Magnetics Design

Primary Constraints:Peak Flux Density (B field) in the core : Bmax (T or Wb/m2)

Core losses Saturation

Peal Current density in the windings : Jmax (A/m2)Resistive lossesHeat

Wire Cross Section: Acond (m2)

Wire with cross section Acond , carrying Irms amperes of current, has current density :

rms

cond

IJ

A

If we have a limit on J < Jmax, then we must choose a wire gauge with :

rmscond

max

IA

J

Core Window Area: AW (window through which all windings must pass)

AW AW

Torroid E Cores

Fill Factor kw, the fractional part of the window actually occupied by conductor cross sections.

Applying our previous constraint to Acond,

N Turns of wire with cross section Acond

w w condk A NA

rmsw

w max

NIA

k J

For transformers with multiple windings, the different windings must accommodate different currents, thus they must have different gauges (different Acond):

1 ,1 2 ,2 ,w w cond cond y cond yy

k A N A N A N A

However, maximum current density must not be exceeded for any winding, thus:

,y rms yy

ww max

N I

Ak J

Inductor Core Cross Section: Acore

The core cross section must accommodate the peak induced flux without exceeding the maximum allowable flux density.

ˆcore

max

AB

For an inductor, the peak flux is proportional to peak current and inductance:

ˆˆ LI

N

ˆcore

max

LIA

NB

Transformer Core Cross Section: Acore

In a transformer, the flux density may not exceed +/- Bmax over one AC

operating cycle at the worst case operating condition.

pri pri cm

pri pri s

V V kT

N N f

Let T be the maximum excitation time for the primary. This will be at most, half the period, or kc /fs , where kc is parameter reflecting the operating duty cycle of a power converter, typically 0.5 for worst case flux density.

Faraday’s Law tells us that the change in magnetizing flux density will be :

Therefore, the change in flux density is:

This places a lower limit on the core cross section:

pri cmmax

core pri s core

V kB

A N f A

c pricore

s pri max

k VA

f N B

**For sinusoidal AC operation, a value of kc = 0.225 is appropriate; Vpri is RMS.

Area ProductWe now have expressions for the minimum core window and cross section, as functions of maximum ratings and operating parameters, which tells us how big the device must be. These expressions are used to express a useful magnetic core design parameter we call the Area Product:

, ,

, ,

y rms y c y rms yc pri y y

p core min w mins pri max w max w s max max

N I k V Ik V

A A Af N B k J k f B J

For Transformers

, ,

ˆˆrms rms

p core min w minmax w max w max max

NI LIILIA A A

NB k J k B J

For Inductors

Quiz

Draw the waveform, and compute the peak and RMS values for a 0.75 A P-P triangular wave riding on a DC value of 5.0 A.

2 22 0.75 2 0.75

5 25 25.0469 5.005 5 Amp123

5 0.75 2 5.375 Amp

RMS

peak

I

I

Core SelectionFrom vendor data, we select a core with Ap, Aw, and Acore which meet our minimum requirements.

cy y

core max s

kN V

A B f

For a transformer, the number of turns for each winding is computed using :

For an inductor, the number of turns is computed using:

ˆ

core max

LIN

A B

The inductor air gap length is computed to provide the appropriate reluctance in the magnetic path for the desired inductance with the chosen number of turns:

Since Ag ~ Acore, and 0 << core, the gap length can be approximated as:

20 0g g

g corecore core

N A Al l

L A

20 core

g

N Al

L

Design Example: InductorDesign Specs

L=100 HIDC = 5 AI = 0.75 A p-pBmax = 0.25 T = 0.25 x 10-6 W/mm2

Jmax = 6.0A/mm2

fs = 100kHzkw = 0.5

4

4

-6

ˆ 10 5.375 53587mm

0.5 0.25x10 6.0rms

pw max max

LIIA

k B J

2 2

2 2 211.3 5.5 97.44 mm4 4 4core

F HA

22 5.5 18 11.3 36.85 mm2W

C FA D

497.44 36.85 3950 mmp core WA A A

4

6

ˆ 5.375 x 1022 Turns

0.25x 10 97.44max core

LIN

B A

25.00.83 mm

6.0rms

condmax

IA

J

2 1020

4

22 4 x 10 97.440.59 mm

10core

g

N Al

L

This would be 18 gauge wire, which is pretty stiff. The author suggests 5 strands of 25 gauge (0.162 mm2)

Design Example: TransformerDesign Specs

v1 = v2 = v3 = 30 vIRMS = 2.5 ABmax = 0.25 T = 0.25 x 10-6 W/mm2

Jmax = 5.0A/mm2

fs = 100kHzkw = 0.5kc = 0.5

,41800 mm

c y rms yy

pw s max max

k V I

Ak f B J

2 2263.9 mm

4 4core

F HA

22 29.2 mm2W

C FA D

22.50.5 mm

6.0rms

condmax

IA

J

1 2 3 -6 5

0.530 10 Turns

63.9 0.25 x 10 10c

core max s

kN N N V

A B f

This would be 20 gauge wire. The author suggests 3 strands of 25 gauge (0.162 mm2)