Report

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 Quality Factor Quality Factor Q is dimensionless Quality factor of an inductor (Imax) 2 = 2 2π 2π = ω= → = ω 2 2 2π π = = = 2 ω ω Q=(ωL)/R 2 = 2 Please note that Q is also equal to Q=Im(Z)/Re(Z) Quality factor of Parallel RL circuit Q=Im(Z)/Re(Z) || ω Z=+= +ω = ω(−ω) 2+ ω 2 Q=ωL(Rp)2/(ω2L2Rp)=Rp/ωL Quality factor of a Capacitor 2 = 2 2π 2π = ω= → = ω 2 2 π 2π = = = 2 ω ω Q=ωCR 2 = 2 Z is the impedance of parallel RC Please note that Q is also equal to Q=Im(Z)/Re(Z) 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) Low-Pass RC Filter High-Pass Filter ωlpf= ωhpf = 2 LPF+HPF ωlpf= ωhpf LPF+HPF (Magnified) Resistor Removed Design Intuition Circuit Quality Factor Q=3.162/(5.129-1.95)=0.99 Mathematical Analysis Transfer Function of a Bandpass Filter Resonant frequency Cutoff Frequency Bandwidth Calculation = ω 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 = ω Q=4.79 Q=0.49 Design Constraints • Specs – Resonant Frequency: 2.4 GHz – RS=50 Ohms – RL=Infinity • List Q, C & L = ω Values Q C L 0.5 0.663 pF 6.63 nH 1 1.326 pF 3.315 nH 10 13.26 pF 331.5 pH 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 Implement the Inductor http://www-smirc.stanford.edu/spiralCalc.html 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 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) Summary Series to Parallel Conversion for RL Circuits 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/(14.96 Ω+50 Ω)=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 Derivation Previous Design Values Q C L 0.5 0.663 pF 6.63 nH 1 1.326 pF 3.315 nH 10 13.26 pF 331.5 pH 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) C1/C2 R’S C1 C2 1 200 Ω 26.52 pF 26.52 pF 2 450Ω 39.78 pF 19.89 pF 3 800Ω 53.04 pF 17.68 pF Include the Effect of Parasitic Resistor