Using igbt in boost converter

Hi,

I am currently designing a dc/dc boost converter to get the output of a solar panel (output is 17.7V) up to 100V.

to get the duty ratio i have done the calculation:

k= 1 - Vs/Vo = 1 - (17.7/100) = 0.823

I want to use an igbt for the switch, and i am getting confused about the timing calculations....

I want the igbt to switch at 20kHz so i have done the calculation:

Tp = 1/f = 50us

On the datasheet of the igbt it states the following typical values:

turn on delay time - 25ns
rise time - 17ns
turn off delay time - 215ns
fall time - 13.2ns

This totals 270.2ns which i will need to add to the Ton time to work out the total time it takes to completely turn on and off the switch.

This is the part where i am getting confused....what sets how long the switch is on for? i.e. Ton

Is the calculation below correct?

using Ton = K.Tp
Ton + 270.2ns = 0.823 x 50us
Ton = 40.88us

Basically i need to know if this is correct as these values are used to calculate the L and C values in the converter.

Any help on this would be greatly appreciated.

Thanks in advance.

20Khz? Thats a pretty Low Value For a SMPS. Most run at 100 Khz to 1Mhz to make parts small and efficient.

There are other formulas to figure out C out and L based on Freq.
https://www.daycounter.com/LabBook/BoostConverter/Boost-Converter-Equations.phtml

if your thinking about trying to get to AC voltages, forget about it. Your better off charging a battery and using an inverter.