Oznog

Having an inductor saturate would be bad news. The current will spike without much limit. So it might increase to 300mA, saturate, and the current suddenly surges up to the 1 amp threshold almost instantly and the 9910 cuts it off before it goes even higher. As soon as the gate is turned off, the inductor current drops back to 300mA almost immediately. Thus the LED may only see a 300mA average current if the inductor saturates at 300mA.

Inductor saturation current goes down as the inductor heats up. That's why knowing the actual saturation current is important rather than just seeing it work once at room temp.

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technogeek

How would I know/calculate saturation current? I know the max current of the device is 0.7A, DC R is 1.5ohm, 1.8mH, ferrite core......

That's all the datasheet lists on that inductor...

Oznog

Max current usually indicates the point at which the device may overheat. It's a matter of I2R losses vs dissipation. Saturation current is generally higher (but this is not a rule) so I'd guesstimate you would be ok but that's not saying much without data. But if the max current is 700mA and you're asking for 700mA continuous by design, you have no safety margin. You need an inductor with a better current rating.

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technogeek

Indeed, it's not my first choice, but it's what I had for this 0.7A supply... The next one will be a 1.4A supply with an inductor rated 2A+, but this is just a "test of theory" to work all the bugs out so to speak. (and there have been a lot as you've seen)

justDIY

don't forget to include ripple in your inductor capacity too...

AN-H48 page 3 has lots of math for calculating just how big an inductor you'll need, but in summary Iout + 30% = 700mA * 0.3 = 210 + 700 = 1 amp minimum on the inductor.

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technogeek

They were using 115% * Iout for the peak inductor current rating?

justDIY

they seem to use it either way ... one datasheet says +15%, the other +30% (115 - 130) ... so I'd aim for 130% unless space savings of a smaller inductor is a big concern.

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Oznog

No. Ripple is NOT a factor. If you set the 9910 threshold at 500mA, the 9910 will shut off at 500mA. If the inductance is too low for the job and ripple is significant, it will lower the average current but the peak current remains fixed at 500mA by the sense resistor on the 9910.

But you DO want a significant safety margin between the peak current and the saturation current. As I say saturation point is highly variable. And another factor is that the inductance value goes down as you get anywhere near the saturation point. The inductor will act like a smaller inductance than indicated in the spec.

Hard to say, but as a rough guess I'd go with an inductor with a saturation 50% greater than what you intend to use.

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technogeek

I put on an 18pF and 10k on the current sense, and the HV9910 self destructed....

I changed the load into a 30 ohm resistor, and measured the current to start at ~0.8A and it decayed (along with an increasing pitch in the whine) to ~0.7A.... but then it just stopped working. The FET was getting quite hot for even the short couple second test runs I was doing, so I thought I killed it.. But I changed it out and still don't have anything. Everything else tests okay, so the 9910 must have bit the dust.

*sigh*

Why does such a simple circuit have so many problems? And what did I do? The IC was getting warm, but nothing that concerned me... This is the FET I'm using-

http://www.fairchildsemi.com/ds/FQ/FQPF11N40C.pdf

Is it possible the gate capacitance or charge is too high for the IC?

justDIY

smps = voodoo

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technogeek

Ahh, so THAT'S the reason I never wanted to touch these things before!!

Oznog

The HV9910 has none of the stability problems associated with most smps.

The gate capacitance could not have fried the HV9910.

I would be almost certain you did not wire this up correctly. Do you have a cap on Vref? Is C1 large enough to avoid excessive ripple?

You put the resistor between the Source and 9910 current sense pin, and then the cap between the current sense pin and ground, right?

10k * 18pF is 180nS, this is less than the specified 300nS.

Oh yeah. The RC network does present a minimum for the "on" time. For example, if your inductor is at 1 amp and you have a 1 amp threshold, the 9910 will still take like 300nS to turn off, during which time the inductor current increases. At 100KHz, that would be a min duty of 1/33. This will generally only show up on high ratios, as a guideline, a 1:33 voltage ratio. Driving a 7v off a 240vdc supply for example.

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technogeek

Source -> 10k -> current sense pin and cap to ground.

It would not surprise me if I did not wire it correctly... I forgot to tie pin 5 to the other Vref pins anyways. Vref has a 2.2uF cap as specified. However it worked for a little while.... C1 was what was spec'd for 0.7A, 68uF I think.

I just downloaded the demo board manual, and it uses an IRF840 in the D2pak, which makes me think it shouldn't be putting out a lot of heat! With how quick it heated up I'd say it was close to 5 watts...

180nS is more smoothing than nothing...

Oznog

Well something is wrong. Unless the 9910 is totally malfunctioning (bad Vdd/Vref) it should either turn the MOSFET on or off. The one you specify is rds-on of 0.5ohms, so it would be 1/4W at 1 amp.

I used it to drive a Lamina LED off of 14.6V, at a couple of amps. I used a transistor in an SO-8 pkg ("hexfet" type, low voltage MOSFET) and it ran totally cool.

I remember raising the question of the current sense pin impedance. I mean if you use a 100M ohm as part of the RC it probably won't read anything due to the leakage current draining the signal voltage away. IIRC he did say the pin's input impedance is pretty high, I think 10k should be fine.

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technogeek

Interesting........