Analysis of the
SallenĆKey Architecture
July 1999 Mixed Signal Products
Application
Report
SLOA024A
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iii
Analysis of the Sallen-Key Architecture
Contents
1 Introduction 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Generalized Circuit Analysis 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 Gain Block Diagram 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Ideal Transfer Function 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Low-Pass Circuit 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Simplification 1: Set Filter Components as Ratios 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Simplification 2: Set Filter Components as Ratios and Gain = 1 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Simplification 3: Set Resistors as Ratios and Capacitors Equal 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Simplification 4: Set Filter Components Equal 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Nonideal Circuit Operation 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Simulation and Lab Data 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 High-Pass Circuit 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1 Simplification 1: Set Filter Components as Ratios 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Simplification 2: Set Filter Components as Ratios and Gain=1 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Simplification 3: Set Resistors as Ratios and Capacitors Equal 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Simplification 4: Set Filter Components as Equal 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Nonideal Circuit Operation 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Lab Data 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Summary and Comments About Component Selection 13. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Figures
iv
SLOA024A
List of Figures
1 Basic Second Order Low-Pass Filter 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Unity Gain Sallen-Key Low-Pass Filter 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Generalized Sallen-Key Circuit 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Gain-Block Diagram of the Generalized Sallen-Key Filter 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Low-Pass Sallen-Key Circuit 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Nonideal Effect of Amplifier Output Impedance and Transfer Function 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 Test Circuits 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 Effect of Output Impedance 8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 High-Pass Sallen-Key Circuit 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 Model of High-Pass Sallen-Key Filter Above fc 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 High-Pass Sallen-Key Filter Using THS3001 12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 Frequency Response of High-Pass Sallen-Key Filter Using THS3001 12. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Analysis of the Sallen-Key Architecture
James Karki
ABSTRACT
This application report discusses the Sallen-Key architecture. The report gives a general
overview and derivation of the transfer function, followed by detailed discussions of
low-pass and high-pass filters, including design information, and ideal and non-ideal
operation. To illustrate the limitations of real circuits, data on low-pass and high-pass
filters using the Texas Instruments THS3001 is included. Finally, component selection is
discussed.
1 Introduction
Figure 1 shows a two-stage RC network that forms a second order low-pass filter.
This filter is limited because its Q is always less than 1/2. With R1=R2 and C1=C2,
Q=1/3. Q approaches the maximum value of 1/2 when the impedance of the
second RC stage is much larger than the first. Most filters require Qs larger than
1/2.
()( )
11C1R1C2R2C1Rs1C2R2C1Rs 1
Vi
Vo 2++++
=
VO
C1
R2
VIC2
R1
Figure 1. Basic Second Order Low-Pass Filter
Larger Qs are attainable by using a positive feedback amplifier. If the positive
feedback is controlled—localized to the cut-off frequency of the filter—almost any
Q can be realized, limited mainly by the physical constraints of the power supply
and component tolerances. Figure 2 shows a unity gain amplifier used in this
manner. Capacitor C2, no longer connected to ground, provides a positive
feedback path. In 1955, R. P. Sallen and E. L. Key described these filter circuits,
and hence they are generally known as Sallen-Key filters.
VO
C1
R2
VIC2
R1 +
–
Figure 2. Unity Gain Sallen-Key Low-Pass Filter
The operation can be described qualitatively:
•At low frequencies, where C1 and C2 appear as open circuits, the signal is
simply buffered to the output.
•At high frequencies, where C1 and C2 appear as short circuits, the signal is
shunted to ground at the amplifier’s input, the amplifier amplifies this input to
its output, and the signal does not appear at Vo.
•Near the cut-off frequency , where the impedance of C1 and C2 is on the same
order as R1 and R2, positive feedback through C2 provides Q enhancement
of the signal.
Generalized Circuit Analysis
2
SLOA024A
2 Generalized Circuit Analysis
The circuit shown in Figure 3 is a generalized form of the Sallen-Key circuit,
where generalized impedance terms, Z, are used for the passive filter
components, and R3 and R4 set a non-frequency dependent gain.
R3
VO
R4
VI
Z4
Z2Z1 Vn
Vf
Vp
Z3
+
–
Figure 3. Generalized Sallen-Key Circuit
To find the circuit solution for this generalized circuit, find the mathematical relationships
between Vi, Vo, Vp, and Vn, and construct a block diagram.
KCL at Vf:
Vf
ǒ
1
Z1
)
1
Z2
)
1
Z4
Ǔ+
Vi
ǒ
1
Z1
Ǔ)
Vp
ǒ
1
Z2
Ǔ)
Vo
ǒ
1
Z4
Ǔ
KCL at Vp:
Vp
ǒ
1
Z2
)
1
Z3
Ǔ+
Vf
ǒ
1
Z2
Ǔå
Vf
+
Vp
ǒ
1
)
Z2
Z3
Ǔ
Substitute Equation (2) into Equation (1) and solve for Vp:
Vp
+
Vi
ǒ
Z2Z3Z4
Z2Z3Z4
)
Z1Z2Z4
)
Z1Z2Z4
)
Z2Z2Z4
)
Z2Z2Z1
Ǔ)
Vo
ǒ
Z1Z2Z3
Z2Z3Z4
)
Z1Z2Z4
)
Z1Z2Z4
)
Z2Z2Z4
)
Z2Z2Z1
Ǔ
KCL at Vn:
V
ǒ
1
R3
)
1
R4
Ǔ+
Vo
ǒ
1
R4
Ǔå
Vn
+
Vo
ǒ
R3
R3
)
R4
Ǔ
2.1 Gain Block Diagram
By letting: a(f) = the open-loop gain of the amplifier, b
+ǒ
R3
R3
)
R4
Ǔ
,
c
+
Z2Z3Z4
Z2Z3Z4
)
Z1Z2Z4
)
Z1Z2Z3
)
Z2Z2Z4
)
Z2Z2Z1 ,
d
+
Z1Z2Z3
Z2Z3Z4
)
Z1Z2Z4
)
Z1Z2Z3
)
Z2Z2Z4
)
Z2Z2Z1 ,
and Ve = Vp – Vn, the generalized Sallen-Key filter circuit is represented in
gain-block form as shown in Figure 4.
(1)
(2)
(3)
(4)
Generalized Circuit Analysis
3
Analysis of the Sallen-Key Architecture
Ve a(f) VO
b
VIc
d
+
–
+
Figure 4. Gain-Block Diagram of the Generalized Sallen-Key Filter
From the gain-block diagram the transfer function can be solved easily by
observing, Vo = a(f)Ve and Ve = cVi + dVo – bVo. Solving for the generalized
transfer function from gain block analysis gives:
Vo
Vi
+ǒ
c
b
Ǔȧ
ȧ
ȡ
Ȣ
1
1
)
1
a
ǒ
f
Ǔ
b
*
d
b
ȧ
ȧ
ȣ
Ȥ
2.2 Ideal Transfer Function
Assuming a(f)b is very large over the frequency of operation, 1
a(f)b
[
0, the ideal
transfer function from gain block analysis becomes:
Vo
Vi
+ǒ
c
b
Ǔȧ
ȡ
Ȣ
1
1
*
d
b
ȧ
ȣ
Ȥ
By letting 1
b
+
K, c
+
N1
D, and d
+
N2
D, where N1, N2, and D are the
numerators and denominators shown above, the ideal equation can be rewritten
as:
Vo
Vi
+ȧ
ȡ
Ȣ
K
D
N1
*
K
@
N2
N1
ȧ
ȣ
Ȥ
. Plugging in the generalized impedance terms gives the
ideal transfer function with impedance terms:
Vo
Vi
+
K
Z1Z2
Z3Z4
)
Z1
Z3
)
Z2
Z3
)
Z1
ǒ
1
*
K
Ǔ
Z4
)
1
(5)
(6)
(7)
Low-Pass Circuit
4
SLOA024A
3 Low-Pass Circuit
The standard frequency domain equation for a second order low-pass filter is:
HLP
+
K
*ǒ
f
fc
Ǔ
2
)
jf
Qfc
)
1
Where fc is the corner frequency and Q is the quality factor. When f<<fc
Equation (8) reduces to K, and the circuit passes signals multiplied by a gain
factor K. When f=fc, Equation (8) reduces to –jKQ, and signals are enhanced by
the factor Q. When f>>fc, Equation (8) reduces to
*
K
ǒ
fc
f
Ǔ
2, and signals are
attenuated by the square of the frequency ratio. With attenuation at higher
frequencies increasing by a power of 2, the formula describes a second order
low-pass filter.
Figure 5 shows the Sallen-Key circuit configured for low-pass:
Z1
+
R1, Z2
+
R2, Z3
+
1
sC1 ,
Z4
+
1
sC2, and K
+
1
)
R4
R3 .
R3
VO
R4
C1
R2
VI
C2
R1 +
–
Figure 5. Low-Pass Sallen-Key Circuit
From Equation (7), the ideal low-pass Sallen-Key transfer function is:
Vo
Vi (Ip)
+
K
s2(R1R2C1C2)
)
s(R1C1
)
R2C1
)
R1C2(1
*
K))
)
1
By letting
s
+
j2
p
f, fc
+
1
2
p
R1R2C1C2
Ǹ
, and Q
+
R1R2C1C2
Ǹ
R1C1
)
R2C1
)
R1C2(1
*
K),
equation (9) follows the same form as Equation (8). With some simplifications,
these equations can be dealt with efficiently; the following paragraphs discuss
commonly used simplification methods.
(8)
(9)
Low-Pass Circuit
5
Analysis of the Sallen-Key Architecture
3.1 Simplification 1: Set Filter Components as Ratios
Letting R1=mR, R2=R, C1=C, and C2=nC, results in: fc
+
1
2
p
RC mn
Ǹ
and
Q
+
mn
Ǹ
m
)
1
)
mn(1
*
K). This simplifies things somewhat, but there is
interaction between fc and Q. Design should start by setting the gain and Q based
on m, n, and K, and then selecting C and calculating R to set fc.
Notice that K
+
1
)
m
)
1
mn results in Q = ∞. With larger values, Q becomes
negative, that is, the poles move into the right half of the s-plane and the circuit
oscillates. Most filters require low Q values so this should rarely be a design issue.
3.2 Simplification 2: Set Filter Components as Ratios and Gain = 1
Letting R1=mR, R2=R, C1=C, C2=nC, and K=1 results in: fc
+
1
2
p
RC mn
Ǹ
and
Q
+
mn
Ǹ
m
)
1. This keeps gain = 1 in the pass band, but again there is interaction
between fc and Q. Design should start by choosing the ratios m and n to set Q,
and then selecting C and calculating R to set fc.
3.3 Simplification 3: Set Resistors as Ratios and Capacitors Equal
Letting R1=mR, R2=R, and C1=C2=C, results in: fc
+
1
2
p
RC m
Ǹ
and
Q
+
m
Ǹ
1
)
2m
*
mK . The reason for setting the capacitors equal is the limited
selection of values in comparison with resistors.
There is interaction between setting fc and Q. Design should start with choosing
m and K to set the gain and Q of the circuit, and then choosing C and calculating
R to set fc.
3.4 Simplification 4: Set Filter Components Equal
Letting R1=R2=R, and C1=C2=C, results in: fc
+
1
2
p
RC and Q
+
1
3
*
K. Now
fc and Q are independent of one another, and design is greatly simplified although
limited. The gain of the circuit now determines Q. RC sets fc—the capacitor
chosen and the resistor calculated. One minor drawback is that since the gain
controls the Q of the circuit, further gain or attenuation may be necessary to
achieve the desired signal gain in the pass band.
Values of K very close to 3 result in high Qs that are sensitive to variations in the
values of R3 and R4. For instance, setting K=2.9 results in a nominal Q of 10.
Worst case analysis with 1% resistors results in Q=16. Whereas, setting K=2 for
a Q of 1, worst case analysis with 1% resistors results in Q=1.02. Resistor values
where K=3 leads to Q=∞, and with larger values, Q becomes negative, the poles
move into the right half of the s-plane, and the circuit oscillates. The most
frequently designed filters require low Q values and this should rarely be a design
issue.
Low-Pass Circuit
6
SLOA024A
3.5 Nonideal Circuit Operation
The previous discussions and calculations assumed an ideal circuit, but there is
a frequency where this is no longer a valid assumption. Logic says that the
amplifier must be an active component at the frequencies of interest or else
problems occur. But what problems?
As mentioned above there are three basic modes of operation: below cut-off,
above cutoff, and in the area of cutoff. Assuming the amplifier has adequate
frequency response beyond cut-off, the filter works as expected. At frequencies
well above cut-off, the high frequency (HS) model shown in Figure 6 is used to
show the expected circuit operation. The assumption made here is that C1 and
C2 are effective shorts when compared to the impedance of R1 and R2 so that
the amplifier’s input is at ac ground. In response, the amplifier generates an ac
ground at its output limited only by its output impedance, Zo. The formula shows
the transfer function of this model.
VO
R2
VIR1
Zo
Vo
Vi
+
1
R1
R2
)
R1
Zo
)
1Assuming Zo<<R1
Vo
Vi
[
Zo
R1
Figure 6. Nonideal Effect of Amplifier Output Impedance and Transfer Function
Zo is the closed-loop output impedance. It depends on the loop transmission and
the open-loop output impedance, zo: Zo
+
zo
1
)
a(f)b, where a(f)b is the loop
transmission. The feedback factor, b, is constant—set by resistors R3 and
R4—but the open loop gain, a(f), is dependant on frequency . With dominant pole
compensation, the open-loop gain of the amplifier decreases by 20 dB/dec over
the usable frequencies of operation. Assuming zo is mainly resistive (usually a
valid assumption up to 100 MHz), Zo increases at a rate of 20 dB/dec. The
transfer function appears to be a first order high-pass. At frequencies above
100 MHz (or so) the parasitic inductance in the output starts playing a role and
the transfer function transitions to a second order high-pass. Because of stray
capacitance in the circuit, at higher frequency the high-pass transfer function will
also roll off.
3.6 Simulation and Lab Data
A Sallen-Key low-pass filter using the Texas Instruments THS3001 shows the
effects described above. The THS3001 is a high-speed current-feedback
amplifier with an advertised bandwidth of 420 MHz. No particular type of filter (i.e.,
Butterworth, Chebychev, Eliptic, etc.) was designed. Choosing Z1=Z2=1kΩ,
Z3=Z4=1nF, R3=open, and R4=1kΩ results in a low-pass filter with fc=159 kHz,
and Q=1/2.
Simulation using the spice model of the THS3001 (see the application note
Building a Simple SPICE Model for the THS3001
, SLOA018) is used to show the
expected behavior of the circuit. Figure 7 shows the simulation circuits and the
lab circuit tested. The results are plotted in Figure 8.
Low-Pass Circuit
7
Analysis of the Sallen-Key Architecture
Figure 7 a) shows the simulation circuit with the spice model modified so that the
output impedance is zero. Curve a) in Figure 8 shows the frequency response as
simulated in spice. It shows that without the output impedance the attenuation of
the signal continues to increase as frequency increases.
Figure 7 b) shows the high-frequency model shown in Figure 6 where the input
is at ground and the output impedance controls the transfer function. The spice
model used for the THS3001 includes the complex LRC network for the output
impedance as described in the application note. Curve b) in Figure 8 shows the
frequency response as simulated in spice. The magnitude of the signal at the
output is seen to cross curve a) at about 7 MHz. Above this frequency the output
impedance causes the switch in transfer function as described above.
Figure 7 c) shows the simulation circuit using the full spice model with the
complex LCR output impedance. Curve c) in Figure 8 shows the frequency
response. It shows that with the output impedance the attenuation caused by the
circuit follows curve a) until it crosses curve b) at which point it follows curve b).
Figure 7 d) shows the circuit as tested in the lab, with curve d) in Figure 8 showing
that the measured data agrees with the simulated data.
VO
R4 = 1 kΩ
VIR2 = 1 kΩR1 = 1 kΩ
C2 = 1 nF
C1 = 1 nF
THS3001
a) Spice – Zo = 0Ωb) Spice – HF Model
Zo
c) Spice – Zo = LCR
Zo
d) Lab Circuit
+15 V
–15 V 100 Ω
VIVO
THS3001
C2 = 1 nF
R1 = 1 kΩ
R2 = 1 kΩ
C1 = 1 nF R4 = 1 kΩ
R2 = 1 kΩR1 = 1 kΩ
C1 = 1 nF VO
VI
C2 = 1 nF
THS3001
R4 = 1 kΩ
+
–
+
–
+
–
R2 = 1 kΩR1 = 1 kΩ
C1 = 1 nF
R4 = 1 kΩ
THS3001
+
–
C2 = 1 nF
VO
VI
Figure 7. Test Circuits
Low-Pass Circuit
8
SLOA024A
–100
–90
–80
–70
–60
–50
–40
–30
–20
–10
0
10
1E+04 1E+05 1E+06 1E+07 1E+08 1E+09
f – Frequency – Hz
b) Spice HF Model a) Spice ZO = 0Ω
c) Spice ZO = LRC
d) Lab Data
VI
VO/– dB
Figure 8. Effect of Output Impedance
High-Pass Circuit
9
Analysis of the Sallen-Key Architecture
4 High-Pass Circuit
The standard equation (in frequency domain) for a second order high-pass is:
HHP
+*
K
ǒ
f
fc
Ǔ
2
*ǒ
f
fc
Ǔ
2
)
jf
Qfc
)
1
When f<<fc, equation (10) reduces to
*
K
ǒ
f
fc
Ǔ
2. Below fc signals are attenuated
by the square of the frequency ratio. When f=fc, equation (10) reduces to –jKQ,
and signals are enhanced by the factor Q. When f>>fc, equation (10) reduces to
K, and the circuit passes signals multiplied by the gain factor K. With attenuation
at lower frequencies increasing by a power of 2, equation (10) describes a second
order high-pass filter.
Figure 9 shows the Sallen-Key circuit configured for high-pass:
Z2
+
1
sC2 ,Z1
+
1
sC1 ,Z3
+
R1, Z4
+
R2, and K
+
1
)
R4
R3 .
R3
VO
R4
C1
R2
VIC2
R1
+
–
Figure 9. High-Pass Sallen-Key Circuit
From equation (7), the ideal high-pass transfer function is:
Vo
Vi (hp)
+
K
1
s2
ǒ
R1R2C1C2
Ǔ)
1
s
ǒ
1
R1C1
)
1
R1C2
)ǒ
1
*
K
Ǔ
R2C1
Ǔ)
1
with some manipulation this becomes
Vo
Vi (hp)
+
K
ǒ
s2(R1R2C1C2)
Ǔ
s2(R1R2C1C2)
)
s(R2C2
)
R2C1
)
R1C2(1
*
K)
)
1)
By letting
s
+
j2
p
f, fo
+
1
2
p
R1R2C1C2
Ǹ
,andQ
+
R1R2C1C2
Ǹ
R2C2
)
R2C1
)
R1C2(1
*
K),
equation (11) follows the same form as equation (10). As above, simplifications
make these equations much easier to deal with. The following are common
simplifications used.
(10)
(11)
High-Pass Circuit
10
SLOA024A
4.1 Simplification 1: Set Filter Components as Ratios
Letting R1=mR, R2=R, C1=C, and C2=nC, results in:
fc
+
1
2
p
RC mn
Ǹ
and Q
+
mn
Ǹ
n
)
1
)
mn(1
*
K). This simplifies things
somewhat, but there is interaction between fc and Q. To design a filter using this
simplification, first set the gain and Q based on m, n, and K, and then select C and
calculate R to set fc.
Notice that K
+
1
)
n
)
1
mn results in Q=∞. With larger values, Q becomes
negative—that is the poles move into the right half of the s-plane and the circuit
oscillates. The most frequently designed filters require low Q values and this
should rarely be a design issue.
4.2 Simplification 2: Set Filter Components as Ratios and Gain=1
Letting R1=mR, R2=R, C1=C, and C2=nC, and K=1 results in:
fc
+
1
2
p
RC mn
Ǹ
and Q
+
mn
Ǹ
n
)
1. This keeps the gain=1 in the pass band,
but again there is interaction between fc and Q. To design a filter using this
simplification, first set Q by selecting the ratios m and n, and then select C and
calculate R to set fc.
4.3 Simplification 3: Set Resistors as Ratios and Capacitors Equal
Letting R1=mR, R2=R, and C1=C2=C, results in:
fc
+
1
2
p
RC m
Ǹ
and Q
+
m
Ǹ
2
)
m(1
*
K). The reason for setting the capacitors
equal is the limited selection of values compared with resistors.
There is interaction between setting fc and Q. Start the design by choosing m and
K to set the gain and Q of the circuit, and then choose C and calculating R to set
fc.
4.4 Simplification 4: Set Filter Components as Equal
Letting R1=R2=R, and C1=C2=C, results in: fc
+
1
2
p
RC and Q
+
1
3
*
K.
Now fc and Q are independent of one another, and design is greatly simplified.
The gain of the circuit now determines Q. The choice of RC sets fc—the capacitor
should chosen and the resistor calculated. One minor drawback is that since the
gain controls the Q of the circuit, further gain or attenuation may be necessary
to achieve the desired signal gain in the pass band.
Values of K very close to 3 result in high Qs that are sensitive to variations in the
values of R3 and R4. For instance, setting K=2.9 results in a nominal Q of 10.
Worst case analysis with 1% resistors results in Q=16. Whereas, setting K=2 for
a Q of 1, worst case analysis with 1% resistors results in Q=1.02. Resistor values
where K=3 leads to Q=∞, and with larger values, Q becomes negative. The most
frequently designed filters require low Q values, and this should rarely be a design
issue.
High-Pass Circuit
11
Analysis of the Sallen-Key Architecture
4.5 Nonideal Circuit Operation
The previous discussions and calculations assumed an ideal circuit, but there is
a frequency where this is no longer a valid assumption. Logic says that the
amplifier must be an active component at the frequencies of interest or else
problems occur. But what problems?
As mentioned above there are three basic modes of operation: below cut-off,
above cutoff, and in the area of cut-off. Assuming the amplifier has adequate
frequency response beyond cutoff, the filter works as expected. At frequencies
well above cut-off, the high frequency (HS) model shown in Figure 10 is used to
show the expected circuit operation. The assumption made here is that C1 and
C2 are effective shorts when compared to the impedance of R1 and R2. The
formula shows the transfer function of this model, where a(f) is the open loop gain
of the amplifier and b is the feedback factor. The circuit operates as expected until
1
a(f)b is no longer much smaller than 1. After which the gain of the circuit falls off
with the open loop gain of the amplifier. Because of practical limitations, designing
a high-pass Sallen-Key filter results in a band-pass filter where the upper cutoff
frequency is determined by the open loop response of the amplifier.
VO
R4
VI
R2
R1
R3
+
–
Vo
Vi
+
1
b
ȧ
ȡ
Ȣ
1
1
)
1
a
ǒ
f
Ǔ
b
ȧ
ȣ
Ȥ
Figure 10. Model of High-Pass Sallen-Key Filter Above fc
4.6 Lab Data
A Sallen-Key high-pass filter using the Texas Instruments THS3001 shows the
effects described above. The THS3001 is a high-speed current-feedback
amplifier with an advertised bandwidth of 420 MHz. No particular type of filter (i.e.,
Butterworth, Chebychev, Eliptic, etc.) was designed. Choosing Z1=Z2=1kΩ,
Z3=Z4=1nF , R3=open, and R4=1kΩ, results in a high-pass filter with fc=159 kHz,
and Q=1/2. Figure 11 shows the circuit, and Figure 12 shows the lab results. As
expected, the circuit attenuates signals below 159 kHz at a rate of 40dB/dec, and
passes signals above 159 kHz with a gain of 1 until the amplifier’s open loop gain
falls to around unity between 300 MHz and 400 MHz. The slight increase in gain
seen just before 300 MHz is due to gain peaking in the amplifier. Setting R4 to a
higher value reduces this, but also reduces the overall bandwidth of the amplifier.
High-Pass Circuit
12
SLOA024A
VO
R4 = 1 kΩ
VI
R2 = 1 kΩ
R1 = 1 kΩ
C1 = 1 nF THS3001
+15 V
–15 V 100 Ω
C2 = 1 nF
Figure 11. High-Pass Sallen-Key Filter Using THS3001
Lab_Data
–50
–40
–30
–20
–10
0
10
1.00E+04 1.00E+05 1.00E+06 1.00E+07 1.00E+08 1.00E+09
f – Frequency – Hz
VI
VO/– dB
Figure 12. Frequency Response of High-Pass Sallen-Key Filter Using THS3001
Summary and Comments About Component Selection
13
Analysis of the Sallen-Key Architecture
5 Summary and Comments About Component Selection
Theoretically, any values of R and C that satisfy the equations may be used, but
practical considerations call for component selection guidelines to be followed.
Given a specific corner frequency, the values of C and R are inversely
proportional—as C is made larger, R becomes smaller and vice versa.
In the case of the low-pass Sallen-Key filter, the ratio between the output
impedance of the amplifier and the value of filter component R sets the transfer
functions seen at frequencies well above cut-off. The larger the value of R the
lower the transmission of signals at high frequency. Making R too large has
consequences in that C may become so small that the parasitic capacitors
—including the input capacitance of the amplifier—cause errors.
For the high-pass filter, the amplifier’s output impedance does not play a parasitic
role in the transfer function, so that the choice of smaller or larger resistor values
is not so obvious. Stray capacitance in the circuit, including the input capacitance
of the amplifier, makes the choice of small capacitors, and thus large resistors,
undesirable. Also, being a high-pass circuit, the bandwidth is potentially very
large and resistor noise associated with increased values can become an issue.
Then again, small resistors become a problem if the circuit impedance is too small
for the amplifier to operate properly.
The best choice of component values depends on the particulars of your circuit
and the tradeoffs you are willing to make. General recommendations are as
follows:
•Capacitors
– Avoid values less than 100 pF.
– Use NPO if at all possible. X7R is OK in a pinch. Avoid Z5U and other low
quality dielectrics. In critical applications, even higher quality dielectrics
like polyester, polycarbonate, mylar, etc., may be required.
– Use 1% tolerance components. 1%, 50V, NPO, SMD, ceramic caps in
standard E12 series values are available from various sources.
– Surface mount is preferred.
•Resistors
– Values in the range of a few hundred ohms to a few thousand ohms are
best.
– Use metal film with low temperature coefficients.
– Use 1% tolerance (or better).
– Surface mount is preferred.
14
SLOA024A
