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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. - PowerPoint PPT Presentation
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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
Quality factor of an inductor
(Imax)
= =ω=→ =
Q=(ωL)/R
Please note that Qis also equal to Q=Im(Z)/Re(Z)
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=→ =
Q=ωCR
Please note that Qis also equal to Q=Im(Z)/Re(Z)
Z is the impedanceof parallel RC
Quality factor of a Capacitor in Series with a Resistor
Q=1/(ωCRS)
Please note that Qis also equal to Q=Im(Z)/Re(Z)
Z is the impedanceof series RC
Low-Pass RC Filter
High-Pass Filter
lpf= pf
𝐿=𝑅2𝐶
LPF+HPF
lpf= pf
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
𝑄=ω𝑜𝑅𝐶
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
𝑄=ω𝑜𝑅𝐶
Design Constraints
• Specs– Resonant Frequency: 2.4 GHz– RS=50 Ohms
– RL=Infinity
• List Q, C & L
𝑄=ω𝑜𝑅𝐶
ValuesQ 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
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 resistancereduces 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 circuitsIdentical at all frequencies, but we can makethese 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 OhmsLp=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 is14.96Ω/(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 ValuesQ 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 & UnknownsKnowns: • RS=50 Ohms• CT=13.26 pFUnknowns:• 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
