Buck Converter Inductance Dimensioning

Hello everyone,

i'm developing a buck converter for a solar application.

For such, i've been following "Buck-Converter Design Demystified" Paper but i wanted to know if i am missing something in my calculation.

So the Values for the system are:
VinMax = 600V
VinMin = 400V
Vmpp = 480V
PinMax = 7,6kW

VoutMax = 380V
VoutMin = 280V
Fsw = 15kHz

So, following the paper logic, i've arrived to a 4,7mH for my worst case scenario :
LIR = 0.1 ; VinMax = 600; Vout=380V;

Do i need to have other concerns when dimensioning a inductance for this kind of converter and this power values?

Thank you all for your time.

Hi
Current is important, Around 300uH is the inductance required


L min(H) = (Vin max – Vout)*(1 -- Vout/Vin max) / 1.4*I out * F
Iout = output current in A
F switching frequency in Hz

smijesh said:

Hi
Current is important, Around 300uH is the inductance required


L min(H) = (Vin max – Vout)*(1 -- Vout/Vin max) / 1.4*I out * F
Iout = output current in A
F switching frequency in Hz

Click to expand...

Hi,

The Max Current i've considered is around 28A (P/Vmin).
So my dimensioning is way higher than what i really need?

Way such a big difference between your method and result and the one i've used?

Thnak you.

Sorry small correction
L min(H) = (Vin max – Vout)*( Vout/Vin max)*(1/F)*(1/I out * LIR)
Iout = output current in A
F switching frequency in Hz
LIR is the constant (Percentage of ripple)


**broken link removed**

We get 1.5mH
Hope it is helpful

smijesh said:

Sorry small correction
L min(H) = (Vin max – Vout)*( Vout/Vin max)*(1/F)*(1/I out * LIR)
Iout = output current in A
F switching frequency in Hz
LIR is the constant (Percentage of ripple)


**broken link removed**

We get 1.5mH
Hope it is helpful

Click to expand...

Using that same formula i get different values. What values did you use to get to that inductance?

But other than the inductance value, and make sure that the saturation current of the inductance is higher than the peak current, do i need to have any further concerns?

Thanks.

I considered 19A if we get 1mH when 28A

You have to take care the saturation current when designing the inductor
or make very small air gap the core to avoid saturation

smijesh said:

I considered 19A if we get 1mH when 28A

Click to expand...

considering the 20A for the 380V output i still get a much higher value.
I'll redo my calculations to see if i'm missing something.

Thanks

Since I like to experiment with simulations, here are results I get.

PinMax = 7,6kW

Fsw = 15kHz

VinMax = 600; Vout=380V

Click to expand...

Consider interleaving two or more buck converters, since your other thread raises a doubt about building an inductor which can handle upwards of 30A.

BradtheRad said:

Since I like to experiment with simulations, here are results I get.

Consider interleaving two or more buck converters, since your other thread raises a doubt about building an inductor which can handle upwards of 30A.

Click to expand...

Hi,

Thanks for the results.
As i said, with my calculations the inductor value is a bit higher.

Im considering that. Using 2 interleaved bucks.

but still, i want to know how to dimension for each case .

thank you for your help.