Electronic – the current limit/saturation of series inductors

Amendment following question correction

Two informative things in the MAX1771 spec: -

If you know the inductance (150uH) and you know the maximum on time (12us) and you know the minimum switch off time (2.8us) you can take a stab at calculating how much power you can transfer.

In 20usec, how much energy can you put into the inductor? In 20us + 2.8us you can assume that energy may be transferred pretty much totally (near, better than 90% I'd say). The cycle repeats every 22.8us i.e. 43,860 times a second.

What are the joules per second - this can be calculated now.

OK in 20us the inductor current will rise to V dt/L because V= L di/dt. Plug the numbers in and I get I to be 12V x 20us / 150uH = 1.6A.

What is the energy? It's 0.5 x L x I^2 = 192uJ. But this happens 43,860 times a second therefore the power is 8.42W.

Your circuit cannot deliver a power greater than 8.42W and if I took losses and only 90% energy transfer it looks more like 7.5W.

You want 300V across 9kohm and that is an output power of 10W - you don't just want an inductor that doesn't saturate you want a lower value inductor. Why? A lower value inductor will attain a higher current and it's current squared that gives you the energy to push to the output.

100uH - current peaks at 2.4A and energy in one cycle is 288uJ. Power is therefore up to 12.6W.

This would be my recommendation - try 100uH and if it's still a little bit light on power (due to inefficiences) try a better FET with less-than the present 0.7ohm on resistance. Bear in mind it's really tricky to estimate how much energy stored in the inductor can be thrown across to the output so the inductor may indeed have to be lower that 100uH and I feel you are starting push the chip to its limitations and you may need to reconsider a chip upgrade.

1. N87 is straight-up ferrite material, not distributed air gap like powder iron types of material. Just because it's in toroidal form doesn't mean it's distributed-gap material - N87 in a toroid will saturate the same way as N87 in an E-core. There's nothing wrong with using straight-up ferrite for a boost inductor, so long as you gap it (more on this later). The fact that it's in toroidal form means you can't gap it. You might want to switch to Kool-Mu if you want to stick with a toroidal form factor.

2. Deriving the core inductance from the \$A_L\$ factor is reasonably good, so long as you have sufficient turns on the core and can keep the winding reasonably uniform and in as few layers as possible. Bear in mind that \$A_L\$ can vary by +/- 25% on some cores.

3. Boost inductors carry both the magnetizing current and energy for the load (that will be stored magnetically and delivered during the off-time.) Once the converter starts operating in continuous conduction mode (when the inductor current never goes to zero), it's even worse since you start operating on a B-H curve that doesn't reset to zero. (Bmax is still Bmax, but you now have a DC offset on which Bpeak is riding.) These are the reasons that the inductor needs air gapping - the core won't be able to handle any significant DC current without saturation otherwise.

4. I'm not sure I understand your test circuit. Both ends of the inductor are essentially clamped to 5V, meaning that the two capacitors (C1 and C2) contribute nothing to the simulation. If your real boost converter is arranged in this manner, it isn't a boost converter and will never work. L1 needs to release its stored energy through D1 to the load, which can never happen when D1 and the load is connected as shown. The only connection between input and output must be through L1 and D1. I would also put R1 in the source of Q1 and do a single ground-referenced measurement instead of a mathematical construction. (L1 will only saturate when Q1 is on, so measuring it when Q1 is off is irrelevant.)