Show all necessary steps making sure to answer all of the qu

Show all necessary steps, making sure to answer all of the questions.

Thank you.

A boost converter operates at the following quiescent point: V_g 28 V, V = 48 V, P_toad = 150 W, f_s = 100 kHz. Design the inductor for this converter. Choose the inductance value such that the peak current ripple is 10% of the dc inductor current. Use a peak flux density of 0.225 T, and assume a fill factor of 0.5. Allow copper loss equal to 0.5% of the load power, at room temperature. Use a ferrite PQ core. Specify: core size, air gap length, wire gauge, and number of turns.

Solution

step 1 : Calculating duty cycle

D = Vo-Vs/V0= 4828/48 = 0.42

Input current =Idc=IL= T0/(1-D)= P0/V0/(1-D) = 150/48/(1-0.42) = 5.387 A

I pp = 10% of Idc+idc = 5.925

Irms = ipp/root12 = 1.712

winding resistance = R = Pcu/irms2 = 0.75/ (Irms )^{2 =} 0.225 ohm

now Inductance L = V0D\'T / ripple current

= 28 * (1-0.42) *10 micro second / (0.538)

= 301 micro henrry

Determining core size

Core size = Kg >= L²I_max²resistivity / (B_max² *R *K_u )10⁸

= 0.098

from PQ ferrite data chart

kg >= .098 nearest value is for PQ26/25

hence A_c = 1.18

W_A= 0.503

MLT=5.62

L_m=5.55

Determinign air gap length

Lg = 4pi*10^-7 *L*I_max² / B_max2A_{c =} 4 *3.14 *10^-7 *301*10^-6*5.925²/ (.225)² *1.18 = 0.222 mm

Number of turns

n = L*I_max / B_max *A_c 10⁴

= 301 microH*5.925/0.225*1.18 *10⁴

⁼ 67