15J Examples

IDDA

IJV 13 a

15 V

Fig. 15.8 Muliiple-output full-bridge isoLiied buck converter example.

Area

- = VDT, 0

- V.

Fig. 15.9 Transforiner wavefonns, full-bridge convener example,

--jv+ -isv

caused by diode forward voltage drops, tme finds that the desired transformer tuins ratios fli.n.riy are 110:5:15. A ferrite EE consisting of Magnefics, Inc. P-mateiiai is to be used in this exiimpie; at 75 It Hz, this material is described by the foiiowiiig parameters: = 7,6 W/rcm 3 = 2,6. A fill factor of K = 0.25 is assumed in this isolated multiple-output application. Total power loss of P, - 4 W, or appi-oxiinateiy 0.5% of the lo;id power, is alitjwed. Copper wire, having a resistivity of fl = 1.724 10 * Si-4;m, is to be used.

The applied primary voit-seconds are

к, = D-!\V = Ш.75)(б.б7 psec )(L60 V) = 800 V- usee The primary rms cirrreut is

The 5 V secondary whidhigs carry nns tiirrent

/D - 5.7 A

/j = i/5vl + 7:> = 66.1A

The 15 V secondary windings carry rms current

The totaJ nns winding current, referred to the ptiniary, is

= 14.4 A

The core size is evaluated using Eq. (15.19):

, (1.724-10-(8(1(1- lf-)( 14,4)(7.6)1

4(0.25)(4)( *

= 0.00937

(15.35)

(15.36)

(15.37)

(15.38)

(15,39)

(15.40)

The EE core data of Appendix D lists the EE40 core with A: = 0.0118 for P = 2.7, Evaluation of Eq. (15.16) shows that = 0,010B for this core, when P = 2.6. In any event, EE40 is the smallest standard EE core size having A < 0,00937. The peak ac flux density is found by evaluation ofEq. (15.20), using the geometrical data for the EE40 core:

AS =

s (l.724 1Q-)(800.10-)(14.4) (S.5)

2(0.25)

(1.1)(1.27)(7.7) (2,б)(7:б)

(15.41)

= 023 Tesla

This flux density is less than the saturation flux density of approximately 0.35 Tesla. The primary turns are determined by evaluation ofEq. (15.21):

4 (800.10 2(U.23)(1.27) = 13.7 turns

(15,42)

The secondary turns are fouud by evaluation of Eq. (15.22). It is desired that the transformer have a 110:5:15 turns ratio, and hence

;i2 = =0.62tJms flS.43)

Пз=-!?!, = 1.87 turns Ц5Л4)

In practice, we might select Mj - 22, = 1, and = 3. This would lead to a reduced with reduced core loss and increased copper loss. Since theresulting iff is suboptimal, the total power loss will be increased. According to Et], (15.3), the pealt ac flux density for the EE40 core will be

The resulting core and copper loss can be computed using Et]s. (15.1) and (15.7):

/V = O-6H0-lt3) (l.27)(7.7) = {).47 W (15.4f.)

(1.724.0-°)(800.](Г)-(14.4) ] ,

4(0.25) (l,l)(l.27) (0.143)- (l-5.47)

= 5.4W

Hence, the total power loss wojld be

r = Ff + P. :).9W (15-48)

Since this is 50% greater than the design goiil of 4 W, it is necessaiy to increase the core size. The next larger EE core is the EE50 core,having of 0.0284. The optimum ac tlux density for this core, given by Eq. (15.3), is AS = 0.14 T; operation at this flux density would require = 12 and would lead to a total power loss of 2.3 W. With ll = 22, calculations similar to Eqs. (15.45) to (15.48) lead to a peak tlux density of АП = 0.08 T. The resulting power losses would then be Pj-, = 0.23 W, P = 3,89 W, P, = 4,12W,

With the EE50 core and = 22, the traction of the available window area allocated to the primary winding is given by Eq. (15.23) as

The fraction ofthe available window area allocated to each half of the 5 V secondary winding should be The fraction ofthe available window area allocated to each half ofthe 15 V secondary winding should be The primary wire area A,i, 5 V secondary wire area А,,з, and 15 V secondary wire areaA, are then given