11. Active Filters and Tuned Circuits
11. Active Filters and Tuned Circuits
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11.1 Active Low - Pass Filters
Butterworth Transfer Function
Active filters: circuits, containing resistors,
capacitors and Op Amps. They are intended for
separating of signals having different
frequencies.
H(f ) =
H0
1+ ( f / f b )
2n
(11.1)
fb – break (cut-off) frequency.
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 gainbandwidth 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.
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.
fb =
1
2π RC
11. Active Filters and Tuned Circuits
(11.2)
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Solution:
Example 11.1 Fourth - Order Low - Pass
Butterworth Filter Design
Design a fourth - order low - pass
Butterworth filter with a cut - off frequency
of 100 Hz. Use the LF411 op amp.
We choose C1=C2=C11=C12=0.1µF and from (11.2)
R1 = R2 = R11 = R12 =
1
1
=
= 15.8kΩ
2π f bC 2π ×100 × 0.1× 10 −6
From table 11.1 K=1.152; K1=2.235. We choose
R4=R14=10kΩ and
R3 = (K − 1)R4 = (1.152 − 1)×10 ×103 = 1.52kΩ
R13 = (K − 1)R14 = (2.235 − 1)×10 ×103 = 12.35kΩ
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 → fb2/f.
H hp ( f ) =
H0
1+ ( f b / f )
2n
(11.3)
Figure 11.8 Sallen-Key high-pass active-filter section.
Figure 11.7 Normalized high-pass Butterworth transfer
functions.
11. Active Filters and Tuned Circuits
Figure 11.9 Bode magnitude plot for the high-pass filter.
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11.3 Active Bandpass Filters
Delyiannis - Friend Bandpass Circuits
f0 =
1
2π C (R1 || R2 )R3
R3
2R1
(11.7)
1
π R3C
(11.8)
H0 =
B=
(11.6)
1/ 2
f 0 1  R3 

= 
Q=
B 2  R1 || R2 
R3 =
Q
π f 0C
(11.10)
R3
2H 0
(11.11)
R3
4Q 2 − 2 H 0
(11.12)
R1 =
R2 =
(11.9)
11. Active Filters and Tuned Circuits
Figure 11.12 Second-order bandpass filter.
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R1 must be large to have high input impedance.
Example 11.3 Bandpass Filter Design
Design a bandpass filter with f0=1 kHz,
B=200 Hz, and H0=10.
From (11.14) R3 = 20R1 and R1 must have
moderate value to keep the value of R3 realistic.
The choice R1 = 10kΩ gives
R3 = 20 R1 = 20 × 10 × 103 = 200kΩ
Solution:
(
R3 200 × 103
R2 =
=
= 2.5kΩ
80
80
)
Q = f 0 B = 1×103 200 = 5
R3 =
Q
1.592 ×10
=
π f 0C
C
(11.13)
R3
R
= 3
2 H 0 20
(11.14)
R3
R3
=
4Q 2 − 2 H 0 80
(11.15)
R1 =
R2 =
−3
1.592 × 10 −3 1.592 × 10 −3
=
= 7.96nF
C=
200 × 103
R3
We select C = 8.2nF and then calculate again the
values of R1, R2 and R3.
Figure 11.13 Bandpass filter designed in Example 11.3.
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11.4 The Series Resonant Circuit
Resonant Frequency and Quality Factor
Resonant frequency
1
LC
ω0 =
f0 =
Figure 11.17 Series resonant circuit.
I=
Vi
jω L + 1 / ( jω C ) + R
(11.16)
Vo = RI
(11.17)
jω RVi
Vo =
− ω 2 L + jω R + 1 / C
Av ( jω ) =
Vo
jω R
=
Vi − ω 2 L + jω R + 1 / C
11. Active Filters and Tuned Circuits
(11.20)
1
(11.21)
2π LC
Quality factor
Q=
ω0 L
(11.22)
R
At the resonance
(11.18)
(11.19)
ω0 L =
1
ω 0C
(11.23)
Another formula for the quality factor
1
Q=
ω 0CR
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(11.24)
8
Voltage transfer ratio, expressed with Q-factor
and resonant frequency
Av ( jω ) =
j (ω / ω 0 )
2
Q 1 − (ω / ω 0 ) + j (ω / ω 0 )
[
]
(11.25)
|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.
Figure 11.18 Voltage transfer function for the series resonant circuit.
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Circuit Bandwidth
B = fH − fL =
f H f L = f 02
f0
Q
(11.28)
(11.29)
If B is relatively small, compared with f0, then
f L ≅ f0 − B / 2
(11.30)
f H ≅ f0 + B / 2
(11.31)
Figure 11.19 Normalized impedance of the series resonant
circuit.
Figure 11.20 Bandwidth and half-power frequencies
for 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:
f0 =
1
2π LC
=
1
= 7.12MHz
2π 5 × 10 ×100 ×10
2πf 0 L 2π × 7.12 × 106 × 5 ×10 −6
=
= 22.4
Q=
R
10
f 0 7.12 × 106
=
= 318kHz
B=
Q
22.4
−6
−12
B
318 ×103
6
= 6.96MHz
f L ≈ f 0 − = 7.12 ×10 −
2
2
318 ×103
B
6
= 7.28MHz
f H ≈ f 0 + = 7.12 × 10 +
2
2
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11.5 The Parallel Resonant Circuit
ω0 L =
1
ω 0C
Q = ω 0 RC
Z ( jω ) = R
Figure 11.25 Parallel resonant circuit.
Z ( jω ) =
1
1 / R + 1 / ( jωL ) + jωC
f0 =
1
2π LC
(11.32)
(11.33)
(11.35)
j (ω / ω 0 )
2
Q 1 − (ω / ω 0 ) + j (ω / ω 0 )
[
]
B = fH − fL =
(11.36)
f0
Q
f H f L = f 02
f L ≅ f0 − B / 2
Q=
R
ω0 L
11. Active Filters and Tuned Circuits
(11.34)
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f H ≅ f0 + B / 2
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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.
Vo = − g mVgs Z ( jω )
Av =
11. Active Filters and Tuned Circuits
Vo
= − g m Z ( jω )
Vin
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(11.57)
(11.58)
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Neutralization
Input Impedance
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.
11. Active Filters and Tuned Circuits
Figure 11.48 Cneut is used to cancel the feedback through Cgd.
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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

1 
 V1 − jω CV2 = 0
 jω C − j
ω L1 


(g m − jω C )V1 +  jω C − j

(11.62)
1
1
+  V2 = 0
ω L2 R 
(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:

1 
 jω C − j

L
ω
1

Figure 11.50 Hartley oscillator.
V1
+ jω C (V1 − V2 ) = 0
jω L1
V
V
g mV1 + jω C (V2 − V1 ) + 2 + 2 = 0
jω L2 R
11. Active Filters and Tuned Circuits
( g m − jω C )
(11.60)
(11.61)
(− jωC )

1
1
 jω C − j
+ 
ω L2 R 

C C
1 
 + − 2
 +
 L1 L2 ω L1 L2 
=0
(11.64)
 ωC

1
−
+ ωCg m  = 0
j 
 R ωRL1

(11.66)
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C C
1 
 + − 2
 +
 L1 L2 ω L1 L2 
 ωC

1
−
+ ωCg m  = 0
j 
 R ωRL1

(11.66)
From zeroing of the real part of (11.66):
ω=
1
C (L1 + L2 )
(11.68)
From zeroing of the imaginary part of (11.66):
gm =
11. Active Filters and Tuned Circuits
L2
RL1
(11.71)
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11.10 Crystal - Controlled Oscillators
The 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 11.53a Colpitts oscillator.
11. Active Filters and Tuned Circuits
Figure 11.58 Pierce oscillator results from the Colpitts
oscillator (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.
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