Coupled Inductors and Transformers

Transcription

1 Cupled nductrs and Transfrmers Self-nductance When current i flws thrugh the cil, a magnetic flux is prduced arund it. d d di di v= = = dt di dt dt nductance: = d di This inductance is cmmnly called self-inductance, because it relates the vltage induced in a cil by the time-varying current in the same cil.

2 Mutual nductance = + d d di di v = = = dt di dt dt d d di di v M = = = dt di dt dt Reverse directin: = + d d di di v = = = dt di dt dt d d di di v M = = = dt di dt dt Mutual inductance is the ability f ne inductr t induce a vltage acrss a neighbring inductr.

3 Fr linear magnetic cupling: M = M = M Determining dt markings: - Arbitrarily mark ne terminal f ne cil (D) - Assign a current int the dtted terminal - Use the right-hand rule t determine the directin f magnetic flux - Arbitrarily pick ne terminal f the secnd cil (A) and assign a current int this terminal - Use the right-hand t determine the directin f magnetic flux - Cmpare the directin f the tw fluxes. - f they have the same directin, place a dt n the selected terminal (A) f the secnd cil. - f the fluxes have ppsite directins, place the dt at the ther terminal (B) f the 3 secnd cil.

4 Dt Cnventin: f a current enters the dtted terminal f ne cil, the reference plarity f the mutual vltage in the secnd cil is psitive at the dtted terminal f the secnd cil. 4

5 Energy Calculatins Energy stred in magnetically cupled cils: w= i + i Mii Since the stred energy cannt be negative i + i - Mii ³ 0 T cmplete the square, add and subtract ii ( ) i ( ) - i + ii - M ³ 0 - M³ 0 r M Cefficient f cupling k k M = 0 k 5

6 Example Calculate the phasr currents. and. Mesh : - + (- j4+ j5) - j3 = 0 r j - j3 = ( + j6) Mesh : - j3 + (+ j6) = 0 r = = (- j4) j3 j4.04 Substituting: j(- j4) - j3 = r = =.9e A 4- j -j63.43 j4.04 -j49.39 Substituting: = (- j4) = 4.47 e.9e A = 3.0e A 6

7 Find the steady-state expressins fr the currents i g and i when v g = 00 cs 0,000t. Find the cupling cefficient. Example Phasr circuit: 00 (5 j0 ) g j5 0 g (0 j5) A (5 j0 ) j5g 0 5 A i ( t) 8.03cs(0,000t56.3 ) A g i () t 5cs(0,000) t A k M (0.00)(0.00) 7

8 Example 3 Determine the cupling cefficient and the energy stred in the cupled inductrs at t =.5 s. M Cupling cefficient: k = = =

9 Mesh : 0 = (4- j4 + j4) - j r 0 = - j () Mesh : 0 =- j + ( + j) r = ( + j) () Substituting () int (): (- j3) = = 0 (- j3) =.78e A = 3.93e A n the time dmain: i = 3.93cs(t+.3 ) A i =.78cs(t ) A At t=.5s, t= 3rad = 7.9 i i = 3.93cs( ) A =-3.94 A =.78cs( ) A =-.85A The ttal energy stred in the cupled inductrs is given by W = 0.5 ( i ) ( i ) -0.5 M( ii ) = 0.5 H( A) H( -.85A) -0.5 H(3.94A.85A) = 9.85J 9

10 inear Transfrmers Find the input and utput impedances f the transfrmer. = ( Z + R + j) - jm S S jm 0 =- jm + ( R + j + Z ) = + + Z = Z = -Z S ab in S R j Z in M R j Z = Z R j Similarly, Z Z R j cd M R j Z = ut = S 0

11 Example 4 Find a) the input impedance f the transfrmer and b) the Thevenin equivalent wrt terminals at. M æ 00 ö Zin = R+ j+ = 00 j R+ j+ Z ç 00 + j j500 çè ø æ 00 (900 j900) ö + j ç ( j ) = = èç ø M æ 00 ö Zut = R+ j+ = 00 j R+ j+ Z ç S 00 + j j00 çè ø æ 00 (700 j3700) ö - ç ( ) = 00 + j600 + = j4.6 = Z èç ø Th

12 Example 4 cnt d te: The pen-circuit vltage will be j00 times the value f j79.9 = = 79.67e ma (700+ j3700) Th - j79.9 j0.7 = j e ma=95.6e

13 Example 5 Find the input impedance and the current frm the vltage surce. Z (4 8) é ( 3 ) ( ù in = + j + ê j - j + + j ) ú ë û = (4+ j8) = 8.58e j j58.05 ( j ) The current frm the vltage surce is = Z = e in =.65e A j

14 4 deal Transfrmer inear Transfrmer 0,, R R k deal Transfrmer Prperties: Hw can we analyze a circuit with an ideal transfrmer where there is n cil impedance and n mutual impedance? ; j M j j M j

15 deal Transfrmer = r = = r = f the cil vltages and are bth psitive r negative at the dt-marked terminal, use a plus sign. Otherwise, use a negative sign. f the cil currents and are bth directed int r ut f the dt-marked terminal, use a minus sign. Otherwise, use a plus sign. 5

16 nput mpedance :n = where n = n = n Z = =, = Z Z = Z in in n n Transfrmers can be used fr impedance matching. Z in æ ö =ç ç ç çè ø Z Cmplex Pwer * ( ) * * S= = n = = S n n an ideal transfrmer, the cmplex pwer supplied t the primary cil is delivered t the secndary cil withut lss. 6

17 Example 6 The surce vltage in the phasr dmain circuit is 5 k. Find the amplitude and phase angle f and. 3 s 5 0 j Z+ç ç ç Z çè ø ( ) ( j ) ( ) j36.87 = = A= 4+j3 A = 5e A æ ö ( ) ( ) ( ) = s - Z= 5, j j3 A= 37, 000- j8,500 j4.39 =- = ( j40) =868.5e 5 j e =5 j6.87 e A = = Z j ( 4-4.4) 7

18 Example 7 An ideal transfrmer is rated as 400/0, 9.6 ka, and has 50 turns n the secndary side. Calculate: (a) the turns rati, (b) the number f turns n the primary side, (c) the current ratings fr the primary and secndary windings. 0 (a) n = = = (b) n= 0.05 = = = 000 turns 0.05 (c) S = = = 9.6 ka 9600A 9600 = = A = 4A A 9600 = = A = 80A 0 8

19 Example 8 deal transfrmers are als used in pwer transmissin frm a generating plant t a residence. Find the turns rati fr each f the ideal transfrmers. 0k 0k 0k 69k 69k 3.8k 3.8k 0 n n n n 3 4 n n 0.34 n 0. 5 n

20 Example 9 n the ideal transfrmer circuit, find and the cmplex pwer supplied by the surce. The (6 - j4) impedance can be reflected t the primary: Z = + (6- j4) 6 = (3- j.5) in j6.57 = 00 (3- j.5) = 9.8e A j6.57 j6.57 =- n=- 9.8e A 4 =-7.454e A 0 ( - )( ) j90 j6.57 j6.57 =- j4 = 4e e A = 78.9e ( )( j ) * * j6.57 = =- j = e = e S A.98 ka 0

21 Example 0 The variable resistr is adjusted until maximum average pwer is delivered t it. Find the value f the resistr and the pwer delivered t it. Hw much pwer is assciated with the ideal transfrmer? Calculate c : c c 0 0; (0) 080 rms

22 Calculate sc and R Th : sc ; sc R 6A Th A rms sc 680A 0 35 Pwer analysis: A ; 3A 380 A rms rms P P S (840)(0.75A) 630 W; P60 60 (0.75A) W 0 (3.75A) 8.5 W; P 35 (3A) 35 W 0 35 P P P P P 0(!) Transfrmer S

23 Equivalent Circuit The T-equivalent circuit is mst cmmnly used fr transfrmers. é ù é j jmùé ù = úê ê ú ê jm j úê ú ë û ë ûëû By cmparisn: ( + ) é ù é j j ù é ù ê ú a c c = ê ú jc j( b c) ê ú ë û ê + ú ëû a ë = - M = -M b c = M û 3

24 Example Determine the T-equivalent circuit f the linear transfrmer. = - M = 0H- H = 8H a = - M = 4H- H = H b c = M = H 4