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Does anyone see anything inherently wrong with this design?

J

JJ

Jan 1, 1970
0
Hi all,

I've been working for the last few months (in my spare time) on a
design which will fire copious amounts of current (albeit for a short
period of time) through a coil.

Now, I can't go into too much detail here, but I can say this:

The current through the coil induces a massive (high speed) movement
of a rod through the coil. (Not unlike a "coil gun" only on a much
larger scale.)

I posted the schematic in a.b.s.e. (Under the name "High Side Firing
Circuit".)

My question is: Does anyone see anything inherently wrong with my
layout? I'm a self edumacated circuit designer, and I'm simply
looking for feedback (in any regard) from the regs here pertaining to
the circuit.

I realize that diodes D1 and D2 need to be higher rated in the real
world: I only used those part numbers in the schematic because they
were the best choices (among the limited selection) in the software.

Pope?, John?, Anyone?

v/r,
JJ Richard

P.S. I haven't visited S.E.D. in years. Whatever happened to Robert?
Did he keep his promise of never posting to SED again or has he
changed his name or....?
 
R

Robert Baer

Jan 1, 1970
0
JJ said:
Hi all,

I've been working for the last few months (in my spare time) on a
design which will fire copious amounts of current (albeit for a short
period of time) through a coil.

Now, I can't go into too much detail here, but I can say this:

The current through the coil induces a massive (high speed) movement
of a rod through the coil. (Not unlike a "coil gun" only on a much
larger scale.)

I posted the schematic in a.b.s.e. (Under the name "High Side Firing
Circuit".)

My question is: Does anyone see anything inherently wrong with my
layout? I'm a self edumacated circuit designer, and I'm simply
looking for feedback (in any regard) from the regs here pertaining to
the circuit.

I realize that diodes D1 and D2 need to be higher rated in the real
world: I only used those part numbers in the schematic because they
were the best choices (among the limited selection) in the software.

Pope?, John?, Anyone?

v/r,
JJ Richard

P.S. I haven't visited S.E.D. in years. Whatever happened to Robert?
Did he keep his promise of never posting to SED again or has he
changed his name or....?

I do not know where that schematic is, but there are a number of
things to consider.
One can explosively vaporize wire with sufficent current; take a 10KV
capacitor, say about 1uF or so fully charged, and (remotely and
carefully) put a number 22 or finer wire across it ("who shot at me"?).
So, reduce that impulse current you implied to unser an amount tomake
the wire glow.
Well you still have a large amount of current.
Take two car jumper wires used for boosting a battery, string them out
parallel and next to each other, and short them together on the far end.
Then connect one to the - side of a fully charged 12V heavy duty truck
battery.
Then carefully and remotely, for a brief amount of time, connect the
other wire to the + side of the battery.
Hope nobody was near those wires, because they are going to *JUMP*
like all hell!
Now, of that 1000 amps (roughly) is OK with you, that is fine, BUT the
damn wire in the damn coil should damn well better be mechanically
restrained or the coil is guaranteed to fail in a few uses!
Are you now getting a dim picture of the possible magnitude of the
forces?
 
J

John Smith

Jan 1, 1970
0
Hi all,

I've been working for the last few months (in my spare time) on a
design which will fire copious amounts of current (albeit for a short
period of time) through a coil.

Now, I can't go into too much detail here, but I can say this:

The current through the coil induces a massive (high speed) movement
of a rod through the coil. (Not unlike a "coil gun" only on a much
larger scale.)

I posted the schematic in a.b.s.e. (Under the name "High Side Firing
Circuit".)

My question is: Does anyone see anything inherently wrong with my
layout? I'm a self edumacated circuit designer, and I'm simply
looking for feedback (in any regard) from the regs here pertaining to
the circuit.

I realize that diodes D1 and D2 need to be higher rated in the real
world: I only used those part numbers in the schematic because they
were the best choices (among the limited selection) in the software.

Pope?, John?, Anyone?

v/r,
JJ Richard



How did the simulation turn out? Were you happy with it?
 
W

Winfield Hill

Jan 1, 1970
0
JJ Richard wrote...
Hi all,

I've been working for the last few months (in my spare time) on a
design which will fire copious amounts of current (albeit for a short
period of time) through a coil. [snip] I posted the schematic in
a.b.s.e. (Under the name "High Side Firing Circuit".)

My question is: Does anyone see anything inherently wrong with my
layout? ... I realize that diodes D1 and D2 need to be higher rated
in the real world: I only used those part numbers in the schematic
because they were the best choices (among the limited selection) in
the software.

Yes, certainly, which means they can't be Schottky types!

Let's examine your FET circuit. First, the 4N25 is a bit wimpy for
driving the high capacitance of FET gates. With your 14mA LED drive
you can only expect about 3mA (min) out of this 20% CTR optocoupler.
Consider, the huge 120N25 parts have Ciss = 7700pF. The formula
dV/dt = i/C tells us you'll have a t = 5.5V 14.4nF / 3mA = 26us delay
to get the FET gates up to turn-on voltage, and the Qgd = 180nC spec
tells us it'll take t = q/i = 360nC/3mA = 120us for the drain-voltage
switching transition. This is pretty slow, and will mean excessive
die heating for the FETs if high drain currents are involved.

Let's examine your inductor's current. The formula dI/dt = V/L tells
us that with 200V applied by the FET switch the current rate-of-rise
will be 200V/300uH = 0.67A/us, or about 80A in 120us if the FETs are
fully on. We can mash some numbers together, like P = 1/2 80A 200V
and get a possible 8000W of FET dissipation during switching.

How high will your current go? I see you've picked very large FETs,
with Ron = 33 milliohms at 200A and 75C die temp. But that doesn't
mean you'll get I = 200V / 0.017 = 11760A of current! Nosirree, BoB!
First, the FETs may well current limit somewhere under 500A. Second,
you'll have the high series resistance of your capacitor bank, and of
your inductor. I don't see either represented in your model.

BTW, may I suggest IGBTs for your high-voltage high-current pulsing?
Some time ago I made a 1200V 200A pulse generator (that's 250kW) with
just two small IGBTs. They have amazing maximum-current capability
and are much easier to drive than FETs. If you're working above 100V
and don't need faster than 0.15us risetime, IGBTs are the way to go.

You can pick up very nice high-power IGBTs on eBay, e.g. CM600HA-24H,
which can deliver 1200A peak. A 1200V rating means you can use high-
voltage capacitor banks, to improve dI/dt. Bolt-style connections
help you to keep your wiring resistance down. And very low transient
thermal resistance (high internal heat capacity) means that they can
withstand high-current abuse for a longer time than smaller parts.

1.2kV 1.2kA = 1.4MW, but for impressive high-current high-voltage
(100kA 20kV = 2000MW) inductor action, see Bert Hickman's quarter-
shrinker web pages, http://205.243.100.155/frames/shrinker.html
P.S. I haven't visited S.E.D. in years. Whatever happened to Robert?

Some say he changed his name to Fred Bloggs, but I'm skeptical.

Thanks,
- Win

whill_at_picovolt-dot-com
 
J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that Winfield Hill
about 'Does anyone see anything inherently wrong with this design?', on
Wed, 31 Dec 2003:
Some say he changed his name to Fred Bloggs, but I'm skeptical.

ISTM that FB is sane but irascible. Robert didn't meet both of those
criteria.
 
R

Rick

Jan 1, 1970
0
Winfield Hill said:
JJ Richard wrote...
Hi all,

I've been working for the last few months (in my spare time) on a
design which will fire copious amounts of current (albeit for a short
period of time) through a coil. [snip] I posted the schematic in
a.b.s.e. (Under the name "High Side Firing Circuit".)

My question is: Does anyone see anything inherently wrong with my
layout? ... I realize that diodes D1 and D2 need to be higher rated
in the real world: I only used those part numbers in the schematic
because they were the best choices (among the limited selection) in
the software.

Yes, certainly, which means they can't be Schottky types!

Let's examine your FET circuit. First, the 4N25 is a bit wimpy for
driving the high capacitance of FET gates. With your 14mA LED drive
you can only expect about 3mA (min) out of this 20% CTR optocoupler.
Consider, the huge 120N25 parts have Ciss = 7700pF. The formula
dV/dt = i/C tells us you'll have a t = 5.5V 14.4nF / 3mA = 26us delay
to get the FET gates up to turn-on voltage, and the Qgd = 180nC spec
tells us it'll take t = q/i = 360nC/3mA = 120us for the drain-voltage
switching transition. This is pretty slow, and will mean excessive
die heating for the FETs if high drain currents are involved.

Let's examine your inductor's current. The formula dI/dt = V/L tells
us that with 200V applied by the FET switch the current rate-of-rise
will be 200V/300uH = 0.67A/us, or about 80A in 120us if the FETs are
fully on. We can mash some numbers together, like P = 1/2 80A 200V
and get a possible 8000W of FET dissipation during switching.

How high will your current go? I see you've picked very large FETs,
with Ron = 33 milliohms at 200A and 75C die temp. But that doesn't
mean you'll get I = 200V / 0.017 = 11760A of current! Nosirree, BoB!
First, the FETs may well current limit somewhere under 500A. Second,
you'll have the high series resistance of your capacitor bank, and of
your inductor. I don't see either represented in your model.

BTW, may I suggest IGBTs for your high-voltage high-current pulsing?
Some time ago I made a 1200V 200A pulse generator (that's 250kW) with
just two small IGBTs. They have amazing maximum-current capability
and are much easier to drive than FETs. If you're working above 100V
and don't need faster than 0.15us risetime, IGBTs are the way to go.

You can pick up very nice high-power IGBTs on eBay, e.g. CM600HA-24H,
which can deliver 1200A peak. A 1200V rating means you can use high-
voltage capacitor banks, to improve dI/dt. Bolt-style connections
help you to keep your wiring resistance down. And very low transient
thermal resistance (high internal heat capacity) means that they can
withstand high-current abuse for a longer time than smaller parts.

1.2kV 1.2kA = 1.4MW, but for impressive high-current high-voltage
(100kA 20kV = 2000MW) inductor action, see Bert Hickman's quarter-
shrinker web pages, http://205.243.100.155/frames/shrinker.html
P.S. I haven't visited S.E.D. in years. Whatever happened to Robert?

Some say he changed his name to Fred Bloggs, but I'm skeptical.

Thanks,
- Win

whill_at_picovolt-dot-com

Have people mostly switched over to IGBT devices from thyratrons, or is that
still a few years away?

The last time I looked, you couldn't get a IGBT device (well, affordable)
over about 4000 volts. If you want
to cascade them, you have to build a pulse transformer or something, right?
 
W

Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
... Nosirree, BoB! *

BoB = Bank of Boston, Boston's largest venerable old bank,
which was rename BankBoston after it absorbed BayBank, and
which was subsequently taken over by Fleet Bank (itself
having previously taken over both Bank of New England and
Shawmut Bank). The combination was named Fleet Boston, and
has now been taken over by Bank of America, which will keep
its name while absorbing most of New England's banking and
becoming one of the country's largest banks.

And I still won't be able to write a check in Pounds, Euros,
Yen, or etc. Sheesh!

Thanks,
- Win

whill_at_picovolt-dot-com
 
J

John Fields

Jan 1, 1970
0
Winfield Hill wrote...

BoB = Bank of Boston, Boston's largest venerable old bank,
which was rename BankBoston after it absorbed BayBank, and
which was subsequently taken over by Fleet Bank (itself
having previously taken over both Bank of New England and
Shawmut Bank). The combination was named Fleet Boston, and
has now been taken over by Bank of America, which will keep
its name while absorbing most of New England's banking and
becoming one of the country's largest banks.

And I still won't be able to write a check in Pounds, Euros,
Yen, or etc. Sheesh!

---
Who cares?

Write it in dollars and it'll be accepted anywhere in the world.

;^) <--- Please note smiley...
 
S

Steve J. Noll

Jan 1, 1970
0
Winfield Hill wrote...

BoB = Bank of Boston, Boston's largest venerable old bank,
which was rename BankBoston after it absorbed BayBank, and
which was subsequently taken over by Fleet Bank (itself
having previously taken over both Bank of New England and
Shawmut Bank). The combination was named Fleet Boston, and
has now been taken over by Bank of America, which will keep
its name while absorbing most of New England's banking and
becoming one of the country's largest banks.

And I still won't be able to write a check in Pounds, Euros,
Yen, or etc. Sheesh!

Thanks,
- Win

whill_at_picovolt-dot-com

For the longest time I wondered why I saw on TV on the walls of
east coast baseball stadiums billboards advertising an enema.
I wouldn't have thought it such a popular product.
(We don't have Fleet Bank here in SoCal.)

Steve J. Noll | Ventura California |
| The Used High-Tech Equipment Dealer Directory
| http://www.big-list.com
| The Peltier Device Information Site:
| http://www.peltier-info.com
 
W

Winfield Hill

Jan 1, 1970
0
John Fields wrote...
Who cares?
Write it in dollars and it'll be accepted anywhere in the world.

Not by your average overseas eBay seller, electronic-component
company, or distributor, etc., who I suppose have to pay a fee
to their bank to make the deposit, or who may just be pissed
off at us these days. For a while I maintained a bank account
in London, but that was before eBay came along, so I let it go.

Thanks,
- Win

whill_at_picovolt-dot-com
 
J

John Fields

Jan 1, 1970
0
John Fields wrote...

Not by your average overseas eBay seller, electronic-component
company, or distributor, etc., who I suppose have to pay a fee
to their bank to make the deposit, or who may just be pissed
off at us these days. For a while I maintained a bank account
in London, but that was before eBay came along, so I let it go.
 
J

JJ

Jan 1, 1970
0
JJ Richard wrote...
Hi all,

I've been working for the last few months (in my spare time) on a
design which will fire copious amounts of current (albeit for a short
period of time) through a coil. [snip] I posted the schematic in
a.b.s.e. (Under the name "High Side Firing Circuit".)

My question is: Does anyone see anything inherently wrong with my
layout? ... I realize that diodes D1 and D2 need to be higher rated
in the real world: I only used those part numbers in the schematic
because they were the best choices (among the limited selection) in
the software.

Yes, certainly, which means they can't be Schottky types!

Let's examine your FET circuit. First, the 4N25 is a bit wimpy for
driving the high capacitance of FET gates. With your 14mA LED drive
you can only expect about 3mA (min) out of this 20% CTR optocoupler.
Consider, the huge 120N25 parts have Ciss = 7700pF. The formula
dV/dt = i/C tells us you'll have a t = 5.5V 14.4nF / 3mA = 26us delay
to get the FET gates up to turn-on voltage, and the Qgd = 180nC spec
tells us it'll take t = q/i = 360nC/3mA = 120us for the drain-voltage
switching transition. This is pretty slow, and will mean excessive
die heating for the FETs if high drain currents are involved.

I'm currently using the 4N25 to fire the FETs, but the FETs are on the
low side of the design as it is now (they provide a ground for the
coil when switched on). The guy that I'm working with on this project
doesn't like the fact that there are wires with 200V on them at all
times out there in the real world (the coil is exposed to the outside
world), so that's why I'm changing the design to put the FETs on the
high side. This way, the 200V is kept safely inside the circuit
enclosure and pulsed out only when the circuit is fired.

Do you think I should go with a beefier coupler even though the FETs
aren't getting warm at all during operation? Am I damaging them
internally due to the slow (120uS) switching?
Let's examine your inductor's current. The formula dI/dt = V/L tells
us that with 200V applied by the FET switch the current rate-of-rise
will be 200V/300uH = 0.67A/us, or about 80A in 120us if the FETs are
fully on. We can mash some numbers together, like P = 1/2 80A 200V
and get a possible 8000W of FET dissipation during switching.
How high will your current go? I see you've picked very large FETs,
with Ron = 33 milliohms at 200A and 75C die temp. But that doesn't
mean you'll get I = 200V / 0.017 = 11760A of current! Nosirree, BoB!
First, the FETs may well current limit somewhere under 500A. Second,
you'll have the high series resistance of your capacitor bank, and of
your inductor. I don't see either represented in your model.

The DCR of the coil is 0.2 ohms, and the ESR of C1 is 0.014 ohms.
That, coupled with the FET Rds(on) as well as wiring losses should
result in ~600+ amps peak through the coil. I don't have a current
clamp with which to measure it, but I suspect that it's pretty close.

The arrangement as it is now works well in that it moves a 1" diameter
steel rod a distatnce of 10.5" in 24mS. But again, that is with the
FETs on the low side. I have not yet built the circuit that we're
discussing now.
BTW, may I suggest IGBTs for your high-voltage high-current pulsing?
Some time ago I made a 1200V 200A pulse generator (that's 250kW) with
just two small IGBTs. They have amazing maximum-current capability
and are much easier to drive than FETs. If you're working above 100V
and don't need faster than 0.15us risetime, IGBTs are the way to go.

You can pick up very nice high-power IGBTs on eBay, e.g. CM600HA-24H,
which can deliver 1200A peak. A 1200V rating means you can use high-
voltage capacitor banks, to improve dI/dt. Bolt-style connections
help you to keep your wiring resistance down. And very low transient
thermal resistance (high internal heat capacity) means that they can
withstand high-current abuse for a longer time than smaller parts.

I thought about going with IGBTs, but they're too expensive. The
120N25's are only $14 each (cheaper in bulk). DigiKey has IGBTs that
will work, but the price range is $80 to $176! EBay is not feasible
since this is a product as oppossed to a one of a kind gadget.

Also, I'm limited to 200V, since any higher voltage will result in a
much larger capacitor. The capacitor I'm using now (C1) is already
3"X6", and size is an issue in this design. As it is, I'm going to be
keeping a 200V cap charged to 195V at all times, but the charging
circuit is uP controlled/monitored.
1.2kV 1.2kA = 1.4MW, but for impressive high-current high-voltage
(100kA 20kV = 2000MW) inductor action, see Bert Hickman's quarter-
shrinker web pages, http://205.243.100.155/frames/shrinker.html

Some say he changed his name to Fred Bloggs, but I'm skeptical.

I'll keep that in mind. :)

Thanks,
- Win

whill_at_picovolt-dot-com


Thanks for the reply Win.

v/r,
JJ
 
J

JJ

Jan 1, 1970
0
I do not know where that schematic is, but there are a number of
things to consider.
One can explosively vaporize wire with sufficent current; take a 10KV
capacitor, say about 1uF or so fully charged, and (remotely and
carefully) put a number 22 or finer wire across it ("who shot at me"?).
So, reduce that impulse current you implied to unser an amount tomake
the wire glow.
Well you still have a large amount of current.
Take two car jumper wires used for boosting a battery, string them out
parallel and next to each other, and short them together on the far end.
Then connect one to the - side of a fully charged 12V heavy duty truck
battery.
Then carefully and remotely, for a brief amount of time, connect the
other wire to the + side of the battery.
Hope nobody was near those wires, because they are going to *JUMP*
like all hell!
Now, of that 1000 amps (roughly) is OK with you, that is fine, BUT the
damn wire in the damn coil should damn well better be mechanically
restrained or the coil is guaranteed to fail in a few uses!
Are you now getting a dim picture of the possible magnitude of the
forces?


You better believe it! :)

Yes, the coil is tightly wound and restrained. What amazes me is the
force that the ensuing magnetic field has on a steel bar positioned at
the mouth of the coil when the current shoots through! It's downright
violent!

The high current pulse only lasts 15mS or so, and the coil is wound
with 14AWG wire, so vaporizing it isn't an issue. In fact, I'm amazed
that the traces on my home-made PCB didn't vaporize (although I did
keep them short and beef them up with solder).

Thanks for the reply; see the thread that I'm having with Win for more
information.

JJ
 
J

JJ

Jan 1, 1970
0
How did the simulation turn out? Were you happy with it?

Yes, it simulates fine, but we all know that simulations don't always
tell the whole story. :)

Look at the thread that I'm having with Win for more details.

Also, someone made what I believe to be a negative post in a.b.s.e
concerning my placing the schematic there. Was I not supposed to do
that? I thought that's what a.b.s.e was for. ??

Tks,
JJ
 
F

Fritz Schlunder

Jan 1, 1970
0
I'm currently using the 4N25 to fire the FETs, but the FETs are on the
low side of the design as it is now (they provide a ground for the
coil when switched on). The guy that I'm working with on this project
doesn't like the fact that there are wires with 200V on them at all
times out there in the real world (the coil is exposed to the outside
world), so that's why I'm changing the design to put the FETs on the
high side. This way, the 200V is kept safely inside the circuit
enclosure and pulsed out only when the circuit is fired.

Do you think I should go with a beefier coupler even though the FETs
aren't getting warm at all during operation? Am I damaging them
internally due to the slow (120uS) switching?


I fully agree with Mr. Hill's assessment regarding 120us being much slower
than desireable. Really though the situation looks even worse than that
though.

In fact. After looking at the circuit I am led to believe that the circuit
would be an excellent way to celebrate the new year. That is, power it up
as the clock strikes twelve and it should make for an impressive fireworks
display to ring in the new year.

Move that resistor R1 from gate to ground to gate to source of the two
MOSFETs. I think this is a very critical and mandatory change. Imagine
what happens when you try to turn on the MOSFETs. The source rises up to
200V above ground potential. In order for the MOSFET to be on the gate
should be at 10V higher than the source. So in other words the gate is at
210V above ground potential when everything is on. But wait a minute. R1
is between gate and ground. So lessee here. 210V/22k = 9.5mA. Wait a
minute. The only source of current for the gate is through the
optoisolator. But Winfield Hill already determined with a CTR of around 20%
it will only conduct something like 3mA (under bad conditions, the 4N25 will
likely have somewhat better CTR in most cases). Hmm...

3mA < 9.5mA. That is, there is no way this is going to work right. The
drain cannot reach 200V or else the gate resistor will suck away too much
juice, and the gate will no longer keep rising even though bootstrapped.

The solution to this problem is quite simple. Just move the resistor from
gate to ground to gate to source of the MOSFETs.

Still though, the drive is much too weak and too stressful on the MOSFETs.
Maybe it works, but it could be much better without much extra trouble.
Using a much higher CTR optocoupler and using a much smaller value resistor
would be two easy ways to make things better.

I suggest getting a very high gain optocoupler (maybe a photodarlington, but
be weary of the extra voltage loss), and then using a gate-source pull down
resistor of something like 1k.
 
J

JJ

Jan 1, 1970
0
I fully agree with Mr. Hill's assessment regarding 120us being much slower
than desireable. Really though the situation looks even worse than that
though.

In fact. After looking at the circuit I am led to believe that the circuit
would be an excellent way to celebrate the new year. That is, power it up
as the clock strikes twelve and it should make for an impressive fireworks
display to ring in the new year.

Move that resistor R1 from gate to ground to gate to source of the two
MOSFETs. I think this is a very critical and mandatory change. Imagine
what happens when you try to turn on the MOSFETs. The source rises up to
200V above ground potential. In order for the MOSFET to be on the gate
should be at 10V higher than the source. So in other words the gate is at
210V above ground potential when everything is on. But wait a minute. R1
is between gate and ground. So lessee here. 210V/22k = 9.5mA. Wait a
minute. The only source of current for the gate is through the
optoisolator. But Winfield Hill already determined with a CTR of around 20%
it will only conduct something like 3mA (under bad conditions, the 4N25 will
likely have somewhat better CTR in most cases). Hmm...

3mA < 9.5mA. That is, there is no way this is going to work right. The
drain cannot reach 200V or else the gate resistor will suck away too much
juice, and the gate will no longer keep rising even though bootstrapped.

The solution to this problem is quite simple. Just move the resistor from
gate to ground to gate to source of the MOSFETs.

Still though, the drive is much too weak and too stressful on the MOSFETs.
Maybe it works, but it could be much better without much extra trouble.
Using a much higher CTR optocoupler and using a much smaller value resistor
would be two easy ways to make things better.

I suggest getting a very high gain optocoupler (maybe a photodarlington, but
be weary of the extra voltage loss), and then using a gate-source pull down
resistor of something like 1k.


Fritz,

Thanks for the reply.

Excellent idea on relocating R1 to go gate to source! Under the
scenario you described, I can see how instead of firing completely, it
could oscillate on/off rapidly, and at those switching currents,
well..maybe the firecrackers popping outside tonight would mask the
smoke and flames. :) Thanks for the heads up!

Two questions, however:

1. Would it not be prudent to then also place a high value (~47K)
resistor in parallel with D2 (for ground reference) to prevent
latch-up when the coil is removed from the circuit?

2. If I lower the value of R1 to 1K, (after relocating it) it will
then draw ~12V/1K = 12mA from C2 (depending on battery voltage) during
the firing sequence, which is more than what I started with. Why not
keep it at 22K?

I will also definitely look at a better opto.

Thanks again for the input, I'm glad I decided to let the SED Heads
chew on it prior to building it!

v/r,
JJ
 
J

JJ

Jan 1, 1970
0
I fully agree with Mr. Hill's assessment regarding 120us being much slower
than desireable. Really though the situation looks even worse than that
though.

In fact. After looking at the circuit I am led to believe that the circuit
would be an excellent way to celebrate the new year. That is, power it up
as the clock strikes twelve and it should make for an impressive fireworks
display to ring in the new year.

Move that resistor R1 from gate to ground to gate to source of the two
MOSFETs. I think this is a very critical and mandatory change. Imagine
what happens when you try to turn on the MOSFETs. The source rises up to
200V above ground potential. In order for the MOSFET to be on the gate
should be at 10V higher than the source. So in other words the gate is at
210V above ground potential when everything is on. But wait a minute. R1
is between gate and ground. So lessee here. 210V/22k = 9.5mA. Wait a
minute. The only source of current for the gate is through the
optoisolator. But Winfield Hill already determined with a CTR of around 20%
it will only conduct something like 3mA (under bad conditions, the 4N25 will
likely have somewhat better CTR in most cases). Hmm...

3mA < 9.5mA. That is, there is no way this is going to work right. The
drain cannot reach 200V or else the gate resistor will suck away too much
juice, and the gate will no longer keep rising even though bootstrapped.

The solution to this problem is quite simple. Just move the resistor from
gate to ground to gate to source of the MOSFETs.

Still though, the drive is much too weak and too stressful on the MOSFETs.
Maybe it works, but it could be much better without much extra trouble.
Using a much higher CTR optocoupler and using a much smaller value resistor
would be two easy ways to make things better.

I suggest getting a very high gain optocoupler (maybe a photodarlington, but
be weary of the extra voltage loss), and then using a gate-source pull down
resistor of something like 1k.


Fritz,

I was just browsing Digi-Key and DL'd the datasheet for the 4NX optos,
and I see the 4N35 has a minimum CTR of 100% at 10V. Now, if I put
20mA into it, and get 19.5mA out (.5mA still goes through R1), that
would reduce my switching time from 120uS to ~18uS. (Assuming I
relocate but keep R1 at 22K.)

Is this fast enough? Or is that still too slow?

Tks,
JJ
 
J

JJ

Jan 1, 1970
0
Fritz,

I was just browsing Digi-Key and DL'd the datasheet for the 4NX optos,
and I see the 4N35 has a minimum CTR of 100% at 10V. Now, if I put
20mA into it, and get 19.5mA out (.5mA still goes through R1), that
would reduce my switching time from 120uS to ~18uS. (Assuming I
relocate but keep R1 at 22K.)

Is this fast enough? Or is that still too slow?

Tks,
JJ


Scratch that. I think MOC207 is a better choice. Looks like it will
keep its CTR at or above 100% over a wider range of temperatures.
What do you think?

Tks,
JJ
 
J

JJ

Jan 1, 1970
0
Fritz,

Thanks for the reply.

Excellent idea on relocating R1 to go gate to source! Under the
scenario you described, I can see how instead of firing completely, it
could oscillate on/off rapidly, and at those switching currents,
well..maybe the firecrackers popping outside tonight would mask the
smoke and flames. :) Thanks for the heads up!

Two questions, however:

1. Would it not be prudent to then also place a high value (~47K)
resistor in parallel with D2 (for ground reference) to prevent
latch-up when the coil is removed from the circuit?

2. If I lower the value of R1 to 1K, (after relocating it) it will
then draw ~12V/1K = 12mA from C2 (depending on battery voltage) during
the firing sequence, which is more than what I started with. Why not
keep it at 22K?

I will also definitely look at a better opto.

Thanks again for the input, I'm glad I decided to let the SED Heads
chew on it prior to building it!

v/r,
JJ

Fritz,

Never mind about question 2. I just figured out that the reason for
the 1k value is to speed up switch-off time.

I'm going to make R1 1K, and go with the MOC207 with 23mA input.. I'm
also going to increase C2's value to 100uF so that the differential
gate/source voltage stays above 10V throughout the 15mS firing cycle.

I believe this means that the turn-on switching time will be
360nC/17.5mA = 21uS.

Also, I'm still wondering about question 1 above.

Sorry for the spaghetti of posts. And thanks again for the wake up
call. :)

v/r,
JJ
 
R

R.Legg

Jan 1, 1970
0
The current through the coil induces a massive (high speed) movement
of a rod through the coil. (Not unlike a "coil gun" only on a much
larger scale.)

The inductance of your load will alter dramatically if there's
physical movement in magnetic material present in the coil.

As this is a one-shot, the use of semiconductors in the switches and
capacitor charging circuitry is probably unneccessary elaboration.

Simply charging C1, then contacting the L1 coil terminal temporarily,
will likely prove just as effective. If the charging current is small,
you could leave this switch normally-closed to prevent voltages
building up in the unattended circuit.

RL
 
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