Impedance Transformation
Topics
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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
𝐼𝑚𝑎𝑥
𝑅
𝑃𝑑𝑖𝑠𝑠 =
2
2π
2π
𝑑𝐼
𝐿 𝑑𝑡 𝐼 𝑑𝑡=
ω= 𝑇 → 𝑇 = ω
2
2
𝐼𝑚𝑎𝑥
𝑅 2π 𝐼𝑚𝑎𝑥
𝑅π
𝐸𝑑𝑖𝑠𝑠 = 𝑃𝑑𝑖𝑠𝑠𝑇 =
=
2 ω
ω
Q=(ωL)/R
2
𝐿𝐼𝑚𝑎𝑥
𝐿𝐼 𝑑𝐼= 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)
𝑅 ||𝑠𝐿
𝑅 𝑗ω𝐿
Z=𝑅𝑃+𝑠𝐿=𝑅 𝑝+𝑗ω𝐿 =
𝑝
𝑝
𝑅𝑝𝑗ω𝐿(𝑅𝑝−𝑗ω𝐿)
𝑅𝑝 2+ ω𝐿 2
Q=ωL(Rp)2/(ω2L2Rp)=Rp/ωL
Quality factor of a Capacitor
2
𝑣𝑚𝑎𝑥
𝑃𝑑𝑖𝑠𝑠 =
2𝑅
2π
2π
𝑑𝑣
𝐶 𝑑𝑡 𝑣 𝑑𝑡=
ω= 𝑇 → 𝑇 = ω
2
2 π
𝑣𝑚𝑎𝑥
2π 𝑣𝑚𝑎𝑥
𝐸𝑑𝑖𝑠𝑠 = 𝑃𝑑𝑖𝑠𝑠𝑇 =
=
2𝑅 ω
ω𝑅
Q=ωCR
2
𝐶𝑣𝑚𝑎𝑥
𝐶𝑣 𝑑𝑣= 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
𝐿 = 𝑅2𝐶
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
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 = ω𝑜𝑅𝐶
Q=4.79
Q=0.49
Design Constraints
• Specs
– Resonant Frequency: 2.4 GHz
– RS=50 Ohms
– RL=Infinity
• List Q, C & L
𝑄 = ω𝑜𝑅𝐶
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