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effects of switching polarity on an electromagnet

W

Winfield Hill

Jan 1, 1970
0
[email protected] wrote...
Hi Win,

Thanks for the generous offer. I did manage to measure my magnet's
inductance, albeit indirectly. I fed 15VAC@60Hz 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.

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

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.
. 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, ...

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.
but I'm not sure how to calculate the startup time.
Is it just t = L (Im/V)?

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

Cheeky blighter that I am, I'll accept for both of us.
Have fun, and please do report back.
 
T

Tony Williams

Jan 1, 1970
0
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.

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.
 
W

Winfield Hill

Jan 1, 1970
0
Tony Williams wrote...
5J per switchover at 10 switchovers per second is 50W
dissipation in the TVS chain.

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

Very nice. I especially like the 5Hz plots.
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.

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
.. | | |
.. '---| said:
Dissipations in all components is quite low.

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.
 
T

Tony Williams

Jan 1, 1970
0
Winfield Hill said:
SFAICS in your first plot the current passes through zero
in about 15ms, which is a little faster than the 20ms
predicted from the resonant frequency.

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.
I suppose that's due to the lossy elements?

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.

One modest issue, the 1000uF has to be a bipolar cap.
. + +
. ---+---|(---+---)|---+--- both caps 1000uF elec
. | | |
. '---|<|--+--|>|---'

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

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.

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.
 
W

Winfield Hill

Jan 1, 1970
0
Tony Williams wrote...
Eh? I shall have to go away and think about that one.

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.
 
G

Graham Grindlay

Jan 1, 1970
0
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.".

This is great, Tony. Thanks!
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.

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:

http://www.controlledautomationinc.com/Documents/Coils.html

Thanks,
Graham
 
T

The Phantom

Jan 1, 1970
0
Tony Williams wrote...

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.

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.
 
T

Tony Williams

Jan 1, 1970
0
Graham Grindlay said:
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.

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.
 
W

Winfield Hill

Jan 1, 1970
0
The Phantom wrote...
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.

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

The Phantom

Jan 1, 1970
0
The Phantom wrote...

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

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.
 
F

Fred Bloggs

Jan 1, 1970
0
Tony said:
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.

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

Jim Thompson

Jan 1, 1970
0
The OP is posting from mit.edu....too funny. He must be from the English
department.

Except there ISN'T an English Department

Probably a visiting physics major from Harvard ;-)

...Jim Thompson
 
F

Fred Bloggs

Jan 1, 1970
0
Jim said:
Except there ISN'T an English Department

Probably a visiting physics major from Harvard ;-)

...Jim Thompson

MIT=Melbourne Institute of Technology
 
J

Jim Thompson

Jan 1, 1970
0
MIT=Melbourne Institute of Technology

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
 
F

Fred Bloggs

Jan 1, 1970
0
Jim said:
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

Maybe they changed the name : http://www.mit.edu.au/
 
W

Winfield Hill

Jan 1, 1970
0
Fred Bloggs wrote...
The OP is posting from mit.edu....too funny. He must be from
the English department.

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.
 
T

Tony Williams

Jan 1, 1970
0
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.

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.
 
W

Winfield Hill

Jan 1, 1970
0
Tony Williams wrote...
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.

Looks good, Tony!
 
T

The Phantom

Jan 1, 1970
0
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.

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.
 
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