Low-Sensitivity Sallen-Key Filter Design with the HP-41C Programmable Calculator

December 21, 2008

This program, which I originally wrote for the HP-67 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:

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, R_F, 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 R_F and R_G). 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 R_F:

Description	Keystrokes	Display
Use engineering notation	ENG 2	0.00  00
Enter f	500       a	500.  00
Enter Q	2       b	2.00  00
Enter H	1       c	1.00  00
Enter C	10 EEx CHS 9       d	10.0 -09
Enter RF	47 EEx 3       e	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 R_G is displayed as zero, meaning it can be omitted, the value of R_F will not matter any more, and R_F 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:

Description	Keystrokes	Display
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:

Description	Keystrokes	Display
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:

Description	Keystrokes	Display
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

Line	Instruction	Comments
01♦	LBL “SK”
02♦	LBL a	Enter and store f
03	STO 11
04	RTN
05♦	LBL b	Enter and store Q
06	STO 12
07	RTN
08♦	LBL c	Enter and store H (gain)
09	STO 13
10	RTN
11♦	LBL d	Enter and store C (capacitor scale)
12	STO 00
13	RTN
14♦	LBL e	Enter and store RF
15	STO 04
16	RTN
17♦	LBL A	Low-pass filter: RG,C1,C2,R1,R2,R3
18	CF 00
19	GTO 00
20♦	LBL C	High-pass filter: RG,C1,C2,R1,R2
21	SF 00
22♦	LBL 00	Forward solution
23	RCL 12
24	2.2
25	×
26	.9
27	−
28	RCL 12
29	.2
30	+
31	÷	(2.2Q-0.9)/(Q+0.2)
32	1
33	x≤y?
34	x↔y
35	STO 15	K = max(1,(2.2Q-0.9)/(Q+0.2))
36	RCL 13
37	x>y?
38	STO 15	K = max(H,1,(2.2Q-0.9)/(Q+0.2))
39	RCL 15
40	÷	H/K
41	x=0?
42	1	Use α = 1 if H/K = 0 (because H was 0)
43	FS? 00
44	1	Always use α = 1 for a high-pass filter
45	STO 14	α = H/K (always 1 for a high-pass filter or H = 0)
46	RCL 04
47	RCL 15
48	1
49	−
50	x≠0?
51	÷	RG = RF/(K-1) if K ≠ 1, or zero if K = 1
52	R/S	Display RG and let user change it
53	STO 05
54	.1	Initialize n to √0.1;
55	√x
56	XEQ 07	n(1+√(1+4Q2(1+n2)(K-1)))/(2Q(1+n2))
57	RCL 08	√0.1 was stored here by subroutine 7
58	x≤y?
59	x↔y	max(n, n(1+√(1+4Q2(1+n2)(K-1)))/(2Q(1+n2)))
60	FS? 00	High-pass filter?
61	XEQ 06	2nQ/(1+√(1+4Q2(K-1-n2)))
62	STO 08
63	RCL 00
64	RCL 08
65	÷	C1 = C/n
66	R/S	Display C1 and let user change it
67	STO 06
68	RCL 08
69	RCL 00
70	×	C2 = nC
71	R/S	Display C2 and let user change it
72	STO 07
73	RCL 06
74	×
75	√x	√(C1C2)
76	XEQ 04	Multiply by 2πf and take reciprocal
77	STO 09	N = 1/(2πf√(C1C2))
78	RCL 07
79	RCL 06
80	÷
81	√x	n = √(C2/C1)
82	XEQ 03	2nQ/(1+√(1+4Q2(K-1-n2))) or n(1+√(1+4Q2(1+n2)(K-1)))/(2Q(1+n2))
83	STO 08
84	RCL 09
85	×
86	STO 03	(R1||R3) = nN
87	RCL 14
88	÷	R1 = (R1||R3)/α
89	R/S	Display R1 and let user change it
90	STO 01
91	RCL 09
92	RCL 08
93	÷	R2 = N/n
94	R/S	Display R2 and let user change it
95	STO 02
96	RCL 03
97	1
98	RCL 14
99	−	( 1-α (R1||R3) )
100	x≠0?
101	÷	R3 = (R1||R3)/(1-α) if α ≠ 1, or zero otherwise
102	R/S	Display R3 and let user change it
103	STO 03
104	RTN
105♦	LBL 04	Multiply by 2πf and take reciprocal
106	RCL 11	f
107	×
108♦	LBL 01	Multiply by 2π and take reciprocal
109	2
110	×
111	π
112	×
113	1/x
114	RTN
115♦	LBL 03	Compute either 2xQ/(1+√(1+4Q2(K-1-x2))) or x(1+√(1+4Q2(1+x2)(K-1)))/(2Q(1+x2))
116	FS? 00	High-pass filter?
117	GTO 07
118♦	LBL 06	Subroutine to compute 2xQ/(1+√(1+4Q2(K-1-x2)))
119	STO 08	Save x for later use
120	RCL 12
121	2
122	×	( 2Q x )
123	×	( 2xQ )
124	LASTx	( 2Q 2xQ )
125	x2	( 4Q2 2xQ )
126	RCL 15
127	1
128	−
129	RCL 08
130	x2
131	−	( K-1-x2 4Q2 2xQ )
132	×	( 4Q2(K-1-x2) 2xQ )
133	XEQ 09	( 1+√(1+4Q2(K-1-x2)) 2xQ )
134	÷
135	RTN
136♦	LBL 07	Subroutine to compute x(1+√(1+4Q2(1+x2)(K-1)))/(2Q(1+x2))
137	STO 08	Save x in register 8 for later use both by this subroutine and the caller
138	XEQ 08	( 2Q 1+x2 )
139	x2
140	×	( 4Q2(1+x2) )
141	RCL 15
142	1
143	−	( K-1 4Q2(1+x2) )
144	×	( 4Q2(1+x2)(K-1) )
145	XEQ 09	( 1+√(1+4Q2(1+x2)(K-1)) )
146	RCL 08
147	×	( x(1+√(1+4Q2(1+x2)(K-1))) )
148	RCL 08
149	XEQ 08	( 2Q 1+x2 x(1+√(1+4Q2(1+x2)(K-1))) )
150	×	( 2Q(1+x2) x(1+√(1+4Q2(1+x2)(K-1))) )
151	÷
152	RTN
153♦	LBL 08	Subroutine to populate stack with 2Q 1+x2
154	x2
155	1
156	+
157	RCL 12
158	2
159	×
160	RTN
161♦	LBL 09	Subroutine to compute 1+√(1+x)
162	1
163	+
164	√x
165	1
166	+
167	RTN
168♦	LBL E	Compute actual f, Q, and Gain
169	XEQ 02	R1 or (R1||R3)
170	RCL 06
171	×
172	STO 08	Save R1C1 for use in calculating Q
173	RCL 02
174	RCL 07
175	×
176	STO 09	Save R2C2 for use in calculating Q
177	×
178	√x
179	XEQ 01	Multiply by 2π and take reciprocal
180	STO 10	Save actual f for use in calculating Q
181	R/S	Display actual f
182	RCL 09	( R2C2 )
183	RCL 08	( R1C1 R2C2 )
184	FS? 00	High-pass filter?
185	x↔y	( R2C2 R1C1 )
186	1
187	RCL 15
188	−
189	×
190	+
191	XEQ 02	R1 or (R1||R3)
192	RCL 07
193	×
194	+
195	RCL 10	Recall actual frequency
196	×
197	XEQ 01	Multiply by 2π and take reciprocal
198	R/S	Display actual Q
199	RCL 04
200	RCL 05
201	x≠0?
202	÷
203	1
204	+
205	XEQ 02	R1 or (R1||R3)
206	×
207	RCL 01
208	÷
209	RTN	Return actual Gain
210♦	LBL 02	Return either R1 or (R1||R3)
211	RCL 01
212	FS? 00	High-pass filter?
213	RTN	Return just R1
214	1/x
215	RCL 03
216	x≠0?	R3 exists? (0 means not)
217	1/x
218	+
219	1/x
220	RTN

Registers and Flags

Register	Use
00	C – capacitor scale
01,02,03	R1, R2, R3 – filter resistors
04,05	RF, RG – feedback resistors
06,07	C1, C2 – capacitors
08	n – variable used during computation
09	N – variable used during computation
11	f – cutoff frequency
12	Q – filter quality
13	H – overall gain
14	α – input attenuator gain
15	K – internal gain
10	Temporary register

Flag	Meaning
00	High-pass filter

Revision History

2008-Dec-21 — 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 (now Texas Instruments) Application Note OA-27, Kumen Blake, 1996

3. Low-Sensitivity, Highpass Filter Design, National Semiconductor (now Texas Instruments) Application Note OA-29, Kumen Blake, 1996

                    

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