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Impedance Transformation. Topics. Quality Factor Series to parallel conversion Low-pass RC High-pass RL Bandpass Loaded Q Impedance Transformation Coupled Resonant Circuit Recent implementation, if time permits. Quality Factor. Quality Factor. Q is dimensionless.
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Topics • Quality Factor • Series to parallel conversion • Low-pass RC • High-pass RL • Bandpass • Loaded Q • Impedance Transformation • Coupled Resonant Circuit • Recent implementation, if time permits
Quality Factor Q is dimensionless
Quality factor of an inductor (Imax) ω=→= == Please note that Q is also equal to Q=Im(Z)/Re(Z) Q=(ωL)/R
Quality factor of Parallel RL circuit Q=Im(Z)/Re(Z) Z== Q=ωL(Rp)2/(ω2L2Rp)=Rp/ωL
Quality factor of a Capacitor ω=→= == Z is the impedance of parallel RC Please note that Q is also equal to Q=Im(Z)/Re(Z) Q=ωCR
Quality factor of a Capacitor in Series with a Resistor Z is the impedance of series RC Please note that Q is also equal to Q=Im(Z)/Re(Z) Q=1/(ωCRS)
High-Pass Filter lpf=pf
LPF+HPF lpf=pf
Circuit Quality Factor Q=3.162/(5.129-1.95)=0.99
Transfer Function of a Bandpass Filter Resonant frequency
Equivalent Circuit Approach At resonant frequency, XP=1/(ωoCp)
Effect of the Source Resistance Q=3.162/(0.664)=4.76
Effect of the Load Resistor 6 dB drop at resonance due to the resistive divider. Q=3.162/(7.762-1.318)=0.49 The loading will reduce the circuit Q.
Summary Q=0.99 Q=4.79 Q=0.49
Design Constraints • Specs • Resonant Frequency: 2.4 GHz • RS=50 Ohms • RL=Infinity • List Q, C & L
Values • Specs: • Resonant Frequency: 2.4 GHz • RS=50 Ohms • RL=Infinity
Design Example Q=2.4/(2.523-2.286)=10.12 BW=237 MHz
Resistance of Inductor • R=Rsh(L/W) • Rsh is the sheet resistance • Rsh is 22 mOhms per square for W=6um. • If the outer diameter is 135 um, the length is approximately 135um x4=540 um. • R=22 mOhms x (540/6)=1.98 Ohms Q=(ωL)/R=(2π2.4G0.336 nH)/1.98 Ω=2.56
Include Resistor In the Tank Circuitry Q=2.427/(3.076-1.888)=2.04 Inclusion of parasitic resistance reduces the circuit Q from 10.
Series to Parallel Conversion We have an open at DC! We have resistor RP at DC! It is NOT POSSIBLE to make these two circuits Identical at all frequencies, but we can make these to exhibit approximate behavior at certain frequencies.
Derivation QS=QP
RP QS=1/(ωCSRS)
Cp QS=1/(ωCSRS)
Resistance of Inductor • R=Rsh(L/W) • Rsh is the sheet resistance • Rsh is 22 mOhms per square for W=6um. • If the outer diameter is 135 um, the length is approximately 135um x4=540 um. • R=22 mOhms x (540/6)=1.98 Ohms Q=(ωL)/R=(2π2.4G0.336 nH)/1.98 Ω=2.56 Rp=RS(1+QSQS)=1.98 Ohms(1+2.56x2.56)=14.96 Ohms Lp=LS(1+1/(QSQS))=331.5 pH(1+1/2.56/2.56)=382.08 nH
Insertion Loss Due to Inductor Resistance At resonant frequency, voltage divider ratio is 14.96Ω/(14.96Ω+50Ω)=0.2303 Convert to loss in dB, 20log10(0.23)=-12.75 dB
Use Tapped-C Circuit to Fool the Tank into Thinking It Has High RS
Previous Design Values • Specs: • Resonant Frequency: 2.4 GHz • RS=50 Ohms • RL=Infinity
Design Problem • Knowns & Unknowns • Knowns: • RS=50 Ohms • CT=13.26 pF • Unknowns: • C1/C2 • R’S
Calculations • CT=C1/(1+C1/C2) • C1=CT(1+C1/C2)