For Peak Inductor Current (\(I_{L(pk)}\)):
\[ I_{L(pk)} = \frac{2 \times 1.414 \times P_O}{\eta \times V_{AC(LL)}} \]
For Inductance (\(L_P\)):
\[ L_P = \frac{t \left(\frac{V_O}{\sqrt{2}} - V_{AC(LL)}\right) \times \eta \times V_{AC(LL)}^2}{\sqrt{2} \times V_O \times P_O} \]
For Switch On-Time (\(t_{on}\)):
\[ t_{on} = \frac{2 \times P_O \times L_P}{\eta \times V_{AC(LL)}^2} \]
Po: Required converter output power.
Vo: Desired output voltage.
\(\eta\): Efficiency of the converter.
Vacll: Minimum AC RMS line voltage.
Lp: Inductance.
t{on}: Switch on-time.
Let's assume the following values:
Step 1: Calculate Peak Inductor Current (\(I_{L(pk)}\)):
\[ I_{L(pk)} = \frac{2 \times 1.414 \times 80}{0.92 \times 90} = 2.74 \text{ A} \]
Step 2: Calculate Inductance (\(L_P\)):
\[ L_P = \frac{40 \times 10^{-6} \left(\frac{230}{\sqrt{2}} - 90\right) \times 0.92 \times 90^2}{\sqrt{2} \times 230 \times 80} = 832.4 \mu H \]
Step 3: Calculate Switch On-Time (\(t_{on}\)):
\[ t_{on} = \frac{2 \times 80 \times 93.47 \times 10^{-6}}{0.92 \times 90^2} = 0.0174 \text{ ms} = 17.4 \mu s \]
