- #1
NexusN
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Homework Statement
I am struggled by a problem encountered in studying SMPS, on reading the book 'Switching Power Supply A-Z',
for a passage talking about the effect of increase in load current of a DC-DC converter on the inductor size,
from the ripple factor formula, r=(delta I)/I_L,
with increased I_L, we will need to also double the delta I to keep the value of ripple factor constant, and we know that
delta I = (Et)/L, for Et keeping constant, the only way is to half the existing inductance of the inductor, then we can get delta I doubled.
From the above, the energy to be stored by the inductor, represented by E= 1/2 (L*I^2), where the I here is the peak current passing through the inductor,
The resulted E of doubled load current will be a doubled E, as L is half-ed while I is doubled.
The book then concluded that :
"Therefore, we can generally state that the size of the inductor is proportional to the
load current."
This confused me a little bit, didn't we half the inductance of the inductor?
How come we finally get a inductor of larger size?
Reading from Wikipedia, I found that a longer inductor would provide a lower inductance(more accurately less windings per unit length), is the length the factor increased here?
So my question is, what exactly the size of an inductor depends on?
The inductance? Or the Energy storing ability of the inductor?
If that is the latter, would you mind giving me the formula describing it?
Thank you.