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Saturation Flux Density

This section talks about saturation flux density, how to calculate it and how to avoid running into it in GDT designs.

 

What Is It?

Saturation flux density is how much magnetic flux the magnetic core can handle before becoming saturated and not able to hold any more. This depends on several factors including ferrite type, temperature and electrical and magnetic conditions on the transformer.

When the ferrite saturates, the transformer no longer acts like an inductor with a linear increase in current over time. Rather, the magnetic field cannot increase further and current is limited by the source impedance of the power supply and the resistance of the transformer wire. This leads to very large currents and blown devices. This is a Bad Thing.

Ferrite typically saturates at a flux density (B) about 0.3 Tesla but this depends on temperature and the ferrite type. A typical design might have as a target a B = 0.25T at 125ºC which gives plenty of operating margin to the limits.

 

Calculations

The formula for flux density is

B = ( V x t ) / ( N x Ae )

where:
B = flux denstiy in Tesla
V = applied Voltage to the winding in Volts
t = time that Volts V is applied for in seconds
Ae = cross sectional area of the core in m^2 - obtainable from the core datasheet

If this were an inductor carrying a DC component, the DC current would play a significant part in calculating the saturation.

The main thing to take from this equation is that for a given core and frequency there is a minimum number of turns that can be used without saturating the core. So, to avoid saturation, use more than the minimum number of turns.

 

Examples

We have a 50% duty cycle, 100kHz drive signal @ 12V peak. Core has a XSA of 50mm2 (= 50.10-6 m2) What is the minimum number of turns assuming a maximum B = 0.2T

A 100kHz drive signal gives a total period of

t = 1 / f
= 1 / 100e+3
= 10 us

And at a 50% duty cycle, this gives an on time of 10 us / 2 = 5 us. Plugging the numbers in to the formula we get a minimum number of turns of:

N = ( V x t ) / ( B x Ae )
= ( 12 x 5e-6 ) / ( 0.2 x 50e-6 )
= 60 / 10
= 6 turns

The V x t product is referred to as Volt-seconds and is the voltage that can be applied for a certain amount of time. In the above case, the volt seconds were 60 V.us (not V/us, we are multiplying!). So how long can we apply 18V to this core, assuming the same amount of turns and same flux density?

t = 60 V.us / 18V
= 3.3 us