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**Second Order Active Filters Based on Bridged-T Networks**

Develop second order active filters based on op amps. Assume ideal op amps, i.e. No input current Negligible voltage drop between the inputs, i.e. infinite gain Use feedback network with R’s and C’s General form of the filter function is: Consider case where we use a particular RC combination called the bridged-T network. Gives well recognized filters, e.g. low pass, high pass, bandpass, notch, etc. Other RC combinations possible, but give less useful (more complicated frequency dependent) filter functions. Will analyze bridged-T networks to show they give second order filters. Can get different filters based on where input is connected into the bridged-T network. Bridged – T Network Vi I-≈ 0 Vo V-≈ 0 A + Bridged – T network Ch. 12 Active Filters Part 2

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**Injecting the Input Signal**

Where do we inject the input signal ? Need to inject the signal at a point where it does not change the poles (expression for t(s) ). We will inject the signal at the bottom of R4 which was at ground. Since we find the poles by turning off the signal source when we use the Gray-Searle technique, this means that we will not be modifying the poles (as long as we take into account the source resistance Rs). Take the output off the op amp output. Where we inject the signal determines what type of filter we get ! Rs Vi VS Vo Ch. 12 Active Filters Part 2

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**Injecting the Input Signal – Source Resistance Problem**

When injecting the input signal, we need to take into account the finite source resistance. Can divide R4 into two parallel resistors. For any α > 0 and < 1, we always get We can easily correct for Rs by subtrac-ting Rs from the needed value for R4/ This approach provides more flexibility in taking Rs into account, especially when R4 is small. If we don’t take Rs into account, the poles (expression for t(s) ) will be altered and the filter design will not meet the specifications. Rs Vi VS Vo Rs Vi VS Vo Ch. 12 Active Filters Part 2

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**Injecting the Input Signal - Bandpass Filter**

Ii I-= 0 V-= 0 Analysis to do: Find T(s) = Vo(s)/Vi(s) Assume op amp is ideal: Input currents are virtually zero. Gain is very large so voltage difference between inputs is nearly zero. Inverting terminal (-) grounded since noninverting (+) is grounded. Filter’s transfer function Ch. 12 Active Filters Part 2

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**Second Order Filter has Bandpass Form**

Circuit analysis gives Recall the form of second order filter is So our filter is a bandpass filter since a2 = 0, a0 = 0 where Finally, recall form of transfer function for bridged-T network was t(s) So the numerator of t(s) is in fact the same as the denominator of T(s), so the poles of T(s) are the same as the zeros of t(s)! T(dB) 0 dB -3 dB 0 Ch. 12 Active Filters Part 2

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**How to Design a Specific Bandpass Filter**

Transfer function for bandpass filter is Given the filter specifications (0 and 0 /Q), we can determine the R’s and C’s. We can also use the desired transmission at 0 to determine the size of . We can get gain [T(0))>1] at if so desired. T(dB) 0 dB -3 dB 0 Ch. 12 Active Filters Part 2

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**Example - Bandpass Filter Design**

R and C calculations Three specifications Five elements to specify: R3 R4 C1 C2 α Conventional approach Set two equal, e.g. C1 = C2 Pick a convenient size for them C1 = C2 = 5 nF Define ratio m = R3/R4 ; R3 = R Simplify equations for o and Q Solve for two parameters: R & m Calculate R3 & R4. Filter specifications: T(dB) 0 dB -3 dB 0 Solution Ch. 12 Active Filters Part 2

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**Bandpass Filter Example**

T(dB) 0 dB -3 dB 0 To get unity transmission (0 dB) at 0, we set T(ωo) = 1. From previous analysis we know C1 = C2 = 5 nF and R3 = 40 kΩ and R4 = 0.1 kΩ so peak value of T(ωo) is given by So if Rs = 5K, we can account for it by making R4/α = 15 K Ch. 12 Active Filters Part 2

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**Bandpass Filter Example - Poles**

Transfer function for bandpass filter Poles of bandpass filter So this bandpass filter is a special form of a feedback amplifier. Ch. 12 Active Filters Part 2

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**Complementary Transformation**

Can we get a different type of filter from this bridged-T network? YES. Need a new injection point. How do we find a new injection point if the grounded end of R4 appears to be our only option? Have to apply a complementary transformation to redraw the circuit with a new injection point. Complementary transformation consists of: 1. Nodes of feedback network originally grounded (c) should be reconnected to op amp output. 2. Nodes of feedback network originally connected to op amp output (b) should be reconnected to ground. 3. Input terminal connections of the op amp should be interchanged. 4. Other input shorted to op amp output instead of being grounded. Complementary Transformed Circuit Original = Note: This is not grounded! Ch. 12 Active Filters Part 2

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**Complementary Transformation of Bridged-T Network**

Reconnect bottom of R4 to op amp output. Remove connection of C1 and R3 from op amp output and connect them to ground. Change connection at op amp input to noninverting (+). Connect inverting (-) input directly to output. Now have two new points (bottom of C1 and R3) that can be used for injection of the input signal to get a new type of active filter. Injection at the bottom of C1 gives a high pass filter. High pass filter Ch. 12 Active Filters Part 2

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**High Pass Filter V-=V0 I+ 0 Vo V1 V+V0 I4 I2 Ii I3 Vi**

Analysis to do: Find T(s) = Vo(s)/Vi(s) Assume op amp is ideal: Input currents are virtually zero. Gain is very large so voltage difference between inputs is nearly zero. Inverting terminal (-) connected to output so noninverting (+) is nearly at V0 since op amp gain A is large. Ch. 12 Active Filters Part 2

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**High Pass Filter T(dB) 0 Circuit analysis gives**

General form of second order filter is So a2= 1 and a0 = a1= 0 Recall form of t(s) As expected the numerator of t(s) is the same as the denominator of T(s), so the poles of T(s) are the same as the zeros of t(s)! T(dB) 0 dB Q(dB) 0 Ch. 12 Active Filters Part 2

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**High Pass Filter Design**

T(dB) 0 dB Q(dB) 0 * Given the filter specifications (0 and 0 /Q), we can determine the R’s and C’s. * Two specifications, four parameters so follow same convention and set C1 = C2 and pick a convenient value, say 5 nF. * Define ratio of resistances m = R3/R4 and set R3 = R. Ch. 12 Active Filters Part 2

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**High Pass Filter - Example**

T(dB) 0 dB Q(dB) NOTE 40 dB/dec 0 Filter specifications: 0 = 1x103 rad/s, Q = 0.5 Poles of high pass filter NOTE: since two poles coincide get falloff at 40 dB/dec below ωo. Ch. 12 Active Filters Part 2

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**Other Active Filters Based on Bridged-T Feedback Networks**

Second bridged-T feedback network similar to first, but with the resistors and capacitors interchanged. Transfer function t(s) is similar. Method of analysis is the same. Find T(s) = Vo(s)/Vi(s) Get it in standard form From specifications (o and Q), determine R’s and C’s. NOTE: Need to pick R1 = R2 and define m = C1/C2 to solve in this case. Rs Vi V0 Ch. 12 Active Filters Part 2

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**Other Active Filters Based on Bridged-T Feedback Networks**

Can use the complementary transformation to get other injection points and other filter types. Ch. 12 Active Filters Part 2

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**Butterworth Filters Low Pass Filter High Pass Filter**

Second order filters Can be low or high pass. Provide improved performance: No peak near band edge that is seen for other filters, i.e. it is maximally flat unlike other second order filters which give the shape shown below Falloff for Butterworth filter is steeper, i.e. 40 dB/dec rather than 20 db/dec for passive RLC filters. High Pass Filter Ch. 12 Active Filters Part 2

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**Low Pass Butterworth Filter**

IC2 Vo Vo Vi V12 IR2 IR1 IC1 General form for biquadratic filter This has form for a low pass biquadratic filter Ch. 12 Active Filters Part 2

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**Low Pass Butterworth Filter Design**

T(dB) 0 dB Q(dB) NOTE 40 dB/dec 0 * Given the filter specification (0), we can determine the R and C. * One specification, two parameters – R and C * Pick a convenient value, say C = 5 nF. * Calculate R from C and ωo. Ch. 12 Active Filters Part 2

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**More Complicated Second Order Active Filters**

Thomas-Tow Biquad Filter Second order filters can employ: More feedback loops. More op amps Can achieve better performance. Added complexity In design In implementation Can achieve different filter types by the choices of capacitances and resistances. Ch. 12 Active Filters Part 2

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**Second Order Biquadratic Filter**

Kerwin-Huelsman-Newcomb (KHN) Biquad Filter NOTE: Here is a general purpose filter that can be used to achieve any one of five filtering functions depending upon where the output is taken. Ch. 12 Active Filters Part 2

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**Second Order Active Filters**

Antoniou Inductor Replacement Biquad Filter Ch. 12 Active Filters Part 2

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**Second Order Active Filters**

Ch. 12 Active Filters Part 2

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**Other Higher Order Active Filters**

Can incorporate other combinations of resistors and capacitors in the feedback network. But may not give a simple filter function. Can use op amps with feedback and other circuit configurations. NOTE: This is a third order filter since it incorporates three capacitors. This will give a 60 dB/dec falloff ! T(dB) 0 dB Q(dB) NOTE 60 dB/dec 0 From G. Parker, Applied Microwave and Wireless, p.74 Jan 1997. Ch. 12 Active Filters Part 2

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**Other Higher Order Butterworth Filters**

Third Order Filter Fourth Order Filter Ch. 12 Active Filters Part 2

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**Summary of Active Filters**

Can use op amps and simple RC feedback networks to implement a variety of second order filters such as: Band pass, High pass, Low pass, Notch filter As an example, we examined use of the bridged-T network of two resistors and two capacitors. Illustrated how to analyze such networks to obtain the transfer function T(s) which describes the filter’s characteristics. Analysis used: Characteristics of ideal op amp Kirchoff’s Laws Also analyzed Butterworth filter Similarly used the characteristics of ideal op amp and Kirchoff’s Laws Finally, pointed out third and higher order filters can also be designed using op amps to get superior filter performance. Ch. 12 Active Filters Part 2

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