Greetings of the day to all,
i am beginner in power electronics and i wanted to ask how can we calculate the value of inductor and filter capictor for simple 48 to 12 V buck operation?
thanks

 

hi m,
Look at this link and PDF.
E
https://passive-components.eu/how-to-choose-the-right-inductor-for-dc-dc-buck-applications/

 

I recommend you use the MC34063A, for learning (40V, so it's not enough for you're 48, but you can put it on breadboard to learn). It is cheap, ubiquitous and well documented-- and it will do BUCK or BOOST conversion. In fact, EEVBlog has a fantastic video discussing the calculations described in the datasheet, which will give you, a far, far better understanding of what goes into buck/boost converter calculations and outcomes.

This may be useful for you, irregardless of which BUCK/BOOST converter you use to calculate efficiencies:

http://www.jmest.org/wp-content/uploads/JMESTN42352977.pdf

https://file.scirp.org/pdf/EPE_2013073111131714.pdf

and a particularl good Application Note on the theory and operation of the MC34063A showing calculations, waveforms, and how a signal is processed through it to achieve a desired output. While this may not be the right BUCK converter for you, the understanding these PDFs provide will help you with virtually any that you use. Even a cursory reading of them.

https://www.onsemi.com/pub/Collateral/AN920-D.PDF

 

Don't use the MC34063, apart from being inefficient, noisy and as old as the hills, it's only rated at 40V.
There are loads of better devices.
See https://www.ti.com/power-management...ching-regulators/step-down-buck/overview.html for more than enough choice!

Calculating the inductance is not an exact science (which is a good thing, because inductors are not particularly accurate and the inductance varies with the current)
Decide on how much ripple current you want. This is your choice. Between 10% and 40% of the maximum output current is a good place to start - more ripple current = more ripple voltage for the same value of capacitance.
Then:
V/L=dI/dt
and, for a square wave:
V/L = ΔI/Δt
Δt is the on time of the converter. Or, if it is a fixed frequency, Δt=(Vout/Vin)*(1/f)
ΔI is the ripple current you decided earlier
L is (obviously) the value of the inductance
and V is the voltage across the inductor (Vin - Vout)

You can repeat the calculation where Δt is the off time
ΔI and L are still the same
and V is the output voltage (i.e. the voltage across the inductor when the switch is OFF)
and you will get the same result.

As for the capacitor, ripple current rating is more important than capacitance. By the time you have found a capacitor with enough voltage and ripple current rating, the odds are that it will have enough capacitance.