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Impedance Transformation

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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

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Quality Factor

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Quality Factor

Q is dimensionless

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Quality factor of an inductor

(Imax)

= =

=

2

=

=

= =

2

2

=

Q=(L)/R

Please note that Qis also equal to Q=Im(Z)/Re(Z)

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Quality factor of Parallel RL

circuit

Q=Im(Z)/Re(Z)

Z=||

+=

+=

()

+

Q=L(Rp)2/(2L2Rp)=Rp/L

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Quality factor of a Capacitor

= =

=

2

=

=

= =

2

2

=

Q=CR

Please note that Qis also equal to Q=Im(Z)/Re(Z)

Z is the impedanceof parallel RC

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Quality factor of a Capacitor in

Series with a Resistor

Q=1/(CRS)

Please note that Qis also equal to Q=Im(Z)/Re(Z)

Z is the impedanceof series RC

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Low-Pass RC Filter

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High-Pass Filter

lpf= hpf

= 2

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LPF+HPF

lpf= hpf

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LPF+HPF (Magnified)

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Resistor Removed

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Design Intuition

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Circuit Quality Factor

Q=3.162/(5.129-1.95)=0.99

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Mathematical Analysis

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Transfer Function of a Bandpass

Filter

Resonant frequency

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Cutoff Frequency

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Bandwidth Calculation

=

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Equivalent Circuit Approach

At resonant frequency, XP=1/(oCp)

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Effect of the Source Resistance

Q=3.162/(0.664)=4.76

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Effect of the Load Resistor

6 dB drop at resonance due tothe resistive divider.

Q=3.162/(7.762-1.318)=0.49

The loading will reduce the circuit Q.

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Summary

Q=0.99

Q=4.79

Q=0.49

=

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Design Constraints

Specs

Resonant Frequency: 2.4 GHz

RS=50 Ohms

RL=Infinity

List Q, C & L

=

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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

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Design Example

Q=2.4/(2.523-2.286)=10.12

BW=237 MHz

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Implement the Inductor

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http://www-smirc.stanford.edu/spiralCalc.html

http://www-smirc.stanford.edu/spiralCalc.htmlhttp://www-smirc.stanford.edu/spiralCalc.htmlhttp://www-smirc.stanford.edu/spiralCalc.htmlhttp://www-smirc.stanford.edu/spiralCalc.html

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Resistance of Inductor

R=Rsh(L/W)

Rshis the sheet resistance

Rsh is 22 mOhms per square for W=6um.

If the outer diameter is 135 um, the length isapproximately 135um x4=540 um.

R=22 mOhms x (540/6)=1.98 Ohms

Q=(L)/R=(22.4G0.336 nH)/1.98 =2.56

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Include Resistor In the Tank

Circuitry

Q=2.427/(3.076-1.888)=2.04

Inclusion of parasitic resistancereduces the circuit Q from 10.

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Series to Parallel Conversion

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Series to Parallel Conversion

We have anopen at DC!

We have resistor RPat DC!

It is NOT POSSIBLE to make these two circuitsIdentical at all frequencies, but we can makethese to exhibit approximate behavior at certain frequencies.

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Derivation

QS=QP

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RP

QS=1/(CSRS)

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Cp

QS=1/(CSRS)

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Summary

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Series to Parallel Conversion for

RL Circuits

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Resistance of Inductor

R=Rsh(L/W)

Rshis the sheet resistance

Rsh is 22 mOhms per square for W=6um.

If the outer diameter is 135 um, the length isapproximately 135um x4=540 um.

R=22 mOhms x (540/6)=1.98 Ohms

Q=(L)/R=(22.4G0.336 nH)/1.98 =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

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Insertion Loss Due to Inductor

Resistance

At resonant frequency, voltage divider ratio is14.96/(14.96 +50 )=0.2303

Convert to loss in dB, 20log10(0.23)=-12.75 dB

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Use Tapped-C Circuit to Fool the

Tank into Thinking It Has High RS

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Derivation

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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 R

L

=Infinity

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Design Problem

Knowns & UnknownsKnowns: RS=50 Ohms

CT=13.26 pFUnknowns:

C1/C2 RS

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Calculations

CT=C1/(1+C1/C2)

C1=CT(1+C1/C2)

C1/C2 RS C1 C2

1 200 26.52 pF 26.52 pF

2 450 39.78 pF 19.89 pF

3 800 53.04 pF 17.68 pF

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I l d th Eff t f P iti

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Include the Effect of Parasitic

Resistor