I am designing a boost converter. I know maybe the specs I want would suit better in another topology like Half-Bridge or Full-Bridge but the boost topology is simpler and inductor winding is simpler too I think.

My boost converter should be:

  • 14V input voltage
  • 15 to 30V output voltage
  • 10A output current so the output power will be 30*10 = 300W
  • 300KHz switching frequency

I am using this application note from Texas Instruments and I am getting this results:

  • Inductor value: 2.5uH
  • Max switch current: 45A

Is it possible inductor has such a low value? Is the peak current really that high? I am planning to use an IRFZ44 MOSFET that specifies a Pulsed Drain Current of 200A, is that the parameter I have to look at or should I calculate Irms and look at Continuous Drain Current?

Also I was looking at this core but the Al value is so big I doesn't event need a turn around the core to get 2.5uH! So I can't use this core?

Also I would like to know how to calculate the inductor to ensure I am not exceeding the Maximum Flux Density. I have a book called Power Supply Cookbook but it seems to focus more on Half-Bridge and Full-Bridge inductor design than Boost design.

I am not an expert so there are a lot of things I don't know any guidance, book, links would be really appreciated! Thanks!


2 Answers 2


Power out = power in minus losses. Power out is 300 watts therefore power in will be about 330 watts. At 14 volts and 330 watts, current in will be 23.6 amps but this is average current. The peak current will depend on the duty cycle and looking at it simplistically, for a 50% duty cycle the peak current might be about 2 x 23.6 amps = 47 amps worst case (borderline discontinuous).

Ideally you need to pick an inductor that doesn't saturate and one with a gap is going to work better. A solid ferrite core will tend to start saturating at a flux density of about 0.4 teslas. It might have a relative permeability of (say) 1000. Using B = \$\mu H\$ allows this to be estimated: -

H = \$\dfrac{0.4}{4\pi \times 10^{-7} \times 1000}\$ = 318 A.t/m

You need to know the mean length of the magnetic field and that is found in the data sheet for the ferrite you are using. Let's say it is 100mm, this means your maximum ampere-turns is 31.8 and of course this would be unsuitable for your application because you can immediately see that it is likely your amperage and one turn is going to heavily saturate the core.

So, if you put a gap in it might reduce the effective permeability to (say) 200 and this allows more 5x H field (5x more current or 5x more turns) but inductance is related to turns squared so there is a net gain because although the \$A_L\$ value for the gapped core has gone down by a factor of 5 you only need \$\sqrt5\$ more turns to recover the inductance.

  • \$\begingroup\$ According to the application note my duty should be about 75%. So a toroidal core is not the best option here? I should use an EE core with gap? How I calculate the inductor to avoid core saturation? \$\endgroup\$
    – Andres
    Commented May 24, 2014 at 20:03
  • \$\begingroup\$ If N30 material has a intial permeability of 4300 how I calculate the relative permeability? H=1200 in N30 but μ has to vary with some parameter like current right? Otherwise there is no way I can't modify B to avoid saturation. \$\endgroup\$
    – Andres
    Commented May 24, 2014 at 20:22
  • \$\begingroup\$ Does my latest answer modification help? Initial permeability is usually quoted as a relative permeability. \$\mu\$ is material, not current dependent. \$\endgroup\$
    – Andy aka
    Commented May 24, 2014 at 20:23
  • \$\begingroup\$ My core datasheets says nothing about magnetic field length :/ mouser.com/catalog/specsheets/epcos_B64290A0084X830.pdf So if my μ=4300 and B=0.4, H becomes 0.000093? Such low value? \$\endgroup\$
    – Andres
    Commented May 24, 2014 at 20:28
  • \$\begingroup\$ Toroids are not going to work for this - look at RM12 ferrite core sets and recognize it will need to be gapped. \$\endgroup\$
    – Andy aka
    Commented May 24, 2014 at 20:32

Is it possible inductor has such a low value?

Well, let's see. Assume a 50% duty cycle so the switch is on \$\frac{1}{2}\frac{1}{300kHz} = 1.67\mu s\$.

During this time, there is 14V across the inductor. Thus, the rate of change of current is

$$\frac{di_L}{dt} = \frac{V_L}{L} = \frac{14V}{2.5\mu H} = 5.6 \frac{A}{\mu s}$$

So, over the time the switch is on, the inductor current will increase by

$$\Delta i_L = 5.6 \frac{A}{\mu s} \cdot 1.67 \mu s = 9.34 \mathrm A$$

From such a "back of the envelope" calculation and the required output current, this sounds reasonable (in the ballpark) to me.

I'm not an expert on boost converter design but I did want to address this particular question to give you an idea of how to do a 'sanity check' yourself.

  • \$\begingroup\$ then add in the average current the inductor must carry delivered to the load on top of the increase due to switching. \$\endgroup\$
    – Marla
    Commented May 24, 2014 at 20:42

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