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4hv.org :: Forums :: Tesla Coils
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Increasing QCW streamer length

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BSVi
Tue Jul 15 2014, 07:06AM
BSVi Registered Member #1637 Joined: Sat Aug 16 2008, 04:47AM
Location: Kiev, Ukraine
Posts: 83
Steve Conner
You are right, P=U^2/Z. Its seems increasing U is really the best way to add length.

As for losses, they are almost independent of voltage as bride is soft-switched (half-zvs, half-zcs). But they are directly proportional to current.

I also have 450v bus capacitors, so with PFC I could add even more to streamer length.

Are you using any sort of Class-DE arrangement with deadtime and snubber capacitors?
I can say so, but without explicit deadtime generator or capacitors. Deadtime is formed by high gate R (10 Ohm at the gates + 1.8 Ohm in primary of GDT). Gate voltage rise looks pretty exponential, fall is, of course, forced by diodes. As for capacitors, frequency is high and output capacitance of IGBTs is large enogth to acheive ZVS.

My power stage is as simple as this:

1405406093 1637 FT164671 Fe1b8f


A QCW with IGBTs has huge switching losses
They are not as huge as you tell :) Conduction loses are major source of loses as I think (because of soft switching). 150A*2V (C-E drop) is 300W. This 300W is spread betwen two paralleled devices (of course, with some error). 150W/die is well within specs even with heatsink at 100 celsius and even for CW operation.

Heatshik heats up to 40 celsius at 1bps without any cooling with ribs facing down. I think there is enough headroom with this thermal design to increase BPS to my target of 5.


I'd guess the reason for better performance at larger coupling to be the OCD to kick in later.
No OCD doesn't trigger at all. You all know the universal soulution in tesla coil building: If OCD kicks in, simply increase it's treshold.
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Dr. Dark Current
Tue Jul 15 2014, 09:23AM
Dr. Dark Current Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
BSVi wrote ...

As for capacitors, frequency is high and output capacitance of IGBTs is large enogth to acheive ZVS.

Why would frequency have anything to do with the turn-off energy loss? It's specified in the data sheet for a given turn-off CURRENT, and is, of course, independent of the switching frequency.

At hundreds of kHz, even for the fastest IGBTs, the switching power losses largely dominate the conduction losses.

So, I'm sorry, but your argument is invalid. tongue
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BSVi
Tue Jul 15 2014, 09:58AM
BSVi Registered Member #1637 Joined: Sat Aug 16 2008, 04:47AM
Location: Kiev, Ukraine
Posts: 83
Why would frequency have anything to do with the turn-off energy loss?
Maybe I'm wrong, but:
Higher frequency->lower primary inductance->higher primary current (~V/L)->more inductive energy (~LI^2)->Easier to charge body capacitance (which independent of freq) druing deadtime.

Of course, my thoughts applys to this topology only.

the switching power losses largely dominate the conduction losses
This coil operates in soft switching mode, so switching loses are not dominant (in my opinion, I don't actually measured them). Of course, they are not non-existent, but heatsink temperature is pretty close to what I calculated it should be taking into accuunt conduction loses only.
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Dr. Dark Current
Tue Jul 15 2014, 10:36AM
Dr. Dark Current Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
The inductance of the primary winding or energy stored in it has absolutely no effect on the switching loss, my friend. Only the turn-off current is what matters and the datasheet values simply hold true.

Your biggest enemy here is the transient thermal imepdance, not a steady-state (DC) dissipation. The temperature ripple of the die should be small. The transistor will work for a while even if it is large (lets say above 30 °C), but it WILL fail sooner or later.
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Steve Conner
Tue Jul 15 2014, 12:17PM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Yes, if you turn off an IGBT at a given current, without a snubber capacitor across it, you will get the switching loss given in the datasheet for that current.
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Uspring
Tue Jul 15 2014, 12:29PM
Uspring Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711
I'd guess the reason for better performance at larger coupling to be the OCD to kick in later.
No OCD doesn't trigger at all. You all know the universal soulution in tesla coil building: If OCD kicks in, simply increase it's treshold.
Oh, and have to mention - this coil has very high coupling - 0.39. Changing it to 0.2 decreases perfomance by about a half. Another idea to try is to increase coupling even more.
Decreasing coupling usually leads to a higher primary Q, i.e. more current for a given bus voltage. I'd expect more input power because of this, leading to longer streamers. Even if current didn't increase (and didn't drop), the input power level would stay the same. So where does all the power go, if not into the streamer?
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Steve Conner
Tue Jul 15 2014, 01:07PM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
I believe the increase in primary Q is equal to the decrease in coupling, in other words the effective amp-turns of MMF seen by the secondary doesn't change.

If the OCD isn't triggering, then on loosening the coupling, the spark length stays the same and the primary current goes up.

If the OCD is triggering, then the primary current stays the same (as the OCD is controlling it) and the spark length goes down.

If the primary circuit is lossy, the increase in Q won't be enough to counterbalance the decrease in coupling, and the spark length will go down.
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Uspring
Tue Jul 15 2014, 03:45PM
Uspring Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711
Qpri, as by the theoretical equation, can vary in dependence of k somewhere between 1/k^2 or not at all. That hinges on the tuning conditions, Qsec, etc.

My argument was based on power conservation and, as you point out losses in the primary play an important role. I imagine the loss resistance Rl and the coupled in resistance Rc (the one implied by the Qpri equation) to sum up in the primary tank. Lost power would be I^2 * Rl and power transferred to the secondary I^2 * Rc. Rc increases in some way with coupling.

EDIT: The situation is much like a voltage connected to 2 resistors in series.

If Rc < Rl, increasing coupling implies more spark output. If Rc > Rl, it will be the other way around. Since BSVi observed the former, the loss resistance is probably dominant. From this it seems, that winding a thicker primary could increase primary current as the OP wanted and at the same time iimprove efficiency.
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BSVi
Tue Jul 15 2014, 03:56PM
BSVi Registered Member #1637 Joined: Sat Aug 16 2008, 04:47AM
Location: Kiev, Ukraine
Posts: 83
The inductance of the primary winding or energy stored in it has absolutely no effect on the switching loss
Yes, I'm agree that at turn off phase shifted IGBTs experiences hardswitching and has very high power dissapation. I calculated it to be around 800W which is much greater than 150W conduction loses. Thaks for pointing this out. I'll calculate transient thermal response and write result here :)

I believe the increase in primary Q is equal to the decrease in coupling, in other words the effective amp-turns of MMF seen by the secondary doesn't change.
Decrease of coupling decreases streamers. I think coupling has much more effect than Q.
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Uspring
Tue Jul 15 2014, 06:09PM
Uspring Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711
Decrease of coupling decreases streamers. I think coupling has much more effect than Q.
Coupling and primary Q are related. A larger coupling will draw more power from the primary. That will show up as a reduced primary Q.
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