RFID Technologies HF Part I

Transcription

1 RFID Technoogies HF Prt I Diprtimento di Ingegneri de Informzione e Scienze Mtemtiche

2 Ampere s w h(r,t) ĉ d = j(r,t) ˆnds t C brt (, ) = µ hrt (, ) S S d(r,t) ˆnds j(r,t) d(r,t) ds ˆn Ø Biot-Svrt (sttic cse) S C h = h(r) ĉ d I πr ˆφ = I I r C b(r) C d ĉ

3 Mgnetic Fied x z N π < λ 0 5 current constnt ong the wire d H r 00 Mgnetic fied strength H (A/m) R = 55 cm R = 7.5 cm R = cm z -3 Figure 4.3 The pth of the ines of mgnetic fux round short cyindric coi, or conductor oop, simir to those empoyed in the trnsmitter ntennsofinductiveycoupedrfidsystems H = f 0 = 3.56 MHz IN ( ) 3 z d << z < λ 0 π Distnce x (m) Figure 4.4 Pth of mgnetic fied strength H in the ner fied of short cyinder cois, or conductor cois, s the distnce in the x direction is incresed

4 Mgnetic Fied Ø Optim ntenn dimeter Mgnetic fied strength H (A/m) 4 3 x = 0 cm x = 0 cm x = 30 cm.5 A/m (ISO 4443) f 0 = 3.56 MHz Rdius R (m) Figure 4.5 Fied strength H of trnsmission ntenn given constnt distnce x nd vribe rdius R, wherei = AndN =

5 Mgnetic Fied H z = N I b 4π b z z b z Ø Rectngur coi x z y b

6 Mgnetic Fied Ø Rectngur coi Δ = 0 cm Δb = 0 cm I = A f 0 = 3.56 MHz y x b z z (m)

7 Mgnetic Fied Ø Rectngur coi Δ = 0 cm Δb = 0 cm I = A f 0 = 3.56 MHz y x b z z (m)

8 Mgnetic Fied Ø Rectngur coi Δ = 0 cm Δb = 0 cm I = A f 0 = 3.56 MHz y x b d=0.8 m z

9 Mgnetic Fied Ø Rectngur vs. circur P r = P c = 4 m I = A f 0 = 3.56 MHz y x b z z (m)

10 Mgnetic Fied Ø Rectngur vs. circur P r = P c = 4 m I = A f 0 = 3.56 MHz y b x z z (m)

11 Mgnetic Fied Ø Rectngur vs. circur P r = P c = 4 m I = A f 0 = 3.56 MHz y x b d=0.8 m z

12 Eectromgnetic Induction Ø Frdy-Neumnn-Lenz Lw emf = Δφ Δt

13 Eectromgnetic Induction Ø Frdy-Neumnn-Lenz Lw Φ[b(r,t)] = S b(r,t) ˆnds ds ˆn S ε = dφ[b(r,t)] dt C d ĉ Ø Mgnetic fux density vs. Mgnetic Fied b(r,t) = µ 0 µ r h(r,t) Wb/m

14 Sef-Inductnce ε = dψ dt = dψ di di dt L = dψ di = Nφ i Ø Stright wound wire >> d ;d cm d c = L = 0.00 n 4 d c 3 up to VHF 4 beyond VHF [µh]

15 Sef-Inductnce Ø Circur coi with singe turn X d ;d cm L µ 0 µ r n 8 d [H]

16 Sef-Inductnce Ø singe-yer ir-coi (circur) with N turn EQUATION 3: L ( N) = ( µh) where: L ( N) = ( µh) N = number of turns = ength in cm = the rdius of coi in cm DS0070C-pge 0 infinite L = µ 0 N π [H]

17 h where: Sef-Inductnce = men rdius of oop in (cm) d = dimeter of wire in (cm) Figure 8 shows n N-turn inductor of circur coi with mutiyer. Its inductnce is ccuted by: b INDUCTANCE OF AN N-TURN SINGLE LAYER CIRCULAR COIL Ø mutipe-yer ir-coi (circur) with N turn FIGURE 7: b X L A CIRCULAR COIL WITH SINGLE N-turns coi TURN where: L 0.3( N) = ( µh) 6 9h 0b EQUATION 4: 0.3( N) = ( µh) 6 9h 0b = verge rdius of the coi in cm N = number of turns b = winding thickness in cm h = winding height in cm EQUATION 3: b L h ( N) = ( µh) where: N = number of turns = ength in cm = the rdius of coi in cm

18 Inductnce of ft squre coi of rectngur cross section with N turns is ccuted by [] : EQUATION 9: Sef-Inductnce Ø stright thin fim trce where: L = in µh L N og og t w 0 (.44) 0.003N 0.35 = ( t w) t = side ength in inches t = thickness in inches w w = width in inches N = tot number of turns Inductnce of ft squre coi of rectngur cross section with N turns is ccuted by [] : INDUCTANCE OF A FLAT SQUARE COIL L 0.00 n w t EQUATION = 9: w t 3 ( µh) Inductnce of ft squre coi of rectngur cross section with N turns is ccuted by [] : FIGURE 3: SQUARE LOOP INDUCTOR WITH Ø thin-fim A RECTANGULAR squre CROSS oop SECTION EQUATION 9: INDUCTANCE OF A FLAT SQUARE COIL L L w t = n w t 3 w = width in cm t = thickness in cm = ength of conductor in cm N og og t w 0 (.44) 0.003N = 0.94 A ( t w) w where: L = N og t w og 0 (.44) 0.003N ( t w) where: L = in µh = side ength in inches t = thickness in inches w = width in inches L = in µh N = tot number of turns = side ength in inches t FIGURE = thickness 3: SQUARE in inches LOOP INDUCTOR WITH w = width in inches A RECTANGULAR CROSS N = tot number SECTION of turns

19 Sef-Inductnce Ø one-turn reder ntenn b [ ] ) ( ) ( ) ( n ) ( n 4 nh A A L b c c c b b = c b = A b = dimensions re in cm

20 Ohmic osses Ø skin depth Ø AC resistnce δ = π f µσ R AC = σ A ctive A ctive πδ f = frequency µ = permebiity (F/m) = µ ο µ r µ o = Permebiity of ir = 4 π x 0-7 (h/m) µ r = for Copper, Auminum, God, etc = 4000 for pure Iron σ = Conductivity of the mteri (mho/m) = 5.8 x 0 7 (mho/m) for Copper = 3.8 x 0 7 (mho/m) for Auminum = 4. x 0 7 (mho/m) for God = 6. x 0 7 (mho/m) for Siver =.5 x 0 7 (mho/m) for Brss EXAMPLE 4: The skin depth for copper wire t 3.56 MHz nd 5 khz cn be ccuted s: EQUATION 5: δ = πf( 4π 0 7 )( ) = ( m) f = 0.08 ( mm) for 3.56 MHz = 0.87 ( mm) for 5 khz

21 Equivent circuit Ø RL circuit ε ε L = Ri ε L di dt = Ri ε = L di dt Ri ε i R ε L resistnce of the coi Ø Time hrmonic V g = jω LI RI

22 Mutu Inductnce Φ(I ), Ψ(I ) B (I ) I A A Tot fux Ψ (I ) ε = dψ (i ) dt = dψ (i ) di di dt M = dψ (i ) di = N ψ (i ) i M = N i b (i ;r,t) ˆn ds S

23 Mutu Inductnce Ø bsic configurtion x M = N φ (I ) N = µ N π b I 0 ( ) 3 x 0.5 Couping coefficient k = f(x) 0. r r r 3 Ø couping coefficient M = M = M = k L L Couping coefficient k (x) Distnce x (m) Figure 4.0 Grph of the couping coefficient for different sized conductor oops. Trnsponder ntenn: r Trnsp = cm, reder ntenn: r = 0 cm, r = 7.5cm, r 3 = cm

24 Mutu Inductnce Ø Mutu inductnce of two conductor segment j L T = L o M M ( µh) d p k q L T = Tot Inductnce L o = Sum of sef inductnces of stright segments M = Sum of positive mutu inductnces M - = Sum of negtive mutu inductnces Ø The mutu inductnce is positive when the directions of current on conductors re in the sme direction, nd negtive when the directions of currents re opposite directions.

25 Reference [4]. Consider n inductor mde of stright segments s shown in Figure 5. The inductnce is the sum of sef inductnces nd mutu inductnces [4] : M jk, = -- {( M k p ) ( M p M q )} Mutu M Inductnce k q EQUATION 3: = -- {( M j M k ) M q } for p = 0 L T = L o M M ( µh) = -- {( M j M k ) M p } for q = 0 Ø Mutu inductnce of two conductor p segment = for p = q where: k M k p M p (c) L T = Tot Inductnce M j k for p = q = 0 L T = L(d) o M M ( µh) L o = Sum of sef inductnces of stright j nd k in the bove figure re indices of conductor, nd segments d p nd q re the indices of the ength for the difference M = Sum of positive mutu inductncesl p q T = Tot in Inductnce the ength of the two conductors. M - = Sum of negtive k mutu inductncesl o = Sum The of sef bove inductnces configurtion of stright (with prti segments) occurs segments The mutu inductnce is the inductnce tht is between conductors in mutipe turn spir inductor. The resuted from the mgnetic fieds produced by djcent M = Sum mutu of positive inductnce mutu inductnces of conductors j nd k in the bove conductors. The mutu inductnce is positive when M - = Sum configurtion of negtive mutu is: inductnces the directions of current on conductors re in the sme direction, nd negtive when the directions of currents re opposite directions. M k The p = mutu k p F k inductnce EQUATION 34: p between two pre conductors is function of the M ength of where the conductors nd of the geometric men jk, = -- {( M k p k p k p M k q ) ( M p M q )} distnce between them. The mutu inductnce of two conductors F k is ccuted p = n by: d j, k d = -- {( M j M k ) M q } for p = 0 () j, k = -- {( M EQUATION 3: d j, k j M k ) M p } for q = 0 (b) d j, k = M k p M p for p = q (c) M = F k ( nh p ) k p = M k for p = q = 0 (d) () (b) MUTUAL INDUCTANCE CALCULATION d j q where is the ength of conductor in centimeter. F is the mutu inductnce prmeter nd ccuted s: EQUATION 33: F -- d -- d = n -- d d -- If the ength of nd re the sme ( = ), then Eqution 34 (d) is used. Ech mutu inductnce term in the bove eqution is ccuted s foows by using Equtions 33 nd 34: EQUATION 35: M k p = k p F k p

26 Mutu Inductnce Ø Mutu inductnce of two conductor segment 3 Gp L T = L o M M ( µh) 4 where L T = L o M M ( µh) = L o M ( µh) M = 0 since the direction of current on ech segment is opposite with respect to the currents on other segments. L o = L L L L 3 L 4 By soving the sef inductnce using Eqution (8), L = L = 59.8 ( nh) L = L 4 = 59.7 ( nh) L 3 = 8 ( nh) L 0 = 8 ( nh) Negtive mutu inductnces re soved s foows: M = ( M, 3 M, 3 M 4, ) M 4, = F, 4 M 3, = -- ( M 3 M M gp ) M, 3 = -- ( M 3 M M gp ) -- d, 4 -- F , 4 = n d, 4 d , 4 d, 4 F d, = n d 3, d , 3 d, 3 F -- d 3, -- = n d 3, d , 3 d 3, -- d, 3 -- F = n d, 3 d , 3 d, 3 M = F M = F M gp = gp F gp F gp gp -- gp = n d gp, 3 d gp, 3 d, 3 d gp, gp

27 Mutu inductnce Ø Equivent circuit Ø Votge porities (dot convention: dots indicte the direction in which the cois re wound)

28 Mutu inductnce Ø When the reference direction for current enters the dotted termin of coi, the reference pority of the votge tht it induces in the other coi is positive t its dotted termin. Ø Mesh equtions (Kirchhoff's votge w) v g i R L di dt M di dt = 0 i R L di dt M di dt = 0

29 Mutu inductnce Ø Equivent circuit for the tg coi B (i ) i M i R u M L L L u R L u L i R L Ø Time hrmonic u i R L di dt M di dt = 0 U = jωmi (R jωl )I