Something doesn't sound right. You speak of "ground clip". If you're using a ground connection with a short wire, your measurement is invalid when dealing with switching power supplies. You need to, at a minimum, connect the probe ground sleeve directly as possible to the common of your circuit.
I've blown out the gates of mosfets, at low voltages and currents, because the body diode had step-recovery behavior. I've seen SRD effects in integrated synchronous buck switchers, too. This is classic:
That's an LM3102. The spikes trashed opamps all over the board.
So I'd expect that people tweak doping profiles in things like body diodes to achieve soft recovery. Slow and safe.
-- John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation
Nice high Q ring there :)
Jamie
It's a beautiful waveform.
For the first 15 ns, the lower mosfet of the synchronous switcher is on.
From 15 to 35, both fets are off, and the output inductor current is flowing through a body diode into ground.
At 35 ns, the upper fet turns on, but the body diode, in reverse recovery, fights it, and the voltage stays low. Lots of current is building up.
At 38 ns, the body diode switches off abruptly, in step-recovery mode. Twang.
-- John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation
Well, no. I've never probed a circuit so bad that >50%, even >10% of the visible signal was due entirely to the ground clip (verification, moving the wire around in space, using different ground points). At worst, this slightly changes the >50MHz squigglies, only by phase and amplitude give or take 10% (or more around a switching device, if you even have a point to clip onto in such a place).
If you're looking at high precision settling time (better than 1%), or specifically the low level transients, you're bound to see something, but that's kind of a gimme, and doesn't compare with the amplitude of the actual switching nodes.
Even probing an avalanche pulse generator (tr < 2ns), I don't see much difference -- yes, it definitely has a lot more bounce and drool with four inches of ground lead (whereas grounding the probe collar to the circuit yields a signal identical to the proper BNC-coupled output), but it doesn't appreciably attenuate the really high frequency stuff. You can still see the general wave shape, rise/fall time is not ludicrously impaired.
When I first built an avalanche pulse generator, I was pleasantly surprised. After reading fear-mongering recommendations like yours, I thought the blip was going to simply disappear altogether with the clip. Turns out that's not the case! Even the cheap 10x probes that I've seen still work pretty well up there.
On a related subject, it's easy to build an avalanche pulse generator on a solderless breadboard, and I mean with junkbox parts -- long squiggly leads and everything, not even any attempt to make it nice. The overall pulse looks more like a gong, of course.
To get good clean pulses, layout is more critical. For example, the first avalanche generator I built was along the lines of AN-94, point-to-point around a connector. Worked okay, but drool amounted to maybe 5-10% of the initial pulse (residual charge, reflections, that sort of thing). I tried again, with a flatter layout over a ground plane, deadbug style, still leaded components, with similar results. Finally, I scored a PCB by hand and used SMT components; at last, this got a lone pulse with low rattle (
Obvious solution for this -- go ahead and let it short out the supply for those 5 or so nanoseconds. Put a big fat inductor in the supply line (which at this rate might be an inch or two of trace over a ground plane cutout). VCC droops and dI/dt is limited, at the very least reducing the rate of recovery, and reducing the peak amplitude if it still snaps. Finally, a little extra capacitance on the switching node slows the transition, so if it still snaps, at least it snaps between diode and cap, rather than the whole supply rail.
If, after all that, it still insists on snapping, there really isn't anything you can do about it besides shop for a device that you hope doesn't snap.
Monolithic chips are nice and handy, but they really shoot themselves in the foot when they expect to draw power for the controller from the same dirty rail that's doing the switching. I'm guessing most chips would find this supply snubbing approach...distasteful.
Other obvious solutions include intentionally slow switchers. This isn't a bad idea when high high sensitivity and bandwidth are required; see LT AN-70. Most power supplies cheat by reporting their noise within a cutoff of 20MHz or whatever, whereas transients can go out to 100s of MHz. Slower operation prevents those harmonics from being generated in the first place, but efficiency and size are the first things to go. Of course, you've maligned many times about the plurality of supplies required to run everything. Sometimes you just can't win.
Tim
-- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
=20
happen=20
My first suspect for that is a measurement problem. From experience trying to get 100 and 300 ampere pulses through 8 feet of cable (3.5 us and 300 ns wide respectively. Even harder to measure well.
?-)
On a sunny day (Wed, 11 Apr 2012 16:16:12 -0700) it happened John Larkin wrote in :
I was thinking about diode turn on time, and started visualizing the vacuum diode. Clearly in vacuum diode it takes time for the electrons hovering around the cathode to start moving when a large voltage is applied, and reach the anode.
The speed is set by the voltage, and the time by the distance between anode and cathode (in the 'perfect' diode tube).
In such a case, where the turn on time depends on the electron speed, one would expect a faster turn on when a higher voltage is applied. Do you also see that with those diodes?
Or is this idea not correct or does not apply to semiconductors?
Once the charge gets into the drift region and starts moving, you get displacement current--you don't have to wait for the charges to get to the other side, either in a tube or a diode. It's the diffusion delay that you have to deal with.
It's actually the same in a NEA photocathode, where the primary photoelectrons can rattle around for a nanosecond or so before escaping. (Cathodes with positive electron affinity don't show the effect so much--if an electron takes more than a couple of bounces, it loses enough energy that it can't escape. That gives you more speed but lower quantum efficiency.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
On a sunny day (Fri, 13 Apr 2012 08:54:40 -0400) it happened Phil Hobbs wrote in :
Is this correct? 'current' is the motion of electrons (as in Ampere electrons per units of time). So in a tube, when electrons start accelerating in the applied field (because of the voltage), as long as an electron does not reach the anode, no current flows. There was a current required to charge the anode up to the voltage, but that is a different current, and long gone to zero before the first electron hits the anode (you basically pulled electrons from the anode making it positive, or maybe better stuffed electrons into the cathode making it negative). I think you can only refer to 'displacement' once something is displaced.
An electron fills a hole in the semiconductor, and as that is not empty space, atoms are close, that probably is true displacement, electrons move one way resulting in holes moving the other way. But in a tube...?
A different picture: Imagine you statically charge anode to 1000V, and cathode to 0V. With the heater off, in a tube. Now you have the anode voltage, and still no current. The current only starts once the cathode gets heated, so electrons become available. And then those take time to reach the anode, if you could heat the cathode up all at once, even then it would take time.
Or this this view wrong?
See Phil's comment about displacement current in vacuum diodes.
In my 1N914 pics, you can see the higher currents, and faster turn-on, when I apply a bigger voltage step. With 1.2 volts across the diode, it hardly conducts at all for the first 4 ns or so. At 5 volts, it appears to be conducting before the generator rise is settled.
(When I designed the P400 output stage, I limited the slew rate to keep within US export limits. They have recently increased the allowable slew limits, so maybe I'll speed it up some day.)
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators
An external schottky diode helps a lot. It keeps that substrate diode from turning on.
Other synchronous switchers don't do this. National flubbed that one.
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators
in
rons per units of time).
ause of the voltage),
ode making it positive,
pace,
available.
eIf I've got an electron moving towards a metal plate (that's part of some other circuit), then there's an image charge on the metal plate. As the electron moves closer the image charge grows.. this charge comes from the circuit.. it looks like current flow. So the pulse of charge in the circuit doesn't happen all at once when the electron arrives at the plate, but over some time.
(At least that's how I see it.)
George H.
ut
On a sunny day (Fri, 13 Apr 2012 07:35:50 -0700 (PDT)) it happened George Herold wrote in :
Its intersting, I allocated some spare brain cycles to it, as now I am running data acquisition tests on my tritium experiment, and it stores once every hour (it lists everything to the terminal though each second), so I hop from one place to the other, anyways when I started thinking about it it got really complicated fast, relativity got into play too :-)
But this: Sure if you have a cloud of electrons (say 10000) on one electrode missing, and a 10000 surplus on the other, then it would depend on how much actually started moving if you take the interaction between the moving electrons in the tube and the ones on the electrodes. If 1 was moving and 10000 missing, no electron would leave its atom in the wires? Actually I am not sure about that, and there is the finite time c where things interact too. Then I started thinking photo-detector, and that 'photons' do not only affect the electrons, but the electrons must then also affect the photons. All the way back to the source? With time delay? QM.? I stopped here, and do not worry about my thoughts, :-)
On a sunny day (Fri, 13 Apr 2012 07:27:24 -0700) it happened John Larkin wrote in :
Cool.
Did not know they had limits on slewrates for exports too.... Indeed it seems you can gain something here with highter voltages.
Forward recovery looks roughly inductive. See for example,
per units of time).
of the voltage),
making it positive,
available.
Google "displacement current".
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
How about an aftermarket cam for the boy-racer set?
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 845-480-2058 hobbs at electrooptical dot net http://electrooptical.net
Larkin
electrons per units of time).
(because of the voltage),
anode making it positive,
displaced.
I think that is a long ways off from tube diode physics.
space,
become available.
cathode
Alternately, heat the cathode up and apply no anode voltage and get a current.
Start with a hot cathode and a space charge with an electron cloud = between the cathode and anode and calculate from there.
but
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diode
not
=20
That makes sense to me. I can't see getting much forward current through the diode while there is a depleted region in place. Thus collapsing it has to do with size, field and carrier mobility.
?-)
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