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Inductance

This section shows how to calculate inductance given a datasheet for a magnetic core and discusses how much inductance is require for a GDT.

 

Inductance Formulae

Inductance of a certain number of turns on a ferrite core can be calculated by using the AL value found in the datasheet which is given in nano-Henries-per-turn-squared or nH/N^2. The formula for calculating inductance is very simple:

L = AL x N^2

Example

A core with 15 turns on it yields an inductance of 815uH - what is the AL?
Well, 815uH is 815,000nH so:

AL = L / N^2
= 815,000 / 15^2
= 815,000 / 225
= 3622 nH/N^2

So for this same core, how many turns would be needed for an inductance of 4.2mH (4.2e+6 nH)?

N^2 = L / AL
N = SQRT [ L / AL ]
= SQRT [ 4.2e+6 / 3622 ]
= 34.052
= 34 turns

 

Design Tradeoffs

What does the indutance need to be in a GDT? Well, the tradeoff is between two choices

    1. Minimise the the number of turns to reduce inductance and to minimise leakage inductance (making sure you don't saturate the cores)
    2. Increase the number of turns to reduce the magnetising current

 

Magnetising Current

Otherwise known as "magnetizing current" (for our viewers from the other side of the pond) this is the current required to maintain the magnetic field in the transformer. Taking the equation for voltage across an inductor...

V = L * ( dI / dt )

...and rearranging it, we can see the peak magnetizing current is determined by

Imag(pk) = ( Vin * period * duty ) / Lmag

Because the current through an inductor appears as a triangle wave, the rms current is a more meaningful result to us. For a triangle wave, this is given by

Imag(rms) = 0.577 * Imag(pk)

This current, added to the current required to charge and discarge the gate of the MOSFET, is the total current required for the gate drivers to supply.

 

Example Calculation

Assume we have a GDT and driver with the following characteristics: -12V to +12V, 100kHz drive signal with a 50% duty cycle being fed into a GDT with a 1mH winding inductance.

Imag(pk) = ( Vin * period * duty ) / Lmag
= ( [ 12 - -12 ] * [ 1 / 100e+3 ] * 0.5 ) / 1e-3
= ( 24 * 10e-6 * 0.5 ) / 1e-3
= 0.12 A (pk)

Imag(rms) = 0.577 * 0.12
= 70 mA (rms)

If we double the winding inductance to 2mH, this will reduce the magnetising current by half to 35mA. Bearing in mind that the inductance is proportional to the number of turns squared, it is easy to add a few more turns to reduce the magnetising current.

 

Conclusion

We can see that the magnetising current is relatively small but for GDT with less turns it can become significant. Therefore more turns for a higher inductance is preferred.

However, leakage inductance will increase with the number of turns and leakage inductance affects the performance more significantly than magnetising current.

Therefore, select the best winding technique to minimise leakage and use as few turns as you can without having a large magnetising current and without hitting saturation.