Inductor sizing and it energy handling ability (SMPS related)

[NexusN]

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.

Homework Equations

The Attempt at a Solution

 

[MisterX]

I wouldn't call myself a SMPS expert, but here is something regarding at least some switched-mode DC-DC converters:

If you want the converter to be in continuous conduction mode, then the peak to peak inductor current ripple should be less than twice the average inductor current. This is dependent on frequency, as at lower frequencies there will be more time for the inductor current to ramp up or down. Thus the peak to peak current ripple on the inductor will be larger at lower frequencies. The peak to peak inductor current ripple is also dependent on inductance. v = L di/dt, and with v constant during each part of the cycle, di/dt, the slope of the current ramps, will be greater with a lower L.

I suppose there might also be a separate issue of avoiding magnetic saturation to consider.

 

[uart]

Hi Nexus. Your problem is a bit like the classic,

Q. How do you confuse an EE 101 student?

A. You tell them that the power dissipation in a resistor is equal to $R I^2$, and therefore proportional to "R". Then you tell them that the power dissipation is also $V^2 / R$, and therefore inversely proportional to "R".

Of course each equation is true, but it can seem confusing when a quantity depends on several different, but inter-related, variables as in your inductor example.

The solution is simply that,

$$\Phi = \frac {L I}{N}$$

Now at first sight it may seem that you're halving "L" so you're halving Phi.

Hopefully though you'll also notice that you're doubling "I" so at second glance you might think that "Phi" remains constant.

Sadly though you'd still be wrong. Since we are reducing L we will also be reduce N, (which is on the denominator), so Phi will increase*. This means we generally require a larger core and therefore a physically larger inductor.

* You may perhaps argue that we could reduce "L", not by reducing "N", but instead by increasing the air-gap on the same physical sized core. Since we're now requiring larger diameter wire however (for the higher current rating) it is not normally possible to fit the same number of turns. So the core size must increase.

 

[NexusN]

Thank you for your replies, I have more ideas now if it is the case similar to the resistor formula,
it looks the inductance of the inductor relates to the inductor size in a complicated way,
I'd better learn more about how each factors affect the others before I can move on in the SMPS game.

Mister:
Yes, the inductor should be so picked that we can avoid the magnetic saturation of it, while at the moment I don't know the factors that may affect the capability of energy holding of an inductor, and if this capability is related to its inductance.
As from the formula I read from Wiki, increasing length(with constant number of winding) and decreasing cross section area of an inductor both will lower the inductance, while they are affecting the inductor size in the opposite way.

 

[uart]

As from the formula I read from Wiki, increasing length(with constant number of winding) and decreasing cross section area of an inductor both will lower the inductance, while they are affecting the inductor size in the opposite way.

Decreasing cross section area of the inductor (with constant number of windings) will cause it to saturate the core. Phi = LI/N stays constant in this case, yet the core area decreases. Therefore this is not an option. So the only option is to increase the size of the core.

Read my previous post it has the correct solution. Ask questions if you don't understand.

 

[NexusN]

Decreasing cross section area of the inductor (with constant number of windings) will cause it to saturate the core. Phi = LI/N stays constant in this case, yet the core area decreases. Therefore this is not an option. So the only option is to increase the size of the core.

Read my previous post it has the correct solution. Ask questions if you don't understand.

Thank you for your follow-up, with cross section area untouched,
does it mean we can only decrease the inductance of the inductor by increasing the length, thus the size of the inductor to achieve what we want,
and this is the reason the book concluded that the load current is proportional to the inductor size?

 

[uart]

Thank you for your follow-up, with cross section area untouched,
does it mean we can only decrease the inductance of the inductor by increasing the length, thus the size of the inductor to achieve what we want,
and this is the reason the book concluded that the load current is proportional to the inductor size?

Yes, since reducing the area doesn't work (causes saturation) then the only other option is to increase the size of the core. We could increase just the "length"* but we often increase the area too. Realistically that means is a bigger core (more cross section) and a bigger air gap.

* In many cases we use gapped cores and the effective length of an inductor is actually controlled by the air gap (particularly for laminated iron or ferrite). The exception is things like a powdered iron toroid where the "distributed" air gap is integral to the construction and non adjustable.

 

[NexusN]

Yes, since reducing the area doesn't work (causes saturation) then the only other option is to increase the size of the core. We could increase just the "length"* but we often increase the area too. Realistically that means is a bigger core (more cross section) and a bigger air gap.

* In many cases we use gapped cores and the effective length of an inductor is actually controlled by the air gap (particularly for laminated iron or ferrite). The exception is things like a powdered iron toroid where the "distributed" air gap is integral to the construction and non adjustable.

Thank you very much for your help,
things are much clearer now.