- L = 22 μH
- \$t_{on}\$ = 26.11 μs
- I have tried to take the solve the equation for the inductor currents at each state
- But it does not seem right
The problem might be your inductor value of 22 μH
It should be much higher for a \$t_{on}\$ of 26.11 μs (D = 0.7833 at 30 kHz). For instance, with an applied DC voltage of 325 volts, after 26.11 μs, the inductor current would be nearly 386 amps.
Or, your problem might be the stated \$t_{on}\$ time of 26.11 μs.....
Back to (some) basics
For an output power (P) of 5 kW at 1500 volts, the load current is 3.3333 amps. The inductor has to raise 325 volts to 5000 volts at 3.3333 amps so, the power needed to be supplied by the inductor is 3.3333 x 1175 = 3916.67 watts. That's an energy (W) of 130.556 mJ per switching cycle.

So,
$$\text{Energy (W)} = \dfrac{\text{Power (P)}}{\text{Frequency(F)}} = \dfrac{L\cdot I^2}{2} \therefore \text{ }\Longrightarrow I = \sqrt{\dfrac{2\cdot P}{F\cdot L}}$$
But we also know this: -
$$V_{in} = L\cdot \dfrac{dI}{dt}$$
- Where \$dI\$ is the peak current in the inductor (used for calculating inductor energy) and
- \$dt\$ is the ON time duration of the switching cycle (\$t_{on}\$).
So, given that we know that the current ramps up linearly we can say: -
$$V = L\dfrac{I}{t_{on}} \Longrightarrow I = \dfrac{V_{in}\cdot t_{on}}{L}$$
Combining the two formulas to remove \$I\$, we can find \$L\$: -
$$L = \dfrac{V_{in}^2\cdot t_{on}^2\cdot F}{2\cdot P}$$
This results in an inductor value of 275.78 μH.
Sanity check
In 26.11 μs, with 325 volts applied, the current will rise to 30.77 amps. As a sanity check, this is an energy in the 275.78 μH inductor of 130.555 mJ (as previously calculated above).
If you did use 22 μH, the problem would be your \$t_{on}\$ time
Re-arranging the inductance formula above gives us: -
$$t_{on} = \sqrt{\dfrac{2\cdot P\cdot L}{V_{in}^2 \cdot F}}$$
Then, \$t_{on}\$ = 7.375 μs and D = 0.2213.
And the peak current would become 108.9 amps. This peak current is also the current that falls to zero during the energy transfer into the output capacitor.
To calculate the time taken for this current to fall to zero (during the 2nd period of the switching cycle aka \$t_{off}\$) we use \$V_{L}\$ = (1500 - 325) volts, L = 22 μH and this formula: -
$$V_{L} = L\dfrac{I_P}{t_{off}} = L\dfrac{108.9}{t_{off}} \Longrightarrow t_{off} = 22 μH\cdot\dfrac{108.9A}{1175V}$$
Hence, \$t_{off}\$ = 2.039 μs.
Finding conducting period for diode in DCM (theoretically)?
This is the time that the diode is conducting for, 2.039 μs.
Charge time = 7.375 μs, transfer in 2.039 μs and hold for 23.91 μs before the next cycle begins.