11. active filters and tuned circuits · 11. active filters and tuned circuits tlt-8016 basic...
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11. Active Filters and Tuned Circuits TLT-8016 Basic Analog Circuits 2005/2006 1
11. Active Filters and Tuned Circuits
11. Active Filters and Tuned Circuits TLT-8016 Basic Analog Circuits 2005/2006 2
11.1 Active Low - Pass FiltersButterworth Transfer Function
( )( ) n
bff
HfH2
0
/1+= (11.1)Active filters: circuits, containing resistors,
capacitors and Op Amps. They are intended for separating of signals having different frequencies.
Requirements for filters:1. Contain few components.2. Have a transfer function that is insensitive to component tolerances.3. Place modest demands on the op amp’s gain-bandwidth product, output impedance, slew rate, and other specifications.4. Be easily adjusted. 5. Require a small spread of component values.6. Allow a wide of useful transfer functions to be realized.
fb – break (cut-off) frequency.
Figure 11.1 Transfer function magnitude versus frequency for low-pass Butterworth filters.
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Sallen - Key Circuits
Figure 11.2 Equal-component Sallen-Key low-pass active-filter section.
RCfb π2
1= (11.2)
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Solution:We choose C1=C2=C11=C12=0.1µF and from (11.2)
Example 11.1 Fourth - Order Low - Pass Butterworth Filter DesignDesign a fourth - order low - pass Butterworth filter with a cut - off frequency of 100 Hz. Use the LF411 op amp.
kΩ8.15101.01002
12
16121121 =
×××===== −ππ Cf
RRRRb
From table 11.1 K=1.152; K1=2.235. We choose R4=R14=10kΩ and
( ) ( )( ) ( ) kΩ35.1210101235.21
kΩ52.110101152.113
1413
343
=××−=−=
=××−=−=
RKR
RKR
Figure 11.3 Fourth-order Butterworth low-pass filter.
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11.2 Active High - Pass Filters
Transformation of low-pass transfer function to high-pass transfer function: f → fb
2/f.
( )( ) n
b
hpf/f
HfH2
0
1+= (11.3)
Figure 11.8 Sallen-Key high-pass active-filter section.
Figure 11.7 Normalized high-pass Butterworth transfer functions. Figure 11.9 Bode magnitude plot for the high-pass filter.
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11.3 Active Bandpass Filters
Delyiannis - Friend Bandpass Circuits
( ) 3210 2
1RR||RC
fπ
= (11.6)
1
30 2R
RH = (11.7)
CRB
3
1π
= (11.8)
21
21
30
21
/
R||RR
BfQ
== (11.9)
CfQR
03 π= (11.10)
0
31 2H
RR =
02
32 24 HQ
RR−
=
(11.11)
(11.12)Figure 11.12 Second-order bandpass filter.
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R1 must be large to have high input impedance.
From (11.14) R3 = 20R1 and R1 must have moderate value to keep the value of R3 realistic.
The choice R1 = 10kΩ gives
Example 11.3 Bandpass Filter DesignDesign a bandpass filter with f0=1 kHz, B=200 Hz, and H0=10.
Solution: kΩ20010102020 313 =××== RR
kΩ5.280
1020080
33
2 =×
==RR( ) 5200101 3
0 =×== BfQ
nF96.71020010592.110592.1
3
3
3
3
=××
=×
=−−
RC
C.
CfQR
3
03
105921 −×==
π(11.13)
We select C = 8.2nF and then calculate again the values of R1, R2 and R3.
2023
0
31
RHRR == (11.14)
80243
02
32
RHQ
RR =−
= (11.15)
Figure 11.13 Bandpass filter designed in Example 11.3.
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11.4 The Series Resonant CircuitResonant Frequency and Quality Factor
LC1
0 =ω
Resonant frequency
(11.20)
LCf
π21
0 = (11.21)Figure 11.17 Series resonant circuit.
( ) RCj/LjVI i
++=
ωω 1Quality factor(11.16)
RLQ 0ω
= (11.22)(11.17)RIVo =
At the resonance
C/RjLRVjV i
o 12 ++−=
ωωω
CL
00
1ω
ω =(11.18) (11.23)
( )C/RjL
RjVVjA
i
ov 12 ++−
==ωωωω Another formula for the quality factor(11.19)
CRQ
0
1ω
= (11.24)
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|Av| = 0 if ω = 0; |Av| → 0 when ω→ ∞. For frequencies between 0 and ∞ |Av| > 0 and has a maximum at ω = ω0.
The circuit is a bandpass filter.
Voltage transfer ratio, expressed with Q-factor and resonant frequency
( ) ( )( )[ ] ( )0
20
0
1 ωωωωωωω
/j/Q/jjAv +−
= (11.25)
Figure 11.18 Voltage transfer function for the series resonant circuit.
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Circuit Bandwidth
QfffB LH
0=−=2
0fff LH =
(11.28)
(11.29)
If B is relatively small, compared with f0, then
(11.30)20 /BffL −≅
(11.31)20 /BffH +≅
Figure 11.20 Bandwidth and half-power frequencies for the series resonant circuit.
Figure 11.19 Normalized impedance of the series resonant circuit.
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Exercise 11.7 Find resonant frequency, Q, the bandwidth, and half - power frequencies of a series resonance circuit having L=5 µH, C=100 pF, and R=10 Ω.
Solution:
MHz28.72
103181012.72
MHz96.62
103181012.72
kHz3184.221012.7
4.2210
1051012.722
MHz12.7101001052
12
1
36
0
36
0
60
660
1260
=×
+×=+≈
=×
−×=−≈
=×
==
=××××
==
=×××
==
−
−−
Bff
Bff
QfB
RLfQ
LCf
H
L
ππππ
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11.5 The Parallel Resonant Circuit
CL
00
1ω
ω =
(11.35)RCQ 0ω=
( ) ( )( )[ ] ( )0
20
0
//1/
ωωωωωωω
jQjRjZ
+−= (11.36)Figure 11.25 Parallel resonant circuit.
( ) ( ) CjLj/R/jZ
ωωω
++=
111
(11.32)
QfffB LH
0=−=
20fff LH =
LCf
π21
0 = (11.33)
20 /BffL −≅
LRQ0ω
= (11.34) 20 /BffH +≅
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Figure 11.26 Normalized impedance of the parallel resonant circuit.
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11.8 Tuned Amplifiers
Tuned amplifiers have resonant circuits at their inputs or outputs or both and they amplify in a narrow bandwidth.
Figure 11.43 Tuned amplifier.
( )ωjZVgV gsmo −= (11.57)
( )ωjZgVVA m
in
ov −== (11.58)
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NeutralizationInput Impedance
Figure 11.48 Cneut is used to cancel the feedback through Cgd.Figure 11.47 Parallel input resistance and reactance for the tuned amplifier of Figure 11.44. Pay attention to the negative
real part at frequencies below the resonant frequency.
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11.9 LC Oscillators
The Hartley Oscillators
Figure 11.50 Hartley oscillator.
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Frequency of Oscillation and the Minimum Transconductance Requirement
0121
1
=−
− CVjV
LjCj ωω
ω (11.62)
( ) 0112
21 =
+−+− V
RLjCjVCjgm ω
ωω (11.63)
(11.62) and (11.63) form a homogeneous set of linear equations concerning V1 and V2. The solution is nonzero if the system determinant is zero:
( )
( )0
11
1
2
1 =
+−−
−
−
RLjCjCjg
CjL
jCj
m ωωω
ωω
ω(11.64)Figure 11.50 Hartley oscillator.
( ) 0211
1 =−+ VVCjLj
V ωω
(11.60)
011
1212
21
=
+−+
−+ mCg
RLRCj
LLLC
LC ω
ωω
ω
(11.66)( ) 02
2
2121 =++−+
RV
LjVVVCjVgm ω
ω (11.61)
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011
1212
21
=
+−+
−+ mCg
RLRCj
LLLC
LC ω
ωω
ω
(11.66)
From zeroing of the real part of (11.66):
( )21
1LLC +
=ω (11.68)
From zeroing of the imaginary part of (11.66):
1
2
RLLgm = (11.71)
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11.10 Crystal - Controlled OscillatorsThe Piezoelectric Effect
Piezoelectric effect: if we apply an electric field to the plates, forces on the ions in the lattice deform the material.
Piezoelectric effect is reciprocal: if deform the crystal, a voltage appears between the plates.
When used as frequency-determining element, quartz crystal vibrates freely at the desired frequency. The mechanical vibrations result in an ac current in the external circuit.
Figure 11.54 Crystal.
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The Equivalent Circuit of the Crystal
Table 11.2 Parameters of a typical 10-MHz Crystal.
Figure 11.56 Equivalent circuit for a crystal.
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Figure 11.57 Crystal reactance versus frequency.
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Crystal Oscillator Circuits
Figure 11.58 Pierce oscillator results from the Colpittsoscillator (Figure 11.53) if the inductor is replaced by a crystal. The dc blocking capacitor C3 shown in Figure 11.53 can be omitted because the crystal is an open circuit for dc.
Figure 11.53a Colpitts oscillator.