the relationship of soft-start capacitor and output cap

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staric

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I want to know the relationship of soft-start capacitor and output capacitor. Whether it can describe with a equation. Who can give me answer? Thanks.
 

For a simple case of a current mode converter, it is modelled as an error amp and a power GM output stage.

For the very crude soft start where you clamp the output of the error amp to the voltage at SS, you can think of the soft start as a triangle ramp of current, proportional to Vss. It is this triangle ramp of current that fills the output cap, so you can just use I=C dV/dt where I is the current proportional to the clamped error amp.

For example, if error amp output ramps from 0-1v forcing the converter output current to ramp from 0-3A in the same time, Power Gm = 3A/V and the total energy delivered to the output is the integral of the current delivered to the output cap.

Rather than a general equation, just use simpler equations to model each part - it will be a good exercise for you to do this.

To get the soft start ramp you will use I=C dV/dt with Iss and Css. Multiply this voltage by power Gm to get the output current, then use I=C dV/dt again to get the output voltage. Your constants of proportionality are Iss, Css, Gmp and Cout.
 

But I have ever read one boost dc-dc datasheet of MAXIM. Calculate Css to be:
Css=21*10^(-6)*Cout*(Vout^2-Vin*Vout)/(Vin*Iinrush-Iout*Vout)
I cannot understand this equation. Who can tell me?
 

here you choose the inrush current you want, they have taken out the time dependence. What part number is it?
 

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