# effects of switching polarity on an electromagnet

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

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

Thanks,
Graham

2. ### Phil HobbsGuest

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 WescottGuest

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. ### Winfield HillGuest

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. ### MarkGuest

using a relay to switch at 10Hz does not seem like a good idea...
Mark

6. ### Tony WilliamsGuest

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 GriseGuest

Back-to-back zeners across the coil?

Good Luck!
Rich

8. ### Rich GriseGuest

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. ### Winfield HillGuest

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.

.. 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 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. ### Tony WilliamsGuest

[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. ### Winfield HillGuest

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 HillGuest

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. ### Tony WilliamsGuest

[/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. ### Winfield HillGuest

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 HillGuest

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. ### Tony WilliamsGuest

[/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. ### Winfield HillGuest

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. ### Winfield HillGuest

wrote...
Tony and I had fun tossing around rapid magnetic-field shutoff
and reversal ideas, indulging ourselves, but Graham, we still
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