Lecture 14
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EE 359 Electronic Circuits
Filters and Tuned Amplifiers Second order LCR resonator Second order Active Filters based on Inductor Replacement
Signal Generators Oscillators
October 19, 2007
1
EE 359 Electronic Circuits
HW
Due today 12.7, 12.9,12.28, 12.34 Review October 26 Quiz#2 October 29
October 19, 2007
2
EE 359 Electronic Circuits
Second Order LCR Resonator
• LCR Resonator V0 1 1 = = I Y 1 / sL + sC + 1 / R s/C = 2 s + s (1 / CR ) + 1 / LC
ω 02 = 1 / LC ω 0 / Q = 1 / CR ω 0 = 1 / LC Q = ω 0CR
• Excite parallel LCR resonator – Determine natural modes (I.e. poles of response function)
October 19, 2007
3
EE 359 Electronic Circuits
Transmission Zeros LPF
• Where to inject iput signal Vi ? T ( s) =
V0 ( s ) Z 2 ( s) = Vi ( s) Z1 ( s) + Z 2 ( s )
• Transmission zeros are when Z2(s) are zero (if Z1
October 19, 2007
4
EE 359 Electronic Circuits
Realization of Low Pass Function
• Transmission zeros are when series impedance becomes infinite. 1 sL → ∞ ⇒ s → ∞ ⇒ →0 sC + 1 / R V0 Z2 Y1 1 / sL T ( s) = = = = Vi Z1 + Z 2 Y1 + Y2 1 / sL + sC + 1 / R
October 19, 2007
5
EE 359 Electronic Circuits
High Pass Filter
• Use the series capacitor. Transmission zero at s=0 (dc), shunt inductor is a tranmission zero at s=0. V0 a2 s 2 T ( s) = = 2 Vi s + s( ω 0 / Q ) + ω 02
October 19, 2007
6
EE 359 Electronic Circuits
Bandpass Filter
• Zero at s=0 due to shunt inductor, s=infinity due to shunt capacitor. T ( s) =
V0 s (1 / CR ) = 2 Vi s + s (1 / CR ) + 1 / LC
October 19, 2007
7
EE 359 Electronic Circuits
Inductance Simulation
• Use opamps instead of inductor!
October 19, 2007
8
EE 359 Electronic Circuits
Low Pass Filter
V0 ( s ) Z 2 ( s) T ( s) = = Vi ( s ) Z1 ( s ) + Z 2 ( s )
V0 Z2 Y1 1 / sL T ( s) = = = = Vi Z1 + Z 2 Y1 + Y2 1 / sL + sC + 1 / R October 19, 2007
9
EE 359 Electronic Circuits
High Pass Filter
•
V0 a2 s 2 T ( s) = = 2 Vi s + s( ω 0 / Q ) + ω 02
October 19, 2007
10
EE 359 Electronic Circuits
Bandpass Filter
•
T ( s) =
October 19, 2007
V0 s (1 / CR ) = 2 Vi s + s (1 / CR ) + 1 / LC 11
EE 359 Electronic Circuits
Low Pass Notch filter
•
October 19, 2007
12
EE 359 Electronic Circuits Second-order narrow-band bandpass filter by transforming a Second-order narrow-band first-order low-pass filter bandpass filter by transforming a first-order low-pass filter. (a) Pole of the first-order filter in the s-plane. (b) Applying the transformation s = p + jω0 and adding a complex conjugate pole results in the poles of the secondorder bandpass filter. (c) Magnitude response of the firs-order low-pass filter. (d) Magnitude response of the second-order bandpass filter.
October 19, 2007
13
EE 359 Electronic Circuits Poles and the frequency response of a fourth-order stagger-tuned narrow-band bandpass amplifier by transforming a second-order lowpass .
October 19, 2007
14
EE 359 Electronic Circuits
SIGNAL GENERATORS /OSCILLATORS
A positive-feedback loop is formed by an amplifier and a frequency-selective network In an actual oscillator circuit, no input signal will be present
October 19, 2007
15
EE 359 Electronic Circuits
Oscillator-frequency stability
October 19, 2007
16
EE 359 Electronic Circuits
Limiter Ckt Comparator
October 19, 2007
17
EE 359 Electronic Circuits
Wien-bridge oscillator
without amplitude stabilization .
October 19, 2007
18
EE 359 Electronic Circuits
Wien bridge w/ Amp. Stabil.
limiter used for amplitude control.
October 19, 2007
19
EE 359 Electronic Circuits
Alternate Wien bridge stabil.
October 19, 2007
20
EE 359 Electronic Circuits
Phase Shift Oscillator
October 19, 2007
21
EE 359 Electronic Circuits
Phase Shift. Osc. W/ Stabil.
October 19, 2007
22
EE 359 Electronic Circuits
Quad Osc. Circuit
October 19, 2007
23
EE 359 Electronic Circuits
Active Tuned Osc.
October 19, 2007
24
EE 359 Electronic Circuits
OPAMP based Tuned Amp. Osc.
October 19, 2007
25
EE 359 Electronic Circuits
Colpitts and Hartley Oscillators
October 19, 2007
26
EE 359 Electronic Circuits
Equiv. Ckt
To simplify the analysis, neglect Cm and rp Consider Cp to be part of C2, and include ro in R.
October 19, 2007
27
EE 359 Electronic Circuits
Collpits Oscillator
October 19, 2007
28
EE 359 Electronic Circuits
Piezzoelectric Crystal
October 19, 2007
29
EE 359 Electronic Circuits
Pierce Oscillator
CMOS inverter as an amplifier.
October 19, 2007
30
EE 359 Electronic Circuits
Bistable Operation
October 19, 2007
31
Filters and Tuned Amplifiers Second order LCR resonator Second order Active Filters based on Inductor Replacement
Signal Generators Oscillators
October 19, 2007
1
EE 359 Electronic Circuits
HW
Due today 12.7, 12.9,12.28, 12.34 Review October 26 Quiz#2 October 29
October 19, 2007
2
EE 359 Electronic Circuits
Second Order LCR Resonator
• LCR Resonator V0 1 1 = = I Y 1 / sL + sC + 1 / R s/C = 2 s + s (1 / CR ) + 1 / LC
ω 02 = 1 / LC ω 0 / Q = 1 / CR ω 0 = 1 / LC Q = ω 0CR
• Excite parallel LCR resonator – Determine natural modes (I.e. poles of response function)
October 19, 2007
3
EE 359 Electronic Circuits
Transmission Zeros LPF
• Where to inject iput signal Vi ? T ( s) =
V0 ( s ) Z 2 ( s) = Vi ( s) Z1 ( s) + Z 2 ( s )
• Transmission zeros are when Z2(s) are zero (if Z1
October 19, 2007
4
EE 359 Electronic Circuits
Realization of Low Pass Function
• Transmission zeros are when series impedance becomes infinite. 1 sL → ∞ ⇒ s → ∞ ⇒ →0 sC + 1 / R V0 Z2 Y1 1 / sL T ( s) = = = = Vi Z1 + Z 2 Y1 + Y2 1 / sL + sC + 1 / R
October 19, 2007
5
EE 359 Electronic Circuits
High Pass Filter
• Use the series capacitor. Transmission zero at s=0 (dc), shunt inductor is a tranmission zero at s=0. V0 a2 s 2 T ( s) = = 2 Vi s + s( ω 0 / Q ) + ω 02
October 19, 2007
6
EE 359 Electronic Circuits
Bandpass Filter
• Zero at s=0 due to shunt inductor, s=infinity due to shunt capacitor. T ( s) =
V0 s (1 / CR ) = 2 Vi s + s (1 / CR ) + 1 / LC
October 19, 2007
7
EE 359 Electronic Circuits
Inductance Simulation
• Use opamps instead of inductor!
October 19, 2007
8
EE 359 Electronic Circuits
Low Pass Filter
V0 ( s ) Z 2 ( s) T ( s) = = Vi ( s ) Z1 ( s ) + Z 2 ( s )
V0 Z2 Y1 1 / sL T ( s) = = = = Vi Z1 + Z 2 Y1 + Y2 1 / sL + sC + 1 / R October 19, 2007
9
EE 359 Electronic Circuits
High Pass Filter
•
V0 a2 s 2 T ( s) = = 2 Vi s + s( ω 0 / Q ) + ω 02
October 19, 2007
10
EE 359 Electronic Circuits
Bandpass Filter
•
T ( s) =
October 19, 2007
V0 s (1 / CR ) = 2 Vi s + s (1 / CR ) + 1 / LC 11
EE 359 Electronic Circuits
Low Pass Notch filter
•
October 19, 2007
12
EE 359 Electronic Circuits Second-order narrow-band bandpass filter by transforming a Second-order narrow-band first-order low-pass filter bandpass filter by transforming a first-order low-pass filter. (a) Pole of the first-order filter in the s-plane. (b) Applying the transformation s = p + jω0 and adding a complex conjugate pole results in the poles of the secondorder bandpass filter. (c) Magnitude response of the firs-order low-pass filter. (d) Magnitude response of the second-order bandpass filter.
October 19, 2007
13
EE 359 Electronic Circuits Poles and the frequency response of a fourth-order stagger-tuned narrow-band bandpass amplifier by transforming a second-order lowpass .
October 19, 2007
14
EE 359 Electronic Circuits
SIGNAL GENERATORS /OSCILLATORS
A positive-feedback loop is formed by an amplifier and a frequency-selective network In an actual oscillator circuit, no input signal will be present
October 19, 2007
15
EE 359 Electronic Circuits
Oscillator-frequency stability
October 19, 2007
16
EE 359 Electronic Circuits
Limiter Ckt Comparator
October 19, 2007
17
EE 359 Electronic Circuits
Wien-bridge oscillator
without amplitude stabilization .
October 19, 2007
18
EE 359 Electronic Circuits
Wien bridge w/ Amp. Stabil.
limiter used for amplitude control.
October 19, 2007
19
EE 359 Electronic Circuits
Alternate Wien bridge stabil.
October 19, 2007
20
EE 359 Electronic Circuits
Phase Shift Oscillator
October 19, 2007
21
EE 359 Electronic Circuits
Phase Shift. Osc. W/ Stabil.
October 19, 2007
22
EE 359 Electronic Circuits
Quad Osc. Circuit
October 19, 2007
23
EE 359 Electronic Circuits
Active Tuned Osc.
October 19, 2007
24
EE 359 Electronic Circuits
OPAMP based Tuned Amp. Osc.
October 19, 2007
25
EE 359 Electronic Circuits
Colpitts and Hartley Oscillators
October 19, 2007
26
EE 359 Electronic Circuits
Equiv. Ckt
To simplify the analysis, neglect Cm and rp Consider Cp to be part of C2, and include ro in R.
October 19, 2007
27
EE 359 Electronic Circuits
Collpits Oscillator
October 19, 2007
28
EE 359 Electronic Circuits
Piezzoelectric Crystal
October 19, 2007
29
EE 359 Electronic Circuits
Pierce Oscillator
CMOS inverter as an amplifier.
October 19, 2007
30
EE 359 Electronic Circuits
Bistable Operation
October 19, 2007
31
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