Jonathan Roderick
Experiment #3
Op-Amps continued
Introduction:
In this lab you will use the op-amp to achieve more advanced filtering techniques. In the previous lab, you might
not of realized it until you had completed the lab, you built a first order low and high pass filter. In this lab you will
learn the basic concepts of a specific second order filter, called the Sallen and Key filter. There will be three types
of filters presented; low pass, band pass and high pass. Second order filters are used because they produce better
roll-off in the cut-off bands, this will be discussed later. The Sallen and Key filter is just one more of many
applications of the good ‘ol op-amp.
Theory:
Active Filters:
As you witnessed in the previous lab, a filter is a circuit that lets certain frequencies pass and blocks other. This
selective nature can be done two ways, either with passive filters or with active filters. Passive filters completely
comprised of passive elements; namely resistors, capacitors and/or inductors. Active filters use active devices, i.e.
an op-amp, to filter out unwanted signals.
Active filters have the following advantages over passive filters.
-Gain and frequency adjustment and tuning.
-No inductors (reduces cost and size).
-No loading effects.
Some disadvantages of active filters.
-Bandwidth limitations
-Fabrication tolerances
-Can only respond to a specific range of signal magnitudes.
Figure 3.3 show the performance of an ideal low-pass, band-pass, and high pass circuit. Active filters can be
classified as; low-pass, high-pass, band-pass, notch, or all pass circuit. These circuits are all used for different
purposes, but this lab will focus on the design of low, band and high pass Sallen and Key Filters. In the exercise #1
of the previous lab, you built and tested a high and low pass circuit. These filters were two first order active circuits.
Sallen and Key filter implement a second order topology that results in better performance than the previous filters
explored in the previous lab. To be sure, the Sallen and Key topologies are also more complicated.
Ideal Lowpass Filter Response
passband
passband
Magnitude
Magnitude
Magnitude
passband
Frequency
Ideal Highpass Filter Response
Ideal Bandpass Filter Response
Frequency
Figure 3.3 Graph of an ideal (a) low pass, (b) band-pass, and (c) high-pass filter output.
1
Frequency
Unfortunately almost nothing in circuit design is ideal. Fig. 3.4 shows the typical behavior of a low pass, band-pass,
and high-pass filter. The first difference between ideal and real filters that you can observe is roll-off. Real filters
do not have the extreme cut-off between the pass-band and reject-band transitions. One might question how the
performance of different filters is compared. This is why key criterion like the 3dB point(s) have been developed.
The 3dB point is defined as the frequency where the magnitude of the gain has dropped by a factor of the square
root of 2.
Lowpass Filter characteristics
Highpass Filter Response
0.6
0.4
0.2
0.0
1.0
Magnitude (Normalized)
Passband
Magnitude (Normalized)
Magnitude (Normalized)
0.8
Bandpass Filter Response
1.0
1.0
0.8
Passband
0.6
0.4
0.2
0.0
Frequency (Log)
0.8
Passband
0.6
0.4
0.2
0.0
Frequency (Log)
Frequency (Log)
Figure 3.4 Graph of a typical (a) low-pass, (b) band-pass, and (c) high-pass filter output.
The limitation of the first order circuits is the roll-off in the cut-off band. First order low or high pass filters
experience a – 20db/decade roll-off in the cut-off regions. A –20dB/decade roll-off is insufficient for some
applications, so a higher order filter is used. A second order filter generates a –40dB/decade. In general, the higher
the order of the circuit, the better the roll-off between the stop and pass bands. These results can be best visualized
using Bode Plots.
Second-Order low-pass filter:
A general transfer function of a second order low-pass filter is
H (s) =
H ( 0) ∗ ω o2
ω 
s 2 +  o  s + ωo2
Q
(3.1)
where H(0) is the dc gain, ωo is the pole frequency, and Q is the quality factor. The damping factor, ξ, can be
calculated from Q= 1/(2ξ). The poles of this transfer function are
p1 , p2 = −
ω0
±
2Q
ω 
j  0  4Q2 − 1
 2Q 
Figure 3.5 shows a schematic diagram of a low-pass Sallen and Key filter.
2
(3.2)
C2
Vs
R1
R3
+
Vo
C4
_
Rx
Ry
Figure 3.5 Low pass Sallen and Key filter.
The gain of the circuit, A, is controlled by the feedback associated with the op-amp, namely Rx and Ry. The
transfer function, pole frequency (ωo ), and Quality factor are given by




A




 Vo ( s )  
R1R3C2C4 



 =


V
(
s
)


1
1
1
A
1
 s 
2
 +
+
+
−
 s + s 


 R3C4 R1C2 R3C2 R3C4  R1R3C2C4  (3.3)
ω0 =
1
R1R3C2C4
(3.4)
 RC
R1C4
RC 
Q= 3 4 +
+ (1 − A) 1 2 
R3C2
R3C4 
 R1C2
3
−1
(3.5)
Magnitude (Normalized)
Low-pass Sallen and Key Filter
1.0
0.8
-40dB/dec
3dB
0.6
0.4
0.2
0.0
ω0
(rad/s)
Figure 3.6 A graphed output of the low pass Sallen and Key filter.
Second-Order high-pass filter:
A general transfer function of a second order high-pass filter is
H (s) =
H ( j∞) ∗ s 2
ω 
s 2 +  o  s + ωo2
Q
(3.6)
where H(j∞) is the high frequency gain (A), ωo is the pole frequency, and Q is the quality factor. The damping
factor, ξ, can be calculated from Q= 1/(2ξ). The poles of this transfer function are
p1 , p2 = −
ω0
±
2Q
ω 
j  0  4Q2 − 1
 2Q 
(3.7)
Figure 3.7 shows a schematic diagram of a high-pass Sallen and Key filter, while 3.8 shows a typical output of a the
high-pass filter.
4
R2
Vs
C1
C3
+
Vo
R4
_
Rx
Ry
Figure 3.7 High-pass Sallen and Key filter.
The transfer function, pole frequency (ωo ), and Quality factor are given by




2
 Vo ( s )  
s A


 =


V
(
s
)


1
1
1
A
1
 s 
2
 +
+
+
−
 s + s


 R2C1 R4C3 R4C1 R2C1  R2 R4C1C3 
ω0 =
1
R2 R4C1C3
(3.9)
 RC
R2C1
R2C3
RC 
Q= 4 3 +
+
−A 4 3 
R4C3
R4C1
R2C1 
 R2C1
−1
The output of a typical Sallen and Key high-pass filter is shown in figure 3.8.
5
(3.10)
(3.8)
Magnitude (Normalized)
High-pass Sallen and Key Filter
1.0
Passband
0.8
3dB
0.6
40dB/dec
0.4
0.2
0.0
ω0
(rad/s)
Figure 3.8 A graphed output of the Sallen and Key high-pass filter.
Second-Order band-pass filter:
A general transfer function of a second order band-pass filter is
ω 
H ( j ω0 ) 0  s
Q
H (s) =
ω 
s 2 +  o  s + ωo2
Q
(3.11)
where H(jω 0 ) is the frequency gain (A) at ωo , ωo is the center frequency, and Q is the quality factor. The damping
factor, ξ, can be calculated from Q= 1/(2ξ). The poles of this transfer function are
p1 , p2 = −
ω0
±
2Q
ω 
j  0  4Q2 − 1
 2Q 
(3.12)
Figure 3.9 shows a schematic diagram of a band-pass Sallen and Key filter, while 3.10 shows a typical output of a
band-pass filter.
6
R2
R1
C3
Vs
+
C5
R4
Vo
_
Rx
Ry
Figure 3.9 Band-pass Sallen and Key filter.
The transfer function, pole frequency (ωo ), and Quality factor are given by


 Vo ( s )  

 =
 Vs ( s )   s 2 + s 1 + 1 + 1

 RC R C R C

 1 5
2 5
4 5

 sA 



 R1C5 

1
A 
1  1 1 
+
 + 
+
−
R4C3 R2C5  R4C3C5  R1 R2   (3.13)
R1
R2
R1R4C3C5
1+
ω0 =
7
(3.14)
Magnitude (Normalized)
Sallen and Key Band-pass Filter
1.0
-20dB/dec
0.8
3dB
0.6
+20dB/dec
0.4
0.2
0.0
ω1
ω0
ω2
(rad/s)
Figure 3.10 A graphed output of the band-pass Sallen and Key filter.
  R1
 R4C3
R1C3
R1C5
+
+
 1 + (1 − A) 
R2
R4C5
R4C3
 R1C5
Q = 

R
1+ 1

R2







−1
(3.15)
Unlike the Sallen and Key high and Low-pass filters, the band-pass rolls off at –20dB/decade. The reason for the
difference is that the pass-band filter has two reject two bands of signals. One pole is used to get a 20bB/decade
roll-up to the center frequency, and the other pole is used to get a –20dB/decade roll-off after the center frequency.
Another way to realize a band-pass is to cascade a low and high-pass filter from experiment #2. Figure 3.11 shows a
schematic diagram of a high and low-pass filter used to get a band-pass effect. A typical response of this circuit is
very similar to the Sallen and Key filter, but the overall circuit is large and requires two op-amps. There are
limitations to this cascade approach. Drawbacks include limitation on Q, power consumption, and others.
The transfer function, Quality factor, and center frequency (fc) of this band-pass filter are:
H ( s) =
Q=
A(ω 0 / Q ) s
s + (ω 0 / Q ) s + ω 02
2
ω0
fc
=
BW
fh − fl
8
(3.16)
(3.17)
fc =
Vs
f l − f h (3.18)
C1
+
R1
R2
+
_
Vo
C2
_
Rx
Rx2
Ry
Ry2
Figure 3.11 A schematic diagram of a cascade band pass filter.
Magnitude ( Normalized)
The typical response of the band-pass filter pictured in figure 3.11 is shown below.
1.0
0.8
3dB
0.6
0.4
0.2
0.0
fl
fc
9
fh
Hz
Reference reading
1)
David Johns & Ken Martin. Analog integrated Circuit Design. John Wiley & Sons, Inc.,
New York, 1997.
2)
Paul R. Gray & Robert G. Meyer. Analysis and Design of Analog Integrated Circuits.
John Wiley & Sons, Inc., New York, 1993.
3)
P. Allen, B Blalock, & S. Milam. The Circuits and Filters Handbook , Chapter 75. CRC
Press Inc., 1995
4)
John Choma, Jr. EE348 lecture notes. University of Southern California. Spring 2001.
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Pre-lab Exercise
1)
You are to design three filters in HSpice using the information given above. Here are the specs for the
bands
a. The first band is a low pass filter that has 3-dB frequency at 1k Hz. The filter should have a low
frequency gain of 1 and a Q=0.5.
b.
The second band is a band-pass filter. It should have a center frequency of 3k Hz, a gain of 2 and
a Q=2.
c.
The third band is a high pass filter. It should have a 3dB frequency at 6k Hz, a gain of 1 and a
Q=0.5.
,
You may choose any R and C values you wish and use any methods that were presented here or that you
have learned in class to complete this pre-lab. Hint: Choose values of R and C that make your life easier by
simplifying the equations. Use known values of capacitors that are readily available in the lab, then solve
for the corresponding resistance values. Depending on what is available in your lab, matching a needed
resistance value can sometimes be a lot easier than finding a required capacitance value.
You will need to turn in a schematic diagram, all parameter calculations, and SPICE simulations to prove
your design works.
2)
Are there any differences between your calculations and your SPICE results? If there are, try to exp lain
why or give any deductive reasoning or insight? Tune or change any parameters to get with in ±10% of the
specifications.
3)
What were the main difficulties in this pre-lab? How did you solve or overcome them?
4)
Is there any potential stability problems with any of these filters that would limit their application or use?
(Hint: Is there any potential stability issues, etc., with any of the filters?) Under what conditions (or
parameter values) may the filter become under damped? (Hint: What is the value of the damping factor, ζ?)
5)
Choose a filter and purposely design it so it is under damped? How do you accomplish this? Simulate it in
SPICE and verify your conclusion in part 4. How does an under damped system behave compared to the
over damped or critically-damped system?
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Lab Exercise
1)
Build the filters you designed in the pre-lab. Sweep the frequency of a 100mV sine wave source and
verify that you filter meets the specifications by recording the magnitude of the output. A good strategy
might be to collect two data points every 1K HZ, then go back and collect data points at smaller intervals
at sudden transition points. This will save time and give you enough data to accurately plot your results
(using excel, or any other equivalent program) for you report. Did your results in lab match your HSpice
simulations within ±10%? Why or why not?
Lowpass
Bandpass
Frequency
2)
Volatge
Frequency
Highpass
Voltage
Frequency
Voltage
If the filter don’t meet the specifications given, then tune your circuit so that you filters perform within
±10%. Verify to your instructor that they work.
12