figure 4, respectively. The Op Amps in these circuits are wired to work as voltage followers. The resistance in the feedback loop is used if drifting bias currents cause excessive drift in the quiescent value of the output. Otherwise, this resistance is replaced with a short circuit. To generalize, I can say that a low-pass (LP) filter, for example, passes frequencies below the cutoff frequency and attenuates frequencies above the cutoff frequency. The high-pass (HP) does the opposite. These first-order filters give a 20 dB/decade of roll-off, but in most applications this is inadequate. An ideal curve is very difficult to achieve, but improvements are quite practical. By cascading two first-order filters, I can have a second-order as shown in figures 5 and figure 6. Figure 5 shows a second-order LP, and figure 6 shows a second-order HP. Combining the two second-order filters should yield a fourth-order band-pass (BP) filter having a roll-off slope of 40 dB/decade as shown in figure 2.

 

 

 

 

Procedure:

            To assign the proper values to the components in the circuit, I had to solve the circuit by hand and come up with the transfer function. To simplify the work, I divided the stages and solved each one separately.

  1. First stage of the LP filter of figure 3:

 

Vo(CS) = Vi/(Ri + 1/(CS))

Vo/Vi = 1/( (CS)(Ri + 1/(CS)) )

Vo/Vi = 1/(CRiS + 1)

Vo/Vi = ( 1/(CRiS) ) / ( S + 1/(CRi) )

 

  1. Second stage of the LP filter has the same transfer function.
  2. Cascading the two means the output of the first stage is the input of the second stage, and the total transfer function is the product of the two (see figure 5):

 

Vo/Vi = ( 1/(CRiS) ) / ( S + 1/(CRi) ) * ( 1/(CRiS) ) / ( S + 1/(CRi) )

Vo/Vi = ( 1/(C2Ri2) )/( S2 + ( 2CRi/(C2Ri2) )S + 1/(C2Ri2) )

\ωo = (1/(C2Ri2))1/2

 

  1. First stage of the HP filter of figure 4:

 

Vo/Ri = Vi/(Ri + 1/(CS))

Vo/Vi = Ri/(Ri + 1/(CS))

Vo/Vi = RiS/(RiS + 1/C)

Vo/Vi = S/(S + 1/(CRi))

 

  1. Second stage of the HP filter has the same transfer function.
  2. Cascading the two means the output of the first stage is the input of the second stage, and the total transfer function is the product of the two (see figure 6):

 

Vo/Vi = S/(S + 1/(CRi)) * S/(S + 1/(CRi))

Vo/Vi = S2/( S2 + ( 2CRi/(C2Ri2) )S + 1/(C2Ri2) )

\ωo = (1/(C2Ri2))1/2

 

Then I cascaded the two filters as shown in figure 7. The transfer function of this filter is the product of the two cascaded filters since the output of the first is the input of the second. Multiplying the two, I came up with the following transfer function:

 

H(S) = Vo/Vi = (1/C2R2)S2 / [S2 + (4/(CR))S3 + (6/(C2R2))S2 + (4/(CR))S + 1/(C4R4)]

 

Please note that I have set capacitors equal, and the resistors too for simplicity. Finally I set the cutoff frequency for both LP and HP equal to the specified center frequency. Because the LP and HP filters are cascaded, they will block all frequencies except for the center frequency. The parameters were calculated as follows:

 

Let C = 10 nF

ωo = (1/(C2Ri2))1/2

ωo = 1/(CRi)

Ri = 1/(Cωo) = 1/(10*10-9*2*π*200)

Ri ≈ 79.6 KΩ

 

Results:

            Simulating the circuit in PSpice using the calculated parameters resulted in the frequency response shown in figure 8. The poles and zero of this system is shown in figure 9.

 

Conclusion:

            This report showed that the filter transfer function can be expressed as the ratio of two polynomials in S. I think the system can be improved by taking more care in choosing the values of the components. The degree of the denominator polynomial is the filter order. And the roots of the denominator polynomial are the poles. I think to obtain a highly selective response, the poles are to be complex and occur in conjugate pairs except for one real pole when the order is odd. The capacitor plays the main role in this kind of systems because its impedance varies with frequency.

            This report showed that the designed filter meets the specifications, but can be improved. Note that the output has no load resistance, and this is one reason why the amplitude of the transfer function is low. I left the load without a resistance assuming that the next circuit to which this filter is to be connected to will make the load. I think this filter can be improved by amplifying the output and make the amplification to be variable so that the user can adjust the amplification.