3. Determine number of winding 1 turns

Vw ((j4 (14.54)

Here, is expressed in Tesla and is expressed in ctn .

3. Determine number of secondary turns Use Che desired turns ratios:

(14.55)

Magnetizing inductance, rctcircd to witiding 1 (H)

Allowed total copper loss (W>

Winding fiU factor

Maximum operating flux density B (T)

The core ditnensions are expressed in ctn:

Core cross-sectional area A. (cm)

Core window area (cm)

Mean length per turn MLT (cm)

The use of centitneters rather than meters requites that appropriate factors be added to the design equations.

1. Deuiminc core size

p4C,1(,s , (l4.52)

Choose a core which is large enough to satisfy this inequality. Note the values of A W and MLT for this core. The resistivity p of copper wire is 1.724- 10 * i2-cm at room tetnpetature,and 2.3 10 fl-cm at 100°C.

2. Determine air gap length

(li-llO ( ,) (14.53)

Here, is expjes.sed in Tesla, is expressed in cm, and is expressed iu tneters. The permeability of free space is /(д = 471 10 H/m, This value is approximate, and neglects fringing flux and other non-idealities.

n,ht

5. Evaluate wire sizes

(14.57)

Select wire with bare copper area le.ss than or etjuai to the.se vahies. An .American Wire Gauge table is included in Appendix D.

14.4 EXAIVIPLES

14.4.1 Coupled Inductor for a Two-Output Forward Converter

As a first exanrple, let us consider the design of coupled inductors for the two-output forward converter illustrated in Fig. 14.12. This element can be viewed as two filter inductors that are wound on the same core. The turns ratio is chosen to be the same as the ratio of the output voltages. The magnetizing inductance performs the function of filtering the switching harmonics for both outputs, and the inagnetizing current is equal to the sum of the reflected winding currents.

At the nominal full-load operating point, the converter operates in the continnons conduction mode with a duty cycle of £> = 0.35. The switching frequency is 200 kHz. .At this operatitig poiut, it is desired that the ripple in the inagnetizing current have a peak magnitude equal to 20% of the dc component of magnetizing current.

The dc component of the magnetizing current / is

= (4A).g(2A) C-S)

= 4.86 A

The magnetizing current ripple Ai can be expressed as

4. Evaluate fraction of window area allocated to each winding

I = 200 kHz

turns

Output 1 28 V 4A

Output 2

=P У2 $ 12V

------1 ,,

Coupled inductor model

Fig, 14.12 TwOOutpj: forward converter example: (a) circuit sctiematic, (b) coupled inductoi model inserted into conveiter secondary-side circuit, (c) matienzing ourrenl and voltage waveforms of coupled indtictor, referred to winding 1.

Since we want Л!/ to be equal to 20% of 7, we should choose LfUS follows:

2Ai,

(28 У)(1-0.35Х5Ц5) ~ 2(4,86 AX20%) = 47 pH

The peak magnetizing current, referred to winding 1, is therefore

/M = f + AW = S.83A

(14,59)

(14.60)

(14.61)

Since the current ripples of the winding currents are small compared to the respective dc components, the