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  • Low-Sensitivity Sallen-Key Filter Design with the HP-67 Programmable Calculator

    This program, a part of my series of analog electronic design programs for the HP-67 programmable calculator, addresses the problem of designing second order single op-amp low- and high-pass filters using the Sallen-Key topology.

    The Sallen-Key filter topology has the advantage of using a minimum of components. The simplest Sallen-Key filters use only two resistors and two capacitors. Additional resistors may be added for input attenuation (low-pass only) and gain adjustment. The following schematics illustrate these generalized Sallen-Key circuits:

    Generalized Sallen-Key low-pass filter
    Generalized Sallen-Key low-pass filter

    Generalized Sallen-Key high-pass filter
    Generalized Sallen-Key high-pass filter

    The equations governing the low-pass filter are as follows,

    where f is the filter's cut-off frequency, Q is its "quality", and H is its gain at the cut-off frequency. The corresponding equations for the high-pass filter are:

    The mathematically inclined will notice that in each case, there are three equations in either nine or ten variables. Thus there is no single "right" solution. At least six or seven of the variables have to be decided arbitrarily (f, Q, H, and three or four component values), at which point the remaining variables (component values) can be solved for. An article by Texas Instruments suggests a number of simplifications to help one choose component values, but this just adds the complication of which simplification to choose.

    I recently came across a pair of application notes by National Semiconductor which gives a procedure for designing Sallen-Key filters to minimize the effect of component value tolerances on the performance of the filter. A side-effect of this procedure is to reduce the number of inputs to five: f, Q, H, RF, and R, where the latter is simply an indication of the magnitude of resistor values desired for R1, R2, and R3. The procedure then dictates how all the other values are chosen, even adjusting for available "real-world" values part way through the solution.

    The program presented here implements this procedure, with some minor changes:

    1. Instead of asking for a desired resistor magnitude, R, the program asks for a capacitor magnitude, C, since in my experience, the capacitors drive the design.

    2. The formula given for internal gain variable K in the procedure (please refer to the application note) seems to have been derived empirically, and has a jump at Q = 1.1. To make the formula simpler to implement, I modified it slightly to:

      The graphs of the original (pink) and revised (blue) formulae show the difference. Testing has shown that the resulting solution is generally at most one real-world capacitor value increment different (with corresponding changes in resistor values of course).

      Also, if H > K, then K is set equal to H, since otherwise it will not be possible to achieve the desired gain.

    For the low-pass circuit, the desired gain, H, is achieved by a combination of input attenuation, α (controlled by R1 and R3), and the internal gain, K, of the filter (controlled by RF and RG). The division of this gain between the two stages depends on H and Q and is chosen to minimize the sensitivity of the circuit to component tolerances. For example, for H = 1 and Q = 2, the attenuator gain is α = 0.629 and the internal gain is K = 1.59, for a net gain of H = αK = 1.

    The high-pass circuit has no attentuation stage, so H must be at least 1, and higher for Q > 0.917. If too low a value is entered for H, it is increased as necessary to make the circuit solvable.

    For either circuit, to achieve a result with minimum component sensitivity without regard to gain, set H = 0. The program will automatically choose the optimal value for K (and thus H). The resulting gain will be output when determining the performance.

    Using the Program

    First type in the program and save it, or read it from a previously recorded magnetic card. The card should be labelled as follows:

    LOW-SENSITIVITY SALLEN-KEY FILTER DESIGN
    f Q H C RF
    LP→RGC1C2R1R2R3   HP→RGC1C2R1R2   f,Q,H

    Filter Design from Specifications

    Example: Design a 500Hz low-pass unity-gain filter with a Q of 2, using capacitors in the 10nF range, and a 47kΩ resistor for RF:

    DescriptionKeystrokesDisplay
    Use engineering notation  h  ENG 
     DSP 2 
     0.00  00 
    Enter f 500 
      f 
     500.  00 
    Enter Q 2 
      f 
     2.00  00 
    Enter H 1 
      f 
     1.00  00 
    Enter C 10 EEx CHS 9 
      f 
     10.0 -09 
    Enter RF 47 EEx 3 
      f 
     47.0  03 
    Compute RG  A  79.5  03 
    Enter real-world RG and compute C1 82 EEx 3 
     R/S 
     31.6 -09 
    Enter real-world C1 and compute C2 33 EEx CHS 9 
     R/S 
     3.16 -09 
    Enter real-world C2 and compute R1 3.3 EEx CHS 9 
     R/S 
     15.4  03 
    Enter real-world R1 and compute R2 15 EEx 3 
     R/S 
     95.9  03 
    Enter real-world R2 and compute R3 100 EEx 3 
     R/S 
     26.1  03 
    Enter real-world R3 27 EEx 3 
     R/S 
     27.0  03 

    Notes

    If a resistor is to be omitted (open circuit), this program displays a value of zero for the resistance. This is different than some of my other programs, which display a "large" value representing infinity.

    When the value for RG is displayed as zero, meaning it can be omitted, the value of RF will not matter any more, and RF can be replaced by a direct connection.

    Filter Performance from Chosen Components

    During the calculation of the solution above, we've entered real-world values in response to each computed value. The real-world values of C1 and C2 are used when computing the values of R1, R2, and R3. However, the real-world values of each of those resistors does not affect the computed value of the remaining ones. Thus, the final filter may not perform exactly as specified. To find out how it does perform, follow these steps:

    DescriptionKeystrokesDisplay
    Compute resulting f E  491.  00 
    Compute resulting Q R/S  1.86  00 
    Compute resulting H R/S  1.01  00 

    A High-Pass Example

    Using the parameters already entered for the low-pass filter above, determine the components for a high-pass filter:

    DescriptionKeystrokesDisplay
    Compute RG for high-pass filter C  79.5  03 
    Enter real-world RG and compute C1 82 EEx 3 
     R/S 
     31.4 -09 
    Enter real-world C1 and compute C2 33 EEx CHS 9 
     R/S 
     3.18 -09 
    Enter real-world C2 and compute R1 3.3 EEx CHS 9 
     R/S 
     9.59  03 
    Enter real-world R1 and compute R2 10 EEx 3 
     R/S 
     97.0  03 
    Enter real-world R2 100 EEx 3 
     R/S 
     0.00  00 

    Now determine the predicted actual performance:

    DescriptionKeystrokesDisplay
    Compute resulting f E  482.  00 
    Compute resulting Q R/S  1.96  00 
    Compute resulting H R/S  1.57  00 

    Notice that H is higher than the specified unity gain. This is because the filter is not possible to construct with unity gain when Q = 2. The smallest possible gain is H = 1.59 (which due to real-world components, has become Q = 1.89 and H = 1.57). To achieve H = 1, you will need either a pre-attenuator with low output impedance, or a post-attenuator with high input impedance.

    Program Listing

    LineInstructionComments
    001♦  LBL a Enter and store f
    002 STO A  
    003 RTN  
    004♦  LBL b Enter and store Q
    005 STO B  
    006 RTN  
    007♦  LBL c Enter and store H (gain)
    008 STO C  
    009 RTN  
    010♦  LBL d Enter and store C (capacitor scale)
    011 STO 0  
    012 RTN  
    013♦  LBL e Enter and store RF
    014 STO 4  
    015 RTN  
    016♦  LBL A Low-pass filter: RG,C1,C2,R1,R2,R3
    017 CF 0  
    018 GTO 0  
    019♦  LBL C High-pass filter: RG,C1,C2,R1,R2
    020 SF 0  
    021♦  LBL 0 Forward solution
    022 RCL B  
    023 2  
    024 .  
    025 2  
    026 ×  
    027 .  
    028 9  
    029 −  
    030 RCL B  
    031 .  
    032 2  
    033 +  
    034 ÷ (2.2Q-0.9)/(Q+0.2)
    035 1  
    036 x≤y?  
    037 x↔y  
    038 STO E K = max(1,(2.2Q-0.9)/(Q+0.2))
    039 RCL C  
    040 x>y?  
    041 STO E K = max(H,1,(2.2Q-0.9)/(Q+0.2))
    042 RCL E  
    043 ÷ H/K
    044 x=0?  
    045 1 Use α = 1 if H/K = 0 (because H was 0)
    046 F? 0  
    047 1 Always use α = 1 for a high-pass filter
    048 STO D α = H/K (always 1 for a high-pass filter or H = 0)
    049 RCL 4  
    050 RCL E  
    051 1  
    052 −  
    053 x≠0?  
    054 ÷ RG = RF/(K-1) if K ≠ 1, or zero if K = 1
    055 R/S Display RG and let user change it
    056 STO 5  
    057 . Initialize n to √0.1
    058 1  
    059 √x  
    060 GSB 7 n(1+√(1+4Q2(1+n2)(K-1)))/(2Q(1+n2))
    061 RCL 8 √0.1 was stored here by subroutine 7
    062 x≤y?  
    063 x↔y max(n, n(1+√(1+4Q2(1+n2)(K-1)))/(2Q(1+n2)))
    064 F? 0 High-pass filter?
    065 GSB 6 2nQ/(1+√(1+4Q2(K-1-n2)))
    066 STO 8  
    067 RCL 0  
    068 RCL 8  
    069 ÷ C1 = C/n
    070 R/S Display C1 and let user change it
    071 STO 6  
    072 RCL 8  
    073 RCL 0  
    074 × C2 = nC
    075 R/S Display C2 and let user change it
    076 STO 7  
    077 RCL 6  
    078 ×  
    079 √x √(C1C2)
    080 GSB 4 Multiply by 2πf and take reciprocal
    081 STO 9 N = 1/(2πf√(C1C2))
    082 RCL 7  
    083 RCL 6  
    084 ÷  
    085 √x n = √(C2/C1)
    086 GSB 3 2nQ/(1+√(1+4Q2(K-1-n2))) or n(1+√(1+4Q2(1+n2)(K-1)))/(2Q(1+n2))
    087 STO 8  
    088 RCL 9  
    089 ×  
    090 STO 3 (R1||R3) = nN
    091 RCL D  
    092 ÷ R1 = (R1||R3)/α
    093 R/S Display R1 and let user change it
    094 STO 1  
    095 RCL 9  
    096 RCL 8  
    097 ÷ R2 = N/n
    098 R/S Display R2 and let user change it
    099 STO 2  
    100 RCL 3  
    101 1  
    102 RCL D  
    103 − ( 1-α (R1||R3) )
    104 x≠0?  
    105 ÷ R3 = (R1||R3)/(1-α) if α ≠ 1, or zero otherwise
    106 R/S Display R3 and let user change it
    107 STO 3  
    108 RTN  
    109♦  LBL 4 Multiply by 2πf and take reciprocal
    110 RCL A f
    111 ×  
    112♦  LBL 1 Multiply by 2π and take reciprocal
    113 2  
    114 ×  
    115 π  
    116 ×  
    117 1/x  
    118 RTN  
    119♦  LBL 3 Compute either 2xQ/(1+√(1+4Q2(K-1-x2))) or x(1+√(1+4Q2(1+x2)(K-1)))/(2Q(1+x2))
    120 F? 0 High-pass filter?
    121 GTO 7  
    122♦  LBL 6 Subroutine to compute 2xQ/(1+√(1+4Q2(K-1-x2)))
    123 STO 8 Save x for later use
    124 RCL B  
    125 2  
    126 × ( 2Q x )
    127 × ( 2xQ )
    128 LSTx ( 2Q 2xQ )
    129 x2 ( 4Q2 2xQ )
    130 RCL E  
    131 1  
    132 −  
    133 RCL 8  
    134 x2  
    135 − ( K-1-x2 4Q2 2xQ )
    136 × ( 4Q2(K-1-x2) 2xQ )
    137 GSB 9 ( 1+√(1+4Q2(K-1-x2)) 2xQ )
    138 ÷  
    139 RTN  
    140♦  LBL 7 Subroutine to compute x(1+√(1+4Q2(1+x2)(K-1)))/(2Q(1+x2))
    141 STO 8 Save x in register 8 for later use both by this subroutine and the caller
    142 GSB 8 ( 2Q 1+x2 )
    143 x2  
    144 × ( 4Q2(1+x2) )
    145 RCL E  
    146 1  
    147 − ( K-1 4Q2(1+x2) )
    148 × ( 4Q2(1+x2)(K-1) )
    149 GSB 9 ( 1+√(1+4Q2(1+x2)(K-1)) )
    150 RCL 8  
    151 × ( x(1+√(1+4Q2(1+x2)(K-1))) )
    152 RCL 8  
    153 GSB 8 ( 2Q 1+x2 x(1+√(1+4Q2(1+x2)(K-1))) )
    154 × ( 2Q(1+x2) x(1+√(1+4Q2(1+x2)(K-1))) )
    155 ÷  
    156 RTN  
    157♦  LBL 8 Subroutine to populate stack with 2Q 1+x2
    158 x2  
    159 1  
    160 +  
    161 RCL B  
    162 2  
    163 ×  
    164 RTN  
    165♦  LBL 9 Subroutine to compute 1+√(1+x)
    166 1  
    167 +  
    168 √x  
    169 1  
    170 +  
    171 RTN  
    172♦  LBL E Compute actual f, Q, and Gain
    173 GSB 2 R1 or (R1||R3)
    174 RCL 6  
    175 ×  
    176 STO 8 Save R1C1 for use in calculating Q
    177 RCL 2  
    178 RCL 7  
    179 ×  
    180 STO 9 Save R2C2 for use in calculating Q
    181 ×  
    182 √x  
    183 GSB 1 Multiply by 2π and take reciprocal
    184 ST I Save actual f for use in calculating Q
    185 R/S Display actual f
    186 RCL 9 ( R2C2 )
    187 RCL 8 ( R1C1 R2C2 )
    188 F? 0 High-pass filter?
    189 x↔y ( R2C2 R1C1 )
    190 1  
    191 RCL E  
    192 −  
    193 ×  
    194 +  
    195 GSB 2 R1 or (R1||R3)
    196 RCL 7  
    197 ×  
    198 +  
    199 RC I Recall actual frequency
    200 ×  
    201 GSB 1 Multiply by 2π and take reciprocal
    202 R/S Display actual Q
    203 RCL 4  
    204 RCL 5  
    205 x≠0?  
    206 ÷  
    207 1  
    208 +  
    209 GSB 2 R1 or (R1||R3)
    210 ×  
    211 RCL 1  
    212 ÷  
    213 RTN Return actual Gain
    214♦  LBL 2 Return either R1 or (R1||R3)
    215 RCL 1  
    216 F? 0 High-pass filter?
    217 RTN Return just R1
    218 1/x  
    219 RCL 3  
    220 x≠0? R3 exists? (0 means not)
    221 1/x  
    222 +  
    223 1/x  
    224 RTN  

    Registers and Flags

    RegisterUse
     0 C - capacitor scale
     1,2,3 R1, R2, R3 - filter resistors
     4,5 RF, RG - feedback resistors
     6,7 C1, C2 - capacitors
     8 n - variable used during computation
     9 N - variable used during computation
     A f - cutoff frequency
     B Q - filter quality
     C H - overall gain
     D α - input attenuator gain
     E K - internal gain
     I Temporary register

    FlagMeaning
     0 High-pass filter

    Revision History

    2008-Dec-12 — Initial release.

    References

    1. Analysis of the Sallen-Key Architecture (Rev.B), Texas Instruments Application Report SLOA024B, James Karki, 2002

    2. Low-Sensitivity, Lowpass Filter Design, National Semiconductor Application Note OA-27, Kumen Blake, 1996

    3. Low-Sensitivity, Highpass Filter Design, National Semiconductor Application Note OA-29, Kumen Blake, 1996

    Other HP Calculator Programs

    I've written programs for many of the HP calculators calculators in my collection. You may be interested in some of these:


     

     
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    Last updated Monday May 25, 2009. E-mail Stefan

     

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