LT3757 gripe

The data sheet indicates that you can supply your own power to the Vint pin, if you expect the average load on Vint to exceed that normally anticipated (

A direct connection between Vint and an external regulated supply should remove loading effects on the internal 7V5 regulator.

The supply sequencing shouldn't allow INTVcc to exceed Vin by more than a schottky diode drop.

Have you changed the chip lately?

RL

I use the LT3757 a lot, mostly with a 10-12V hard rail like you did. Seldom with the internal 7V rail because that isn't sufficient for standard level FETs and makes even logic level ones sluggish. It never had any of the problems you described, always pulls hard to zero. It doesn't even need a gate driver supply to do that.

Now don't get upset, but have you checked the ground pad underneath? Is it really connected?

What is the reason for all those resistors and caps at the gate? I'd replace them with ... nothing, just a direct connection driver to gate.

Yes. In fact, I changed from the 3757 to the 3757A, with no change in what I was seeing.

Good thought -- yes, I believe it's connected. I replaced the chip recently and the thing showed solder on the pad.

The network was there to try to make the gate pull a bit negative on turn- off. As I said, I've seen such things work well in similar circumstances in the past.

One of the things I managed to do was to make the heat spreading work well -- so _everything_ gets hot. I'm wishing I had a thermal imager.

A finger makes a pretty good differential temperature sensor! But an imager is a fabulous tool. They are getting affordable.

My HV supply has a bunch of copper pours to cool things; the biggest pour cools the transformer.

I could have optimized things to get the efficiency up a bit, but it didn't matter in this application. I did breadboard it, so there were no surprises when we did the PC board.

That 2SK4177 fet turned out to be awful; too much Rds-on, too much capacitance. FDD7N20 works a lot better; 600 mohms on, only 5 nC gate charge, 5 pF feedback capacitance.

Did you try to make the resistors larger and the cap smaller? You need only very little energy and 110ohms during on-phases burns a lot of energy. Not sure if the LT3757 or most others are really rated to do that.

It might be better to use a small Schottky diode or two forward, a high value resistor from gate to ground and a small cap across the diode(s). That way the gate never really pulls above Vdrive mins the diode drop(s), leaving the pedestal available for slight negative pull during the transition.

I should try that. Frankly, I started with that but it was going to be hard to hack onto the board (yes, bad reason, I know -- but there you are).

A diode from the gate drive output to the gate? In place of the 10 ohm resistor in Tim's pic. (If so why would you want that?)

George H.

Keeping in mind that it's kind of shot-gunning the problem, the theory (borne out to some extent with simulation) is that the gate is not being pulled low forcefully enough to turn the FET off promptly, causing excessive switching losses.

Putting a diode or resistor in there causes the gate to not rise up to the full voltage supplied by the gate drive, which, in turn, means that the gate is actually pulled a bit negative when the gate driver pulls low.

At the time that I learned the trick it saved our bacon -- the circuit that my colleague was working on turned from a clever but expensive device to scatter burning bits of transistor around the lab into a real live switching amplifier.

Can you post a scope trace? drive out and Vgs?

RL

Thank you for this comment. I've burned my fingertips so often on electronics and model airplane motors that they're fairly insensitive to temperature, so the "differential finger temperature measurement" doesn't work for me.

However -- I dredged up some thermistors from a junk drawer (for a never- realized airspeed indicator project). It looks like it is the coil that's getting hot, after all (D'oh). So I have to stop griping about the chip.

I don't think the inductor is saturating per se - I don't see that characteristic upwards curl in the current waveform that makes one smell burnt electronics in the mind's nose. However, I suppose the coil could be exhibiting excessive hysteresis losses.

I'm using a Bourns SRP1270

I don't see anything about core losses on the data sheet -- is there any way to tell (other than just by knowing ahead of time) what the losses are? If there was some alternate part I could slap in there it would be lovely. If not, I'd at least like to know if there's some way of knowing what the losses will be so that I can simulate them or calculate them.

Thanks.

Yeah I got that much. The goal is to get charge out of the gate... or put less in to start with.

Yeah this is where my confusion comes in. You've got a diode between the gate and the driver, so how is the driver pulling any charge out at all. Or is it because the capacitor is there...? You make the diode bypass cap bigger than the gate C?

George H.

You make the bypass cap significantly bigger than the gate capacitance, yes. Initially the gate voltage follows the gate driver, but over time it drops by one diode drop or so, thereby pulling it lower than ground at the bottom.

The problem may be my coil -- see my reply to J.L. elsewhere in the thread.

Omega sells stick-on thermocouples, very thin, almost foil elements on a band-aid sort of sticky fabric. Some of the cheap DVMs accept t/c inputs, so that combination is good for investigating hot spots.

Are you not using the current sense resistor? It does not appear to be optional. It's not in the diagram, but I'd bet that it's providing a predictive signal to make the circuit stable with minimal external parts.

Get rid of the extra parts on the gate.

Tim keeps referring to this circuit as a speedup for the switching edges. I think the added capacitor will speedup the switching compared to just having the resistors, but otherwise this is like the scope probe compensation circuit and will give an even frequency response which the resistor only circuit won't. Otherwise it is better to just not have any circuit at all, no? With 110 ohm resistive load, the assumption is the output resistance is low in comparison, otherwise this circuit just loads down the output.

Same here but mostly because of mountain biking where the insides of the hands become calloussed. I find that the outside of the 2nd digit of fingers works well. But be careful, with a 1uH inductor operated this way it could hurt ... phssss ... OUCH.

Manufacturers rarely spec core losses anymore these days. But 1uH at