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

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 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