Sync buck

This is the rising edge coming out of an LMR23625 sync buck making 3.3 V @ 1.6 A from 5V. The data are from a Tek TDS 694C with a P6249 FET probe.

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Impressive. The chip and the inductor both run very cool.

Now I just have to figure out the EMI issue. ;)

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Phil Hobbs
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That waveform looks a lot like this:

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There are distinct phases: bottom fet on

fet off and a substrate diode catching the current

upper fet on and the diode fighting the rise diode running out of charge and a step-recovery snap.

That switcher trashed opamps all over the board. The SRD step was brutal; it challenges most scopes and probes. The current must have been awesome.

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John Larkin

This has to share a small board with an LPC804 MCU that has two 12-bit ADC channels. I may have to disable the regulator during measurements, but the ADC is pretty fast (2 us) so that won't cause too many headaches.

That little MCU is a pretty sweet device actually--it's a Cortex M0+, has a 12-bit ADC and a 10-bit DAC, comes in 20-TSSOP, and is half the price of an ATmega of the same size.

Cheers

Phil Hobbs

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Phil Hobbs

This has to share a small board with an LPC804 MCU that has two 12-bit ADC channels. I may have to disable the regulator during measurements, but the ADC is pretty fast (2 us) so that won't cause too many headaches.

That little MCU is a pretty sweet device actually--it's a Cortex M0+, has a 12-bit ADC and a 10-bit DAC, comes in 20-TSSOP, and is half the price of an ATmega of the same size.

Cheers

Phil Hobbs

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Phil Hobbs

It might be prudent to use a switcher that doesn't do that; junctions everywhere will rectify fast spikes. Are there any opamps? They could be trashed for a lot longer than 2 us.

How can people sell stuff like that?

I like TPS54302. It's quiet, but low frequency so needs a relatively big inductor.

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John Larkin

Not on this board. Just two switchers, a CPU, a USB serial chip, and a few blinkenlights. It's stacked on top of the TEC driver board, bottom-to-bottom, and in a metal box, which will help. The TEC board has some op amps---a TLC272 and a couple of LM358s. The actual sensitive stuff is at the other end of a cable. So hopefully this isn't too sporty.

The spec I'm most concerned about is noise in DC-2 MHz, so a 2.1 MHz switcher has a lot of charm. I looked at the LMZM33063 as well, which has an integrated inductor, but for some reason it can't produce an output near its input voltage, whereas the LMR23625 can. It also doesn't run at 2 MHz.

Cheers

Phil Hobbs

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Phil Hobbs

If "figure out" == "why is the EMI so bad", the answer is fairly easy. OTOH, if "figure out" == "now what do I do", well, it depends. What's your layout look like? Synchronous bucks are a little picky but if it's done right there usually isn't much of a problem, particularly with the more recent TI parts. Some of the older ones are unfixable but the new ones have really nice packages and control the switch nodes really well.

Reply to
krw

Impressive indeed, something must clearly be wrong with it.

Is the regulator in a light load condition, where it auto-selects from PWM to PFM? Note that there are 2 chip versions available in the SO8 package - one of them always has forced PWM operation, the other may switch to PFM depending on load conditions. In the WSON package only the auto-select version is available (hopefully you have space for SO8 on the board so you can try out the other configuration).

Note also that there is a distinctive area on your scope trace just before the snap - starting some 3.2 ns before the dashed line. In this area, the high side MOSFET is already on, (probably pulling a large current) and "fighting" some heavy load.

That heavy load could either be the low side MOSFET or a diode reverse-recovery characteristic. The voltage level is clearly not at Vdd/2 midscale, but with the low side MOSFET being typically lower in Rdson than the high side one, either case could still be realistic.

If it's the transistor, that would mean TI has royally screwed up the chip design and got the shoot-through prevention timing the wrong way, but I think the culprit here is more likely the diode.

Note that in a PFM operation, there is a long time immediately prior to the switch-on of the high side FET, where both FETs are off, hiving plenty of opportunity for the low side substrate diode to conduct. That can precharge the diode and make it snap into step recovery when the polarity changes, hence my question on what exact mode it was in.

Even in PWM, there should still be a "dead time" between the MOSFET "on" states in order to prevent cross-conduction. During this time the diode will conduct and will likely get sufficiently "pre-charged".

Note that if it's really a chip timing mess-up with cross-conduction, that is not fixable on the board level - in this case the chip model needs to go and never be used again! But if it's "only" the diode, it can be helped by way of an external low Vf Schottky diode in parallel.

A MOSFET substrate diode can be prevented from "charging up" and going into a step recovery by clamping its forward voltage low enough that it does not get into full conduction.

A target below about 0.5 V should be good enough, and there are diodes like a PMEG3050EP (Nexperia) with a nice low 360 mV forward drop at 5 A current that should clamp away any misbehaving silicon junction.

You can try to parallel such a diode externally from SW to PGND pin of the switching regulator, and from SW to VCC for good measure (who knows if the high side FET diode is also misbehaving in a similar way).

Best to route the PCB trace from the inductor such that it first goes to the diodes and from there to the switching regulator, but to try things out, one can just dead-bug the diode over the chip with as short as practical wire connections.

Regards Dimitrij

Reply to
Dimitrij Klingbeil

Not screwed up, or not necessarily at least. Zero deadtime/interleave is preferred, as long as you have the supply inductance and damping to deal with it.

A very short transient implies the switch node capacitance and loop inductance are very small, giving the high frequency; but not small enough given the commutation rate of one thing or another. (It could very well be body diode recovery at work here.)

The unfortunate part about integrated regulators is that they are usually powered from the high side drain pin, making it impractical to impossible to introduce supply impedance there.

It's another one of those things that seems convenient (saving pins and space) but is actually retarded once you know you need it. See also: internally compensated regulators, dominant pole compensated op-amps, etc. :(

This is the remaining option, downside being, with switches this fast, it's difficult to impossible to place such a diode close enough (low enough Lstray), plus you incur the additional Cjo which is significant at these frequencies and/or switch rates.

Tim

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Reply to
Tim Williams

Nope, it's generating a continuous 1.6A in CCM.

I'm pretty sure it isn't gross shoot-through, because the period is less than 500 ns, and the chip runs very cool even at two thirds of full load.

Interesting idea, thanks. I'll look at diodes to see if there's one with low enough capacitance to be a help there. Otherwise I might try ~20 nH in series with the input.

Cheers

Phil Hobbs

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Phil Hobbs

Did people see my post above? The problem is apparently step-recovery of a substrate diode that conducts during the mosfet non-overlap time.

Adding an external schottky helped my case some, but not enough.

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John Larkin

Yes, many times. That's been a favorite example for, what, half a decade? (I wouldn't mind if you'd put it on a webpage proper; it always seems to move around, first back when you had an FTP, then through however many variations of Dropbox it's been through...)

Anyway, it's not necessarily step recovery. It could just be that fast.

Probable proof that it is just that fast...

In any case, the solution is the same: slow commutation, dampen the reactance, and filter what's left.

Sadly these aren't always possible on an integrated reg, which goes back to Dimitrij's point: "the chip model needs to go and never be used again!"

Tim

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Reply to
Tim Williams

I did the measurement on an eval board--after a couple of unpleasant surprises I now make it a rule never to use a switcher chip I haven't had a chance to check out in person.

I'll wire up a DIP op amp with a gain of 100 and wave it around the switcher to see if it misbehaves.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
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Phil Hobbs

That's really surprising from TI (recently). Again, I don't know that part but their newer switchers in the new packages have been very well behaved, IME. Their eval boards are usually done right, too. It's sorta important for sales.

Good idea but there are conducted issues (same planes, etc.), too.

Reply to
krw

Phil had the problem, and I answered him. I don't post the waveform unless someone brings up the subject. The SRD spikes can require a board spin.

The waveform, and the extraordinary ESD effects, suggest a high-current SRD snap. The insane sub-ns rise time of the output node is a small fraction of the fall time. Most half-bridge switchers have a faster fall time. It would be about impossible to make a mosfet switch that fast.

I don't think any external filter will kill that current/voltage spike.

We can agree on that.

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John Larkin   Highland Technology, Inc   trk 

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John Larkin

The actually sensitive things are on different boards.

Cheers

Phil Hobbs

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Phil Hobbs

Particularly at the supply pin, make sure that the filter (if any) does not have any appreciable undamped inductance. Thing is, during the time when both the low side diode and the high side FET conduct, they act as an effective short circuit. The duration is not long (here some 3.2 ns) but still, with the shoot-through being higher than the subsequent load current, it's going to charge the inductance. And when the SRD finally snaps off, the inductance can quickly overshoot the chip's VCC supply.

If you make the inductance small, it will charge up and overshoot the VCC rail. If you make it big, the VCC will sag and wreak havoc in the chip's internal logic functions by starving the chip's power supply.

It may be more effective to use an intentionally lossy filter element in the VCC line, like a ferrite bead. If an inductor is needed, at least it can be parallel-damped with a resistor just low enough R to keep it at least critically damped (or slightly overdamped) at SRF.

Now, of course, an intentionally lossy filter element is not exactly the best efficiency improvement measure known to mankind :)

Anyway, at least the parallel low Vf diode could be worth testing. It can sometimes work, but the details depend on each case. At least, I once used that approach to quieten the SRD snap-off of a MOSFET in a synchronous rectifier circuit of a power supply - in that instance it killed the SRD effect more or less completely. The risetime was still fast, but with the diode in place it was a different kind of fast. It wast just rising quickly like other signals can rise quickly (in a steady way), but not that nasty "precharge the SRD, blast it with a near short circuit and dump the whole energy in ps time" variety.

Note however that this approach only works well with diodes that can maintain a Vf well below silicon junction conduction even at full peak current levels to happen there. This typically means a flat lead-less surface mounted Schottky with something like 0.3 V typical Vf that can maintain below 0.5 V at a peak of several A. At least the circuit in question seems to be powered from a modest VCC of about 5 V, it should be possible to find a suitable part to try. Also both PGND and SW pins on the chip package are conveniently oriented "towards" the same edge with nothing between them - a nice place to put a part in parallel.

Reply to
Dimitrij Klingbeil

Never underestimate the noise spreading capabilities of a common ground lead ;)

Reply to
Dimitrij Klingbeil

That helps a lot. It makes any necessary isolation a lot easier.

Reply to
krw

It occurrs to me that we may not have a plain SRD, but a drift step-recovery (Grehkov) diode effect in a high-frequency switcher. The short diode ON time enhances the ultimate step-recovery rise time.

The IC designers could maybe do some diffusion tricks to make the substrate diode softer. A minority of synchronous switchers have this ghastly rise time.

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John Larkin   Highland Technology, Inc   trk 

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John Larkin

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