# How can the charge time/startup time for a boost generator be calculated?

I've been reading about boost converters and how to do calculations with them, but I haven't found equations/an equation for determining the startup (?) time for them (where the load has a high resistance value so it can be approximated as being capacitive). This being the time it takes for the average voltage curve to reach the high value.

One way I was thinking of for finding this was if the energy of the inductor is transferred to the capacitor during each cycle and by determining the di/dt of the inductor, the peak current of the inductor can be determined and thus the energy that the capacitor.
So for a boost generator with an inductor with inductance L=10^-6 Henries and a output load capacitor with a capacitance of 10^-6 Farads, the equation Vl=L*(di/dt), with Vl=10 volts would yield (di/dt)=10/L=10^-7 Amps/second (assuming just a linear current rise). Assuming a duty cycle of 95%, the current from the inductor flows through to charge the capacitor during the off time, which is for 10^-6 seconds if the frequency is 50 khz, which would mean that (dt)=10^-6 seconds, and thus (di)=10 amps. If the peak current through the inductor is 10 amps, then the energy stored is (1/2)(LI^2)=0.5(100)(10^-6)=5(10^-5) Joules, and the energy that flows into the capacitor would be found by setting that equal to 1/2(C)(V^2), and solving for V, the inductor would charge the capacitor by 10 volts each cycle. Then to get the time for the final voltage you could calculate how many increments there would have to be to reach the final output voltage. The simulation I ran seems to approximate this, at least during part of the time period:

The problem is that as time goes on, the voltage the capacitor is charged by during a given cycle is reduced over time. That's where I get stuck:

Sorry if my understandings is off with this subject but I'm trying to understand this with the math that I know as of this point.

$$\text{Energy} = \dfrac{1}{2}\cdot C\cdot V^2$$
Of course after the first energy transfer event, the capacitor starts to acquire terminal voltage (call it $$\V_i\$$) and the next energy transfer lifts if from $$\V_i\$$ to a level a little bit higher. But each subsequent dose produces a slightly less increase in voltage as per the energy equation of a capacitor being related to voltage squared.