Looking for pulse-rated zener.

This is not too arduous a task. Look at 'zenamics' and similar - generally they are 'fast' zeners made for emc purposes. Why a zener and not a diode?

Assuming you have an input supply rated at/programmed to your required current then you might just stick a mosfet on the other end of you coils and switch it off. 15A with 8uH in 1uS is 120V so a 200V avalanche rated device will do it.

Read a datasheet....

That's a 200V avalanche rated device with 40mR Rdson (@25C) so your 15A gives 9W continuous loss, you'll need a heatsink. The extra mJ per 5 seconds isn't likely to hurt it... Have a look at the transient thermal impedance curves.

If you've managed to figure out what sort of coils you need then you have a physics mind and should be able make some sense of the information given.

I just grabbed that one as an example. You might find something more suitable.

Feel free to ask again if it doesn't make sense.

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( and these things are only sold in bulk. I don't need 12,000 or 25,000

Another alternative is to use a power supply and clamp the voltage to it with a fast high current diode. Fast high current diodes are easy to obtain, they're in every SMPS, used for a rectifier. The power supply could be something as simple as a battery, however it could be derived from the existing supply with a little ingenuity. Schottky rectifiers might be fast enough, also. I'm not sure what the zener voltage is from the above part number, so I can't say what the PS might be.

Please

This is a very easy job for ordinary parts, as I'll show below, but first there're a few other consideration you need to account for to get 1us turnoff. One is parallel capacitance. To get < 1us turnoff time you'll need > 250kHz resonant frequency (1/4 cycle = 1us), and you can calculate C < 1 / (2pi f)^2 L = 50nF. The total capacitance of your coil, cable, switching FET and clamping zener diode must be less than 50nF... but that's easy. OK, one down!

For capacitances of much less than 50nF, you can use the dI/dt = V/L formula to get V = LI/t = 120 volts to rid a 8uH coil of 15A in 1us. Let's say you choose a 150V zener. This means you'll need to choose a 200V power MOSFET or IGBT to switch your 15 amps. Hmm, that takes a medium to large FET, such as Fairchild's FQA34N20 or FQP34N20.

Mouser stocks the whole series,

One the front page of the FQA34N20 or FQP34N20 datasheet, you'll see a parameter called Single-Pulse Avalanche Energy with a spec of 640mJ. The note tells us this is for an 830uH inductance with 34A through it interrupted to fly back to the in-excess of 200V breakdown of the FET, and lose its energy in avalanche. So your a '34N20 type MOSFET can eat your puny 1mJ for breakfast, and come back asking for more!

The note also says the starting junction temperature is 25C.

If you continuously run 15A through a '34N20 with its 75-milli-ohm max ON resistance you'll dissipate 17 watts. I'm sure you'll have a good big heat sink, but lets still assume the junction temperature rises by say 50 degrees. This increases Ron by 1.4x, see figure 8, increasing the worst-case dissipation to 24 watts. But, although this reduces the Single-Pulse Avalanche Energy with a spec by 100/150 = 426mJ, it's still a far far higher capability than your 1mJ requirement.

[Parenthetical note, a '34N20 could drop up to 1.6V at 15A, so an IGBT switch, or two '34N20 in parallel might be a better choice...]

Yep, see the '34N20 Repetitive Avalanche Energy spec of 21mJ. For careful avalanche-energy capability calculations, you should turn to the Transient Thermal Response Curves, figure 11.

You need a TVS ( transient voltage suppresor ).

Errr..... like PKE something ( don't have data in front of me sadly ).

Widely available.

Graham

You need 120 volts or so, so why not series a string of, say, ordinary

1-watt zeners? That spreads the power dissipation and reduces capacitance.

Zeners are pretty tough for short pulses.

Or how about just an r-c snubber? You're going to need something to kill the ringing anyhow. Roughly 30 ohms and 8 nF would overshoot to about 400 volts and be close to critically damped.

John

One thing to consider: Allowing current to flow after the switch opens will mean that the magnetic field will continue to exist while it collapses, and if current continuse to flow,the magnetic field will reverse. Using a zener or equivalent across the coil as some kind of snubber would decrease the time that current flows during flyback time, but would allow current to flow when the polarity of the inductor reverses. In all cases the inductor will resonate with its own self-capacitance and other external capaacitances; the frequency and "Q" will change as shunt and/or series resistances change (zener conducting forward, zener conducting reverse, etc) making for a non-linear energy damping system. So, first you need to determine if it is OK to allow the magnetic field to reverse in polarity, and then determine the best way to dump or transfer the stored energy in that time while doing the best to ensure that energy dump / transfer does not continue after desired point (ie: magnetic field reversal). The stored energy can be dissipated in resistances (switching device losses, inductor resistance and resistors) as well as transferred into another inductor or a capacitor (which then is discharged at leisure).

"Mark Becker" wrote in message news: snipped-for-privacy@stowetel.com... | Hello - (hope this is the right place .. pardon the cross-post)... | | I was recently asked to come up with a circuit for rapidly turning off a | modest magnetic field (about 100 gauss) in a microsecond or less. Did a | little math, some thinking, and came up with a Helmholtz coil pair that | seems to do the right thing. Approximate inductance of the coil pair is | 8 uH. | | Turning the field off in under a microsecond was the difficult part.. | but a hint in a reference suggested placing a zener across the | inductors. This was tried with a much lower current (and a much lower | field) and indications are that this will work. However... | | Generating a 100 gauss field requires about 15 amperes. Turning off | that field by circulating the current through a zener means the zener | must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less | than a microsecond. The repetition rate is once every 5 seconds, maybe | a little faster if the research turns the frequency up. | | I have found only one maker of pulse-rated zener diodes.. and while I | was able to get a few as samples, now a small quantity is needed (

Why a zener? Does that give you a constant energy/time graph? Why is that good?

The energy has to go someplace. Where else can you dump it? Can the coil itself take the heat? What if you use a traditional relay-damping diode? 15 A diodes are easy to get.

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I think you missed what the OP's looking for. He said he already _has_ a solution, to his satisfaction. Pls reread above.

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1 mJ or 1 MJ? The former should be no challenge at all.

If you're content to just dissipate the energy at the end of each cycle, and it's only 1 mJ, then you can dump it into a Transorb. They amount to a sort of Zener which is designed for power absorption rather than for a nice sharp Zener knee. They are still available in lots of voltage ratings and you can string them together if you need higher voltages. [You didn't specify the voltage you needed.]

Otherwise, as someone else here implied, you can drive the coils with a partial H-bridge, and when you turn off the active transistors in the H-bridge the coil current will buck backwards into the power supply capacitors with a voltage drop equal to the PS voltage. If you want a faster cutoff than this you can add Transorbs into the buck leg so that the voltage is the sum of the PS voltage and the Transorb voltage(s).

-

----------------------------------------------- Jim Adney snipped-for-privacy@vwtype3.org Madison, WI 53711 USA

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(snip)

I think you are laboring under a misconception, here. The coil inductive voltage reverses the moment the current starts to decrease (V=L*(di/dt)), but the magnetic field polarity does not change till the current changes direction. Once the current passes through zero and the coil (and zener and switch) stray capacitance that is sitting at zener voltage starts to dump current into the coil as the capacitance discharges back toward zero volts, then the magnetic field polarity will also pass through zero and reverse.

look at TVS diodes. (Transient Voltage Diodes); they are designed to work very fast and clamp a good load.

i meant to say Transient Voltage Suppressers. i dont know where my head was at when i wrote that! :)

Err... Take a look at the current and voltage waveforms in a flyback system. Turn on the switch, inductive current increases in the standard R/L form. When the switch is opened, the magnetic field starts to collapse; the waveform across the inductor is square-ish in most practical circuits. During that time, the current goes rapidly to zero as the flyback voltage pulse rises; roughly remains near zero at the flattish top, and then goes negative as the flyback voltage pulse drops to zero. However, the voltage would continue to decrease and go negative (L-C oscillations), but the switch (FET) internal diode conducts, allowing the coil current to continus to flow. The voltage pulse seen has the *same* polarity as the supply. In a standard flyback scheme, the negative current waveform is mirror-image of the ramp-like charging time. This remains to be a fairly close picture of operation waveforms, even if the core of the inductor saturates to some extent.

This is incorrect. The current goes "rapidly" to zero _after_ the flyback voltage pulse rises, during the flattish top...

John is correct, the current shouldn't reverse, excepting perhaps for a small amount of ringing. Perhaps, when you speak of mirror image, you're thinking of a reversal of the rate-of-change in the flyback coil current, rather than the current polarity, per se?

In the rapid magnetic-field collapse system we're talking about, using MOSFET or TVS avalanche to absorb the coil's energy, the avalanching junction will stop conducting the instant the current drops to zero (the physics of avalanche doesn't have any reverse- recovery time). There will be a small amount of current reversal (and ringing) due to discharging the system capacitance from the avalanche voltage level back to the supply V, but it'll be small compared to the 15A magnetizing current.

For example, a 34n20 FET's capacitance is 400pF at 150V, a 1.5kW 150V TVS is 100pF, and 1 meter of cable is another 100pF. The resonant frequency with 8uH will be 2.3MHz. The energy stored in 600pF of capacitance at 130V (assume Vs = 20V supply for the coil) pushes the inductor to a peak reversal of i = V sqrt(C/L) = 1.1A, and the voltage to +20V -130V = -110V after a T/2 time of 220ns, assuming nothing else in the path. However, the FET's intrinsic body diode and the TVS diode will prevent the voltage from going below ground, bringing things to a stop after only 130ns, limiting the ringing to the 20V supply Vs, and the peak reversal current to about -600mA, only 4% of the original 15A.

Since the magnitude is only 4% and it only lasts a few hundred ns, it's not useful to characterize this as "current reversal."

A 2pi f L = 100-ohm resistor paralleled with the inductor would nicely damp the 2.3MHz ringing, and reduce the magnitude of the single ring as well, to say 2%. It would need to be a 3W part, etc., unless a 0.01uF series capacitor was added to stop any DC current. We call this R-C network a snubber.

--
 Thanks,
    - Win

Hello -

First, I wish to thank everyone participating in this discussion. While I'm an EE, it has been a long time since some of this has crossed my mind.. and the technology has improved considerably. The discussion has poked a few embers.. Thank you.

the inductor?

Using the effective series resistance of the inductor does not dissipate the energy fast enough. For 8 uH and a series resistance of something like .050 ohms, tau is on the order of 160 microseconds.

The zener has the feature of placing an (almost) constant voltage across the coil during field collapse. Ideally, superposition applies.. and I(t) = I(0) - Integral(v(t), t)/L. Faster than exponential decay.

Think I looked at this when first thinking about it.. and wasn't happy at the response time. I'll go and look again. I am also not familiar with "avalanche mode" transient absorbers and will stare at those as well.

The current circuit (pun not intended) uses a pair of IRFB18N50K MOSFETs and a pair of series-connected BZW03D100 100V across the MOSFETs. Hmm.. maybe the MOSFETs should be across the coils with a series fast diode..

Another poster suggested using the MOSFET intrinsic diode as the zener clamp. I stared at that one.. is this feasible (I'm still climbing out of using BJTs) ? I like the idea of the snubber better.. but have to think a little more on it.

Especially as now the researcher wants the magnetic field to enclose a larger volume. The coils for THAT assembly work out to about 400 uH .. and still need 15A.

Your time is *really* appreciated.

Mark