Coil-gun Project, questions I couldnt fine answers to
PlayNice, Mon Dec 10 2012, 09:15PMHey everyone, I would like to build a coil gun and had some questions I can't quite find answers to. I just want to understand the concept better before I start.
My understanding is high Henry/slower coil is better in lower stages of the gun (first few stages), and low Henry/faster coil in higher stages of the gun right?
Lets just say someone was making a multi-stage (say 100 stages) coil gun but "HAD TO" make all coils 100% identical, same size/turns, same gauge wire...
What would an experienced person choose? High H/slow coils, or low H/fast coil?
Wouldn't a bunch of low Henry/fast amp rise coils give more velocity in the end? Or would a bunch of high Henry/slow amp rise coils give more velocity in the end? Which of the 2 options would be more efficient? Is it even possible to know?
Power supply would be 3, 12v batteries in series so 36v, @ about 100amps.
Thanks everyone
Re: Coil-gun Project, questions I couldnt fine answers to
Turkey9, Wed Dec 12 2012, 03:17AM
The reason that a larger inductance is generally better for the first few stages is that the projectile is moving slower. Slower projectile means a longer pulse. However, this is when using capacitors. An LC circuit will have a certain time constant, or how long it takes to go through a single period. In your case, you don't have a capacitor so the only factor your inductance will change is the rise time of the pulse. You want the rise time to be as fast as possible. So in short, I would make all your coils have a pretty small inductance, say around 10uH. That's 10E-6 Henry.
Turkey9, Wed Dec 12 2012, 03:17AM
The reason that a larger inductance is generally better for the first few stages is that the projectile is moving slower. Slower projectile means a longer pulse. However, this is when using capacitors. An LC circuit will have a certain time constant, or how long it takes to go through a single period. In your case, you don't have a capacitor so the only factor your inductance will change is the rise time of the pulse. You want the rise time to be as fast as possible. So in short, I would make all your coils have a pretty small inductance, say around 10uH. That's 10E-6 Henry.
Re: Coil-gun Project, questions I couldnt fine answers to
PlayNice, Wed Dec 12 2012, 06:30AM
Thank you for your reply. Just what I needed to know. If I make my coils small inductance like you said say 20uH, at 100a 36v will the first coil even get my 3.5gram projectile moving though? Also I am considering using a capacitor but mainly just to collect the flyback voltage to make my setup more efficient (not sure if that's what that will actually do). PS what do you think about my circuit? BTW I know nothing about designing a circuit, I'm pretty much a beginner so I have a lot of questions. After the IGBT turns off, how long of a pulse is the flyback voltage compared to the IGBT pulse? Will this cap that I added help my coilgun? What about the size of the cap?
Here is what I have so far:
PlayNice, Wed Dec 12 2012, 06:30AM
Thank you for your reply. Just what I needed to know. If I make my coils small inductance like you said say 20uH, at 100a 36v will the first coil even get my 3.5gram projectile moving though? Also I am considering using a capacitor but mainly just to collect the flyback voltage to make my setup more efficient (not sure if that's what that will actually do). PS what do you think about my circuit? BTW I know nothing about designing a circuit, I'm pretty much a beginner so I have a lot of questions. After the IGBT turns off, how long of a pulse is the flyback voltage compared to the IGBT pulse? Will this cap that I added help my coilgun? What about the size of the cap?
Here is what I have so far:
Re: Coil-gun Project, questions I couldnt fine answers to
Turkey9, Thu Dec 13 2012, 01:13AM
Remember that the strength of the magnetic field is proportional to the current through the coil. So yes, 100A will be enough to move the projectile.
Your circuit will not collect the reverse voltage on the inductor. Really, it is best for a beginner to put the diode in anti-parallel with the coil and short any of the reverse current. Trying to recycle it is just too complicated for your first design.
As for the capacitor, it will help provide a kick of current, but once it drains the battery will be supplying all the current. Another consideration you need to make is how to turn off the IGBT in time for the projectile not to experience any suck-back.
Turkey9, Thu Dec 13 2012, 01:13AM
Remember that the strength of the magnetic field is proportional to the current through the coil. So yes, 100A will be enough to move the projectile.
Your circuit will not collect the reverse voltage on the inductor. Really, it is best for a beginner to put the diode in anti-parallel with the coil and short any of the reverse current. Trying to recycle it is just too complicated for your first design.
As for the capacitor, it will help provide a kick of current, but once it drains the battery will be supplying all the current. Another consideration you need to make is how to turn off the IGBT in time for the projectile not to experience any suck-back.
Re: Coil-gun Project, questions I couldnt fine answers to
PlayNice, Thu Dec 13 2012, 02:33AM
Ok so I can still keep the cap to help give it more kick at the begging of a pulse. My IGBT has a antiparallel diode in it, is that enough? And well my IGBT will be powered by a MOFSET 4amp driver, driver will be powered by a very fast LED opto-switch circuit from a RPM counter, so I am assuming here that my IGBT will have super fast ON/OFF.
Important question about the flyback voltage. How fast is the flyback voltage? Lets say that my IGBT has a ON pulse of 1ms, is the flyback voltage pulse just as long so 1ms? Or is the flyback voltage way faster since it way higher voltage? Also if my circuit is 40v, what voltage is the flyback 400v? 1000v?
I need to know how fast the flyback voltage is so that I can then move my opto switch bit away from the coil to compensate for the flyback and try to avoid suckback.
PlayNice, Thu Dec 13 2012, 02:33AM
Ok so I can still keep the cap to help give it more kick at the begging of a pulse. My IGBT has a antiparallel diode in it, is that enough? And well my IGBT will be powered by a MOFSET 4amp driver, driver will be powered by a very fast LED opto-switch circuit from a RPM counter, so I am assuming here that my IGBT will have super fast ON/OFF.
Important question about the flyback voltage. How fast is the flyback voltage? Lets say that my IGBT has a ON pulse of 1ms, is the flyback voltage pulse just as long so 1ms? Or is the flyback voltage way faster since it way higher voltage? Also if my circuit is 40v, what voltage is the flyback 400v? 1000v?
I need to know how fast the flyback voltage is so that I can then move my opto switch bit away from the coil to compensate for the flyback and try to avoid suckback.
Re: Coil-gun Project, questions I couldnt fine answers to
Yandersen, Thu Dec 13 2012, 04:44AM
Flyback voltage is so fast, that noone can catch it - don't even try! :)))
Yandersen, Thu Dec 13 2012, 04:44AM
Flyback voltage is so fast, that noone can catch it - don't even try! :)))
Re: Coil-gun Project, questions I couldnt fine answers to
Turkey9, Thu Dec 13 2012, 06:02AM
You're not going to get a force from the flyback voltage. Don't move your optical sensor to compensate.
Voltage is the derivative of current through your inductor. The reason such a large voltage is created is that you cut the current off suddenly. This creates a step whose derivative is theoretically infinity. Of course, in real life when you cut the current off it isn't a real step. So the flyback voltage is proportional to how fast you stop the current flowing.
Turkey9, Thu Dec 13 2012, 06:02AM
You're not going to get a force from the flyback voltage. Don't move your optical sensor to compensate.
Voltage is the derivative of current through your inductor. The reason such a large voltage is created is that you cut the current off suddenly. This creates a step whose derivative is theoretically infinity. Of course, in real life when you cut the current off it isn't a real step. So the flyback voltage is proportional to how fast you stop the current flowing.
Re: Coil-gun Project, questions I couldnt fine answers to
PlayNice, Thu Dec 13 2012, 07:49AM
Oh ok, I was under the impression that flyback voltage would create some pull force too. So what do I do to stop the osculation of fly-back v? Is my antiparallel diode inside the IGBT enough?
PlayNice, Thu Dec 13 2012, 07:49AM
Oh ok, I was under the impression that flyback voltage would create some pull force too. So what do I do to stop the osculation of fly-back v? Is my antiparallel diode inside the IGBT enough?
Re: Coil-gun Project, questions I couldnt fine answers to
Turkey9, Fri Dec 14 2012, 09:04AM
I always like to put my diodes in antiparallel with the coil so that no current flows back into the capacitor.
Turkey9, Fri Dec 14 2012, 09:04AM
I always like to put my diodes in antiparallel with the coil so that no current flows back into the capacitor.
Re: Coil-gun Project, questions I couldnt fine answers to
PlayNice, Sun Dec 16 2012, 02:49AM
Why not let the flyback current collect back in the cap? I though if you let he flyback to into the coil it would make the system more efficient?
PlayNice, Sun Dec 16 2012, 02:49AM
Why not let the flyback current collect back in the cap? I though if you let he flyback to into the coil it would make the system more efficient?
Re: Coil-gun Project, questions I couldnt fine answers to
Yandersen, Sun Dec 16 2012, 07:00AM
The normal coilgun must return unused energy (that is what you keep calling "flyback current") back into the cap. Even with a perfect timing and other perfect conditions it is not possible to convert all of cap's energy into the projectile kinetic energy - 8-30% is a maximum of the conversion rate (assuming no suckback and no heat dissipation).
To make coilgun operate efficiently, all unused energy must be returned or reused by subsequent stages. The easiest way is to use 3-element circuit - non-polar cap, thyristor and a coil. All unused energy will stay in a cap after a shot, howhether, cap will be repolarized. You may add a second stage to reuse the energy and return cap into normal polarization. I did it once in my first recuperational coilgun and get 6.6% of efficiency from the first stage and 13% from second one; cap had around 25% of initial energy left after the shot. You may try the same - this design is even simplier than classic one. Just make sure charger is disconnected from a cap before the shot, otherwise it will be fried by repolarized cap.
The second normal design is a halfbridge. Here coil discharges unused energy back into the power source. Saz is an expert in this field, he can say more. :) All I can mention is that halfbridge is pretty complicated for the beginners and operates on transistors, which cost more than thyristors and generally less powerfull (at least price and size compared).
All other designs which burn out unused energy are monkey designs - inefficient and do not inspire hobbyists to learn electronics. You can easily determine those by damper diode used. But being honest, I did one like that long time ago. :)
Yandersen, Sun Dec 16 2012, 07:00AM
The normal coilgun must return unused energy (that is what you keep calling "flyback current") back into the cap. Even with a perfect timing and other perfect conditions it is not possible to convert all of cap's energy into the projectile kinetic energy - 8-30% is a maximum of the conversion rate (assuming no suckback and no heat dissipation).
To make coilgun operate efficiently, all unused energy must be returned or reused by subsequent stages. The easiest way is to use 3-element circuit - non-polar cap, thyristor and a coil. All unused energy will stay in a cap after a shot, howhether, cap will be repolarized. You may add a second stage to reuse the energy and return cap into normal polarization. I did it once in my first recuperational coilgun and get 6.6% of efficiency from the first stage and 13% from second one; cap had around 25% of initial energy left after the shot. You may try the same - this design is even simplier than classic one. Just make sure charger is disconnected from a cap before the shot, otherwise it will be fried by repolarized cap.
The second normal design is a halfbridge. Here coil discharges unused energy back into the power source. Saz is an expert in this field, he can say more. :) All I can mention is that halfbridge is pretty complicated for the beginners and operates on transistors, which cost more than thyristors and generally less powerfull (at least price and size compared).
All other designs which burn out unused energy are monkey designs - inefficient and do not inspire hobbyists to learn electronics. You can easily determine those by damper diode used. But being honest, I did one like that long time ago. :)
Re: Coil-gun Project, questions I couldnt fine answers to
PlayNice, Mon Dec 17 2012, 02:48AM
Thanks a lot for the explanation Yanderson.
So basically I dont think I can use the first design you mention the "3-element" as I would like to keep my power supply (batteries) connected to the caps. And I don't like thyristors as my understanding is they don't turn off and I need them to before the suck-back occurs.
I would like to hear more about the half-bridge design. Especially if the unused energy can be discharged back into the cap or the batteries I'm using for the power source. I am a bit scared of the complicated design tho but I can start small I guess.
PlayNice, Mon Dec 17 2012, 02:48AM
Thanks a lot for the explanation Yanderson.
So basically I dont think I can use the first design you mention the "3-element" as I would like to keep my power supply (batteries) connected to the caps. And I don't like thyristors as my understanding is they don't turn off and I need them to before the suck-back occurs.
I would like to hear more about the half-bridge design. Especially if the unused energy can be discharged back into the cap or the batteries I'm using for the power source. I am a bit scared of the complicated design tho but I can start small I guess.
Re: Coil-gun Project, questions I couldnt fine answers to
Yandersen, Mon Dec 17 2012, 06:37AM
Well, if it is complicated for you to connect a coil to the non-polar cap via thyristor, than just forget about half-bridge - you need to deeply learn electronics for that.
Thyristor (the same as SCR) is a simpliest switching element, cheap and powerfull. Once triggered it will stay conductive up untill all current will go through it. When connecting a coil to the non-polar cap through the SCR, you must give it just one short single pulse. Energy goes from cap to coil, than back into cap and SCR closes after that. And it is not possible to trigger it again because the inversed polarity will not let it become opened. But if you add a second stage connected to that cap via another SCR, but turned upside down, this one may be triggered. It is the simpliest coilgun design and it is few times more efficient than classic one. Here is my two-stage recuperational coilgun schematic:
Sure don't wonna try?
Yandersen, Mon Dec 17 2012, 06:37AM
Well, if it is complicated for you to connect a coil to the non-polar cap via thyristor, than just forget about half-bridge - you need to deeply learn electronics for that.
Thyristor (the same as SCR) is a simpliest switching element, cheap and powerfull. Once triggered it will stay conductive up untill all current will go through it. When connecting a coil to the non-polar cap through the SCR, you must give it just one short single pulse. Energy goes from cap to coil, than back into cap and SCR closes after that. And it is not possible to trigger it again because the inversed polarity will not let it become opened. But if you add a second stage connected to that cap via another SCR, but turned upside down, this one may be triggered. It is the simpliest coilgun design and it is few times more efficient than classic one. Here is my two-stage recuperational coilgun schematic:
Sure don't wonna try?
Re: Coil-gun Project, questions I couldnt fine answers to
PlayNice, Mon Dec 17 2012, 08:13AM
I might try it, looks simple. But I already bough all these polar caps spend big money on them too, sucks that I would have to buy new ones now
PlayNice, Mon Dec 17 2012, 08:13AM
I might try it, looks simple. But I already bough all these polar caps spend big money on them too, sucks that I would have to buy new ones now
Re: Coil-gun Project, questions I couldnt fine answers to
Ash Small, Mon Dec 17 2012, 11:00AM
I've a question here....I read (and posted a link) that they stay conductive until the voltage across them is zero (or negative).
Now, if it feeds an inductor, is not the current 90 degrees out of phase with the voltage? Won't there still be current flowing when the voltage is zero?.....(or have I missed something?)
Ash Small, Mon Dec 17 2012, 11:00AM
Yandersen wrote ...
Once triggered it will stay conductive up untill all current will go through it.
Once triggered it will stay conductive up untill all current will go through it.
I've a question here....I read (and posted a link) that they stay conductive until the voltage across them is zero (or negative).
Now, if it feeds an inductor, is not the current 90 degrees out of phase with the voltage? Won't there still be current flowing when the voltage is zero?.....(or have I missed something?)
Re: Coil-gun Project, questions I couldnt fine answers to
Yandersen, Mon Dec 17 2012, 04:22PM
As any semiconductor device, zero voltage means unconductive state at which flow of the current is not possible. Vice versa is true also - till current runs, voltage drop will be around 1-2V.
Inductor keeps current flow constant, allowing voltage to jump any way possible, while cap keeps voltage potential at the same level allowing infinite discharge current. All this in ideal case, of course.
When cap discharges onto coil through the thyristor, then voltage drop on thyristor is around 2V while current is running. When cap is at 0V, thyristor is at 2V, and coil is at peak of it's current value while maintaining -2V. Running current starts to charge cap again (repolarizing it) and decrease. When current drops to around 0A, voltage on cap is -Vinitial and thyristor closes changing anode-cathode voltage drop from +2V to -Vinitial.
Use an LT spice (I never did as I have a real oscilloscope).
"But I already bough all these polar caps spend big money on them too, sucks that I would have to buy new ones now"
Don't you know that you can make a fake non-polar cap from two electrolithics? Just connect two of the same electrolithics "minus-to-minus" and put protection diodes for each of them and you have a polar cap with 1/2 capacitance. This way you can keep charger connected to one of the caps without risking to burn it during the shot.
Real non-polars polypropylene caps are just better by all characteristics except lower capacitance-to-volume ratio (less energy per volume they store, at least by now - future may be promising).
Yandersen, Mon Dec 17 2012, 04:22PM
As any semiconductor device, zero voltage means unconductive state at which flow of the current is not possible. Vice versa is true also - till current runs, voltage drop will be around 1-2V.
Inductor keeps current flow constant, allowing voltage to jump any way possible, while cap keeps voltage potential at the same level allowing infinite discharge current. All this in ideal case, of course.
When cap discharges onto coil through the thyristor, then voltage drop on thyristor is around 2V while current is running. When cap is at 0V, thyristor is at 2V, and coil is at peak of it's current value while maintaining -2V. Running current starts to charge cap again (repolarizing it) and decrease. When current drops to around 0A, voltage on cap is -Vinitial and thyristor closes changing anode-cathode voltage drop from +2V to -Vinitial.
Use an LT spice (I never did as I have a real oscilloscope).
"But I already bough all these polar caps spend big money on them too, sucks that I would have to buy new ones now"
Don't you know that you can make a fake non-polar cap from two electrolithics? Just connect two of the same electrolithics "minus-to-minus" and put protection diodes for each of them and you have a polar cap with 1/2 capacitance. This way you can keep charger connected to one of the caps without risking to burn it during the shot.
Real non-polars polypropylene caps are just better by all characteristics except lower capacitance-to-volume ratio (less energy per volume they store, at least by now - future may be promising).
Re: Coil-gun Project, questions I couldnt fine answers to
Ash Small, Mon Dec 17 2012, 05:53PM
Now I'm confused. Say the cap is charged to xV, thyristor opens. Current is zero.Voltage on cap falls as cap discharges, current through inductor rises. Current is a maximum when voltage across cap is zero. current continues to flow, gradually decreasing until cap is charged to -xV (assuming no losses), so current is at a maximum when voltage across thyristor is zero. (Voltage and current are 90 degrees out of phase.)
What am I missing? (Apart from the fact that the thyristor will close when the voltage across it is zero, which is when current through it is at a maximum)
From Wikipedia: "A thyristor is a solid-state semiconductor device with four layers of alternating N and P-type material. They act as bistable switches, conducting when their gate receives a current trigger, and continue to conduct while they are forward biased (that is, while the voltage across the device is not reversed)."
(EDIT: Post edited.)
I have two oscilloscopes, but neither has a 'current' setting, they only measure voltage.
Ash Small, Mon Dec 17 2012, 05:53PM
Yandersen wrote ...
As any semiconductor device, zero voltage means unconductive state at which flow of the current is not possible.
As any semiconductor device, zero voltage means unconductive state at which flow of the current is not possible.
Now I'm confused. Say the cap is charged to xV, thyristor opens. Current is zero.Voltage on cap falls as cap discharges, current through inductor rises. Current is a maximum when voltage across cap is zero. current continues to flow, gradually decreasing until cap is charged to -xV (assuming no losses), so current is at a maximum when voltage across thyristor is zero. (Voltage and current are 90 degrees out of phase.)
What am I missing? (Apart from the fact that the thyristor will close when the voltage across it is zero, which is when current through it is at a maximum)
From Wikipedia: "A thyristor is a solid-state semiconductor device with four layers of alternating N and P-type material. They act as bistable switches, conducting when their gate receives a current trigger, and continue to conduct while they are forward biased (that is, while the voltage across the device is not reversed)."
(EDIT: Post edited.)
Yandersen wrote ...
Use an LT spice (I never did as I have a real oscilloscope).
Use an LT spice (I never did as I have a real oscilloscope).
I have two oscilloscopes, but neither has a 'current' setting, they only measure voltage.
Re: Coil-gun Project, questions I couldnt fine answers to
Yandersen, Mon Dec 17 2012, 09:14PM
OMG... Current is measured by a voltage drop on a resistor.
What is forcing thyristor's voltage to some value in a circuit with an inductor? Nothing! Inductor voltage changes any way possible trying to stabilize the current. If thyristor is "jammed" between two caps - that's completely different thing. Voltage drop on a thyristor does not change all the way down to the zero current state when it suddenly closes - only then bias changes from +2V to -xV. This sudden voltage change event is filtered via resistor-cap filter in a two-stage coilgun circuit on the picture above, and works perfectly reliable in my practice. Actually, there are two triggering shots for the second thyristor - when first one opens and when it closes, but first time cap's polarity is not suitable for the second thyristor to open, so it got triggered only after the first SCR relaxes.
Yandersen, Mon Dec 17 2012, 09:14PM
OMG... Current is measured by a voltage drop on a resistor.
What is forcing thyristor's voltage to some value in a circuit with an inductor? Nothing! Inductor voltage changes any way possible trying to stabilize the current. If thyristor is "jammed" between two caps - that's completely different thing. Voltage drop on a thyristor does not change all the way down to the zero current state when it suddenly closes - only then bias changes from +2V to -xV. This sudden voltage change event is filtered via resistor-cap filter in a two-stage coilgun circuit on the picture above, and works perfectly reliable in my practice. Actually, there are two triggering shots for the second thyristor - when first one opens and when it closes, but first time cap's polarity is not suitable for the second thyristor to open, so it got triggered only after the first SCR relaxes.
Re: Coil-gun Project, questions I couldnt fine answers to
Ash Small, Mon Dec 17 2012, 10:32PM
Yep, I have come across that before, thanks for reminding me.
I'm aware that you get voltage spikes on flyback transformers, etc. when the current is suddenly stopped. I can also see that there must be a voltage drop across the thyristor, because it has resistance, but I still don't see how there can be a voltage drop of 2V across the thyristor when the voltage across the cap is zero. I can see that, when the thyristor switches off, and current stops flowing through the inductor, the voltage across the inductor rises to ~ infinity, but the thyristor is already off by then.
I still can't see what happens in the thyristor when the voltage across it is between 2V and -xV. I suppose I'll have to buy a thyristor or two, and 'scope it.
Are you saying that the thyristor switches off when the voltage on the cap drops to 2V?
Ash Small, Mon Dec 17 2012, 10:32PM
Yandersen wrote ...
OMG... Current is measured by a voltage drop on a resistor.
OMG... Current is measured by a voltage drop on a resistor.
Yep, I have come across that before, thanks for reminding me.
Yandersen wrote ...
What is forcing thyristor's voltage to some value in a circuit with an inductor? Nothing! Inductor voltage changes any way possible trying to stabilize the current. If thyristor is "jammed" between two caps - that's completely different thing. Voltage drop on a thyristor does not change all the way down to the zero current state when it suddenly closes - only then bias changes from +2V to -xV. This sudden voltage change event is filtered via resistor-cap filter in a two-stage coilgun circuit on the picture above, and works perfectly reliable in my practice. Actually, there are two triggering shots for the second thyristor - when first one opens and when it closes, but first time cap's polarity is not suitable for the second thyristor to open, so it got triggered only after the first SCR relaxes.
What is forcing thyristor's voltage to some value in a circuit with an inductor? Nothing! Inductor voltage changes any way possible trying to stabilize the current. If thyristor is "jammed" between two caps - that's completely different thing. Voltage drop on a thyristor does not change all the way down to the zero current state when it suddenly closes - only then bias changes from +2V to -xV. This sudden voltage change event is filtered via resistor-cap filter in a two-stage coilgun circuit on the picture above, and works perfectly reliable in my practice. Actually, there are two triggering shots for the second thyristor - when first one opens and when it closes, but first time cap's polarity is not suitable for the second thyristor to open, so it got triggered only after the first SCR relaxes.
I'm aware that you get voltage spikes on flyback transformers, etc. when the current is suddenly stopped. I can also see that there must be a voltage drop across the thyristor, because it has resistance, but I still don't see how there can be a voltage drop of 2V across the thyristor when the voltage across the cap is zero. I can see that, when the thyristor switches off, and current stops flowing through the inductor, the voltage across the inductor rises to ~ infinity, but the thyristor is already off by then.
I still can't see what happens in the thyristor when the voltage across it is between 2V and -xV. I suppose I'll have to buy a thyristor or two, and 'scope it.
Are you saying that the thyristor switches off when the voltage on the cap drops to 2V?
Re: Coil-gun Project, questions I couldnt fine answers to
Yandersen, Mon Dec 17 2012, 11:18PM
What I'm saying is written in black on white in the posts above. From the point the SCR opens and up until it closes voltage drop over it is constant: anode is 1-2V over the cathode, no matter the cap's voltage and polarity. Voltage across inductor changes in the way to maintain that drop.
Yeah, by a couple of 40TPS80 - my recuperational gauss built on them, so I know they are good.
Yandersen, Mon Dec 17 2012, 11:18PM
What I'm saying is written in black on white in the posts above. From the point the SCR opens and up until it closes voltage drop over it is constant: anode is 1-2V over the cathode, no matter the cap's voltage and polarity. Voltage across inductor changes in the way to maintain that drop.
Yeah, by a couple of 40TPS80 - my recuperational gauss built on them, so I know they are good.
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