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effects of switching polarity on an electromagnet

Discussion in 'Electronic Design' started by [email protected], Jan 21, 2006.

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  1. Guest

    Hi All,

    I have an application for which I'd like to be able switch the polarity
    of an electromagnet at farily low frequencies (less than 10Hz, but
    variable). The magnet coil is 3.3 Ohms and my DC power souce provides
    [email protected], so about 8 amps running through the magnet. I'm thinking of
    using a DPDT relay, driven by some digital circuitry, to switch the
    magnet between both polarities.

    I'm wondering if there are any potential issues to be aware of, given
    that this would approximate a bipolar square wave as a power souce.
    Since the polarity is switching, I can't use a shunting diode, but
    maybe a varistor would work to limit the inductive kick when the relay
    switches? Are there other concerns with this design (eddy currents)?

    I'd really appreciate any general advice on how to proceed or ideas
    about alternative designs.

    Thanks,
    Graham
     
  2. Phil Hobbs

    Phil Hobbs Guest

    You need to measure the inductance of the magnet. Switching a big one
    in a short time is going to take a very impressive power supply, because
    V = L * dI/dt. If it's 1 H, and you want to switch it through +- 16 A
    in 10 ms, you're going to need to apply 1*16/.01 or 1.6 kV. It will
    also take fairly heroic measures to avoid the windings arcing over.

    It's really quite easy to kill yourself this way, so be careful.

    Cheers,

    Phil Hobbs
     
  3. Tim Wescott

    Tim Wescott Guest

    Well, maybe 7.3 amps?

    A simple resistor in parallel with the coil will limit the voltage spike
    (and help bleed off current). If you used a 13.2 ohm resistor you'd
    develop about 96 volts when the relay opened, at the cost of a couple of
    amps with the relay closed.

    You could also do this with diodes to each power rail, so when either
    terminal of the coil rose above 24V or fell below ground a diode would
    conduct. This could be handled by a bridge rectifier if you wanted to
    save space -- and you could still put in a series resistor to raise the
    coil voltage above 24V to bring the coil current down.
     
  4. Phil Hobbs wrote...
    OK, Graham, go explore a circuit we call an H-bridge.
    .. __________________________________
    .. | | | |
    .. XXX _|_ _|_ XXX
    .. | /_\ /_\ | |
    .. '---+----, LOAD ,----+----' XXX = switch,
    .. +---o o---+ e.g., a MOSFET
    .. ,---+----' '----+----,
    .. | _|_ _|_ |
    .. XXX /_\ /_\ XXX
    .. _|___|______________________|____|

    These normally have reverse diodes to protect the switches.

    Phil mentioned the issue of rapid switching. If you need this
    capability, add a set of series diodes to limit the current flow
    of each bridge-leg to its primary direction. Add another set of
    diodes with zeners in series, to limit the flyback voltage when
    you want to switch rapidly. See below. Open the turned-ON pair
    of switches (i.e., all the switches open), wait for the flyback
    voltage to settle, then close the other switch pair.

    .. H-bridge, with provision for rapid inductive-load shutoff.
    .. _______________________________________________
    .. | | zener zener | |
    .. XXX '-|>|--|<|--, ,--|>|--|<|-' XXX
    .. | | | |
    .. '------|>|------+ LOAD +---------------' XXX = switch,
    .. +---o o---+ e.g., a MOSFET
    .. ,------|>|------+ +---------------,
    .. | | | |
    .. XXX ,-|<|--|>|--' '--|<|--|>|-, XXX
    .. _|___|_zener________________________zener_|___|

    BTW, the zener voltages must be higher than the supply voltage.
    Make sure the zeners can handle the magnet's energy. Use what we
    call TVS, or Transient Voltage Suppressor zeners, these have slugs
    of copper on each side of the semiconductor die to absorb heat.

    In practice, it may not be useful to use especially high-voltage
    zeners, because they only speed up the shutoff time, whereas the
    H-bridge supply voltage determines the startup time in the other
    direction. If you have a current-regulating bridge, you can use
    higher supply voltages to get faster startup time. But then the
    XXX elements are no longer switches, unless you're using a PWM.
    You can read up on the technology. As you can see, making rapid
    magnetic reversals can become fairly complicated, but it can be
    lots of fun. First educate yourself, and like Phil says, by all
    means, BE CAREFUL! One hand behind your back, tucked into your
    belt, whenever the lids are off and the power is on.
     
  5. Mark

    Mark Guest

    using a relay to switch at 10Hz does not seem like a good idea...
    Mark
     
  6. A relay would be a problem because of arcing of the
    contacts, so it has to be a semiconductor H-bridge.

    A few years ago I was asked to look into ways of
    reversing the current in a 180mH/8A/1.5ohm inductor,
    (5.76 Joules!), as reasonably quickly as possible.

    A quick trial with an H-bridge and diodes "allowing
    the inductive energy to flow back into the supply"
    destroyed an expensive programmable supply... due
    to too much energy from the inductor going back
    into the supply.

    This led to the interesting scheme below.

    D1 Vs, allowed to rise to Vpk.
    12.5V------|>|---+----+----------+----+------+
    | | | | |
    + _|_ _|_ + |
    Sw1/ /_\ D D /_\ /Sw2 |
    + | | + |
    | | | | +|
    +----+--//////--+----+ ===C
    | | R+L | | |
    + _|_ _|_ + |
    Sw3/ /_\ D D /_\ /Sw4 |
    + | | + |
    | | | | |
    0v---------------+----+----------+----+------+

    Assume the R+L is up at current I. When the bridge
    is swapped over the energy stored in L+R is tranferred
    via diodes (D) into capacitor C. The bridge supply
    rises to voltage Vpk. On the way, diode D1 gets back
    biassed to isolate the Vsupply.

    At Vpk, the inductor current is zero and the capacitor
    voltage is at a defined maximum. This takes a quarter
    sine of the resonant period of L+C. Then the direction
    of the energy flow reverses... the C starts to deliver
    current into L. One q-sine later the C is discharged
    and the L now has nearly -I flowing in it. However,
    at the end of the recharge the voltage has fallen below
    Vsupply, so D1 conducts again and holds the current in
    the inductor at -I.

    The approximate sums are relatively simple.

    If T is the time taken for the current to reverse, then
    1/2T = 1/2.pi.sqrt(L.C).

    And if the energy transfer between the L and C is 100%
    then C.Vpk^2 = L.I^2.

    Combining those produces the useful sums, Vpk = L.I.pi/T,
    and C = T^2/L*pi^2.

    So reversing the current in that 180mH/8A inductor in
    10mS would require Vs to swing up to a Vpk of 452V,
    (with C= 56uF). 100mS gives 45V, requiring 5600uF.

    I did play around with a breadboard and the scheme did
    look promising. One early problem was the tolerance of
    big electrolytics and a TVS to clamp Vpk to a safe value
    for the MOSFETs was a swift addition. Unfortunately
    the customer changed his mind about what was wanted.
     
  7. Rich Grise

    Rich Grise Guest

    Back-to-back zeners across the coil?

    Good Luck!
    Rich
     
  8. Rich Grise

    Rich Grise Guest

    What if he got a supermagnet, and epoxied it to some kind of shaft,
    and rotated it 180 degrees each half-cycle?

    Cheers!
    Rich
     
  9. Guest

    Thanks to all who responded to my post. I think that I'm going to give
    Winfield's suggested circuit for rapid switching a shot, but want to be
    sure that I understand his comments regarding the zeners. In my
    application, I am more concerned with the rapidity with which the field
    dies down than I am with how quickly it starts up (although I certainly
    don't want startup to be slow) and I know that simply applying my 24V
    power source yields an acceptable startup time. So, as I understand
    it, to get the quickest shutoff time, I should be using zeners with a
    breakdown voltage just a hair higher than my supply (24V) rather than
    ones with say a 50V rating. Am I correct here?

    I'm also curious as to how long the flyback voltage might take to
    settle in this circuit. Are we talking on the order of a few
    milliseconds here? How would one go about calculating the required
    time?

    Thanks again (and thanks very much for the Art of Electronics,
    Winfield!),
    -Graham
     
  10. wrote...
    Here's my circuit,

    .. H-bridge, with provision for rapid inductive-load shutoff.
    .. _______________________________________________
    .. | | zener zener | |
    .. XXX '-|>|--|<|--, ,--|>|--|<|-' XXX
    .. | | | |
    .. '------|>|------+ LOAD +------|<|------' XXX = switch,
    .. +---o o---+ e.g., a MOSFET
    .. ,---------------+ +---------------,
    .. | |
    .. XXX XXX
    .. _|____________________________________________|

    My circuit can be considerably simplified by keeping the bottom
    FET on during the flyback discharge. You can drive the FETs with
    a bridge driver IC like Intersil's HIP4080A. This part allows
    you to turn off the upper FETs while keeping the bottom one on.
    http://www.intersil.com/cda/deviceinfo/0,1477,HIP4080A,0.html
    During normal operation of the electromagnet the 8A current is
    developed across the 3.3-ohm magnet resistance. When the XXX
    switches are opened the inductive flyback reverses the voltage
    across the magnet terminals, to the supply plus two diodes plus
    a zener. The coil's inductance begins its current fall at the
    dI/dt = V/L rate we're so familiar with. Ignoring the coil's
    resistance we can say the time will be t = L I/V. So clearly
    you want V to be as high as is reasonable. To calculate some
    typical times we'd need to know your magnet's inductance.

    Actually, I liked Tony's circuit the best. It can be simpler,
    and has the advantage you can simply reverse the H-bridge
    switches, rather than opening the first pair for a discharge
    time, before closing the other pair, as my circuit required.
    http://groups.google.com/group/sci.electronics.design/msg/efec70381aaaac1c

    .. D1 Vs, allowed to rise to Vpk.
    .. 12.5V------|>|---+----+----------+----+------+
    .. | | | | |
    .. + _|_ _|_ + |
    .. Sw1/ /_\ D D /_\ /Sw2 |
    .. + | | + |
    .. | | | | +|
    .. +----+--//////--+----+ ===C
    .. | | R+L | | |
    .. + _|_ _|_ + |
    .. Sw3/ /_\ D D /_\ /Sw4 |
    .. + | | + |
    .. | | | | |
    .. 0v---------------+----+----------+----+------+

    When you reverse the set of closed switches (MOSFETs), say 2,3
    instead of 1,4 the current flow through the new FETs is in the
    reverse direction. However, unless the drop across the FET's
    Rds(on) exceeded 0.7 volts (and generally you'd select them so
    it wouldn't) in the normal drain-to-source direction, the drop
    in the reverse direction would be about the same, and the diodes
    shown in the drawing wouldn't conduct. This means the diodes
    don't play a special role, except during the short open time as
    the bridge is reversed. Power MOSFETs have intrinsic diodes as
    a part of their structure, so we don't need to explicitly add
    diodes if we use MOSFET switches. Tony's circuit becomes:

    .. D1 allowed to rise to Vpk <= Vzener
    .. V+ ----|>|---+------------------+--------+-----,
    .. | | | |
    .. |--' '--| | |
    .. |<-, Q1 Q2 ,->| | |
    .. |--+ +--| | \_|_
    .. | | +| C /_\
    .. +------//////------+ === | TVS
    .. | L + Rs | | | units
    .. |--' '--| | \_|_
    .. |--, Q3 Q4 ,->| | /_\
    .. |--+ +--| | |
    .. | | | |
    .. 0v-----------+------------------+--------+-----'

    Ignoring Rs, and equating L and C energies, Vpk = I sqrt L/C,
    or the TVS voltages, whichever is lower.

    For rapid discharge only, with no energy return, I'd probably
    eliminate the capacitor and rely on the TVS power zeners alone.
    This way the discharge is a fast ramp, rather than a half-sine.

    Any way you do it, the result is more simple than my circuit.

    If you want, the energy in the magnet can be returned to the
    power supply's output capacitor.

    .. TVS zeners rises to Vzener + Vs
    .. V+ -+--o--|>|--|>|---+-------+----------+--------,
    .. | | | |
    .. | |--' '--| |
    .. | |<-, Q1 Q2 ,->| |
    .. | |--+ +--| |
    .. +| Cs | --> Im | | C2
    .. === +------//////------+ ===
    .. | | L + Rs | |
    .. | |--' '--| |
    .. | |--, Q3 Q4 ,->| |
    .. | |--+ +--| |
    .. | | | |
    .. 0v--+--o-------------+------------------+--------'

    You might want to use a small C2 to limit the flyback risetime,
    calculate dV/dt = Im / C2, and reduce RFI, etc.

    Be careful, Cs in the power supply has to be large enough to
    absorb the magnet's inductive energy without soaring too much.
    If Cs absorbed all of the inductor's energy (ignoring the TVS)
    its voltage after flyback would be Vs' = sqrt (Vs^2 + LC I^2).

    Hmm, you can add your own external electrolytic cap to be safe.
    In that case add diode D1 and get the best of both worlds, fast
    dI/dt ramp, plus some energy storage and return to the magnet
    in the other direction, saving power and speeding up reversal.

    .. V+ D1 TVS zeners rises to Vzener + Vs + dVcap
    .. o--|>|-+--|>|--|>|---+------------------+--------,
    .. | | | |
    .. | |--' '--| |
    .. | |<-, Q1 Q2 ,->| |
    .. | |--+ +--| |
    .. +| C1 | | | C2
    .. === +------//////------+ ===
    .. | | L + Rs | |
    .. | |--' '--| |
    .. | |--, Q3 Q4 ,->| |
    .. | |--+ +--| |
    .. 0V | | | |
    .. o------+-------------+------------------+--------'


    One big difference between Tony's circuit and mine, my MOSFET's
    voltage ratings are low, just safely above the supply voltage,
    whereas for Tony's scheme the FETs need to be rated at a much
    higher voltage, safely above the maximum flyback. However, with
    your magnet's modest 8A draw, there are plenty of good inexpensive
    high-voltage FETs. Lots of attractive 800V power MOSFETs.

    There are good high-voltage MOSFET driver ICs, like IR's 600V
    ICs, allowing you to use impressively-high flyback voltages.
    http://www.irf.com/product-info/cic/fsgatedriverics.html
    http://www.irf.com/product-info/datasheets/data/ir2111.pdf
    These are only $2.58 each, in stock at DigiKey.
    http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail?Ref=120907&Row=68496&Site=US
    (fix any link wraparound) I've used the IR2113, which is rated
    for 2A gate drive, but the IR2111 is fine. Its higher sinking
    than sourcing gate-drive capability is appealing.

    That just leaves the issue of the TVS diodes, used to maximize
    the discharge ramp. These have to absorb the magnet's energy.
    TVS parts have internal copper slugs for this purpose. Note I
    showed parts in series. The largest practical size may be the
    5kW units, which can absorb 5J of energy in one ms (less if the
    time is shorter, scaled by the square root of time). You can
    purchase low-cost automotive types, e.g., the 5kp26A parts,
    which breakdown at about 30 volts for 8 amps, and wire them in
    series for a higher voltage. E.g., 15 for 450 volts. They'd
    handle 75J, which would be enough if your inductance less than
    L = 2E/I^2 = 2.3H. If it's more, you can use paralleled stacks
    of TVS diodes.

    OK, that's it, I've said enough.
    Hey, thanks, you're very welcome, Graham.
     
  11. [huge snip]
    ISTR Win that Q1/Q2 could be low voltage, but the
    drains of Q3 or Q4, (when OFF), are always going
    to be lifted to the Vpk of C2.

    Lots of forward diode drop in the circuit now. On
    the original single diode circuit I was going to
    try an N-channel MOSFET in parallel with D1,
    wired 'backwards', Source to V+, etc. The MOSFET
    would be normally ON, just gated OFF during the
    flyback/current-reversal (which is whenever the
    voltage on the H-bridge is higher than V+).
     
  12. Tony Williams wrote...
    Good point. Especially if the TVS consists of 15 in series!
    We can easily solve this by moving D1 (which must now be a
    high-voltage diode) and adding one small diode to establish
    the voltage on C1. After a few flybacks the standing voltage
    on C1 will rise to be 15 diodes-drops higher than Vs.

    .. Vs+ D1 rises to Vzener + Vs + 15Vd + dVcap
    .. o-+----|>|------------+------------------+--------,
    .. | TVS zeners | | |
    .. '-|>|-+--|>|--|>|---+ | |
    .. | | | |
    .. | |--' '--| |
    .. | |<-, Q1 Q2 ,->| |
    .. | |--+ +--| |
    .. +| C1 | | | C2
    .. === +------//////------+ ===
    .. | | L + Rs | |
    .. | |--' '--| |
    .. | |<-, Q3 Q4 ,->| |
    .. | |--+ +--| |
    .. 0V | | | |
    .. o-------+-------------+------------------+--------'
    Yes, that's a good addition to save the power lost in D1.
     
  13. Winfield Hill wrote...
    I took a quick look for a sample of 600V low-Ron power MOSFETs,
    and found IR's irfps30n60k. DigiKey stocks these for $11.85.
    http://www.irf.com/product-info/datasheets/data/irfps30n60kpbf.pdf

    The 30n60 is among the larger 600V FETs available, but still
    has Rds(on) = 0.16 ohms typ at 25C, which means they would drop
    1.3V at 8 amps, or 10 watts of dissipation, assuming Tj = 25C.
    Given an estimated 50C of heating, Rds(on) rises to ~ 0.24 ohms,
    for so 1.9V of drop and Pd = 15 watts. That's much worse than
    a diode when considered as a replacement for D1. It also means
    IGBTs can be considered for these tasks. An irg4pc50ud drops
    ~1.1V at 8A when warmed up (about 9 watts), $11.44 at DigiKey.
    http://www.irf.com/product-info/datasheets/data/irg4pc50ud.pdf
     
  14. [/QUOTE]
    I'm always nervous about dissipating forward power in
    TVS devices. It raises the die temperature and reduces
    the transient energy dissipation capacity. So another
    diode across the TVS zeners chain could be useful.

    Things are starting to look a little mucky now though.
    Might be better to put the TVS chain back across C2 and
    forget about using them to recover energy.
     
  15. Tony Williams wrote...
    I'd say there's not much concern with pre-heating, because forward
    current flows through the TVS diodes only long enough to discharge
    C1 to Vs + 15 diode drops, then D1 takes over for the reverse-
    direction steady current. However, there's still a motivation to
    add a diode across the TVS, as you suggest, to discharge C1 closer
    to Vs, recover more energy, and speed the current reversal. Still,
    the whole idea was that by adding the TVS parts, one is trading off
    a fast magnetic-field shutoff with a slow startup in the reverse
    direction, as the O.P. requested.

    .. Vs+ D1 rises to Vzener + Vs + dVcap
    .. o-+----|>|------------+------------------+--------,
    .. | TVS zeners | | |
    .. '-|>|-+--|>|--|>|---+ | |
    .. +----|>|------+ | |
    .. | D2 | | |
    .. | |--' '--| |
    .. | |<-, Q1 Q2 ,->| |
    .. | |--+ +--| |
    .. +| C1 | | | C2
    .. === +------//////------+ ===
    .. | | L + Rs | |
    .. | |--' '--| |
    .. | |<-, Q3 Q4 ,->| |
    .. | |--+ +--| |
    .. 0V | | | |
    .. o-------+-------------+------------------+--------'
    Tony, I'm puzzled about where you suggest putting the TVS chain.

    In the drawing above, C2 is a small capacitor to slightly limit
    the dV/dt as the voltage flys up, to start the rapid-discharge
    cycle. Whereas in your original design C2 was large, grabbing
    the magnet's energy and slowing the discharge to a resonant LC
    half-sine as a penalty, but then speeding up current reversal.
    So we see C2 in the two designs plays quite different roles.

    If we see C1 just as a large BFC cap to play it safe with the
    power supply, forgetting any fast-reversal energy recovery,
    that certainly simplifies the circuit. Except that C1 will be
    potentially much larger than before.

    .. TVS zeners
    .. Vs ,--|>|--|>|---, rises to Vzener + Vs
    .. o-------+----|>|------+------------------+--------,
    .. | D1 | | |
    .. | |--' '--| |
    .. | |<-, Q1 Q2 ,->| |
    .. | |--+ +--| |
    .. +| C1 | | | C2
    .. === +------//////------+ ===
    .. | | L + Rs | |
    .. | |--' '--| |
    .. | |<-, Q3 Q4 ,->| |
    .. | |--+ +--| |
    .. 0V | | | |
    .. o-------+-------------+------------------+--------'

    We also get into an interesting subject, how far are we allowed
    to force the power supply above its intended output voltage?
    I'd venture to say most supplies can be safely raised a few
    volts, maybe 5V. I'd want to see the schematic, but these days
    many manufacturers withold that information, for example Xantrex,
    my favorite high-power lab switching-supply-with-PFC company.
     
  16. Winfield Hill wrote...
    Of course MOSFETs could be used, if a number of 600V parts were
    paralleled, or if a lower TVS flyback voltage was used, allowing
    the use of lower-voltage lower-Ron FETs.

    Also, we should remember Tony's point that Q1 and Q1 can be
    low-voltage switches, that can be a big help when considering
    paralleled FETs, and in helping to reduce the heat-sink size.
    Low-voltage switches have much lower Ron for a given size.
    One issue if IGBTs are used: Unlike MOSFETs, they don't conduct
    current in the reverse direction, which is what's happening in
    the bridge during the flyback. That's why I selected irg4pc50ud
    parts, note the "d" on the end. Most IGBTs come with and without
    the extra parallel diode. Unlike the case with MOSFETs, where the
    parallel diode comes for free as an intrinsic part of the die, in
    IGBTs the diode takes considerable extra die space. If you check
    at DigiKey, you'll see the irg4pc50ud IGBT costs $11.44, whereas
    the irg4pc50u costs $8.68, so the diode adds an extra 32%. Whew!

    Because of the extra cost, manufacturers often give designers a
    choice, "Sir, would like a diode on the side, with your IGBT?"
     
  17. [/QUOTE]
    From a previous drawing of yours Win. C2 as indicated.

    ------------------------------------------------------------
    D1 allowed to rise to Vpk <= Vzener
    V+ ----|>|---+------------------+--------+-----,
    | | | |
    |--' '--| | |
    |<-, Q1 Q2 ,->| | |
    |--+ +--| | \_|_
    | | +| C /_\
    +------//////------+ === | TVS
    | L + Rs | became|C2 | units
    |--' '--| | \_|_
    |--, Q3 Q4 ,->| | /_\
    |--+ +--| | |
    | | | |
    0v-----------+------------------+--------+-----'

    Ignoring Rs, and equating L and C energies, Vpk = I sqrt L/C,
    or the TVS voltages, whichever is lower.

    For rapid discharge only, with no energy return, I'd probably
    eliminate the capacitor and rely on the TVS power zeners alone.
    This way the discharge is a fast ramp, rather than a half-sine.
    ----------------------------------------------------------------
    Yes. The resonant LC half-sine is a solution to
    the OP's original problem of reversing the current.
    How much energy is being recovered, against the
    energy dissipated in the TVS chain? Is there
    an optimum balance between the two?
    It will be an individual thing, depending the voltage
    rating of the output C, on any Overvoltage trips, or
    even crowbars.

    This discussion has highlighted the fact that
    "putting the inductive energy from an H-bridge,
    back into the supply" should not be done casually.
     
  18. Tony Williams wrote...
    Maybe it shouldn't be done at all. For rapid shutoff of
    the field, just a small C2 to control flyback slew rates.
    No C1. Q1 Q2 low-voltage MOSFETs. Q3 Q4 rated for the
    TVS-stack total, plus safety margin, perhaps using IGBTs
    (with internal diode) if Vzener is over 200V. Double up
    on Q3 Q4 if necessary to reduce heat-sink size. Simple.

    .. D1 rapidly rises to Vzener
    .. V+ ----|>|---+------------------+--------+-----,
    .. | | | |
    .. |--' '--| | |
    .. |<-, Q1 Q2 ,->| | |
    .. |--+ +--| | \_|_
    .. | | C2 | /_\
    .. +------//////------+ === | TVS
    .. | L + Rs | | | units
    .. |--' '--| | \_|_
    .. |--, Q3 Q4 ,->| | /_\
    .. |--+ +--| | |
    .. | | | |
    .. 0v-----------+------------------+--------+-----'
     
  19. wrote...
    Tony and I had fun tossing around rapid magnetic-field shutoff
    and reversal ideas, indulging ourselves, but Graham, we still
    are curious about your project, and your inductor. Have you
    had a chance to measure its inductance? If you like, you could
    take a walk over to the Institute and we could measure it for you.
    Then we could put some real-world numbers to all our theorizing.
     
  20. Guest

    Hi Win,

    Thanks for the generous offer. I did manage to measure my magnet's
    inductance, albeit indirectly. I fed [email protected] through the magnet and
    a decade box, R, in series:

    |--------/////////--------|
    | |
    ~ |
    | |
    |-----------R-----------|

    I adjusted R such that the voltage across R and the voltage across the
    magnet were equal. Then, using R's value as the reactance, X_L, I
    solved for L = X_L/(2*pi*60Hz). I get L = 161mH.

    In my application, I definitely can't tolerate field reversal times of
    more than 100ms. In fact it'd be highly preferable to be able to fully
    switch in about 10ms. It's okay if the field shutoff and startup times
    are unequal, but shutoff should be quicker than startup (say 20ms
    shutoff and 80ms startup for the worst-case 100ms scenario). This
    balance becomes less important as the total field reversal time becomes
    shorter.

    So, it seems that for the circuit below, I'd be safe with only two of
    the 5kp26A TVS zeners since the magnet energy is E = (161mH*8A^2)/2 =
    5.15J.

    .. D1 rapidly rises to Vzener
    .. V+ ----|>|---+------------------+--------+-----,
    .. | | | |
    .. |--' '--| | |
    .. |<-, Q1 Q2 ,->| | |
    .. |--+ +--| | \_|_
    .. | | C2 | /_\
    .. +------//////------+ === | TVS
    .. | L + Rs | | | units
    .. |--' '--| | \_|_
    .. |--, Q3 Q4 ,->| | /_\
    .. |--+ +--| | |
    .. | | | |
    .. 0v-----------+------------------+--------+-----'

    This just leaves me with finding an appropriate value for C2 and
    figuring out some startup and shutoff times. It seems as though
    shutoff should be very quick (<1ms) since it just depends on the TVS
    zener ramp, but I'm not sure how to calculate the startup time. Is it
    just t = L (Im/V)?

    Again, Win and Tony, thank you both so much for your help.

    Cheers,
    Graham
     
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