Powering up the RFID chip - Remotely 1. Basic Reader-Tag System Rectifier Logic & Memory Tag Reader 2

Powering up the RFID chip - Remotely

1

Basic Reader-Tag System

Rectifier

Logic & Memory

Tag

Reader

2

Simple Magnetically Coupled Circuit

Vi = Z1.I1 - jM.I2 0 = Z2.I2 - jM.I1

Applying KVL in each loop

Z1’ and Z2’ can be used to representresistors, capacitors etc. as required

Define self-impedance of each loop:Z1 = Z1’ +R1+ jL1 Z2 = Z2’ +R2+ jwL2

Z1’

Z2’

I1

. .+

I2

~ViL1, R1 L2, R2

2

M1

1

i 2

ZZ

I

V Input impedance

Transfer admittance 2

2.1

M1

2.1

M

.2.1

j

i

2

ZZ

ZZ

ZZV

I

General Expressions

Reflected impedance

3

2

M1

1

i 2

ZZ

I

V Input impedance


2.1

M1

2.1

M

.2.1

j

i

2

ZZ

ZZ

ZZV

I

General Expressions

2

M.j

1

2

ZI

I Current Transfer ratio

4

Vi = (R1 + jL1).I1 - jM.I2 0 = (R2 + jL2).I2 - jM.I1

I2

I1

(R2 + jL2).I2= jM.I1

(R1 + jL1).I1

jM.I2

Vi

Example: Inductively Coupled Resistive Circuit (Transformer)

I1

. .+

I2

~ViL1 L2

R1

R2

VoltageCurrentSource voltage

5

I2

I1

(R2 + jL2).I2= jM.I1

jL1.I1

jM.I2

ViVi = jL1.I1 - jM.I2

0 = (R2 + jL2).I2 - jM.I1

I1

. .+

I2

~ViL1 L2

R1 ~ 0

R2

Ideal Transformer


N.k

1

1L

2L.

k

1

2L.1Lk

2L

Mj

2Lj2R

2I

1I

R1 << .L1R2 << .L2k ~ 1

6

Self Quiz

1. Inductively coupled circuit with R1= 1, R2= 2, L1=L2, .L1=200, k= 0.8

If I1= 1A, what is the approximate value of I2? (KVL)

2. If R2 = 1, what is the approximate value of I2?

3. What is approximate input impedance in each case?

4. What is the approximate input impedance if k ~ 1?

7

1. 0.8 A

2. 0.8 A (Same!)

3. (1+ j.72) (Unchanged!)

4. 1

8


2.1

M1

2.1

M

.2.1

j

i

2

ZZ

ZZ

ZZV

I

Effectiveness to drive current through secondary – would like to maximize for effective power transfer

Introduce resonance

Let resonance occur at

~ C2

I1

. .+

I2

Vi

R2

L2L1

R1C1

Self impedances:Z1 = 1/ jC1 +R1+ jL1 Z2 = 1/ jC2 +R2+ jL2

2C.2L

1

1C.1L

10 which is our excitation frequency

CAVEAT: Series resonance for illustration only!9

At we have Z1 =R1, Z2 =R2 and Transfer admittance is

222Q.1Qk1

2Q.1Qk.

2R.1R

j

2R.1R

M1

2R.1R

M

.2R.1R

j

i

2

V

I

Coupling Coefficient %

22Q.1Qk1

2Q.1Qk

Q1=30Q2=40

Peak occurs at 12Q.1Qk Beyond this value of k, Transfer admittance falls!

0.1 1 10 1000

0.1

0.2

0.3

0.4

0.5

10

Self Quiz

Reader and Tag both has Q =25, and each has ESR (effective series resistance ) = 5. The reader is excited by 1V. What is the current in the Tag for k = 1%, 4%, 10% if both primary and secondary tuned to same frequency?

11

Q= 25 R ohm= 5

k k.Q kQ/(1+kQ^2) I amps I^2. R mW

0.01 0.25 0.235294 0.047 11.07

0.04 1 0.5 0.1 50.00

0.1 2.5 0.344828 0.069 23.78

0.16 4 0.235294 0.047 11.07

12

0.1 1 10 1000

0.1

0.2

0.3

0.4

0.5


22Q.1Qk1

2Q.1Qk

Diminishing return – does not help reducing the spacing beyond a certain point

Tight couplingSmall Separation

Weak couplingLarge Separation

Transfer admittance

spacing

~Spacing ↑ => Coupling coefficient ↓

13

Weak Coupling Case

12Q.1Qk If then coupling is weak

2R.1R

M.

2R.1R

2Q.1Qk.j

i

2

V

IThen

0.1 1 10 1000

0.1

0.2

0.3

0.4

0.5

12R.1R

M

In other words

14

Resonant vs. Non-resonant

2.1

Mj

i

2

ZZV

I 1

2.1

M

ZZ

Transfer admittance- general expression

2

2.1

M1

2.1

M

.2.1

j

i

2

ZZ

ZZ

ZZV

I

For weak coupling: =>

2Q.1Q)2jQ1).(1jQ1(2R.1R

)2Lj2R)(1Lj1R(

resonantnon_2

resonant_2

I

I

)2Lj2R)(1Lj1R(

M.j

2.1

Mj

i

2

ZZV

I

For non-resonant situation

2R.1R

Mj

i

2

V

I

For resonant situation

Current increases by Q1.Q2 (Product of loaded Q’s) 15

Effects of Resonance

• Resonance helps to increase current in coupled loop ~1000X

• But it causes strange behavior (reduction of secondary current at close range). Why ?

16

Self Quiz

• The primary coil is tuned to a certain frequency and excited by a voltage source of the same frequency. A secondary coil, also tuned to the same frequency is gradually brought in from far distance. How does the current in the secondary coil behave with changing distance? (qualitative description)

• Two coils each of Q=50 is taken. Current is measured in second coil with and without tuning capacitor (tuned to frequency of excitation). What is the ratio of currents in the two scenarios?

17

Self Quiz

• The primary coil is tuned to a certain frequency and excited by a voltage source of the same frequency. A secondary coil, also tuned to the same frequency is gradually brought in from far distance. How does the current in the secondary coil behave with changing distance?

Increases till k.sqrt(Q1.Q2) = 1, then decreases

• Two coils each of Q=50 is taken. Current is measured in second coil with and without tuning capacitor (tuned to frequency of excitation). What is the ratio of currents in the two scenarios?

50*50 = 2500

18

Self Quiz

• A Reader-tag system has a certain maximum read range determined by current needed to turn on the Tag chip. Q of the tag is halved. How much is the max read range compared to original? [Assume weak coupling]

R2 is doubled (M/R1.R2) halved range halved

19

Vi = [R1 + j(L1-1/C1)].I1 - jM.I2 0 = [R2 + j(L2-1/C2)].I2 - jM.I1

I2

(R2+j.X2).I2= jM.I1

-jM.I2

Inductively Coupled Series Resonant Circuits


~ C2

I1

. .+

I2

Vi

R2

L2L1

R1C1

Excitation at higher than resonant frequency

I1

(R1+j.X1).I1

Phase angle between Vi and I1 may be > or < 0 depending on coupling

~

+

++

20


I2

I1

R2.I2= jM.I1

-jM.I2



~ C2

I1

. .+

I2

Vi

R2

L2L1

R1C1

R1.I1

Vi

Excitation at resonant frequency

21


I2

(R2-j.X2).I2= jM.I1

-jM.I2



~ C2

I1

. .+

I2

Vi

R2

L2L1

R1C1

Excitation at lower than resonant frequency

I1

(R1-j.X1).I1

• Phase angle between Vi and I1 may be > or < 0 depending on coupling

• I1 and I2 flowing in same direction for lossless case 22

Below resonance (capacitive)

Above resonance (inductive)

I1I2I1

I2I1

Resonance (resistive)

1

2

1

2

1

2

I2

+ + +

23

Power Transmission Efficiency

sourcelable fromPower avai

loadipated at Power diss

Rectifier

Logic & Memory

Tag

Reader Equivalent Resistive Load

24

Parallel to Series Transformation

≡RLC

RLs

CsAt a certain frequency

C.RLXC

RLQ

If Q>>1 then:

RL

XCRLs

CCs

2

Example:

f = 13.56 MHzC= 50.0 pF (XC = 235RL = 2000

Cs pF (Exact): 50.7 pFCs pF (Approx): 50.0 pF

RLs (Exact): 27.2 RLs (Approx): 27.6

25

Assuming both Reader and Tag are resonant at excitation frequency

~

C2I1

. .+

I2

Vi

R2

L2L1

R1C1

RLs

Power dissipated at load = |I2|2.RLs

Power available from source = |I1|2.Re(Zin)

2Z

M1ZRe

RLs.

2Z

Mj

)ZinRe(.1I

RLs.2I22

2

2

2

RLs2R

M1R

RLs.

RLs2R

M222

22

Zin

26

1 10 1000

20

40

60

Load resistance Kohm

Pow

er tr

ansf

er e

ffic

ienc

yM = 5

M = 15

For weak coupling, efficiency is maximum when R2 = RLs

22 2C.RL

1

2R

RL↑ => C2 ↓ for given R2Low dissipation chips usually use less tank capacitance 27

Special Case

• Both Reader and Tag are resonant at excitation frequency

L1.C1=L2.C2 = 02

• Weak coupling

R1>> Reflected impedance• Tag is independently matched to load

R2=RLs => Total resistance in Tag = 2R2 = 2RLs• Q of load (XC2/RLs) >> 1

1R2

R

1R.RLs.4

M0 reflect22

reflect2

reflect2

222

2

2

R1I.2

1R.

1R.2

V

RLs

M0

1R.4

VPchip

28

Self Quiz

XC = 200 ohm (C~ 50 pF)

RL = 10Kohm

What is the value of Tag resistance for optimum power transfer at weak coupling?

If XC is changed to 300 ohm, what is the value of Tag resistance for optimum power transfer at weak coupling?

29

Self Quiz

XC = 200 ohm (C~ 50 pF)

RL = 10Kohm

What is the value of Tag resistance for optimum power transfer at weak coupling?

200^2/10e3= 4 ohm [Traces could be too wide for a compact tag!]

If XC is changed to 300 ohm (C~ 33 pF), what is the value of Tag resistance for optimum power transfer at weak coupling?

300^2/10e3= 9 ohm [Compact tag is realistic]

30

Measurement of Resonance Parameters• Resonant frequency• Loaded Q

• Caution:– Maintain weak coupling with

probe loop

Vector Network Analyzer

Sensing Loop

31

Measurement on a Tag attached to curved surface

32

33

Principle of Measurement

0Z1

0Z1M_11

Z

Zs

0Z2

0Z2D_11

Z

Zs

Sensing Loop alone – stored in Memory

Sensing Loop + DUT – ‘Data’

Data – Memory = s11_D - s11_M DUT2 Y.M.

0Z

2

0Z

)12(2

).20Z).(10Z(

)12.(0Z.2

ZZ

ZZ

ZZ

Z1 = R1 + j.L1 Sensing Loop alone

Z2 = R1 + j.L1 + (M)2. YDUT Sensing Loop + DUT

YDUTZ2 - Z2 = (M)2. YDUT

If s-parameter is used

Approximation valid if Z0>> Z1, Z2. error for low values of YDUT

Transmission method is more accurate34

Spectral Splitting

35

0.1 1 10 1000

0.1

0.2

0.3

0.4

0.5


22Q.1Qk1

2Q.1Qk

Tight couplingSmall Separation

Weak couplingLarge Separation

spacing

~ secondarycurrent

Are these phenomena related?

36

I1

. .

I2

L1 L2

R1

V1 V2

+ +R2

M

M

L2-ML1-MR1 R2

V1 V2

+ +

I1 I2

≡

V1= (R1+jL1).I1 + jM.I2V2= (R2+jL2).I2 + jM.I1

~ C2

I1

. .+

Vi

R2

L2L1

R1C1

M

L2-ML1-MR1 R2I1

C2Vi ~C1≡

37

If coupling is NOT weak:

At f=f0:R2+j.[0.(L2-M)-1/(0.C2)] = R2- j0.M

I1 Let:(L1, C1) => f0(L2, C2) => f0i.e.0.L1=1/(0.C1)0.L2=1/(0.C2)

If M~0 (weak coupling), I1 exhibits series resonance behavior determined by L1, C1

Parallel resonance chokes current at f0 [+j.M and –j.M in shunt]

Input is capacitive

If R2 ↑ (Q2↓) => choking ↓

M

L2-ML1-MR1 R2

C2Vi ~C1

L1-MR1I1

Vi ~C1

M

~1/02.M

~02.M2/R2

(0.M)/R2>>1

+j.M -j.M

38

Self Quiz

• Lossless Resonators tuned at f1 and f2. When coupling is increased, at what frequency parallel resonance occurs?

39

Self Quiz

• Lossless Resonators tuned at f1 and f2. When coupling is increased, at what frequency parallel resonance occurs?

• f2 when looking from resonator 1 and vice versa

40

Series resonances

L1-MR1I1

Vi ~C1

Mf<f0‘Odd Mode’

L1-MR1I1

Vi ~C1

Mf>f0‘Even Mode’Occurs when shunt arm is shorted

Series and parallel resonances alternate

Frequency↓=> Shunt arm more and more capacitive

Frequency↑=> Shunt arm less and less capacitive and then more and more inductive

L2-M R2

C2

41

R1=R2=6 ohm L1=L2=2700 nH C1=C2=50 pF

Q1=Q2=38.7 f01=f02=13.7 MHz

Critical coupling = 0.026

Excitation voltage = 1V

2Q.1Q

1kc

10 12 14 16 180

20

40

60

80

100

k=kck=0.1k=0.25k=kc/2

Magnetically Coupled Series Resonators

Frequency MHz

Sec

onda

ry c

urre

nt m

A

13.7

42

Resonances for Lossless Identical resonators

C).ML(

10

C.L

10

ParallelSeries Series

L1=L2=L C1=C2=C R1=R2=0

C).ML(

11

CC C

L-M L-M L-M2M

43

Two NFC Tags ~ equally coupled with Sensing Loop

44

Realistic Situation

R1=R2=6 ohm L1=L2=2700 nH C1=50pF C2= 47pF

Q1=38.7 (at f01) Q2=39.9 (at f02) f01=13.7 MHz f02= 14.1 MHz

Critical coupling = 0.025

Excitation voltage = 1V

10 12 14 16 180

20

40

60

80

100

k=kck=0.1k=0.25k=kc/2

Magnetically Coupled Series Resonators

Frequency MHz

Sec

onda

ry c

urre

nt m

A

13.714.1

45

Excitation Frequency as Parameter

1 10 1000

20

40

60

80

100

13.714.113.914.313.5

Coupling coeff %

Sec

onda

ry c

urre

nt m

A

Significant degradation in weakly coupled region when frequency of excitation is outside the band between resonant frequencies with a little bit improvement in close range

%6.22Q.1Q

1

46

• For two magnetically coupled resonators tuned at same frequency, we observed that parallel resonance occurs above a certain M. To arrive at this we used an equivalent T network for magnetically coupled inductors. How this phenomenon is explained by reflected impedance?

Review Quiz

47

Review Quiz

• For two magnetically coupled resonators tuned at same frequency, we observed that parallel resonance occurs above a certain M. To arrive at this we used an equivalent T network for magnetically coupled inductors. How this phenomenon is explained by reflected impedance?

2ωM

R1

12

Z

Primary current ~ is maximized when Z2 is minimum

Series resonance in secondary => parallel resonance in primary

48

Documents

Powering up the RFID chip - Remotely 1. Basic Reader-Tag System Rectifier Logic & Memory Tag Reader 2