effects of switching polarity on an electromagnet
grin...@gmail.com
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
10A@24V, 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
Phil Hobbs
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
Tim Wescott
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.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Winfield Hill
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.
--
Thanks,
- Win
Mark
Mark
Tony Williams
<grin...@gmail.com> wrote:
> 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 10A@24V, 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.
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.
--
Tony Williams.
Rich Grise
Back-to-back zeners across the coil?
Good Luck!
Rich
Rich Grise
What if he got a supermagnet, and epoxied it to some kind of shaft,
and rotated it 180 degrees each half-cycle?
Cheers!
Rich
grin...@gmail.com
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
Winfield Hill
>
> 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?
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
> 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?
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.
> Thanks again (and thanks very much for the Art of Electronics,
> Winfield!),
Hey, thanks, you're very welcome, Graham.
--
Thanks,
- Win
Tony Williams
Winfield Hill <Winfiel...@newsguy.com> wrote:
[huge snip]
> . 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.
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+).
--
Tony Williams.
Winfield Hill
>
>In article <drlgd...@drn.newsguy.com>,
> Winfield Hill <Winfiel...@newsguy.com> wrote:
>
> [huge snip]
>
>> ... 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.
>
>>. o--|>|-+--|>|--|>|---+------------------+--------,
>>. | | | |
>>. | |--' '--| |
>>. | |<-, Q1 Q2 ,->| |
>>. | |--+ +--| |
>>. +| C1 | | | C2
>>. === +------//////------+ ===
>>. | | L + Rs | |
>>. | |--' '--| |
>>. | |<-, Q3 Q4 ,->| |
>>. | |--+ +--| |
>>. 0V | | | |
>>. o------+-------------+------------------+--------'
>
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-------+-------------+------------------+--------'
> 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+).
Yes, that's a good addition to save the power lost in D1.
--
Thanks,
- Win
Winfield Hill
> Tony Williams 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
--
Thanks,
- Win
Tony Williams
Winfield Hill <Winfiel...@newsguy.com> wrote:
> > Lots of forward diode drop in the circuit now.
> 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-------+-------------+------------------+--------'
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.
--
Tony Williams.
Winfield Hill
> Winfield Hill <Winfiel...@newsguy.com> wrote:
> 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-------+-------------+------------------+--------'
> 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.
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.
--
Thanks,
- Win
Winfield Hill
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.
> 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
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?"
--
Thanks,
- Win
Tony Williams
Winfield Hill <Winfiel...@newsguy.com> wrote:
> > 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.
> Tony, I'm puzzled about where you suggest putting the TVS chain.
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.
----------------------------------------------------------------
> 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.
Yes. The resonant LC half-sine is a solution to
the OP's original problem of reversing the current.
> 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.
How much energy is being recovered, against the
energy dissipated in the TVS chain? Is there
an optimum balance between the two?
> . 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.
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.
--
Tony Williams.
Winfield Hill
>
> How much energy is being recovered, against the energy
Right.
> This discussion has highlighted the fact that "putting
> the inductive energy from an H-bridge, back into the
> supply" should not be done casually.
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-----------+------------------+--------+-----'
--
Thanks,
- Win
Winfield Hill
>
> 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
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.
--
Thanks,
- Win
grin...@gmail.com
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.
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
Winfield Hill
>
> 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.
--
Thanks,
- Win
Tony Williams
<grin...@gmail.com> wrote:
> 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.
--
Tony Williams.
Winfield Hill
>
> Graham wrote:
>
>> 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.
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
. | | |
. '---|<|--+--|>|---'
> 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.
--
Thanks,
- Win
Tony Williams
Winfield Hill <Winfiel...@newsguy.com> wrote:
> 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.
--
Tony Williams.
Winfield Hill
> Winfield Hill wrote:
>
>>
>>. + +
>>. ---+---|(---+---)|---+--- both caps 1000uF elec
>>. | | |
>>. '---|<|--+--|>|---'
>
> 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.
--
Thanks,
- Win
Graham Grindlay
> 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
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.
Tony Williams
Graham Grindlay <grin...@mit.edu> wrote:
> 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.
> http://www.controlledautomationinc.com/Documents/Coils.html
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.
Winfield Hill
Whoa! What are you making, a high-throughput machine gun?
--
Thanks,
- Win
The Phantom
wrote:
>The Phantom wrote...
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.
Fred Bloggs
The OP is posting from mit.edu....too funny. He must be from the English
department.
Jim Thompson
wrote:
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
Fred Bloggs
MIT=Melbourne Institute of Technology
Jim Thompson
wrote:
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".
Fred Bloggs
Maybe they changed the name : http://www.mit.edu.au/
Jim Thompson
wrote:
I guess so. It's been 10 years since I was there.
Winfield Hill
> Tony Williams wrote:
>> Graham Grindlay wrote:
>>
>>> 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.
>>
>>> http://www.controlledautomationinc.com/Documents/Coils.html
>>
>> 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.
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
Tony Williams
Tony Williams <to...@ledelec.demon.co.uk> wrote:
> In article <dsfih...@drn.newsguy.com>,
> Winfield Hill <Winfiel...@newsguy.com> wrote:
> > 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?
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.
Winfield Hill
> Tony Williams <to...@ledelec.demon.co.uk> wrote:
>> Winfield Hill <Winfiel...@newsguy.com> wrote:
>>> 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?
>
> 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!
--
Thanks,
- Win
The Phantom
>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.
>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
Rich Grise
Hey, Win, if you lose your tech, I'm available, if you can afford
to relocate me. ;-)
Cheers!
Rich
Winfield Hill
>
> Graham Grindlay wrote:
>
>> 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.
>
>> http://www.controlledautomationinc.com/Documents/Coils.html
>
> 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.
Graham, we're still looking for an answer to this question.
What are you working on? Maybe we can come up with an
estimate.
--
Thanks,
- Win
Graham Grindlay
> I think Grahammust be at MIT's Media Lab.
Yep, you've got me pegged, Win. I'm a first-year grad student in a
music group at the Media Lab. I've actually got more of a comp. sci.
background, but am just starting to get my feet wet in the fascinating
(and eminently useful) world of electronics. I realize that many of my
questions are probably pretty naive and may seem obvious to the regulars
here on s.e.d and so I particularly appreciate everyone's help and patience.
-Graham
Graham Grindlay
still trying to wrap up a couple of projects from break...
> Graham, we're still looking for an answer to this question.
> What are you working on? Maybe we can come up with an
> estimate.
>
This is actually for a force-feedback drum project that I've been
thinking about recently. The idea is to seat the magnet just behind the
head of a drum, pulse the field direction rhythmically such that it
attracts and repels a permanent magnet embedded in a drum stick tip.
The original motivation was purely creative: see if this arrangement
could be used to produce a sort of 'player drum' effect (e.g. you strike
the drum once, but at the same time the magnet is used to rapidly attact
and then repel the permanent magnet in the drum stick tip, say in the
pattern of a fast triplet). Lately, I've been thinking that, if I can
get this thing to work, it would be interesting to look at pedagogical
applications. If you could 'quantize' a beginning player's sloppy
timing you might affect their learning rate by giving them a feel for
the correctly played rhythm. There are also some musical concepts such
as 'playing behind the beat' that just about impossible for a teacher to
convey verbally, so 'learning by feel' might be a real advantage in some
situations. In any case, this is obviously all very early stage stuff.
>>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 short of it is that, while I'm not sure, I don't expect the
mechanical load to be significant (unless, of course, someone *really*
took a swing at the drum!). I do know the maximum pull of the permanent
magnets (14.2lbs), but this would seem to be largely overwhelmed by the
added force of the player. Does this seem like a reasonable assumption?
Thanks,
Graham
Winfield Hill
>
> This is actually for a force-feedback drum project that I've been
> thinking about recently. The idea is to seat the magnet just behind
> the head of a drum, pulse the field direction rhythmically such that
> it attracts and repels a permanent magnet embedded in a drum stick tip.
> The original motivation was purely creative: see if this arrangement
> could be used to produce a sort of 'player drum' effect (e.g. you strike
> the drum once, but at the same time the magnet is used to rapidly attact
> and then repel the permanent magnet in the drum stick tip, say in the
> pattern of a fast triplet). Lately, I've been thinking that, if I can
> get this thing to work, it would be interesting to look at pedagogical
> applications. If you could 'quantize' a beginning player's sloppy
> timing you might affect their learning rate by giving them a feel for
> the correctly played rhythm. There are also some musical concepts such
> as 'playing behind the beat' that just about impossible for a teacher to
> convey verbally, so 'learning by feel' might be a real advantage in some
> situations. In any case, this is obviously all very early stage stuff.
Very interesting, and highly relevant to your design. I was very
concerned over the issue of coil heating, because 192 watts (8A*24V)
will overheat and destroy a small coil like yours in the end. You
may not notice trouble during short tests as the heat is taken by
the copper's thermal mass, but the equilibrium temperature will get
you. I envisioned that you had a steady maximum field and suddenly
wanted to collapse the field and then reverse it. But if what you
need is short, fast musically-timed field pulses, the average power
dissipation could be acceptable. That's good.
Second, your need for as fast as a 100ms repetition rate may be for
only a few cycles at a time, good, and you may be able to live with
slightly reduced force pulses after the first pulse, which we saw in
Tony's simulations of fast reversals.
I'm wondering if you do want a reversal. The shape of the field
pulse must be a strong factor in the drummer's sensation, surely
in normal use you don't want to interfere too strongly with the
drum stick's natural bounce off the surface. That must be why
you want to collapse the field quickly, to stop the attractive
forces by the time the stick hits the surface. But do you also
want a repulsive force? Perhaps a weakened one, to gain stick
distance for the fast triplet beat you're thinking about, etc?
I also wonder if you want a standing full-strength field before
the drum-beat event? What's the time-frame you envision for the
onset of your attractive field? The natural t = L/R = 53ms time
constant from simply switching your 24V supply seems pretty slow.
You can push single powerful half-sine uni-polar pulses of current
into an electromagnet with a charged bipolar capacitor and an SCR.
. half-sine positive current pulse
. +HV supply ---/\/\------,
. + C1 C2 + | --- coil --- t = pi sqrt L*C
. ,---|(---+---)|---+ 160mH 3-ohms
. ----+---|<|--+--|>|---+--#####--/\/\--, Ip = V sqrt C/L
. gnd |______________|/|________________|
. trigger__/|\| scr
The bipolar capacitor came up earlier in this thread. Since the
value you'll need is high, it's made up from two electrolytics.
As the voltage on initially-charged capacitor C1 approaches zero
halfway through the pulse, the field current reaches its peak.
The coil maintains a declining current flow in the second half
of the pulse, charging C2. When the current reaches zero the
SCR shuts off. C1 and C2 are both equal to C in the equations.
After the pulse you still have most of the energy, but it's in
the second cap, and the full bipolar capacitor has a reversed
voltage (diodes protect the electrolytics during the voltage
reversal). This is another place a high-voltage H-bridge may
be useful, to recover and reuse the energy.
To make a single-cycle bipolar current pulse, replace the SCR
with a TRIAC, and provide gate drive through the first half of
the cycle, plus a bit, then remove it so the TRIAC will turn off
after the second half finishes. A significant portion of the
charge will be back in C1, ready for the next pulse.
If you make C2 larger than C1, the attack will be fast but the
middle portion of the cycle will be longer than the attack. Or
if you make C2 smaller than C1, the current cutoff and reversal
will be faster than the attack and decay, which might be better
for your drumsticks. C2 will need a higher voltage rating.
BTW, you can get TRIACs with current ratings to 6000A. :-)
>>> 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.
This must be a very loosely-coupled magnetic system, so the
electromagnet's inductive Q will not be mechanically reduced.
--
Thanks,
- Win