# 12v to 300v Boost Converter - Ideal Inductor for 9kOhm load

I'm currently working with a MAX1771 chip, the design can maintain a stable voltage up to around 15kOhm and is as follows: 12V to 300V DC-DC Boost : Disregard NMOS and DIDODE Value

The inductor currently being used is a SRN1060-151M, unfourtanately it has a peak current of 1.25A and a saturation current of 1.8A. My guess is that the low saturation and peak current are causing problems with lower loads. I'm rather inexperienced with actually ordering proper parts and mostly ended up getting parts that fit the initial specs (matched values and picked low DC-resistance).

I take it that for a 9kOhm load I would need an inductor that can handle up to 3A. The NMOS is a STD8NM50N which can handle up to 5A and the DIODE is an EGL34G-E3/98GICT-ND.

Can anyone recommend an affordable inductor on DigiKey that can get the job done? Or changes to the circuit that may assist in making the 300V with 9kOhm load possible?

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.

• My apologies, I'm currently aiming for 300V with a 9kOhm load, currently it is stopping at 200V. – diverges Jun 19 '13 at 13:18
• Does it achieve 300V with the 10kohm load? If not what load allows the full 300V? Are you actually using a pot for the feedback control (as per the circuit)? – Andy aka Jun 19 '13 at 14:18
• With 9kOhm it achieves 226V, I've tried it with 50kOhm and it achieves the 300V just fine. POT is a 50kOhm variable resistor and at around 15kOhm of load resistance the voltage starts to drop. – diverges Jun 19 '13 at 14:23
• According to your circuit the POT is 25kohm. I was referring to that POT. But it does sound like a better inductor is what you need. – Andy aka Jun 19 '13 at 14:40
• Yeah POT varies from 10k to 47k, it sets the output voltage which can go from 180V to 360V. Although that means nothing if the load causes the circuit to hit the current limit. I did find an inductor with 1.8A peak current but it is 100uH, hopefully this isn't too much of a problem. – diverges Jun 19 '13 at 16:20