Download presentation

Presentation is loading. Please wait.

Published byEmanuel Collman Modified over 3 years ago

1
Impedance Transformation

2
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

3
Quality Factor

4
Q is dimensionless

5
Quality factor of an inductor (I max ) Q=(ωL)/R Please note that Q is also equal to Q=Im(Z)/Re(Z)

6
Quality factor of Parallel RL circuit Q=Im(Z)/Re(Z) Q=ωL(R p ) 2 /(ω 2 L 2 R p )=R p /ωL

7
Quality factor of a Capacitor Q=ωCR Please note that Q is also equal to Q=Im(Z)/Re(Z) Z is the impedance of parallel RC

8
Quality factor of a Capacitor in Series with a Resistor Q=1/(ωCR S ) Please note that Q is also equal to Q=Im(Z)/Re(Z) Z is the impedance of series RC

9
Low-Pass RC Filter

10
High-Pass Filter

11
LPF+HPF

12
LPF+HPF (Magnified)

13
Resistor Removed

14
Design Intuition

15
Circuit Quality Factor Q=3.162/(5.129-1.95)=0.99

16
Mathematical Analysis

17
Transfer Function of a Bandpass Filter Resonant frequency

18
Cutoff Frequency

19
Bandwidth Calculation

20
Equivalent Circuit Approach At resonant frequency, XP=1/(ω o C p )

21
Effect of the Source Resistance Q=3.162/(0.664)=4.76

22
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.

23
Summary Q=0.99 Q=4.79 Q=0.49

24
Design Constraints Specs – Resonant Frequency: 2.4 GHz – R S =50 Ohms – R L =Infinity List Q, C & L

25
Values QCL 0.50.663 pF6.63 nH 11.326 pF3.315 nH 1013.26 pF331.5 pH Specs: Resonant Frequency: 2.4 GHz R S =50 Ohms R L =Infinity

26
Design Example Q=2.4/(2.523-2.286)=10.12 BW=237 MHz

27
Implement the Inductor

28
http://www-smirc.stanford.edu/spiralCalc.html

29
Resistance of Inductor R=R sh (L/W) – R sh 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

30
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.

31
Series to Parallel Conversion

32
We have an open at DC! We have resistor R P 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.

33
Derivation Q S =Q P

34
RPRP Q S =1/(ωC S R S )

35
Cp Q S =1/(ωC S R S )

36
Summary

37
Series to Parallel Conversion for RL Circuits

38
Resistance of Inductor R=R sh (L/W) – R sh 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 R p =R S (1+Q S Q S )=1.98 Ohms(1+2.56x2.56)=14.96 Ohms L p =L S (1+1/(Q S Q S ))=331.5 pH(1+1/2.56/2.56)=382.08 nH

39
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, 20log 10 (0.23)=-12.75 dB

40
Use Tapped-C Circuit to Fool the Tank into Thinking It Has High R S

41
Derivation

42
Previous Design Values QCL 0.50.663 pF6.63 nH 11.326 pF3.315 nH 1013.26 pF331.5 pH Specs: Resonant Frequency: 2.4 GHz R S =50 Ohms R L =Infinity

43
Design Problem Knowns & Unknowns Knowns: R S =50 Ohms CT=13.26 pF Unknowns: C 1 /C 2 R’ S

44
Calculations C T =C 1 /(1+C 1 /C 2 ) C 1 =C T (1+C 1 /C 2 ) C 1 /C 2 R’ S C1C1 C2C2 1200 Ω26.52 pF 2450Ω39.78 pF19.89 pF 3800Ω53.04 pF17.68 pF

46
Include the Effect of Parasitic Resistor

Similar presentations

Presentation is loading. Please wait....

OK

1 © Unitec New Zealand DE4401 AC R L C COMPONENTS.

1 © Unitec New Zealand DE4401 AC R L C COMPONENTS.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google