A6986I datasheet(39/62 Pages) STMICROELECTRONICS

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Figure 43. Primary (blue) winding current, secondary (pink) winding current and magnetizing current

(black)

ILsec

ILpri

ILpri_pos_peak

ILsec_peak

ILpri_neg_peak

Therefore, setting the current ripple ΔIL, the inductance is so defined:

Lpri = VIN−VOUT_pri∙VOUT_pri

VIN∙fSW∙∆IL

(39)

For example, assuming IOUT_prim = 500 mA, IOUT_sec = 100 mA, N= 5, the current ripple can be set around 30%

of the total current IOUT_pri + N×IOUT_sec = 1A, therefore 300 mA. If VIN = 12 V, VOUT_pri = 5 V and fSW = 500 kHz,

the inductor value should be 19.4 µH (and 18 µH or 22 µH the closest standardized value).

Peak and RMS current

As any inductor, peak and RMS current for each winding must be calculated in order to define saturation and

RMS currents that the transformer should fulfill.

For the primary winding, the equations below are valid:

Ipri_pos_peak = IOUT_pri+IOUT_secN+ΔIpri2

(40)

Ipri_RMS = IOUT_pri+IOUT_secN∙ 1+112∙ ΔIpri∙N

N∙IOUT_pri+IOUT_sec

(41)

For the secondary winding the peak current can change depending on the leakage inductance. In the image

below secondary winding current waveforms with different leakage inductances are simulated. It is evident

how the peak current can significantly vary. Considering the target leakage inductance value for an iso-buck

(recommended up to 1% of the primary inductance), the waveform can be approximated with a sawtooth shape

and the peak and RMS currents can be hence estimated as:

Isec_pos_peak = 2∙IOUT_sec

1−D

(42)

Isec_RMS = Isec_pos_peak∙ 1−D3

(43)

A duty cycle higher than 50-60% significantly increases the peak current in the secondary winding. Furthermore,

this affects the negative peak current at the primary side (see Figure 44).

A6986I

Design of the external components

DS13481 - Rev 2

page 39/62

A6986I Datasheet(PDF) 39 Page - STMicroelectronics