SLYT830 June 2022
Although a baby boost converter can reduce a bulk capacitor’s size and capacitance, minimizing the converter’s footprint will help preserve the original high power density goal. Since a baby boost converter operates in a very short amount of time (an AC dropout event), the peak operational current and voltage stresses will determine your power-stage component selections, instead of continuous power dissipation. At VBulk,min, current stress should be at maximum. Select the boost diode and metal-oxide semiconductor FET (MOSFET) to handle current stress at VBulk,min while rated for VBulk,max. The baby boost inductor needs to handle peak current at VBulk,min.
Equation 2 determines the baby boost inductor inductance:
where VBB is the voltage at CBB, ΔiLBB is the baby boost inductor peak-to-peak ripple current and Fs,BB is the baby boost converter switching frequency.
Since the goal is to minimize the footprint of the inductor, Equation 3 assumes that the peak-to-peak ripple current equals twice the input current at VBulk,min and the maximum output power:
With VBB regulating to 390 V and assuming Fs,BB = 500 kHz, Equation 2 calculates LBB as 7.385 µH.
Because the footprint is a higher design priority than the power dissipation, an inductor core with a higher saturation point is preferable; in the case of the baby boost converter, a powder iron core is a better choice than a ferrite core. The soft-saturation characteristic of the powder iron core [7] makes the design of the baby boost inductor design a bit tricky, however. With the core permeability dropping (inductance dropping) as the current increases, you must ensure the LBB calculated in Equation 2 is the inductance at iLBB peak. Equation 4 estimates the inductance at a given magnetizing field:
where AL is the inductance factor in henry/turns2, µi% is the remaining percentage of the initial permeability at a given magnetizing field, and N is the number of turns applied to the inductor.
Equation 5 expresses the relationship between µi% and the magnetizing field, according to the core manufacturer:
where a, b and c are constant coefficients and H is the magnetizing field.
Assuming the application of a Magnetics 0076381A7 [8] (a Kool Mμ Hƒ core [9]) for the baby boost inductor, the constant coefficients a, b and c would be 0.01, 4.064∙10-7 and 2.131, respectively.
According to Ampere’s law, Equation 6 expresses the relationship between H and N:
where I is the current through the winding and le is the effective magnetic path length in centimeters.
With Equation 2 and Equation 3 calculating LBB, Equation 4, Equation 5 and Equation 6 determine the N needed to achieve the inductance at a given magnetizing field.
It is also possible to estimate N iteratively. Assuming that a given inductor has an inductance that operates at a certain H with a given current, you can use Equation 4, Equation 5 and Equation 6 to evaluate whether the calculated H is close to the assumed H.
For example, if your initial guess is that H = 140 Oe with I = 25 A and the inductor has an inductance of 7.385 µH, Equation 4 calculates µi% at 39.65%. Then, taking the LBB calculated from Equation 2 and Equation 3, along with the µi% calculated into Equation 4, N is then 20.8.
To verify H using Equation 6 with calculated N, you have H = 125.67 Oe. Since there is still error between the guessed H and the calculated H, you can make a second guess about H and calculate H again until the error becomes negligible. You will find the correct turns (operation point) after a couple of iterations. Using the iterative method, H is 108.75 Oe, with N = 18.009. The inductance is 7.385 µH at 25 A.