**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.

## Best Answer

Two inductors in series would still saturate at 10A and also you'd have twice the inductance. Two inductors in parallel would share the current reasonably well but now the net inductance has halved because they are in parallel.

This would mean you choose inductors that are twice the inductance and this has a net benefit because, for a given ferrite size/mass, getting twice the inductance only needs 1.414 (sq root of 2) times the number of turns and flux saturation is driven by amperes x turns.

So you started with 1 x 10uH inductor saturating at 10A and you end up with two parallel inductors of 14.14uH each saturating at 7.07A but because they are in parallel the net saturation current is 14.14 amps.

Small print: Assumes that the two inductors are well-matched i.e. reasonably share the load current and that they don't leak enough flux to each other to make calculations more complicated.