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# Does anyone see anything inherently wrong with this design?

W

#### Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
Let's examine the rather-brief IXYS datasheet. A novice engineer
might take the 120N25's 20-milli-ohm Rds(on) spec at face value,
but at high currents (and the resulting high junction temperatures
during a pulse) it'll be nearly 2x higher for 200A and Tj = 100C,
as shown by figures 4 and 5, or more like 40 milli-ohms. Therefore
200A in a 120N25 will result in a Vds = 8V drop, substantially more
voltage than 4V for a set of the lower-cost IGBTs I suggested. The
single-pulse Transient Thermal Resistance curve, fig 12, gives us
a nice low 0.06C/W for 15ms (the 120N25 has a much larger die than
the 4pc50), so that 200A * 8V = 1600W. This is more dissipation
than the IGBTs see, yet we still get a reasonable Tj rise of 96C.
So you might be able to parallel three 120N25 FETs for single 600A
15ms pulses and keep their junction temp rise under control. But
four FETs would be much safer, given the FET's current sharing
won't be perfect.

Oops, a phrase ended up in the wrong place! Corrected paragraph:

... 200A in a 120N25 will result in a Vds = 8V drop, substantially
more than the 4V drop for a set of the lower-cost IGBTs I suggested.
This results in more dissipation than the IGBTs see. But the two
parts do not handle pulses the same, due to die size. The 120N25's
single-pulse Transient Thermal Resistance curve, fig 12, gives us a
nice low 0.06C/W number for a 15ms pulse (note: the 120N25 has a much
larger die than the 4pc50), so 200A * 8V = 1600W. In the end we get
a more reasonable Tj rise of 96C. You may well be able to parallel
three 120N25 FETs for single 600A 15ms pulses and keep their junction
temp rise under control. But four FETs would be much safer, given
the FET's current sharing won't be perfect.
----

One more point: the above calculations are for conducting dissipation
losses only, and do not include switching losses or dissipation. On
the other hand, the conduction dissipation is overstated because the
current during the entire 15ms pulse duration.

These simple calculations are in the back-of-the-envelope category.
Which, as we say in H&H's AoE, is not to be taken lightly.
Toshiba's popular MG300 or MG400 modules drop about 3.5V at 600A.

A single IGBT transistor from their even more popular (and cheaper)
MG200 parts may work well. A bit higher drop, but still very good.

Thanks,
- Win

whill_at_picovolt-dot-com

W

#### Winfield Hill

Jan 1, 1970
0
JJ wrote...
The arrangement as it is now works well in that it moves
a 1" diameter steel rod a distance of 10.5" in 24ms.

OK, that's 25mph, not very fast. What are you working on?

Thanks,
- Win

whill_at_picovolt-dot-com

J

#### JJ

Jan 1, 1970
0
JJ wrote...

OK, that's 25mph, not very fast. What are you working on?

Thanks,
- Win

whill_at_picovolt-dot-com

Win,

Agreed, 25MPH isn't very impressive on face value. And I wish that I
could further elaborate on the aspects of this design, however, I have
promised the person that I'm working with on this design otherwise.

This design won't remain a secret forever, though, and I will be able
to disclose the details soon enough.

Thanks Win,
JJ

J

#### JJ

Jan 1, 1970
0
No problem.

Don't let this bother you too much. My impression is EE programs in general
don't cover practical and real world useful information very much. I can
only talk from my own exerience with a single EE program (Arizona State
University in my case), but my impression is this is true of most EE
programs. All they want to talk about is mathematics and "theory" type
things. Although knowing math and theory can certainly be helpful, and
definitely isn't hurtful, it isn't usually instantly practical or
applicable. High side MOSFET gate drive techniques is something way too
practical and instantly useful for them to teach (although admittedly
wouldn't be useful for everyone).

The net result is a recent EE graduate can often have dramatically inferior
circuit designing capabilities to any dedicated serious hobbiest.

Yeah my thought is it is better to take your lumps in this stage of
development rather than later. That is... If you build a circuit with
design problems it will most likely explode if you are playing with high
voltages and high currents. When your circuits always explode it can take a
real toll on self esteem, perhaps even a lasting toll. Usually it is very
difficult to learn very much by trial and error when it comes to power
electronics. Usually there are just far too many reasons that could result
in your device failure that it is very difficult to know for sure what
caused the problem. Also, it tends to get quite expensive if you blow away
dozens or more of expensive MOSFETs. If you figure out all of the problems
with a design in the development stage, it can save quite a bit of
money/time/heartache.

MOSFET characteristics and their gate drive requirements is quite a tricky
subject. I've been playing around with them for a long time now but I still
learn new things about them somtimes. But rest assured if you persevere
(and don't skip the research steps) you will get it eventually.

If you build your high power circuits and they work flawlessly and robustly
(especially the first time around), it can have an opposite effect on self
esteem. It always gives me a temporary boost when I make something really
great that works well. It makes me feel powerful, like I know what I am
doing, etc.

Okay. Go and read this document, "Design and Application Guide for High
Speed MOSFET Gate Drive Circuits" by Laszlo Balogh distributed by Texas
Instruments:

http://focus.ti.com/lit/ml/slup169/slup169.pdf

It is a very long document, but extremely useful for anyone interested in
MOSFET gate drive. This is by far the most difinitive resource on this
topic I have seen to date. After reading this document, take a look at the
references section at the end. Read all of the references by International
Rectifier. They are also extremely useful. They can be found on
International Rectifier's web site:

http://www.irf.com

The TI document talks quite a bit about boostrap drive techniques which is
of particular interest to you. Read all of it though, it is very useful.

Good. You will probably learn this stuff faster than it took me (especially
if you read those documents I've suggested). I can tell you it took more
than a few days in my case.

Alright. Well... The new design isn't quite right either, however I can
tell you it is starting to look reminiscent of the bootstrap drive
techniques that are employed inside integrated circuits like the IR2117.

R5 doesn't appear to serve any purpose. D3 has too little voltage blocking
ability (needs to be over 200V capable). Q5 is rated for too little
voltage. When the MOSFETs are on the gate is at about 210V. The base of Q3
should therefore be about one diode drop higher than this. If this is so,
then Q5 would have to be rated for more than 210V. Without doing any
calculations C2 looks like it might be a little smaller than desireable even
if it does work fine (good to have a little design margin). Otherwise it
appears similar to other tried and true boostrap driver designs out there.

Look forward to reading it.

Fritz,

I just got home (I allowed a friend of mine to drag me out to watch
the Sugar Bowl). It's late, and I've been drinking. Soooo, I'm
making a firm decision to defer from making a reply to this thread at
present.

I do plan to read and evaluate all of the posted text that we've
shrared thus far.

Sorry, Fritz. I'm just a little too tipsy to formulate a thinking
reply at this point.

v/r,
JJ

W

#### Winfield Hill

Jan 1, 1970
0
JJ wrote...
Can you tell us more about the capacitor, brand and part number?

A minor point, milliseconds is abbreviated ms, not mS. S is
Siemens, a unit of conductance. Sorry, it just bothers me, sort
of like scratching on a chalkboard, to see mS for time. :>)

You're moving the steel rod through the coil and on past it?
What's the current waveform in the coil...?

You haven't said much about the coil, or the current waveform in
it, but we know it uses #14 wire and has 0.2 ohms of resistance,
so that must be about 79 feet of wire, and weigh about one pound.

Your coil must be just over 1" in diameter to pass the 1" rod,
say 1.25", and it could be say 2" long. Now we can estimate your
coil's inductance using a weird form of Wheeler's equation that
I just whipped up,

.. C^2 where C = wire length, inches
.. L = --------------------- uH/in and d = diameter,
.. 2 pi^2 d (9 + 20 b/d) and b = coil length.

With C = 950 inches, d = 1.25 inches and b = 2 inches, I estimate
that's about 240 turns and I get an inductance of about L = 890uH.
If you wound the coil longer than 2", then the inductance would
be lower, e.g. for 5" length I get 410uH.

Hmm, 890uH is a lot more than the 100uH you used in the drawing.

The coil's circuit time constant is t = L/R = 890uH / 0.25-ohms
= 3.6ms, so we can make some guesses about your current waveform.

You mention switching 200V with the FETs for 15ms. dI/dt = V/L
with 200V gives us 118A/ms, so ignoring resistance, we'd approach
your claimed 600A in about 5ms. Let's estimate a 6ms risetime to
500A. The energy stored in the coil would be E = 0.5 L I^2 = 160J
at 600A. After you turn off the FETs the current will fall with a
3.6ms time constant, or about 7ms to drop to 100A. How do these
estimates square up with your observations?

Thanks,
- Win

whill_at_picovolt-dot-com

W

#### Winfield Hill

Jan 1, 1970
0
Winfield Hill wrote...
You haven't said much about the coil, or the current waveform in
it, but we know it uses #14 wire and has 0.2 ohms of resistance,
so that must be about 79 feet of wire, and weigh about one pound.

Your coil must be just over 1" in diameter to pass the 1" rod,
say 1.25", and it could be say 2" long. Now we can estimate your
coil's inductance using a weird form of Wheeler's equation that
I just whipped up,

C^2 where C = wire length, inches
L = --------------------- uH/in and d = diameter,
2 pi^2 d (9 + 20 b/d) and b = coil length.

With C = 950 inches, d = 1.25 inches and b = 2 inches, I estimate
that's about 240 turns and I get an inductance of about L = 890uH.
If you wound the coil longer than 2", then the inductance would
be lower, e.g. for 5" length I get 410uH.

Hmm, 890uH is a lot more than the 100uH you used in the drawing.

Not even including the effect of the steel rod, which could easily
increase the inductance by 3x or more when fully inside the coil.
The coil's circuit time constant is t = L/R = 890uH / 0.25-ohms
= 3.6ms, so we can make some guesses about your current waveform.

Once again, ignoring the slowing effect of the steel rod, and the
induced-current effects from its motion through the coil.

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
Mon, 5 Jan 2004:
A minor point, milliseconds is abbreviated ms, not mS. S is
Siemens, a unit of conductance. Sorry, it just bothers me, sort
of like scratching on a chalkboard, to see mS for time. :>)

This minor (?) error is extremely prevalent, unfortunately. But you risk
getting a pedantry award from Genome for pointing it out. (;-)

J

#### JJ

Jan 1, 1970
0
No problem.

Don't let this bother you too much. My impression is EE programs in general
don't cover practical and real world useful information very much. I can
only talk from my own exerience with a single EE program (Arizona State
University in my case), but my impression is this is true of most EE
programs. All they want to talk about is mathematics and "theory" type
things. Although knowing math and theory can certainly be helpful, and
definitely isn't hurtful, it isn't usually instantly practical or
applicable. High side MOSFET gate drive techniques is something way too
practical and instantly useful for them to teach (although admittedly
wouldn't be useful for everyone).

The net result is a recent EE graduate can often have dramatically inferior
circuit designing capabilities to any dedicated serious hobbiest.

Yeah my thought is it is better to take your lumps in this stage of
development rather than later. That is... If you build a circuit with
design problems it will most likely explode if you are playing with high
voltages and high currents. When your circuits always explode it can take a
real toll on self esteem, perhaps even a lasting toll. Usually it is very
difficult to learn very much by trial and error when it comes to power
electronics. Usually there are just far too many reasons that could result
in your device failure that it is very difficult to know for sure what
caused the problem. Also, it tends to get quite expensive if you blow away
dozens or more of expensive MOSFETs. If you figure out all of the problems
with a design in the development stage, it can save quite a bit of
money/time/heartache.

MOSFET characteristics and their gate drive requirements is quite a tricky
subject. I've been playing around with them for a long time now but I still
learn new things about them somtimes. But rest assured if you persevere
(and don't skip the research steps) you will get it eventually.

If you build your high power circuits and they work flawlessly and robustly
(especially the first time around), it can have an opposite effect on self
esteem. It always gives me a temporary boost when I make something really
great that works well. It makes me feel powerful, like I know what I am
doing, etc.

Okay. Go and read this document, "Design and Application Guide for High
Speed MOSFET Gate Drive Circuits" by Laszlo Balogh distributed by Texas
Instruments:

http://focus.ti.com/lit/ml/slup169/slup169.pdf

It is a very long document, but extremely useful for anyone interested in
MOSFET gate drive. This is by far the most difinitive resource on this
topic I have seen to date. After reading this document, take a look at the
references section at the end. Read all of the references by International
Rectifier. They are also extremely useful. They can be found on
International Rectifier's web site:

http://www.irf.com

Thanks for those references. I have not had time to sit down and read
them (I work two jobs), but I will look start with them tonight.

The TI document talks quite a bit about boostrap drive techniques which is
of particular interest to you. Read all of it though, it is very useful.

Good. You will probably learn this stuff faster than it took me (especially
if you read those documents I've suggested). I can tell you it took more
than a few days in my case.

Alright. Well... The new design isn't quite right either, however I can
tell you it is starting to look reminiscent of the bootstrap drive
techniques that are employed inside integrated circuits like the IR2117.

R5 doesn't appear to serve any purpose. D3 has too little voltage blocking
ability (needs to be over 200V capable). Q5 is rated for too little
voltage. When the MOSFETs are on the gate is at about 210V. The base of Q3
should therefore be about one diode drop higher than this. If this is so,
then Q5 would have to be rated for more than 210V. Without doing any
calculations C2 looks like it might be a little smaller than desireable even
if it does work fine (good to have a little design margin). Otherwise it
appears similar to other tried and true boostrap driver designs out there.

Ok, thanks again for taking the time to do this.

R5 is there only to keep Q5 in conduction should the uP be accidently
removed with power applied (I frequently remove it to re-program it
during testing), incorrectly programmed, etc. No surprise firing
sequences is was I what thinking when I added R5. When this thing
fires, it is pretty violent.

D3 was an oversight on my part. I meant to change that number to
1N4004. LTSpice doesn't seem to have a large selection to choose
from, and so it's up to me to remember to go back and place the actual
part numbers that will be used in the design.

Ditto with Q5. I had a 300Vceo transistor already picked out for that
one, but again I forgot to update the number.

I usually use CircuitMaker to do my sims, but I saw a thread in here
about LTSpice, so I dl'd it and have been using it instead just out of
curiosity.

Thanks, Fritz.

JJ

J

#### JJ

Jan 1, 1970
0
Well with the resistor between gate and source there would be no way to turn
the MOSFETs on once C2 gets drained. It is important to realize that
MOSFETs only care about their gate to source and drain to source voltages.
If the gate is at the same potential as the source pin, it doesn't matter
where in the circuit the device is or what "ground" is, the channel will be
very high resistance.

Hmm... Well I hate to disappoint you but I liked the other design with the
optoisolator and the previously discussed modifications (with the 300% CTR
optoisolator) much better than the new design. With the old design but
using a 300% CTR optoisolator the turn on time becomes somewhat reasonable
and well within the SOA curve for these devices (figure 12 of the
datasheet):

http://www.ixys.com/98879.pdf

Upon closer inspection of the SOA curves it looks like you might be running
these devices either right on the edge or exceeding the SOA somewhat during
the main discharge on time. It is kind of hard to tell for sure.

This new design is most peculiar. I've never seen anything quite like it.
It looks like it sorta might work (maybe), and I'm guessing you simulated it
with the given parameters and it worked. I certainly would not use it
though. I notice you changed the 300uH inductor into a 1uH inductor. I
don't think this change really makes for a more accurate model of reality.
It is pretty hard to scientifically quantify what the coil will really
behave like since it depends upon so many things that aren't to easy to
specify. The 0.2R DC resistance is guaranteed to be there, but my guess is
the current will intially ramp up very rapidly (although perhaps slowed a
little bit by "leakage inductance") to some fairly high value, but then on
top of that you will also have an increasing current waveform. But further
complicating this is the main capacitor C1 will be decaying and thus the
current would tend to decrease, or at least stop ramping up at such as fast
rate. Meanwhile the projectile starts moving and further interacting
causing other changes. Ultimately I have no idea how much the coil will
behave as a resistor and how much as an inductor, but I doubt 1uH with 0.2R
DC resistance is it. I suspect both the resistive and inductive components
will behave as somewhat larger values, but they might be dynamic.

So... Back to your circuit...

Suppose what happens with your circuit when you first try to turn it on. Q5
turns on and that central node with R2, R3, and R4 goes near ground. Okay.
So Q3 turns on and under the initial conditions there is about 12V minus two
diode drops at the base of Q3. Since the collector node of Q5 is at near
ground potential resistor R4 conducts something like 10V/1.2k = 8.3mA. Okay
not bad. But wait a minute. As the MOSFETs turns on the source rises up to
200V. If that is so, then the top of C2 rises up as well to ~211V above
ground potential. The base of Q3 is one diode drop lower than this at
around ~210V. Even so, the other side of R4 is still at ground potential.
So R4 has about 210V across it. At 1.2k this works out to 175mA at 210V.
Ouch. That is almost 37 watts of dissipation and it is going to start
draining C2 very fast (even if it is 100uF). If C2 loses voltage too fast
(or the C1 doesn't discharge fast enough because the coil's inductance isn't
1uH but more and is slowing things down), then it may start starving the
voltage provided to the gate. If this happens then the MOSFETs enter linear
operation and you've got trouble. Maybe this won't happen and everything
will work fine... But gee... It is nonetheless quite scary and perhaps
stressful on the driving components. What happens during turn off is also
equally peculiar.

Why are you using MOSFETs for this application? Winfield Hill preaches the
benefits of IGBT technology, and although I tend to agree with him about the
greatness of IGBTs in pulse current applications, 200V is right on the edge
where MOSFETs may in some cases still be attractive. The higher the voltage
the more and more attractive the IGBTs get.

Four or five IRG4BC40F devices in parallel might be adequate for this
application and they only cost $3.52 each in ten unit quantities from Digikey. That is cheaper than the two MOSFETs isn't it? Datasheet at: http://www.irf.com/product-info/datasheets/data/irg4bc40f.pdf That's a good point. However, that would make for as many as 10 or so transistors since I'm actually putting two firing circuits onto the PCB. I also have other circuitry to put on there as well. Also, that would complicate the PCB assembly process with respect to the heatsink. Why aren't you using an SCR device instead? By the sounds of the description and energy values it seems like all you are trying to do is rapidly discharge your capacitor bank into the coil. It sounds like you want all the energy available. If this is what you are trying to do, there really is no better choice than an SCR. They have by far the best pulse current handling capability at high voltages of really pretty much any semiconductor switching device (with the possible exception of plain diodes). You can get allot of bang for your buck with SCRs. If this is really all you need something like the S6055R made by Teccor available from Digikey will set you back a paltry$2.49 in 100 unit
quantities (\$3.11 for one). If we look at the datasheet:

http://rocky.digikey.com/WebLib/Teccor/Web Data/S4016N RP.pdf

We notice in figure E6.3 the device can handle something like 760A peak
capacitive discharges where five time constants equals 15ms. Thus perhaps
only one device is needed. If not, then a slightly larger device could be
used.

Of course SCRs have different drive requirements.

But how do I turn the SCR off? Even after C1 discharges, the 12V
potential from the battery will still cause ~60A of current through
the D1/SCR/L1 combo. Actually, during actual firing, C1 doesn't get
down below ~20V after the uP's 15ms pulse ends. (I usually still show
about 20-23V on C1 after firing.)

Thanks Fritz,
JJ

J

#### JJ

Jan 1, 1970
0
Can you tell us more about the capacitor, brand and part number?

A minor point, milliseconds is abbreviated ms, not mS. S is
Siemens, a unit of conductance. Sorry, it just bothers me, sort
of like scratching on a chalkboard, to see mS for time. :>)

Thanks for the correction. Sorry for poking you in the eye so many
times in this thread.

You haven't said much about the coil, or the current waveform in
it, but we know it uses #14 wire and has 0.2 ohms of resistance,
so that must be about 79 feet of wire, and weigh about one pound.

Your coil must be just over 1" in diameter to pass the 1" rod,
say 1.25", and it could be say 2" long. Now we can estimate your
coil's inductance using a weird form of Wheeler's equation that
I just whipped up,

. C^2 where C = wire length, inches
. L = --------------------- uH/in and d = diameter,
. 2 pi^2 d (9 + 20 b/d) and b = coil length.

With C = 950 inches, d = 1.25 inches and b = 2 inches, I estimate
that's about 240 turns and I get an inductance of about L = 890uH.
If you wound the coil longer than 2", then the inductance would
be lower, e.g. for 5" length I get 410uH.

Hmm, 890uH is a lot more than the 100uH you used in the drawing.

The coil's circuit time constant is t = L/R = 890uH / 0.25-ohms
= 3.6ms, so we can make some guesses about your current waveform.

You mention switching 200V with the FETs for 15ms. dI/dt = V/L
with 200V gives us 118A/ms, so ignoring resistance, we'd approach
your claimed 600A in about 5ms. Let's estimate a 6ms risetime to
500A. The energy stored in the coil would be E = 0.5 L I^2 = 160J
at 600A. After you turn off the FETs the current will fall with a
3.6ms time constant, or about 7ms to drop to 100A. How do these
estimates square up with your observations?

Thanks,
- Win

We used the calculator on this page:

http://mgc314.home.comcast.net/opticoil_trans_in.htm

to determine the coil parameters. Initially, we wanted to use the
readily available 12V battery as the firing voltage source. It soon
became apparent that 12V wasn't going to cut it.

We then decided to go with 48V instead. (This came after much
tinkering with rod diameter, material, weight and timing issues.) So,
he had coils made (which I thought was premature) that meet the specs
that the above link gives if you input 48V, 180A, 0.15 ohms of losses,
0.875" I.D. and 2" L. (I was off a bit on the 1" diameter that I
previously specified; I just got the correct diameter from my partner
via phone.)

That works out to be 377uH on the site. (I think the reason that I was
using 300uH was because that's what my brain had stored in memory for
some reason; this circuit has changed a lot since its inception.)

We built the circuit and tested it out and the speed was pretty good.
Then, however, my partner decided that he wanted a longer travel
(10.5" vs. 8"). This meant that we needed more speed than that which
the present coil was capable of giving us at 48V. The problem was/is
that we already have all of these coils lying around.

So, we went back to the site and changed the 48V input to (48*4) =
192V, and the peak current from 180A to (180 * 4) = 720A. The results
you see when you plug those values into the site's calculator indicate
that the inductance remains the same and the expected peak current is
648A.

We plugged these values into the simulator and modeled different
capacitances until it drew a predicted current pulse of just over 600A
(at 2.6ms peak) which decayed to zero in ~15ms.

So, we acquired a 200V, 25mF capacitor (Mallory p/n: CGS253T200X8L),
the 120N25 xsistors, and tested it out. It worked as expected, with
the capacitor discharging to near 0 (~20V) in 15ms. The speed
increase due to the higher voltage (even with the longer rod) was
astounding.

The reason that 15ms has been chosen is that the rod moves so fast
that if the pulse width is increased much beyond that, the opposite
end of the rod encounters kickback effects from the coil. (If I
program even as little as 20ms, kickback begins to occur.)

We have since found a capacitor with a smaller physical size (3" X
5.875") that is rated at 200V/23mF, and are awaiting samples to
arrive.

So, at least in this regard, real world testing does seem to confirm
what the simulation predicted. We have fired and timed this
configuration dozens, if not a hundred times. I don't have a current
clamp-on meter to measure the actual current waveform, however the
capacitor's voltage discharge waveform (as viewed on a TKS storage
scope) does agree with the simulated expectations.

The firing xsistors (the 120N25s) do not even get warm to the touch
(with no heatsink attached) after a firing sequence but, as I've
learned in this thread, that doesn't necessarily mean that they aren't
going to suffer premature deaths due to excessive stresses over time.

Thanks for taking the time to reply, Win.

JJ

W

#### Winfield Hill

Jan 1, 1970
0
JJ wrote...
That's a good point. However, that would make for as many as 10 or so
transistors since I'm actually putting two firing circuits onto the
PCB. I also have other circuitry to put on there as well. Also, that
would complicate the PCB assembly process with respect to the
heatsink.

No, that would be unecessarily-conservative with these IGBTs in a
single-pulse application. Their nice characteristic in this respect
is a much lower Vce compared to FETs Vds, as I showed elsewhere in
this thread. Here it means you can use four in parallel at 600A.

Now that you just described fully discharging the capacitor each time
(the binary drawings have long since scrolled out of view) it's clear
that you don't need to turn off the pulse (the capacitor does that),
so I agree an SCR is the ideal part. You'll probably only need one.
But how do I turn the SCR off? Even after C1 discharges, the 12V
potential from the battery will still cause ~60A of current through
the D1/SCR/L1 combo. Actually, during actual firing, C1 doesn't get
down below ~20V after the uP's 15ms pulse ends. (I usually still show
about 20-23V on C1 after firing.)

An easy answer to that is a different circuit arrangement. I'll leave
the suggestions to Fritz as I no longer have access to your circuit.

Thanks,
- Win

whill_at_picovolt-dot-com

F

#### Fritz Schlunder

Jan 1, 1970
0
An easy answer to that is a different circuit arrangement. I'll leave
the suggestions to Fritz as I no longer have access to your circuit.

Thanks,
- Win

whill_at_picovolt-dot-com

Good question. How do you turn an SCR off? You could try pouring root beer
on it. That may not be a reliable or practical means of commutating it
however. Alternatively you could use some sort of other scheme kind of like
this guy's coil gun uses:

http://www.pskovinfo.ru/coilgun/indexe.htm

His circuit schematic can be found here:

http://www.pskovinfo.ru/coilgun/vcircuit.gif

What if you used say two SCRs, one to discharge most of the capacitor's
energy through the main coil, and then another one that triggers a little
before the 15ms is up that very rapidly discharges what little energy is
left in the capacitor bank into a simple low inductance power resistor. If
you do it right the capacitor voltage might very rapidly decay (much faster
than if it simply flows into the coil which is limited by that "big" 0.2R
resistance), thus freeing the coil to naturally decay.

Hmm... "Naturally decay," eh'? Sounds allot like garbage. I'm sure glad
my profession is electrical engineering instead of garbage man (although for
all the garbage men and women out there reading this don't be offended, it
is a very respectable job that someone has to do). Count your blessings
they say.

SCRs really shine when you don't have to turn them off. I don't really have
enough experience with coilguns to have much useful intuition on what
circuitry produces the best price/performance in this application, although
since the start of this thread I have been inspired to build a coilgun. So
far I have only achieved about 18 m/s velocity with a two stage system
firing an estimated 3.9g mass with an input energy of 54 Joules. The
efficiency is outright pitiful. I'm using a MAC224A8 TRIAC for switching
each stage. Making the storage capacitor excessively big definitely
produces a very significant reduction in output velocity presumably due to
"suck back" effect.

Peace out Mr. JJ and Mr. Hill.

F

#### Fritz Schlunder

Jan 1, 1970
0
But how do I turn the SCR off? Even after C1 discharges, the 12V
potential from the battery will still cause ~60A of current through
the D1/SCR/L1 combo. Actually, during actual firing, C1 doesn't get
down below ~20V after the uP's 15ms pulse ends. (I usually still show
about 20-23V on C1 after firing.)

Thanks Fritz,
JJ

Oh yeah. The dog was humping the leg on the computer table while I was
composing the last post. Naturally I was distracted and forgot about the
12V flowing through D1. Obviously something would be needed to stop that.
A simple low voltage P-channel would probably be adequate here.

Hmm... It just occurred to me. I don't own a dog.

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