Looking for pulse-rated zener.

[Mark Becker]

Hello - (hope this is the right place .. pardon the cross-post)...

I was recently asked to come up with a circuit for rapidly turning off a
modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

Your time is appreciated.

Mark

P.S. Postings to the newsgroup preferred.. however, if you e-mail a
response to this note, the e-mailer at stowetel will challenge. Please
ack the challenge.. it slows down the spammers. Thanks.

 

[R.Lewis]

Hello - (hope this is the right place .. pardon the cross-post)...

I was recently asked to come up with a circuit for rapidly turning off a
modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

Your time is appreciated.

Mark

P.S. Postings to the newsgroup preferred.. however, if you e-mail a
response to this note, the e-mailer at stowetel will challenge. Please
ack the challenge.. it slows down the spammers. Thanks.

This is not too arduous a task.
Look at 'zenamics' and similar - generally they are 'fast' zeners made for
emc purposes.
Why a zener and not a diode?

 

[Genome]

Hello - (hope this is the right place .. pardon the cross-post)...

I was recently asked to come up with a circuit for rapidly turning off a
modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

Your time is appreciated.

Mark

Assuming you have an input supply rated at/programmed to your required
current then you might just stick a mosfet on the other end of you coils and
switch it off. 15A with 8uH in 1uS is 120V so a 200V avalanche rated device
will do it.

Read a datasheet....

http://www.irf.com/product-info/datasheets/data/irfb260n.pdf

That's a 200V avalanche rated device with 40mR Rdson (@25C) so your 15A
gives 9W continuous loss, you'll need a heatsink. The extra mJ per 5 seconds
isn't likely to hurt it... Have a look at the transient thermal impedance
curves.

If you've managed to figure out what sort of coils you need then you have a
physics mind and should be able make some sense of the information given.

I just grabbed that one as an example. You might find something more
suitable.

Feel free to ask again if it doesn't make sense.

DNA

 

[Watson A.Name - \Watt Sun, the Dark Remover\]

Hello - (hope this is the right place .. pardon the cross-post)...

I was recently asked to come up with a circuit for rapidly turning off a
modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

Your time is appreciated.
Mark

Another alternative is to use a power supply and clamp the voltage to it
with a fast high current diode. Fast high current diodes are easy to
obtain, they're in every SMPS, used for a rectifier. The power supply
could be something as simple as a battery, however it could be derived
from the existing supply with a little ingenuity. Schottky rectifiers
might be fast enough, also. I'm not sure what the zener voltage is from
the above part number, so I can't say what the PS might be.

 

[Winfield Hill]

Mark Becker wrote...

I was recently asked to come up with a circuit for rapidly turning
off a modest magnetic field (about 100 gauss) in a microsecond or
less. Did a little math, some thinking, and came up with a Helmholtz
coil pair that seems to do the right thing. Approximate inductance
of the coil pair is 8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much
lower field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in
less than a microsecond.

This is a very easy job for ordinary parts, as I'll show below, but
first there're a few other consideration you need to account for to
get 1us turnoff. One is parallel capacitance. To get < 1us turnoff
time you'll need > 250kHz resonant frequency (1/4 cycle = 1us), and
you can calculate C < 1 / (2pi f)^2 L = 50nF. The total capacitance
of your coil, cable, switching FET and clamping zener diode must be
less than 50nF... but that's easy. OK, one down!

For capacitances of much less than 50nF, you can use the dI/dt = V/L
formula to get V = LI/t = 120 volts to rid a 8uH coil of 15A in 1us.
Let's say you choose a 150V zener. This means you'll need to choose
a 200V power MOSFET or IGBT to switch your 15 amps. Hmm, that takes
a medium to large FET, such as Fairchild's FQA34N20 or FQP34N20.
http://www.fairchildsemi.com/sitesearch/fsc.jsp?command=text&attr1=34N20
Mouser stocks the whole series, http://www.mouser.com

One the front page of the FQA34N20 or FQP34N20 datasheet, you'll see
a parameter called Single-Pulse Avalanche Energy with a spec of 640mJ.
The note tells us this is for an 830uH inductance with 34A through it
interrupted to fly back to the in-excess of 200V breakdown of the FET,
and lose its energy in avalanche. So your a '34N20 type MOSFET can
eat your puny 1mJ for breakfast, and come back asking for more!

The note also says the starting junction temperature is 25C.

If you continuously run 15A through a '34N20 with its 75-milli-ohm max
ON resistance you'll dissipate 17 watts. I'm sure you'll have a good
big heat sink, but lets still assume the junction temperature rises by
say 50 degrees. This increases Ron by 1.4x, see figure 8, increasing
the worst-case dissipation to 24 watts. But, although this reduces
the Single-Pulse Avalanche Energy with a spec by 100/150 = 426mJ,
it's still a far far higher capability than your 1mJ requirement.

[Parenthetical note, a '34N20 could drop up to 1.6V at 15A, so an
IGBT switch, or two '34N20 in parallel might be a better choice...]

The repetition rate is once every 5 seconds, maybe a little faster
if the research turns the frequency up.

Yep, see the '34N20 Repetitive Avalanche Energy spec of 21mJ. For
careful avalanche-energy capability calculations, you should turn to
the Transient Thermal Response Curves, figure 11.

I have found only one maker of pulse-rated zener diodes.. and while
I was able to get a few as samples, now a small quantity is needed
(<100) and these things are only sold in bulk. I don't need 12,000
or 25,000 diodes, not for a research effort.

OK, let's talk zener diodes. For example, how much transient energy
can two ordinary puny little 1n4761 75-volt zeners in series absorb?
http://www.onsemi.com/pub/Collateral/1N4728A-D.PDF
http://www.onsemi.com/pub/Collateral/AN784-D.PDF

Look at figure 5 of the datasheet, and observe the ability of the die
to absorb energy. For example, the single-event curve shows 100W can
be dissipated for 20us, which says E = Pt = 2mJ. This means two 75V
zeners (ON Semi doesn't offer a 150V part in this series) can handle
4mJ. For 1us we'd have to derate that by about sqrt 20 to 0.9mJ.

Now consider, this is a very small part with a small die. It's not a
complete wimp, but it can't begin to handle the energy your MOSFET can.

Let's say you prefer a zener to the FET, where would you turn? Why to
silicon Transient Voltage Suppressors of course! These are just zener
diodes mated to slugs of copper, to aborb lots of energy quickly. For
example, consider the small p6ke type, rated at 600W for 1ms pulses.
These can handle 600W average for 500us, or 300mJ per pulse for such
a long pulse. http://www.onsemi.com/pub/Collateral/P6KE6.8A-D.PDF

For shorter pulses, where the heat doesn't have time to spread
throughout the copper, they're more limited. But the fig 4 shows
they can handle 12kW for 1us, or 12mJ of energy, way more than the
1N4728-series zeners. The p6ke's are available in 150V versions,
p6ke150a, only 26 cents from Mouser. Also, FYI, Vishay makes parts
rated up to 540V.

Now, let's say you had a really serious problem... You can get much
larger versions of these Transient Voltage Suppressors. For example,
I used Fagor 5kp 5kW parts to rapidly absorb and stop the 900A field
current in Lena Hau's stopped-light experiments here at the Institute.
OK, I used 8 in parallel, each with a 0.05-ohm low-inductance ballast
resistor to insure equalizing the current, but that's another story.
See the two 1997 s.e.d. threads, "Living dangerously with 1800 amps."

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

No.

 

[Pooh Bear]

Hello - (hope this is the right place .. pardon the cross-post)...

You need a TVS ( transient voltage suppresor ).

Errr..... like PKE something ( don't have data in front of me sadly ).

Widely available.

Graham

 

[John Larkin]

Hello - (hope this is the right place .. pardon the cross-post)...

I was recently asked to come up with a circuit for rapidly turning off a
modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

Your time is appreciated.

Mark

P.S. Postings to the newsgroup preferred.. however, if you e-mail a
response to this note, the e-mailer at stowetel will challenge. Please
ack the challenge.. it slows down the spammers. Thanks.

You need 120 volts or so, so why not series a string of, say, ordinary
1-watt zeners? That spreads the power dissipation and reduces
capacitance.

Zeners are pretty tough for short pulses.


Or how about just an r-c snubber? You're going to need something to
kill the ringing anyhow. Roughly 30 ohms and 8 nF would overshoot to
about 400 volts and be close to critically damped.

John

 

[Robert Baer]

Hello - (hope this is the right place .. pardon the cross-post)...

I was recently asked to come up with a circuit for rapidly turning off a
modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

Your time is appreciated.

Mark

P.S. Postings to the newsgroup preferred.. however, if you e-mail a
response to this note, the e-mailer at stowetel will challenge. Please
ack the challenge.. it slows down the spammers. Thanks.

One thing to consider:
Allowing current to flow after the switch opens will mean that the
magnetic field will continue to exist while it collapses, and if current
continuse to flow,the magnetic field will reverse.
Using a zener or equivalent across the coil as some kind of snubber
would decrease the time that current flows during flyback time, but
would allow current to flow when the polarity of the inductor reverses.
In all cases the inductor will resonate with its own self-capacitance
and other external capaacitances; the frequency and "Q" will change as
shunt and/or series resistances change (zener conducting forward, zener
conducting reverse, etc) making for a non-linear energy damping system.
So, first you need to determine if it is OK to allow the magnetic
field to reverse in polarity, and then determine the best way to dump or
transfer the stored energy in that time while doing the best to ensure
that energy dump / transfer does not continue after desired point (ie:
magnetic field reversal).
The stored energy can be dissipated in resistances (switching device
losses, inductor resistance and resistors) as well as transferred into
another inductor or a capacitor (which then is discharged at leisure).

 

[Daniel A. Thomas]

| Hello - (hope this is the right place .. pardon the cross-post)...
|
| I was recently asked to come up with a circuit for rapidly turning off a
| modest magnetic field (about 100 gauss) in a microsecond or less. Did a
| little math, some thinking, and came up with a Helmholtz coil pair that
| seems to do the right thing. Approximate inductance of the coil pair is
| 8 uH.
|
| Turning the field off in under a microsecond was the difficult part..
| but a hint in a reference suggested placing a zener across the
| inductors. This was tried with a much lower current (and a much lower
| field) and indications are that this will work. However...
|
| Generating a 100 gauss field requires about 15 amperes. Turning off
| that field by circulating the current through a zener means the zener
| must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
| than a microsecond. The repetition rate is once every 5 seconds, maybe
| a little faster if the research turns the frequency up.
|
| I have found only one maker of pulse-rated zener diodes.. and while I
| was able to get a few as samples, now a small quantity is needed (<100)
| and these things are only sold in bulk. I don't need 12,000 or 25,000
| diodes, not for a research effort.
|
| Questions:
| 1. Anyone know a house selling small quantities of BZW03D100 diodes?
| 2. Anyone know a maker of suitable zeners?
| 3. Do I really need a high-power zener for this?
|
| Your time is appreciated.
|
| Mark
|
| P.S. Postings to the newsgroup preferred.. however, if you e-mail a
| response to this note, the e-mailer at stowetel will challenge. Please
| ack the challenge.. it slows down the spammers. Thanks.

Mark,

One of the methods for creating a power zener is to insert the zener
between drain and gate (or between the base and collector). There is no
need for special pulse rated zeners as the '34N20 takes care of the brunt
of the load.

Of course you'll need to add whatever protection is required for the FET
in the gate circuit and determine your drive circuit...


o
|
|
|
C|
C| 8u
C|
|
|
o---------|
| |
| |
| |
Z |
A | D
| |
R | ||-+
___ | ||<- '34N20
o------|___|------o------||-+
G |
|
| S
|
|
gnd

created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

Hope this helps
Dan Thomas

 

[Hal Murray]

I was recently asked to come up with a circuit for rapidly turning off a

modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

Why a zener? Does that give you a constant energy/time graph?
Why is that good?

The energy has to go someplace. Where else can you dump it?
Can the coil itself take the heat? What if you use a traditional
relay-damping diode? 15 A diodes are easy to get.

 

[Winfield Hill]

Daniel A. Thomas wrote...

One of the methods for creating a power zener is to insert the zener
between drain and gate (or between the base and collector). There is
no need for special pulse rated zeners as the '34N20 takes care of
the brunt of the load.

Of course you'll need to add whatever protection is required for the
FET in the gate circuit and determine your drive circuit...

o
|
|
C|
C| 8u
C|
|
o---------|
| |
Z |
A | D
R | ||-+
___ | ||<- '34N20
o------|___|------o------||-+
G | S
|
gnd

This active-zener method works well with low-voltage power MOSFETs,
such as under 100V, but it's dangerous with high-voltage FETs, 200V
and up, because they have a bad tendency to go into RF oscillation.
This is a high-power RF oscillation at frequencies of 15 to 40MHz,
which is very difficult to damp with external parts such a ferrite
beads, gate resistors, etc. That's because the RF oscillation is
internal to the FET, employing its inductance and self capacitance.
The required linear properties occur whenever a high current flows
while the drain-source voltage is higher than 10 to 20V. The latter
condition causes the FET capacitances to drop to the levels where RF
amplification is efficient.

A second point is that such an active-zener method isn't necessary,
because most power MOSFETs are just as happy dissipating 250mJ in
avalanche mode as in their active power regions. So there's little
reason to use them in a fashion that risks high-power oscillation.
(In the *old* days some FETs had problems with possible gate damage
due to the current pathways during avalanche. To satisfy oneself
that the FET has been designed to avoid this, and is safe to use,
look for an Eas Single-Pulse Avalanche Energy rating on the spec
sheet. Most parts have this rating, including the 34n20 types.)

Moreover, if you carefully calculate the relationship of the Eas
spec, and its associated pulse length, to the maximum power you
can disspate using the Transient Thermal Response Curve, you'll see
that the maximum Eas spec uses the same full thermal capability of
the MOSFET, identical to what you could do in active zener mode.
To calculate the maximum Avalanche Energy capability for durations
other than for the spec'd value, simply use the Transient Thermal
Response Curves with the usual junction-temperature deratings.
This way you can see exactly how much safety margin you have, and
just what the FET's junction temperature is doing at all times.

.. 8A <-- coils
.. ,---UUUUU--UUUUU--- constant-current supply
.. | or supply + ballast R
.. |--'
.. ||<-, MOSFET
.. --/\/\--'|--+ fqaA34n20
.. |
.. gnd

If the breakdown voltage of a FET candidate is too high, e.g. 250V
or more for the 34n20, then simply choose a lower-voltage MOSFET.
For example, a 100V part would breakdown at 120V or higher.

Or, if you're still worried about such an approach, then use a TVS,
and dump the inductor's energy into its massive copper blocks. The
common 1500-watt versions have plenty of capability.

.. 8A <-- coils
.. ,---UUUUU--UUUUU--- constant-current supply
.. |
.. | TVS 1.5ke120A, 1n6297A, etc.
.. +-----|<|----,
.. | / |
.. |--' gnd
.. ||<-,
.. --/\/\--'|--+ fqaA34n20
.. |
.. gnd

 

[Watson A.Name - \Watt Sun, the Dark Remover\]

One thing to consider:
Allowing current to flow after the switch opens will mean that the

magnetic field will continue to exist while it collapses, and if current
continuse to flow,the magnetic field will reverse.
Using a zener or equivalent across the coil as some kind of snubber
would decrease the time that current flows during flyback time, but
would allow current to flow when the polarity of the inductor reverses.
In all cases the inductor will resonate with its own self-capacitance
and other external capaacitances; the frequency and "Q" will change as
shunt and/or series resistances change (zener conducting forward, zener
conducting reverse, etc) making for a non-linear energy damping system.
So, first you need to determine if it is OK to allow the magnetic
field to reverse in polarity, and then determine the best way to dump or
transfer the stored energy in that time while doing the best to ensure
that energy dump / transfer does not continue after desired point (ie:
magnetic field reversal).

I think you missed what the OP's looking for. He said he already _has_
a solution, to his satisfaction. Pls reread above.

 

[Fred Bloggs]

| Hello - (hope this is the right place .. pardon the cross-post)...
|
| I was recently asked to come up with a circuit for rapidly turning off a
| modest magnetic field (about 100 gauss) in a microsecond or less. Did a
| little math, some thinking, and came up with a Helmholtz coil pair that
| seems to do the right thing. Approximate inductance of the coil pair is
| 8 uH.
|
| Turning the field off in under a microsecond was the difficult part..
| but a hint in a reference suggested placing a zener across the
| inductors. This was tried with a much lower current (and a much lower
| field) and indications are that this will work. However...
|
| Generating a 100 gauss field requires about 15 amperes. Turning off
| that field by circulating the current through a zener means the zener
| must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
| than a microsecond. The repetition rate is once every 5 seconds, maybe
| a little faster if the research turns the frequency up.
|
| I have found only one maker of pulse-rated zener diodes.. and while I
| was able to get a few as samples, now a small quantity is needed (<100)
| and these things are only sold in bulk. I don't need 12,000 or 25,000
| diodes, not for a research effort.
|
| Questions:
| 1. Anyone know a house selling small quantities of BZW03D100 diodes?
| 2. Anyone know a maker of suitable zeners?
| 3. Do I really need a high-power zener for this?
|
| Your time is appreciated.
|
| Mark
|
| P.S. Postings to the newsgroup preferred.. however, if you e-mail a
| response to this note, the e-mailer at stowetel will challenge. Please
| ack the challenge.. it slows down the spammers. Thanks.

Mark,

One of the methods for creating a power zener is to insert the zener
between drain and gate (or between the base and collector). There is no
need for special pulse rated zeners as the '34N20 takes care of the brunt
of the load.

Of course you'll need to add whatever protection is required for the FET
in the gate circuit and determine your drive circuit...


o
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C|
C| 8u
C|
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o---------|
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Z |
A | D
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R | ||-+
___ | ||<- '34N20
o------|___|------o------||-+
G |
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gnd

created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de

Hope this helps
Dan Thomas

It will be tough driving the FET into the ohmic region when the forward
bias on the zener clamps Vgd to a diode drop. And you will spend a long
time in the lab taming the all the parasitic effects- to some unknown
degree - unless you plan on spending even more time there.

 

[Jim Adney]

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

1 mJ or 1 MJ? The former should be no challenge at all.

If you're content to just dissipate the energy at the end of each
cycle, and it's only 1 mJ, then you can dump it into a Transorb. They
amount to a sort of Zener which is designed for power absorption
rather than for a nice sharp Zener knee. They are still available in
lots of voltage ratings and you can string them together if you need
higher voltages. [You didn't specify the voltage you needed.]

Otherwise, as someone else here implied, you can drive the coils with
a partial H-bridge, and when you turn off the active transistors in
the H-bridge the coil current will buck backwards into the power
supply capacitors with a voltage drop equal to the PS voltage. If you
want a faster cutoff than this you can add Transorbs into the buck leg
so that the voltage is the sum of the PS voltage and the Transorb
voltage(s).

-

 

[John Popelish]

One thing to consider:
Allowing current to flow after the switch opens will mean that the
magnetic field will continue to exist while it collapses, and if current
continuse to flow,the magnetic field will reverse.

(snip)

I think you are laboring under a misconception, here. The coil
inductive voltage reverses the moment the current starts to decrease
(V=L*(di/dt)), but the magnetic field polarity does not change till
the current changes direction. Once the current passes through zero
and the coil (and zener and switch) stray capacitance that is sitting
at zener voltage starts to dump current into the coil as the
capacitance discharges back toward zero volts, then the magnetic field
polarity will also pass through zero and reverse.

 

[Jamie]

Hello - (hope this is the right place .. pardon the cross-post)...

I was recently asked to come up with a circuit for rapidly turning off a
modest magnetic field (about 100 gauss) in a microsecond or less. Did a
little math, some thinking, and came up with a Helmholtz coil pair that
seems to do the right thing. Approximate inductance of the coil pair is
8 uH.

Turning the field off in under a microsecond was the difficult part..
but a hint in a reference suggested placing a zener across the
inductors. This was tried with a much lower current (and a much lower
field) and indications are that this will work. However...

Generating a 100 gauss field requires about 15 amperes. Turning off
that field by circulating the current through a zener means the zener
must (a) handle a pulse of 15 amps and (b) dissipate nearly 1 mJ in less
than a microsecond. The repetition rate is once every 5 seconds, maybe
a little faster if the research turns the frequency up.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

Questions:
1. Anyone know a house selling small quantities of BZW03D100 diodes?
2. Anyone know a maker of suitable zeners?
3. Do I really need a high-power zener for this?

Your time is appreciated.

Mark

P.S. Postings to the newsgroup preferred.. however, if you e-mail a
response to this note, the e-mailer at stowetel will challenge. Please
ack the challenge.. it slows down the spammers. Thanks.

look at TVS diodes.
(Transient Voltage Diodes);
they are designed to work very fast and clamp a good load.

 

[Jamie]

look at TVS diodes.
(Transient Voltage Diodes);
they are designed to work very fast and clamp a good load.

i meant to say Transient Voltage Suppressers.
i dont know where my head was at when i wrote that!

 

[Robert Baer]

(snip)

I think you are laboring under a misconception, here. The coil
inductive voltage reverses the moment the current starts to decrease
(V=L*(di/dt)), but the magnetic field polarity does not change till the
current changes direction. Once the current passes through zero and the
coil (and zener and switch) stray capacitance that is sitting at zener
voltage starts to dump current into the coil as the capacitance
discharges back toward zero volts, then the magnetic field polarity will
also pass through zero and reverse.

Err...
Take a look at the current and voltage waveforms in a flyback system.
Turn on the switch, inductive current increases in the standard R/L form.
When the switch is opened, the magnetic field starts to collapse; the
waveform across the inductor is square-ish in most practical circuits.
During that time, the current goes rapidly to zero as the flyback
voltage pulse rises; roughly remains near zero at the flattish top, and
then goes negative as the flyback voltage pulse drops to zero.
However, the voltage would continue to decrease and go negative (L-C
oscillations), but the switch (FET) internal diode conducts, allowing
the coil current to continus to flow.
The voltage pulse seen has the *same* polarity as the supply.
In a standard flyback scheme, the negative current waveform is
mirror-image of the ramp-like charging time.
This remains to be a fairly close picture of operation waveforms,
even if the core of the inductor saturates to some extent.

 

[Winfield Hill]

Robert Baer wrote...

Err...
Take a look at the current and voltage waveforms in a flyback system.
Turn on the switch, inductive current increases in the standard R/L form.
When the switch is opened, the magnetic field starts to collapse; the
waveform across the inductor is square-ish in most practical circuits.
During that time, the current goes rapidly to zero as the flyback
voltage pulse rises; roughly remains near zero at the flattish top,

This is incorrect. The current goes "rapidly" to zero _after_
the flyback voltage pulse rises, during the flattish top...

and then goes negative as the flyback voltage pulse drops to zero.
However, the voltage would continue to decrease and go negative
(L-C oscillations), but the switch (FET) internal diode conducts,
allowing the coil current to continus to flow.
The voltage pulse seen has the *same* polarity as the supply.
In a standard flyback scheme, the negative current waveform is
mirror-image of the ramp-like charging time.
This remains to be a fairly close picture of operation waveforms,
even if the core of the inductor saturates to some extent.

John is correct, the current shouldn't reverse, excepting perhaps
for a small amount of ringing. Perhaps, when you speak of mirror
image, you're thinking of a reversal of the rate-of-change in the
flyback coil current, rather than the current polarity, per se?

In the rapid magnetic-field collapse system we're talking about,
using MOSFET or TVS avalanche to absorb the coil's energy, the
avalanching junction will stop conducting the instant the current
drops to zero (the physics of avalanche doesn't have any reverse-
recovery time). There will be a small amount of current reversal
(and ringing) due to discharging the system capacitance from the
avalanche voltage level back to the supply V, but it'll be small
compared to the 15A magnetizing current.

For example, a 34n20 FET's capacitance is 400pF at 150V, a 1.5kW
150V TVS is 100pF, and 1 meter of cable is another 100pF. The
resonant frequency with 8uH will be 2.3MHz. The energy stored
in 600pF of capacitance at 130V (assume Vs = 20V supply for the
coil) pushes the inductor to a peak reversal of i = V sqrt(C/L)
= 1.1A, and the voltage to +20V -130V = -110V after a T/2 time
of 220ns, assuming nothing else in the path. However, the FET's
intrinsic body diode and the TVS diode will prevent the voltage
from going below ground, bringing things to a stop after only
130ns, limiting the ringing to the 20V supply Vs, and the peak
reversal current to about -600mA, only 4% of the original 15A.

Since the magnitude is only 4% and it only lasts a few hundred
ns, it's not useful to characterize this as "current reversal."

A 2pi f L = 100-ohm resistor paralleled with the inductor would
nicely damp the 2.3MHz ringing, and reduce the magnitude of the
single ring as well, to say 2%. It would need to be a 3W part,
etc., unless a 0.01uF series capacitor was added to stop any DC
current. We call this R-C network a snubber.

 

[Mark Becker]

Hello -

First, I wish to thank everyone participating in this discussion. While
I'm an EE, it has been a long time since some of this has crossed my
mind.. and the technology has improved considerably. The discussion has
poked a few embers.. Thank you.

Why a zener? Why not just a fast diode to circulate the current back through the inductor?

Using the effective series resistance of the inductor does not dissipate
the energy fast enough. For 8 uH and a series resistance of something
like .050 ohms, tau is on the order of 160 microseconds.

The zener has the feature of placing an (almost) constant voltage across
the coil during field collapse. Ideally, superposition applies.. and
I(t) = I(0) - Integral(v(t), t)/L. Faster than exponential decay.

[ TVS ]

Think I looked at this when first thinking about it.. and wasn't happy
at the response time. I'll go and look again. I am also not familiar
with "avalanche mode" transient absorbers and will stare at those as
well.

The current circuit (pun not intended) uses a pair of IRFB18N50K MOSFETs
and a pair of series-connected BZW03D100 100V across the MOSFETs. Hmm..
maybe the MOSFETs should be across the coils with a series fast diode..

Another poster suggested using the MOSFET intrinsic diode as the zener
clamp. I stared at that one.. is this feasible (I'm still climbing out
of using BJTs) ? I like the idea of the snubber better.. but have to
think a little more on it.

Especially as now the researcher wants the magnetic field to enclose a
larger volume. The coils for THAT assembly work out to about 400 uH ..
and still need 15A.

Your time is *really* appreciated.

Mark