TWI327402B - Power circuit component parameters design method for compensating the loosely coupled inductive power transfer system - Google Patents

Description

1327402 九、發明說明： 【發明所屬之技術領域】 本發明係關於感應電力傳輸系統之設計方法，更特別十之，伏關於一 種弱耦合感應電力傳輸系統功率補償的電路元件參數設計方法 \ 、1327402 IX. Description of the Invention: [Technical Field] The present invention relates to a method for designing an inductive power transmission system, and more particularly to a method for designing a circuit component parameter for power compensation of a weakly coupled inductive power transmission system.

【先前技徇J 弱耦合感應電力傳輸系統(Loosely coupled Inductive細打丁削命[Previous Technology J weakly coupled inductive power transmission system (Loosely coupled Inductive)

System ’ Lam難t能自1備傳輸至另―設備，而無需直接透過 電力線傳導。因為不必經過插座直接接觸轉送電力，故具有可避免產 生火花或觸電危險的優點。此外，因為沒有接觸之磨損也有益於該設備 壽命之提升。以下列舉數例以說明其一些應用，在礦坑或是石油鑽採等 場所，若是傳統接觸式供電，可能會_接觸不良職生的火花導 致***；X如在編的環境巾’傳統觸式供電較紐生電擊的潛在危 險，另外在氣密式儀器設備内部供電時，傳統接觸式的電能傳輸除了需 要t木用較喊本的連接!!設計外，同時有無法完全確絲紐的危險； 因此採用可分離式隔離感應電力傳輸的技術即可以避免上述的問題發 生0 如圖1所不，為上典型可分離式隔離感應電力傳輸系統之基本架構卜 利用可为離式之高頻變壓器配合前置反流器，以將電能由電源端傳輸 至系統負載側。其中以由諧振反流$ 10、可分離式隔離壓胃20及可控 1/4 30二個部份所組成的架構最廣為採用。 6 1327402 Μ :代表可分離式隔離變壓器20互感 ς :代表可受離變壓器2〇—次側串聯補償電容 C2 :代表可分離式隔離變壓器20二次側並聯補償電容 圪：代表系統之交流電阻負載40 Z2 :代表可分離式隔離變壓器20二次側阻抗 & :代表Z2經由可分離式隔離變壓器20反射至一次側之等效阻抗 :代表系統輸入阻抗 首先由圖2可求出1/_/似(：2與/^之並聯阻抗如下所示。 2 = Rac 1 二 Rac =Rac-JRac2c〇C2 2 心 +1 "<yC2 1 + 风C2 \ + {RacaC2)2 =-^--j—R〇c ^-i — (1) \ + {Rac〇C2f J\ + (Rac&C2f 其次，吾人可利用電路學基本原理求出Z2經由可可分離式隔離變壓器20 反射至一次側之等效阻抗A:System ’ Lam can't be transferred from one device to another without having to transmit directly through the power line. Since there is no need to directly contact the transfer power through the socket, there is an advantage that the risk of sparking or electric shock can be avoided. In addition, wear without contact is also beneficial to the life of the device. The following is a few examples to illustrate some of its applications. In the case of mine pits or oil drilling, if it is a traditional contact power supply, it may cause a smoldering spark to cause an explosion; X, such as the environmental towel in the 'traditional contact power supply' Compared with the potential danger of Newson's electric shock, in addition to the internal power supply of the airtight instrument, the traditional contact type power transmission needs to be connected with the shouting of the wood! In addition to the design, there is a danger that the wire cannot be completely confirmed; Therefore, the technology of detachable isolated inductive power transmission can be used to avoid the above problem. 0 As shown in Figure 1, the basic structure of the typical separable isolated inductive power transmission system can be used as a high-frequency transformer. A pre-flow reverser to transfer electrical energy from the power supply to the system load side. Among them, the architecture consisting of two parts: resonant backflow $10, separable isolation stomach 20 and controllable 1/4 30 is the most widely used. 6 1327402 Μ : Represents the detachable isolation transformer 20 mutual inductance ς : represents the transformer 2 〇 - secondary side series compensation capacitor C2 : represents the separable isolation transformer 20 secondary side parallel compensation capacitor 圪: represents the system AC resistance load 40 Z2 : represents the secondary side impedance of the separable isolation transformer 20 & represents the equivalent impedance of Z2 reflected to the primary side via the separable isolation transformer 20: represents the input impedance of the system. First, 1 /_ / can be obtained from Figure 2. The parallel impedance of (:2 and /^ is as follows. 2 = Rac 1 2 Rac =Rac-JRac2c〇C2 2 heart +1 "<yC2 1 + wind C2 \ + {RacaC2)2 =-^-- j—R〇c ^-i — (1) \ + {Rac〇C2f J\ + (Rac&C2f Secondly, we can use the basic principles of circuit science to find out that Z2 is reflected to the primary side via the cocoa separation transformer 20 Effective impedance A:

ζ {ωΜ)1 r jc〇L2 + Z2 (ωΜ)2 R^+ j{oL2 -X2) (ωΜΥ^-Κω^-Χ,)] R^ + iaL.-X,)2 {ωΜ)2 · (ωΜ)2 -(oL2-X2) R；+{aL2-X2)2 + A~ ^{aL2-X2f 9 1327402 最後由式(2)吾人可求得圖2系統之輸入阻抗又ζ {ωΜ)1 r jc〇L2 + Z2 (ωΜ)2 R^+ j{oL2 -X2) (ωΜΥ^-Κω^-Χ,)] R^ + iaL.-X,)2 {ωΜ)2 · (ωΜ)2 -(oL2-X2) R;+{aL2-X2)2 + A~ ^{aL2-X2f 9 1327402 Finally, we can find the input impedance of the system of Figure 2 by equation (2).

Zr+j&L, j(〇C' = --l— + Xr] a>C' ^Rin +J^ln 由於圖2系統係假設該電力傳輸系統無其他耗摘， (3) 等於輸入之實功率。而輸入複數功率瓦可求得如 因此其輸出實功率將會 Z,Zr+j&L, j(〇C' = --l- + Xr] a>C' ^Rin +J^ln Since the system of Figure 2 assumes that the power transmission system has no other consumption, (3) is equal to the input Real power. And input complex power watts can be obtained as such, so its output real power will be Z,

Rm - ΐΧin RJ +X, ^PP+jQp 因由能量核定律，吾人可以轉輸出到貞載之功率A為 Vp (4) 户s (必，C〗，C2) = 上式中Rm - ΐΧin RJ +X, ^PP+jQp Due to the law of energy, we can convert the output power to the load A as Vp (4) household s (must, C, C2) = above

R RJ+X 2 (5) (ωΜ)% R. in R^+ioL.-Xy 1__ ωα ^22 + (ωΙ^χ^ 本發明之目標即在首先蚊(^以最佳錄祕麟最大的輸出功率 Ps ’此點可利驗積分求極值的方法由下列三充分條件解得之： 1327402 dPs. -0 (6a) dC' dPs . -0 (6b) dC2 dPs_ δω =0 (6c) 將式(5)代入⑹可得下列三方程式： dC, 3R RJ+X 2 (5) (ωΜ)% R. in R^+ioL.-Xy 1__ ωα ^22 + (ωΙ^χ^ The goal of the invention is to be the first mosquito in the first place. The output power Ps 'this point can be used to determine the integral value of the integral method by the following three sufficient conditions: 1327402 dPs. -0 (6a) dC' dPs . -0 (6b) dC2 dPs_ δω =0 (6c) Substituting equation (5) into (6) gives the following three equations: dC, 3

R dC,R2 + Xt. r) = 〇 => dR dC,R2 + Xt. r) = 〇 => d

RR

dC^Rj + Xj) dRdC^Rj + Xj) dR

々—(RX) - 2RJ ·各-2Rin. Xin. & = Q dCx dC, dC' ⑺ dC2 => d々—(RX) - 2RJ ·Each-2Rin. Xin. & = Q dCx dC, dC' (7) dC2 => d

VP R, dC2、Rin2+Xj => dVP R, dC2, Rin2+Xj => d

R dcK) )=〇 o 脅凡 2+()-2(.¾ -2R-X, ax. in ac, dC2 2RX. da ⑻ 1327402 dPs_ δω => => dR dcK) )=〇 o threat 2+()-2(.3⁄4 -2R-X, ax. in ac, dC2 2RX. da (8) 1327402 dPs_ δω =>=> d

d〇}KR：+Xhid K d〇(Rin2+Xin2) dR r) = 〇 δω 3ω Βω ^-R：)^-2RinXm^0 ⑼D〇}KR:+Xhid K d〇(Rin2+Xin2) dR r) = 〇 δω 3ω Βω ^-R:)^-2RinXm^0 (9)

將式(7)、（8)與(9)用商業軟體MATLAB計算，並將其結果重新整理可得 dPs _ 2L,L2k2R〇r(^6+C^+E^2+Rj) = 〇 5C, ~ C,3 - [{L2C2RJ · + (Z22 - 2L2C2RJ)o2 + RJ)2 上式中 4=L'C、L22C22Rj(k2-l) C, =(L2C2RJ -(2-^) E^L^-lL^Rj-LfA： dPs _ -IL^k^Rj · ω2 (^ω6 + 02ω4 + Ε2ω2 -1) dC2=q2-[{L2C2RJ ·ω^ +(L22 -2L2C2RJ)o2 +RJ)2 上式中 4 =ι,2ς24〇2-(^-2^ + 1) £2=L2C2+1^.(2]2) dPs -21^2L . (Ay + C〆 + G3(〇2 - 2RJ) ai = C,2 · ^ · [{L2C2RJ · + (42 - 2L2C2RJ )ω2 + RJ f 12 1327402 上式中 α,ςΛ2)2.(^-2^+ΐ) ^=-V+a2c2+i,Q-2i?j 下 而由式(10)、（11)與(12)最後吾人可得Cl及C如 -1 ω L^k-l^k + l) ω2Ι,(1-^) 2 2 ωΊ (13) (14) k 其中t為可分離式隔離變壓器2〇，其定義如下： Μ 0<k<\ (15) 事實上在上述導親財，是假關2 +元件均為縣元件，沒有其他損 耗，因此吾人亦可由工程設計之觀點來求解^與^如下：Equations (7), (8) and (9) are calculated with commercial software MATLAB, and the results are rearranged to obtain dPs _ 2L, L2k2R〇r(^6+C^+E^2+Rj) = 〇5C , ~ C,3 - [{L2C2RJ · + (Z22 - 2L2C2RJ)o2 + RJ)2 In the above formula 4=L'C, L22C22Rj(k2-l) C, =(L2C2RJ -(2-^) E^L ^-lL^Rj-LfA: dPs _ -IL^k^Rj · ω2 (^ω6 + 02ω4 + Ε2ω2 -1) dC2=q2-[{L2C2RJ ·ω^ +(L22 -2L2C2RJ)o2 +RJ)2 Where 4 =ι,2ς24〇2-(^-2^ + 1) £2=L2C2+1^.(2]2) dPs -21^2L . (Ay + C〆+ G3(〇2 - 2RJ) Ai = C,2 · ^ · [{L2C2RJ · + (42 - 2L2C2RJ )ω2 + RJ f 12 1327402 α,ςΛ2)2.(^-2^+ΐ) ^=-V+a2c2+i Q-2i?j and then by equations (10), (11) and (12), we can get Cl and C such as -1 ω L^kl^k + l) ω2Ι, (1-^) 2 2 ωΊ ( 13) (14) k where t is a separable isolation transformer 2〇, which is defined as follows: Μ 0<k<\ (15) In fact, in the above-mentioned guides, it is a false-off 2 + components are county components, no Other losses, so we can also solve the ^ and ^ from the perspective of engineering design as follows:

首先欲獲得最小輸入伏安容量以降低成本， 為零而求得C,。 吾人可令式(4)中之虛功率込 =^=0 二十 1 (ωΜΥ(ω12~Χ2) 〇ct ii22+(^L2~x2y 13 (16)1327402 => —=ω21^ - iz2+z3)-z4 上式中 χχ = m4L22C22Rac2 - 2o2L\C2Rj + Rac2 + &2L22 Z2 = (o%k2ClRac2-&^L22C22Rac2) Z3 = {-<〇2^ik2(^iKc + 0)1 ^iKc ~ Kc ~ + <y2z,^2) % 其次再將(5)式代入(6b)式可得 2ω%%Ψ〇%(：2-1) (〇4L22C22Rj - 2〇2L2C2Rj + RJ + ω^Ι^γ = 0 (17)First, you want to get the minimum input volt-ampere capacity to reduce the cost, and find C for zero. We can make the virtual power in equation (4) 込=^=0 二十1 (ωΜΥ(ω12~Χ2) 〇ct ii22+(^L2~x2y 13 (16)1327402 => —=ω21^ - iz2+z3 ) -z4 上χχ = m4L22C22Rac2 - 2o2L\C2Rj + Rac2 + &2L22 Z2 = (o%k2ClRac2-&^L22C22Rac2) Z3 = {-<〇2^ik2(^iKc + 0)1 ^iKc ~ Kc ~ + <y2z,^2) % Next, substituting (5) into (6b) gives 2ω%%Ψ〇(:2-1) (〇4L22C22Rj - 2〇2L2C2Rj + RJ + ω^ Ι^γ = 0 (17)

因此由式(17)吾人可得和(6b)相同結果，然後再將所得&值代入(i6) 式中便可得C；最佳值，其結果與(Μ)完全相同。至此吾人^知式(⑽、 (6b)與(6c)的結果，在假設埋想元件條件下即相當於要求輪入伏安容量 為最小值。由於在最佳參數值條件下，=〇或者相當於式(4) 中之込為零，亦即Therefore, from equation (17), we can get the same result as (6b), and then substitute the obtained & value into (i6) to obtain C; the optimum value, the result is exactly the same as (Μ). At this point, the results of our knowledge ((10), (6b), and (6c) are equivalent to the requirement that the wheel volt-ampere capacity be the minimum under the assumption of the buried component. Because of the optimum parameter value, =〇 or Equivalent to zero in equation (4), ie

將式(18)重新整理可得下列#㈤）多項式方程式 Ν(ω) = Α^ω6 + +^ω2'^2=〇 (19) 1327402 上式中 A^mqL^Rj -L^eef^Rj) C^^eqC.Rj+L^L,2-L^RJ-IL^L^RJ-L.L^C,) 尽=(-ι22+2ΐ2ς〇,ς〇 由於(14)知最佳C2值必須滿足β = 1/(Ζ2ί：2)之關係式，因此吾人可將 #(的表示成下列形式The following equation can be obtained by refining equation (18): (5)) Polynomial equation Ν(ω) = Α^ω6 + +^ω2'^2=〇(19) 1327402 In the above formula, A^mqL^Rj -L^eef^Rj C^^eqC.Rj+L^L,2-L^RJ-IL^L^RJ-LL^C,) ==(-ι22+2ΐ2ς〇, ς〇 because (14) knows the best C2 value must Satisfy the relationship of β = 1/(Ζ2ί:2), so we can represent #( in the following form)

Ν{ω) = {ω2--) X (αω2 + δω + ο)χ {άω1 + e& + f). [2C2 (20) 將式(19)與(20)兩式之係數比較吾人可得 ad = (L2^C23-Rj) (21) ae + bd = 0 (22) (―ί;)=^τ^ LULXRJ (23) bf + ce ae + bd Λ --=0 l2c2 (24) ce — af^e + cd Λ r^Rj^ (25) L 2 2 2 〇c \ bf + ce A l2c2 _ u (26) cf =v (27) l2c2 因此由(20)式吾人可由分解之因式求得其它解 15 (28) 1327402Ν{ω) = {ω2--) X (αω2 + δω + ο)χ {άω1 + e& + f). [2C2 (20) Comparing the coefficients of the equations (19) and (20) Ad = (L2^C23-Rj) (21) ae + bd = 0 (22) (―ί;)=^τ^ LULXRJ (23) bf + ce ae + bd Λ --=0 l2c2 (24) ce — Af^e + cd Λ r^Rj^ (25) L 2 2 2 〇c \ bf + ce A l2c2 _ u (26) cf =v (27) l2c2 Therefore, we can solve the factor of decomposition by (20) Other solutions 15 (28) 1327402

αω2 + b& + c = 0 άω2 +εω + f = 〇 由式(28)取正實數解可得％如下 yTa\'(a2+yi^) 「^Qki-a：2) 同理’由式(29)取正實數解可得％如下仞 _—Ε:(α2-在) H 2L2C2Rac(l-k2) 上二解中q、<32及<33之定義如下 at=L2-(k2-l) α2 = (A - L2. A:2 - 2. C2 又2 + C2. 〇2) a,=(L2-L2-k2-C2-R0^k2) a2~(L2-L2 'k2-4-C2-Rj +^-C2-RJ .k2^ (29) (30) (31) 由上述結果可知： (一)該串並聯型可分離式隔離感應電力傳輸系統電力傳輸系統在其最大輸 出功率之充分條件下共有三個可工作於最小輸入伏安容量的工作頻 率’即A、及％。其中叫及％與交流電阻負載凡及耦合 係數&有關，但則相對的保持定值。 (~)通例的情況下之串聯補償電容ς與並聯補償電容q之參數值，可期能 1327402 匹配非接觸型魏器之以及則寺性，以獲得最大輸出功率。 然而，有鐘於前述所得，如公式甚為複雜，因此吾人更進一步特別針 對一特殊應用例子，導演出—較觸單的公式，同時亦有助於對非接觸型 電力傳齡財更進-紅瞭解。轉殊鮮是舢定電阻負載情況下， 若吾人特擇ς與以數，使制2之—摘㈣錄_與二次側並 聯共振頻率恰相等。Ωω2 + b& + c = 0 άω2 + εω + f = 〇 The positive real number solution from equation (28) can be obtained as follows yTa\'(a2+yi^) ”^Qki-a:2) the same as the equation (29) The positive real number solution can be obtained as follows: 仞 _—Ε: (α2- in) H 2L2C2Rac(l-k2) In the upper two solutions, q, <32 and <33 are defined as follows: at=L2-(k2 -l) α2 = (A - L2. A:2 - 2. C2 and 2 + C2. 〇2) a,=(L2-L2-k2-C2-R0^k2) a2~(L2-L2 'k2- 4-C2-Rj +^-C2-RJ .k2^ (29) (30) (31) From the above results, we can see that: (1) The power transmission system of the series-parallel detachable isolated inductive power transmission system is at its maximum output. Under the sufficient conditions of power, there are three operating frequencies that can work at the minimum input volt-ampere capacity 'A, and %. The % is related to the AC resistance load and the coupling coefficient & however, but the relative constant value. (~) In the case of the general case, the value of the series compensation capacitor ς and the parallel compensation capacitor q can be used to match the non-contact type Weier and the temple to obtain the maximum output power. However, there is a clock in the above. If the formula is very complicated, so we go further and specifically for a special application. The director out of the formula that is more than one-touch, and also helps to understand the non-contact type of power-age-age--------------------------------------------------------------- The system 2 - pick (four) record _ and the secondary side parallel resonance frequency are exactly equal.

首先二次側阻抗之共振頻率可由下式求得： im{>4+^//—i-} = 〇 j(〇C2 _K_ 1 + 6>2C22First, the resonance frequency of the secondary side impedance can be obtained by the following formula: im{>4+^//-i-} = 〇 j(〇C2 _K_ 1 + 6>2C22

+ j(aL2 - 〇C,R 2Wq^)} = 0 2 、〇c (32) 上式中，符號係代表"其虛數部分"。吾人由式(Μ)可得+ j(aL2 - 〇C,R 2Wq^)} = 0 2 , 〇c (32) In the above formula, the symbolic system stands for "the imaginary part". I can get it by formula (Μ)

(〇)L2 - =>mL7 —d2 1 + ω2Γ2Κ 2 1 QC =>fi>2 _C2Roc、L2 c,2KX => 6)- _+ l^2^ac2 ~L2 Ί ^7RJl2 由式(33)取正實數解可得叫如下 (33) 1327402 同理由一次側阻抗虛數部分為零，如下式所示(〇)L2 - =>mL7 —d2 1 + ω2Γ2Κ 2 1 QC =>fi>2 _C2Roc, L2 c,2KX => 6)- _+ l^2^ac2 ~L2 Ί ^7RJl2 33) The positive real number solution can be called as follows (33) 1327402. The reason is that the primary side impedance imaginary part is zero, as shown in the following equation.

R i+〇2c22Rj+J〇)Ll+^q)} = °R i+〇2c22Rj+J〇)Ll+^q)} = °

R l + ωR l + ω

ac_TCJFC iyC, )} = 〇 (35) (36) => {〇)Lx--) = 0 吾人可得 λ/Α^ι 今特別針對此固定負載特例，設計參數ς與c2使得一次側共振頻率a 等於其二次側共振頻率咚且令其為叫'如下式： ω, 「呢— iC22RjL2 則式(19)之#(似)在此特殊例子下可表示成下列式子 Ν(ω) - {ω2 (37) ——)·(^ -——)·[α2ω +^ + c2] = 0 L2C2 L}L} => Ν(ω) = Α5ω6 + B5as + C5<y4 + Ό5ω3 + Ε5ω2 + F5a + G5 = 0 (38) 上式中 a. -(L,C) -\-L2C2)-a2 L'C'L2C2 1327402 -(L'C、+ L2C2). b2 LxCxL2C2 ~(Z,C) + L2C2) -c2+a2 L、C、L2C2 L'C'L2C2 C2ac_TCJFC iyC, )} = 〇(35) (36) => {〇)Lx--) = 0 λ/Α^ι I can get this special fixed load special case, design parameters ς and c2 make primary side resonance The frequency a is equal to its secondary side resonance frequency 咚 and is called 'the following formula: ω, 』 — - iC22RjL2 Then #(like) of equation (19) can be expressed as the following formula ω (ω) - {ω2 (37) ——)·(^ -——)·[α2ω +^ + c2] = 0 L2C2 L}L} => Ν(ω) = Α5ω6 + B5as + C5<y4 + Ό5ω3 + Ε5ω2 + F5a + G5 = 0 (38) where a. -(L,C) -\-L2C2)-a2 L'C'L2C2 1327402 -(L'C, + L2C2). b2 LxCxL2C2 ~(Z,C ) + L2C2) -c2+a2 L, C, L2C2 L'C'L2C2 C2

LxCxL2C2LxCxL2C2

同理比較(19)與(38)二式之係數，吾人可求得 a2=L'C'L22C22Rj(l-k2) (39) b2 = 0 (40) C2 = (41) 因此由(38)式吾人可由分解之因式求得其它解 a2〇)2 + b2〇) + c2 = 0 (42) 由式(42)取正實數解可得 1 ^L2C2-(l-k2) (43) 所以式(38)可以重新寫成 •) = ("2、2C2H"2-AClHiy、C2.(1-W = 〇 (44) 因此針對一次側串聯共振頻率與二次側並聯共振頻率恰相等時，吾人可得 (45) ω〇=^~-ω〇 (46)Similarly, comparing the coefficients of (19) and (38) two formulas, we can find a2=L'C'L22C22Rj(l-k2) (39) b2 = 0 (40) C2 = (41) Therefore by (38) We can find other solutions by decomposition factor a2〇)2 + b2〇) + c2 = 0 (42) From the equation (42), we can get 1 ^L2C2-(l-k2) (43) Equation (38) can be rewritten as •) = ("2, 2C2H"2-AClHiy, C2.(1-W = 〇(44) Therefore, when the primary side series resonance frequency is equal to the secondary side parallel resonance frequency, We have (45) ω〇=^~-ω〇(46)

19 (47)1327402 而其對-Κ各點 π 刀州τ以獲件如下簡單公式乃(Ο=5 (〇 事貫上’由式(37)吾人可得 (48 (49) (50)19 (47) 13274002 and its opposite - Κ π 刀 州 τ to obtain the following simple formula is (Ο = 5 (〇事上的 by (37) is available to us (48 (49) (50)

所以由式(51)及(48)吾人可得k -K-C2 同樣地，由通式(13)及(14)吾人可得 (51 (5：Therefore, from the formulas (51) and (48), we can obtain k -K-C2. Similarly, from the general formulas (13) and (14), we can obtain (51 (5:

A (53) 縱上所述，於一次側串聯共振頻率與二次側並聯共振頻率恰相等時 離式隔離感應電力傳輸系統的阻抗匹配補償方法如下： 可分 步驟一：設定一次側串聯共振類率與二次側並聯共振頻率恰相等 20 1327402 步驟二：設定輸入電壓與輪出負載電阻 步驟三：求得二次側並礴補償電容c2 首先由式(53)吾人可得 (55) (55) 1 _(1-^2) L'C' L2C2 因此由式(37)與(54)吾人可得A (53) In the longitudinal direction, the impedance matching compensation method of the isolated isoelectric power transmission system is as follows when the primary side series resonant frequency is equal to the secondary side parallel resonant frequency: Step 1: Set the primary side series resonance class The ratio is equal to the secondary side parallel resonance frequency. 20 1327402 Step 2: Set the input voltage and the wheel load resistance. Step 3: Find the secondary side and compensate the capacitor c2. First, we can get (55) by equation (53). ) 1 _(1-^2) L'C' L2C2 is therefore available to us by equations (37) and (54)

(l-k2) = C2Rj-L2 L2C2 ' C22L2RJ => (C22L2RJ)·(\-k2) = (C2RJ -L2) ·L2C2 ^(C2Rj)-(\-k2) = (C2Rj-L2) ^>L2 = C2Rj-(C2Rj)-(\-k2) ^L2 = C2-Rj-k2(l-k2) = C2Rj-L2 L2C2 ' C22L2RJ => (C22L2RJ)·(\-k2) = (C2RJ -L2) ·L2C2 ^(C2Rj)-(\-k2) = (C2Rj-L2) ^&gt ;L2 = C2Rj-(C2Rj)-(\-k2) ^L2 = C2-Rj-k2

其次由式(37)與(53)吾人可得二次側並聯補償電容(：2的數學關係式 (1-灸2) K)2 =>C2 l2c2 (l-k2) ^Lf-L2 (56) 步驟四：求得一次側串聯補償電容ς 將式(56)代入(55)中吾人可得設計非接觸型變壓器二次側自感Α的數學關 係式 L2=c2-Rj-e 21 1327402 —r (1-^) VW ocSecondly, by the equations (37) and (53), we can obtain the secondary side parallel compensation capacitor (: 2 mathematical relationship (1- moxibustion 2) K) 2 => C2 l2c2 (l-k2) ^Lf-L2 ( 56) Step 4: Find the primary side series compensation capacitor ς Substituting equation (56) into (55), we can obtain the mathematical relationship of the secondary side self-inductance of the non-contact type transformer L2=c2-Rj-e 21 1327402 —r (1-^) VW oc

R 2k2 or *V 以-(1]2) (57) 將式(53)代入(52)中吾人可得設計非接觸型變壓器一次側自感A的數學關係 式 μ \R 2k2 or *V with -(1]2) (57) Substituting equation (53) into (52), we can design the mathematical relationship of the primary side self-inductance A of the non-contact type transformer.

kl、H k"'K Α·(ΐ-^2)Kl, H k"'K Α·(ΐ-^2)

A kKRac Ps{C〇L)^{\~k2) (58) 傷 由式(45)吾人可得設計非接觸型切換式直流電源供應器一次側補償電容q 的數學關係式 =>c, ,K)2· (59) 經由上述四個步驟可將此複雜的技術歸納出一套簡易的設計準則，可以根 據此準則很方便地設計可分離式隔離感應電力傳輸系統的阻抗匹配補償 之功率電路元件參數。圖3與圖4為相應之模擬波形圖。由圖3與圖4可 知本發明於輸入電壓源為弦波且頻率為叫工作時’正如理論所期般其輸 22A kKRac Ps{C〇L)^{\~k2) (58) Injury formula (45) The mathematical relationship of the primary side compensation capacitance q of the non-contact switching DC power supply is designed =>c, , K)2· (59) Through the above four steps, this complex technology can be summarized into a set of simple design criteria, according to which the power of the impedance matching compensation of the separable isolated inductive power transmission system can be conveniently designed. Circuit component parameters. Figure 3 and Figure 4 show the corresponding analog waveforms. 3 and 4, the present invention can be used when the input voltage source is a sine wave and the frequency is called a work.

Claims (1)

1327402 ⑽年㈣修正替換頁 、申請專利範圍： 1. :種=合感應電力傳輪系統功率補償的電路元件參數設計方 (b)該第二電容值之設計值為c = (1-灸2) ’’、、2— (C)該第一電容值之設計值為^ Κ)2·Α5 其中，无為-可分離式隔離變壓器之輕合係數；為該第一自感 與該第一電容之串聯共振頻率；Α為該第一自感之感值；ι2為該 第二自感之感值；以及 該弱麵合感應電力傳輸系統包括： 該可分離式隔離變壓器，具—次顺：_，該— 自感，該二次側具該第二自感； 、Λ第一 該第-電容，串聯該可分離式隔離變壓器之該一次側； 參 該f二電容’並聯該可分離式隔離變之該二次側； -交流電阻負載，並聯該可分離式隔離變壓器之該二次側。 2.如申請專利範圍第i項所述之方法，進一步包括該第 感值由該交流電阻負載…輸出功率與—輸人電壓決定。感之 3‘如申請專利範圍第1項所述之古、丄 、 -非接觸式變壓器。方法，該可分離式隔離變壓器為 4.如申請專利範圍第1項所述之方 流裝置與-直流電阻負載組成。’以、机貞栽係由-整 241327402 (10) Year (4) Amendment replacement page, patent application scope: 1. : = = Inductive power transmission system power compensation circuit component parameter design side (b) The second capacitance value design value is c = (1 moxibustion 2 ) '',, 2 - (C) The design value of the first capacitance value is ^ Κ) 2 · Α 5 where, the light-coincidence coefficient of the non-separable isolation transformer; for the first self-inductance and the first The series resonant frequency of the capacitor; Α is the sense of the first self-inductance; ι2 is the sense of the second self-inductance; and the weak-faced inductive power transmission system comprises: the detachable isolation transformer, with a second-order : _, the self-inductance, the secondary side has the second self-inductance; Λ the first of the first capacitor, the primary side of the separable isolation transformer is connected in series; the f-capacitor is connected in parallel to the separable The isolation is changed to the secondary side; - an alternating current resistance load, in parallel with the secondary side of the separable isolation transformer. 2. The method of claim i, further comprising determining, by the AC resistance load, the output power and the input voltage. Sense 3 'As in the application of patent scope, the ancient, 、, - non-contact transformer. The detachable isolation transformer is composed of a square current device and a DC resistance load as described in claim 1 of the patent application. ',,,,,,,,,,,