How to correctly use the formula for calculating the air-gap in a pulse transformer?

Question

To achieve a certain value of the magnetizing inductance (Lm), we can adjust the transformer air gap.

Air gap calculation formula:

There is no particular information on this, but I assumed that:

u0 = Vacuum permeability (4π×10−7 H/m)
N - primary turns? (x18)
Ae effective area (PQ5050) = 328mm2 (Core datasheet)
Lm = magnetic inductance (my goal - 60uH)

Substituting these values ​​into the formula, I got 2.2 meters of the gap, which is clearly incorrect.

(0.000001257 * 324(18^2) * 0.328) / 0.00006

Next, I tried replacing N with the transformation ratio (n = 1.5), I got 15.46mm of the gap. This is closer to true, but still a very large gap.

(0.000001257 * 2.25(1.5^2) * 0.328) / 0.00006

I also tried to remove the square from N, but this is still a very large gap.

Question: what is my mistake? I assumed that the matter is in N (it is not clear whether this is the number of turns of the primary winding, or this is the transformation ratio (but it is denoted by a small n))

Perhaps there are other formulas with which you can get the Air gap value, the desired value of the magnetizing inductance?

Answer 1

Perhaps there are other formulas with which you can get the Air gap value, the desired value of the magnetizing inductance?

Well, you have the ferrite core-set data sheet and you have the ferrite material spec; 3C95 so let's proceed from there. The important formulas to know are these: -

The effective permeability of a core set when gapped

$$\mu_e = \dfrac{\mu_i}{\mu_i\cdot\dfrac{\ell_g}{\ell_e}+1}$$

• Where \$\mu_i\$ is the ungapped magnetic relative permeability of 3C95 (2530)
• Where \$\mu_e\$ is the gapped magnetic relative permeability of 3C95 (trying to find this)
• Where \$\ell_g\$ is the length of the gap in metres (maybe 1mm or 0.001 metres)
• Where \$\ell_e\$ is the core magnetic length in metres (0.113 metres)

If you plug-in the numbers with a gap of 1 mm, \$\mu_e\$ equals 108.17 and this is used in the next formula for inductance: -

Inductance of the gapped core set

$$L = \dfrac{\mu_0\cdot\mu_e \cdot N^2 \cdot A_e}{\ell_e + \ell_g}$$

• Where \$A_e\$ is the effective cross sectional area in m² (328 mm² or 3.28 x \$10^{-4}\$ m²)
• Where \$N\$ is the number of turns (18)
• Where \$\mu_0\$ is \$4\pi\times 10^{-7}\$
• The formula assumes that gapping has increased the magnetic length of the core - should you in fact "grind" down the centre limb, the total effective length remains as \$\ell_e\$.

So, if you plug in the numbers now (1 mm gap) you get 126.7 μH

Given the above, I reckon you should be able to figure out the reverse process to get the inductance you require (about 2 mm).

Summary

You will also have to double check that you have provided enough of a gap so that the peak magnetic flux density remains significantly less than 400 mT. I usually aim for 200 mT but, in the Cree document that you linked, they appear to be aiming for 120 mT. This is fairly easy to check remembering that \$B = \mu_0\cdot \mu_e\cdot H\$ where \$H\$ is the peak current multiplied by number of turns and then divided by \$\ell_e\$.

Or, \$H = \dfrac{MMF}{\ell_e}\$

Answer 2

Although this question is fairly old and has an accepted answer already, I'd like to post another answer to show where the air gap formula given in the OP comes from and what optimization it has, so that this can be a guide for the future visitors.

Here's the magnetic circuit of a transformer with \$N\$ turns around a gapped core:

^{simulate this circuit – Schematic created using CircuitLab}

Here,

• \$R_c\$ is the magnetic reluctance of the core
• \$R_g\$ is the magnetic reluctance of the air gap
• \$\Phi\$ is the magnetic flux
• \$N \ i(t)\$ is basically the magnetomotive force (MMF)

According to the model above, we can write that

$$ N \ i(t) = \Phi (t) \ (R_c + R_g) \\ R_c = \frac{l_e}{\mu_0\mu_i\ A_e}, \ R_g = \frac{l_g}{\mu_0 \ A_e} \\ $$

• I will not redefine \$l_g\$, \$l_e\$, \$\mu_x\$ as they are already defined in Andy's answer.
• Notice that \$R_g\$ does not contain \$\mu_i\$ because it's not related to core but the air instead.

Using \$v_L(t)=L\ \frac{di(t)}{d(t)}=N\ \frac{d\Phi(t)}{d(t)}\$, we can find the inductance equation from the one above:

$$ L=\frac{N^2}{R_c+R_g} = \left(\frac{\mu_0\mu_i\ A_e}{l_e+\mu_i l_g}\right)\ N^2 $$

Now from this final equation, it becomes easy to find the required air gap knowing \$A_e\$ and \$l_e\$ from the core's datasheet, and \$\mu_i\$ from the material's datasheet.

However, since \$\mu_i \ l_g\$ product will be way larger than \$l_e\$ because \$l_e\$ is in millimetres (thus it has a multiplier of 0.001) and \$\mu_i\$ is a few thousands for ferrites, we can simply neglect \$l_e\$ and simplify the equation above:

$$ L=\left(\frac{\mu_0\ A_e}{l_g}\right)\ N^2 $$

And finally, we obtain the air gap formula:

$$ l_g=\left(\frac{\mu_0\ A_e}{L}\right)\ N^2 $$