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effects of switching polarity on an electromagnet

Sounds good. I'm curious, how big is your coil? What's it like?

You wouldn't want only two of the 26-volt TVS because a 55V flyback would be slower than you're looking for, but you could use a modest number of higher-voltage TVS parts. It's also be nice to drop back to 1.5kW parts, which are more commonly available. For example, you could use six of them in series, for a nice 40% safety margin.

Well, if you were to use 450-volts worth of TVS in series, then you'd get a discharge time I*L/V = 2.8ms, likely faster than you need. If you pick a lower flyback voltage you'll have an easier time finding low Ron switches, which is important to avoid a big fan-cooled heatsink. :-) For example, a 175V flyback would give you a 7.3ms shutoff but allow the use of 200V FETs.

Fairchild's FQA34N20 or ST's STW34NB20 (both in stock at Mouser for about $3.20 each), have Rds(on) of about 0.06 ohms, so they'd dissipate only about 4W, which is a big improvement over what you were facing with 500V parts. I'd even consider using one of these (wired backwards) to replace D1, because their 0.5V drop at 8A is better than you can do with a 250V diode. I'd use good heatsinks on the FETs, you want to keep the junction temperature down to avoid an increase in Rds(on).

To get the 175V flyback, you could use six series 1.5ke24a (also called 1p5ke24a, to eliminate the decimal) TVS, which drop 27-30V at 8A. 24c at Mouser. Six of these are rated to absorb 9J in a 1ms pulse, even more in your 7ms pulse, fine for your 5J energy.

As for C1, it's be wise to limit the flyback risetime to say 0.2ms, limiting dV/dt to a modest 0.9V/us. That means a capacitance of C = I*t/V = 9uF. An ordinary 10uF 250V electrolytic would probably work fine, given your presumably-long interval between events.

The current ramp has a V/L slope as it starts, but the coil's high series resistance makes it tail off in the classic manner you're use to seeing with a charging RC. The time constant is L/R, or 0.16/3 = 53ms. But it'll be within 10% of 8A after about 100ms.

Cheeky blighter that I am, I'll accept for both of us. Have fun, and please do report back.

--
 Thanks,
    - Win
Reply to
Winfield Hill
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5J per switchover at 10 switchovers per second is 50W dissipation in the TVS chain.

I finally got round to LTspice'ing the half-sine full recovery circuit. The LTspice file and waveforms at 100mS and 200mS reversing intervals has been posted to a.b.s.e under the title "Reversing a 161mH/8A inductor.".

The capacitor is 1000uF, chosen so that the peak of the half-sine does not exceed 100V. This allows the use of a Schotty diode and meaty low-voltage MOSFETs.

Dissipations in all components is quite low.

Off out now.... more later on today.

--
Tony Williams.
Reply to
Tony Williams

Ahah, I was wondering how often he was going to run this thing. I see you got the infomation from the first post.

Very nice. I especially like the 5Hz plots.

SFAICS in your first plot the current passes through zero in about 15ms, which is a little faster than the 20ms pedicted from the resonant frequency. I suppose that's due to the lossy elements?

Anyway, that compares to 2.8 and 7.3ms for 450 and 175V zener-limited flyback voltages. OTOH, your circuit rapidly reverses the magnetic field, getting to within 75% of the new current before using up the capacitor's stored energy after another 15ms. That compares to the rather slow 53ms risetime using the 24V supply alone. Although your 15ms shutoff falltime is half the speed for the 175V zener case, if Graham is going to attempt running at 10Hz the faster- flyback scheme's slow reversal risetime will be a killer.

One modest issue, the 1000uF has to be a bipolar cap.

. + + . ---+---|(---+---)|---+--- both caps 1000uF elec . | | | . '---||---'

Agreed, and getting rid of 50W when running a zener or cap + diode clamp at 10Hz is another aspect Graham can weigh in his decision hopper. 'Course, he's already dealing with a 192W heating level in his magnetic coil. I imagine that's a pretty good-sized coil with lots of copper thermal mass, but if he runs it for long the interior will get very hot, and he'll need to consider water cooling, etc.

I didn't look, did you model the MOSFET, capacitor and diode losses? Sounds like your having fun with your old circuit Tony. Too bad they canned that project long ago.

--
 Thanks,
    - Win
Reply to
Winfield Hill

You have a keen eye for an anomoly Win.

The half-sine has an apparent resonant frequency of 17Hz, nowhere near the 12.5Hz of the LC, or the 12.4Hz of the LCR.

Well, if the 3.3 ohm coil resistance is dropped to .001, (and the supply resistance increased to keep it at 8A current), then the quarter-sine risetime certainly goes from 15mS to 20mS.

My simple-minded resonance approach suggested that the coil resistance should only affect the Vpk, not the resonant frequency. It seems to be the other way round... the Vpk drops only slightly, and the resonant frequency has a large change.

Eh? I shall have to go away and think about that one.

The biggest losses for most components will be at the dc (non-reversing) condition.

There are no actual MOSFETs, just generic Spice switches with 0.1 ohm ON resistance. At 8A dc that would be 6.4W per switch. At 5Hz the effective current drops to 7.1A, and each switch is reported to have about 2.5W losses.

At 5Hz the capacitor current is 1.74Arms. An ESR of 0.1 ohms would be 0.3W and LTspice reports 0.33W.

The LTspice 100V/10A Schottky diode has a fwd drop of 0.68V at 8A. At dc that would be 5.4W, and at 5Hz it is reported as 4.3W. Note that this diode has a peak reverse voltage of (Vpk - Vsupply).... in theory. In fact any inductance in the supply cable will spike Vsupply (at the diode) negatively. Some close-in decoupling looks to be prudent.

Note: The 1000uF cap is actually modelled as 1000u in series with 0.1 ohm and 50uH (to represent ESL plus careless wiring). There are two 1uF caps also across the bridge.

To demo a poor 1000uF or careless construction those two caps can be dropped to say 100pF each. There is then an initial spike of about 1.6KV at switchover.

There are actually four Schottky diodes in the bridge to model the body diodes (I was being lazy). In a real bridge these would be a TVS across each MOSFET for close-in over-V protection.

--
Tony Williams.
Reply to
Tony Williams

Oops! I withdraw my comment. An ordinary electrolytic will do the job fine, as the voltage soars and drops back to Vs.

I was distracted by a coil-gun coil design I was working on at the same time yesterday morning, in which the voltage reverses. In that case, two series 1000uF caps would look like 500uF while resting, but as soon as they go to work and current flows, one diode or the other conducts and the network looks like 1000uF. Only one cap works at a time.

--
 Thanks,
    - Win
Reply to
Winfield Hill

This is great, Tony. Thanks!

Any reason why I couldn't use a heavier-duty diode (maybe IR's MUR2020CTPBF 200V 20A part), higer-voltage MOSFETs (say, STMicroelectronics STW34NB20 200V 34A FET), and smaller cap to get a quicker rise and fall in current through the magnet? A 330uF would give T = pi*sqrt(.161*330uF) = 23ms.

I realize that this will all take some fairly serious heat-sinking. From my limited experimenting, the magnet itself doesn't really seem to generate much/any heat though. Of course, I haven't pushed it too hard, but it doesn't seem to be too much of an issue. It came from a place called Controlled Automation and was originally intended for use in a feeder bowl. Their website has a few difficult to see pictures, if you're interested:

formatting link

Thanks, Graham

Reply to
Graham Grindlay

I'm not sure exactly what is going on in your coil-gun circuit, but are you sure that only cap works at a time? Once that circuit has been running, at least one, and usually both, of the caps tend to stay charged, so subsequent applications of current may not only exercise one cap at a time.

See the posting over on ABSE.

Reply to
The Phantom

Yes that would work. The diode will be dissipating nearly 7W though. That would run at about 160V pk across the inductor, provided it will take that.

Do you have a mechanical load on this electromagnet, and do you know how many mechanical watts it will take?

A mechanical load could collapse the magnetic field faster and slow the recharge time. In fact the Q of the magnet+load could be so low that the resonant reversal just will not work.

--
Tony Williams.
Reply to
Tony Williams

Whoa! What are you making, a high-throughput machine gun?

--
 Thanks,
    - Win
Reply to
Winfield Hill

No. I'm not necessarily suggesting the same use you have. Sometimes one needs a high value capacitor to carry AC current and a non-polar capacitor would be the thing to use. If you can't buy one, then you have to make it. I'm just pointing that it behaves in a different way than one might think at first glance.

Reply to
The Phantom

The OP is posting from mit.edu....too funny. He must be from the English department.

Reply to
Fred Bloggs

Except there ISN'T an English Department

Probably a visiting physics major from Harvard ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
           Liberalism is a persistent vegetative state
Reply to
Jim Thompson

MIT=Melbourne Institute of Technology

Reply to
Fred Bloggs

Again, not true, I've lectured there with Willy Sansen doing the CMOS, me doing the bipolar analog... it's RMIT as in "Royal Melbourne Institute of Technology".

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
           Liberalism is a persistent vegetative state
Reply to
Jim Thompson

Maybe they changed the name :

formatting link

Reply to
Fred Bloggs

I guess so. It's been 10 years since I was there.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
           Liberalism is a persistent vegetative state
Reply to
Jim Thompson

What's the matter with you, Fred? The world doesn't revolve around engineers designing things to specs. The world needs visionaries, who come up with off-the-beaten-path creations, and who seek out engineering types for help. I think Graham must be at MIT's Media Lab. Nothing funny about that at all. They must have the whole gamut from Physicists to Musicians to Engineers there. A lot of good stuff has come out of that place. Even though it often gets people pushing the envelope in areas outside their expertise, I have been happy to watch and applaud the Media Lab's approach. I've been told there are lots of copies of The Art of Electronics over there. If so, that could be an impetus driving some of them to seek us out here at s.e.d. That's something we should encourage.

--
 Thanks,
    - Win
Reply to
Winfield Hill

With a Q of 3.6 there should only be about a 1% difference between the resonant frequency of the LC and LCR. So this 5mS apparent error has been bothering me, but I think it can be reasonably explained now.

The LC or LCR quarter-sine calculation is based on the time it takes for the C to go from 0v to Vpk, (Vpk=90v). In fact the C is already precharged by the supply to 28v, so it is already partly up the quarter-sine waveshape.

28v/90v = 0.31, equivalent to 18 degrees, and 20mS*18/90 = 4mS.... not far off the apparent 5mS reduction.

Since Vc returns to 28v, a similar reduction happens on the way down. Which is why the overall half-sine measures about 32mS instead of 40.

--
Tony Williams.
Reply to
Tony Williams

Looks good, Tony!

--
 Thanks,
    - Win
Reply to
Winfield Hill

Since you have been connecting and disconnecting your 24V supply, presumably by simply breaking the connnection, have you looked with a scope to see how high the voltage across the coil gets when you have the very high di/dt of a break of the metallic connection? I wonder if the coil has some built-in voltage limiting device to protect the insulation of the coil. It might be good to know what the value of that voltage limit is, if there is one.

Reply to
The Phantom

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