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I'm in the process of compiling some notes on inductor design. I have good hands-on experience in the field (excuse the pun), but I feel this experience hasn't been totally systematic. I have been reading online literature including design procedures from core manufacturers, but I have a specific question related to the iterative process that typically takes place due to the intrinsic complexity of this design task. It goes like this:

Let's assume you have a working flyback which uses say a gapped ferrite RM8. This particular flyback has specific constant working conditions (input voltage, output voltage, on time, switching frequency, output power, etc.) Let's assume that:

  • The transformer is reset to zero flux density every cycle.
  • The transformer size is driven by temperature rise rather than peak energy storage. The flux density swing and operating frequency are such that the heat drives the size, not I^2*L. The ferrite operates at a comfortable flux density far away from saturation.
  • The core losses in those conditions are whatever P_core.

Our transformer works well, but say we now replace it with an oversized transformer of the same effective permeability, say for example an RM12. As its A_L is higher, we wind it with less turns so that its primary inductance L_1 is preserved, keeping the basic oscilloscope waveforms unaltered, so to say. Naturally, the bigger component will heat up less. However, and this is my question, will that be at higher or lower total core loss?

My thoughts so far are that N decreases as A_L increases (so as to preserve L_1 as described above), l (magnetic path length) increases, peak H decreases as a consequence of all this, peak to peak B decreases consequently, P_V (specific core losses) decreases significantly as a result... but obviously the core volume is significantly higher and, since most of these relationships are non-linear, I'm not sure what happens.

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  • \$\begingroup\$ Once I was told casually that an inductor that doesn't heat up significantly is not only too big but also has higher losses than a smaller, equivalent one would. I'm failing to justify (or refute) this statement. \$\endgroup\$
    – Alex Lopez
    Jul 27, 2021 at 13:59
  • \$\begingroup\$ I'm still unsure what your question is. \$\endgroup\$
    – Andy aka
    Jul 27, 2021 at 14:12
  • \$\begingroup\$ Sorry if the question got hidden in the lenghty text. It's in there, through: "However, and this is my question, will that be at higher or lower total core loss?" \$\endgroup\$
    – Alex Lopez
    Jul 29, 2021 at 6:47

2 Answers 2

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Once I was told casually that an inductor that doesn't heat up significantly is not only too big but also has higher losses than a smaller, equivalent one would. I'm failing to justify (or refute) this statement.

  1. They mean it is overdesigned. Like using a wire AWG so large that you can run 1000A through it but it rises 1C above ambient. Excessive use of material.

  2. Larger inductors also tend not to have more high frequency losses but have lower DC losses so there is a balance.

  3. Just because it heats up less because it is larger doesn't mean it is necessarily dissipating less power; It has more volume to spread that power around which may result in lower temperatures even if the dissipated power is higher.

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  • \$\begingroup\$ It'll be the same magnetization current if inductance is the same (as specified in the question). \$\endgroup\$
    – Andy aka
    Jul 27, 2021 at 14:14
  • \$\begingroup\$ 2. Larger would mean higher Al so for a given set of turns, lower magnitizing current. \$\endgroup\$
    – winny
    Jul 27, 2021 at 16:01
  • \$\begingroup\$ I'll back off on the magnetizing current and someone else can clear it up. \$\endgroup\$
    – DKNguyen
    Jul 27, 2021 at 16:03
  • \$\begingroup\$ Thanks, all. Please read carefully. The number of turns is not give, the inductance is. The question is very specific in an attempt to minimise confusion and subjectivity :) \$\endgroup\$
    – Alex Lopez
    Jul 29, 2021 at 6:39
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I've combined some of the usual equations and it looks like the core loss doesn't really change.

Inductance factor:

$$A_L=\frac{L}{N^2}$$

$$L=N\cdot\frac{\Phi}{I}=N\cdot\frac{B\cdot A_e}{I}$$

$$A_L=\frac{B\cdot A_e}{N\cdot I}$$

$$N\cdot I=H\cdot l$$

$$A_L=\frac{B\cdot A_e}{H\cdot l}=\mu\cdot\frac{A_e}{l}$$

Volume:

$$V=A_e\cdot l$$

As it simplifies math, let's assume that we increase the core volume by a factor of 64. (Obviously this is not an RM8 to RM12 increase, but the math is the same.)

Because the geometry (shape) is kept, so are all the proportions, the magnetic path length will increase x4 and the magnetic material cross-section will increase x16, i.e. all lengths increase by the same amount and the resultant volume increases x64.

The inductance factor will increase x4:

$$A_L=\mu\cdot\frac{A_e}{l}$$

As we require the same inductance as before, the number of turns halves:

$$N=\sqrt\frac{L}{A_L}$$

The magnetic field decreases by a factor of 8:

$$H=\frac{N\cdot I}{l}$$

Because we are far away from saturation, the permeability is still the same, the flux density B decreases by a factor of 8 just as the magnetic field and so does its swing (peak-peak value).

The specific loss P_V is proportional to the square of this swing, so it decreases by a factor of 64.

Therefore, the core loss hasn't changed:

$$P=P_V\cdot V$$

...Obviously you still want your core small for cost reasons, but that's a different story.

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