[SOLVED] Inductor quality value for switching regulators

[Easy peasy]

The duty cycle was defined above as 70% max with load on the booster - which is not too far from 50% i.e. a relatively even pulse arrangement.

most engineers would know that corner cases, e.g. V short or V long pulses, are just that, and need special attention - but even so losses in a buck or boost choke tend to be constant for constant RMS AC ( from 10% to 90% duty cycle ) and constant DC values of thru current, i.e. the average.

Arguing the corner cases does not do much for the OP except perhaps to add confusion ...

 

[Akanimo]

Hi,

What you should do is go to the power loss density versus deltaB chart in the datasheet for the core that you hope to use and use the Empirical Steinmetz equation to determine the core loss for your switching frequency (1.2MHz in your case) and deltaB.

Actually, core loss Pc = k1*(deltaB^beta)*vc
[where vc is the core volume, and both k1 and beta are fitting parameters already considered in the chart].

So just trace core loss density for your switching frequency at your design deltaB. Multiply that value by the core volume to get your core loss at that frequency.

 

[KlausST]

Hi,

Arguing the corner cases does not do much for the OP except perhaps to add confusion ...

I'm not talking about corner cases.
I'm talking about the use cases.
It's not unusual that a CCM driven inductor has a current of 550mA to 650mA (600mA DC plus 100mApp ripple) which is about 600mA RMS.
It's also not unusual hat a DCM driven inductor has a current of 0 to 2500mA ... with a duty cycle that also gives 600mA RMS.

"Q" ... how I've learned it - is defined for DC free sinusoidal waveform....but with a switcher
* the current is neither DC free
* nor sinusoidal.

The OP did not bring "Q" into the discussion, so I was rather confused why one brings "Q" into the discussion.
I try to be objective ... and also just out of curiosity I did an internet search for switcher inductor datasheets. I opend 5 of different manufacturers and checked whether they mention "Q". None of them did. I assume they have a good reason for it.

I also can't remember that I have seen "Q" of inductors mentioned in a switcher application note.

Am I wrong with this:
If "Q" is different for different frequencies (sine waveform), and because it's obvious that the typical swicher waveform is far away from sine ( because of it's overtones and DC) .... isn't it then obvious that one has to take care about the waveform?

But in the end everyone is free to design electronics his own way.
I'm just a friend of standards.
--> Thus I recommend the OP to keep on the good and freely available application/design notes from universities and the big manufacturers of inductors and switcher ICs. If they calculate and explain with "Q" ...then go this way.

Klaus

 

[c_mitra]

I also can't remember that I have seen "Q" of inductors mentioned in a switcher application note.

You are right; Q is meaningful only for a resonant system but for an ideal inductor, Q cannot be defined.

For a real inductor, there is always some losses and that can be used to define a time constant. But that may or may not be useful for a meaningful Q.

As an example, a LCR circuit has a clearly defined absorption spectrum and it is easy to define a Q for the circuit (but not for the individual component L or C or R).

For the same reason, Q cannot be defined at any arbitrary frequency: engineers use FWHM (full width at half maximum) to define the Q but it is strictly defined for a Gaussian waveform.

Form a mechanical engineering point of view, it is a simple case of forced vibration close to the resonant frequency.

Like capacitors giving rise to poles in a complex diagram: it depends on the circuit and not on the individual capacitor.

That is what I have learnt.

 

[KlausST]

Hi,

I totally agree with all of this.

From my experience with designing switching regulators:
I usually choose the inductors with the values given in the datasheets. Especially when they are loaded above 20% of their rating, the calculations were pretty close with reality in the circuit.
But when comparing different inductors at light load I experienced differences between different inductors that I could not explain with the datasheet specifications.
For one application I spent several days exploring the best inductor for the light load condition. The aim was to get longest run time from a battery. I used Excel to calculate the loss with the inputs. (For sure one possible problem was, that I was not able to "calculate" the effect that was caused when the switcher changed to discontinous burst mode)

--> I ended up to simply measure V and I of input as well as output (all clean DC values). This was an easy method to determine the best inductor for battery life....as well as overall loss and overall efficiencies.
The drawback is that one can not differentiate between inductor loss, switcher IC loss, catch diode loss...

Klaus

 

[Zak28]

There should be a distinct parameter or table depicting core losses as a function of magnetizing current and frequency.

Currently most datasheets are insufficient about their material losses.

 

[c_mitra]

But when comparing different inductors at light load I experienced differences between different inductors that I could not explain with the datasheet specifications.

Inductors are basically current device; they do well at high current (unlike capacitors that do well under light load).

For switching regulators, there will an optimum load at which the efficiency will have a peak (high efficiency) but performance will go down on either side.

But it is not a fault of the inductor; it is possible (I have not done any design work) to optimize the design at a given load. Simple switchers are not happy with variable loads.

- - - Updated - - -

Currently most datasheets are insufficient about their material losses.

Please see post #22.

Is really not so difficult to estimate hysteresis loss. For most ferrites, eddy losses can be considered zero.

Most ferrites are poor heat conductors and if there is considerable heat build up over time, the core will *****.

 

[Easy peasy]

I'm not sure C_mitra has a good handle on the real meaning of Q & therefore D. I cannot agree with much of what (s)he has posted.

A low loss coil has a higher Q than a higher loss one - every time ...

Q can be defined for a single part at a single freq, and over a freq band, Q is not limited to any resonance condition

It is called quality factor for a reason

Manufacturers shy away from specifying Q - as it generally points clearly to how poor the total losses are in their parts.

 

[FvM]

I agree that Q is related to power inductor core losses, although it's a small signal quantity. The commonly understood core loss parameter is however watts respectively watts per volume unit for a specific frequency and core flux.

I won't contradict Easy peasy that Q numbers can be used for power inductor comparison. But due to the strong flux to loss dependency (raised to a power of 2 to 3), it's also easy to misinterpret data.

 

[c_mitra]

I'm not sure C_mitra has a good handle on the real meaning of Q & therefore D. I cannot agree with much of what (s)he has posted....

Of course I have a very very limited knowledge. So I looked up other places:

1. Wikipedia is reasonably decent reference. It says

"In physics and engineering the quality factor or Q factor is a dimensionless parameter that describes how underdamped an oscillator or resonator is,[1] and characterizes a resonator's bandwidth relative to its centre frequency.[2] Higher Q indicates a lower rate of energy loss relative to the stored energy of the resonator; the oscillations die out more slowly. A pendulum suspended from a high-quality bearing, oscillating in air, has a high Q, while a pendulum immersed in oil has a low one. Resonators with high quality factors have low damping, so that they ring or vibrate longer."

2. Maxim Integrated is an electronics company: They say

"Glossary Term: Q Factor

Definition

A measure of the quality of a resonant (tank) circuit. A "high-Q" circuit has mostly reactive components (inductive and capacitive), with low resistance. It resonates strongly, with little damping (low loss). A high-Q circuit will have low bandwidth relative to its center frequency (that is, it will have a narrow bandwidth vs frequency curve).

Q = 2 π * (Energy stored / Energy dissipated per cycle)"

My comment: both of them mention resonant circuit- is that a coincidence?- these two definitions are equivalent under some conditions but I do not have the derivations ready...

3. A Stanford university site says the same thing:

"Quality Factor (Q)

The quality factor (Q) of a resonator may be defined as the resonance frequency divided by the resonator bandwidth:

$\displaystyle \zbox {Q \isdefs \frac{\mbox{ResonanceFrequency}}{\mbox{Bandwidth}}} $

where the resonance frequency and bandwidth must be given in the same units (e.g., Hz). The resonance frequency is typically defined as the frequency at which the peak gain occurs. The bandwidth is typically defined as the 3dB-bandwidth, i.e., the bandwidth between the -3dB points (half-power-gain points) straddling the resonance frequency.

We define Q in the context of continuous-time resonators, so that the transfer function $ H(s)$ is the Laplace transform of the continuous impulse-response $ h(t)$ , $ t\in\mathbb{R}$ (instead of the z transform of the discrete-time impulse-response $ h$ , $ n=0,1,2,\ldots$ ). An introduction to Laplace-transform"

I believe this is the most widely followed definition in classical textbooks.

Manufacturers shy away from specifying Q - as it generally points clearly to how poor the total losses are in their parts.

Just a poor personal opinion; they are interested in selling their products and if vendor1's Q is better then vendor2, they would certainly love to say so...

 

[Zak28]

For most ferrites, eddy losses can be considered zero.

There are many ferrite formulations each with their capabilities many ferrites will become very lossy at 1200khz with slight magnetizing current.

Is really not so difficult to estimate hysteresis loss.

They don't need to estimate, they should take a quantity and run them thru tests varying frequency and magnetizing current then have a software plot the results and publish the chart with the accompanying datasheet.

Manufacturers should be doing this perdue the many battery operated devices which require optimum performance and minimal losses.

 

[c_mitra]

There are many ferrite formulations each with their capabilities many ferrites will become very lossy at 1200khz with slight magnetizing current.

Most ferrites have very high electrical resistivity and that means that the eddy losses are very low.

What you are seeing is called hysteresis loss; it is dependent on the area of the hysteresis loop. Most of the ferrites have very low hysteresis loss.

If you want a low loss ferrite for 1200 kHz, you have to search because at high frequency the µ drops off; this has to do with the domain rotations and µ becomes complex with real and imaginary parts.

If you want to use high frequency (>1MHz) you need to reduce the level of magnetization and live with greater dissipation.

I randomly looked up the TDK site https://product.tdk.com/info/en/catalog/datasheets/ferrite_mn-zn_material_characteristics_en.pdf and it does specify all the important parameters quite explicitly.

Note that variations can be large (20-30% difference with published values are not uncommon) and you need to design conservatively.

 

[Easy peasy]

Respectfully:

Most ferrites have very high electrical resistivity and that means that the eddy losses are very low

is not that accurate for power grade ferrites, more accurately MnZn (power ) ferrites have only a few ohm-metre's of resistivity and NiZn ferrites have about x10 more for freq's of 500kHz up ...

 

[c_mitra]

more accurately MnZn (power ) ferrites have only a few ohm-metre's of resistivity and NiZn ferrites have about x10 more for freq's of 500kHz up ...

True.

For a comparison, metals like Ag, Cu and Au (the coinage metals) have more than a million times less resistivity (of the order of 1E-8) ...

These values (resistivity of common ferrites) are comparable to intrinsic semiconductors (Si, Ge etc).

Eddy current losses are proportional to conductivity (reciprocal of resistivity) AND proportional to the square of the frequency.

NiZn ferrites have about x10 more for freq's of 500kHz up...

Resistivity is a DC concept independent of frequency; it is also measured for a DC potential. Frequency is never mentioned in resistivity data.

At high frequency, you will have impedance (that will have an imaginary component).

I do not know at what frequency dispersion effects will become significant; because ferrites are ionic (defect) conductors, it may become appreciable at microwave frequency (exact value I do not know).

 

[Easy peasy]

copper has a DC resistivity - it also has AC resistance that goes up with freq - exactly the same for ferrites - ferrite conductivity goes up with temp - and it is also affected by applied voltage ... ferrites suffer temp rise from self heating due to eddy current losses and I^2R losses due to DC+AC current heating ...

 

[Amadeus]

Hi,
Am I wrong with this:
If "Q" is different for different frequencies (sine waveform), and because it's obvious that the typical swicher waveform is far away from sine ( because of it's overtones and DC) .... isn't it then obvious that one has to take care about the waveform?

Yes, Let's say that is not completely correct.
Most of the calculations that were used are for sine waveform. Why? Because of the Laplacian transform.
How can we solve square waves? or different duty cycles? Square waves can be represented as multiple sine waves at different frequencies, therefore %50 duty cycle SW and %15 duty cycle SW are completely different.

Lots of power elec. books and application notes have calculations that use duty cycles BUT those are "guestimations".
For example; small-angle approximation sin(x) = x, is it true ? no, but it is good enough.

 

[c_mitra]

it also has AC resistance...

I am learning new things. Do you have some reliable references where I can learn more about the AC resistance of copper?

Resistance of metal increases with temp- due to phonon electron scattering. Resistance of insulators and semiconductors decreases with temp - because or carrier generation. The reasons and mechanisms are completely different.

But I give up.

 

[FvM]

I am learning new things. Do you have some reliable references where I can learn more about the AC resistance of copper?

Easy peasy is obviously referring to AC resistance caused by skin and proximity effect of the winding.

 

[schmitt trigger]

Wikipedia has a good explanation:

https://en.wikipedia.org/wiki/Skin_effect

 

[Akanimo]

There is AC resistance actually. That's one of the reason you would prefer 'multifilar' winding or litz wire to solid copper.

- - - Updated - - -

As a matter of fact, you need to calculate for your optimum magnetics wire size. It is natural to think that the bigger the wire cross-section, the lesser the resistance, but in reality it is not so. As frequency goes up, more and more of the inner portion of your wire core becomes useless. Electric current tends to flow at the 'skin' of your conductor thereby creating 'ac resistance'.

- - - Updated - - -

One way to go about this optimization is to use a table of AWG with maximum current and maximum frequency for maximum skin depth.

- - - Updated - - -

Check here for a typical example of such table:
https://www.powerstream.com/Wire_Size.htm