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26 or 18 awg for 315*1650mm secondary (and Balancing MMC and Primary on chosen Secondary in DR)

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Intra

Sat Jun 09 2018, 01:55PM

Registered Member #2694 Joined: Mon Feb 22 2010, 11:52PM
Location: Russia, Volgograd (Stalingrad).
Posts: 97

Uspring wrote ...

Sparks live on power, so it is desirable to input as much energy as the bridge allows to the primary tank.

The primary tank is a mmc?

Uspring wrote ...

The max current in the primary due to arc loading is dependent on the choice of the primary tank impedance.

I have 75pcs of TPC FAG86-9301--C 3.1uF 600Vrms 150Arms in 15S5P configuration which gives 9kV 1.0(3)uF 750Arms and which connected to each other with tin-galvanized 25*6*~128mm copper busses and brass bolts\studs.
How can I calculate or measure primary tank impedance?

Uspring wrote ...

There is an optimal secondary Q for strongest arc loading effects, which is around Q=1/k

JavaTC told me use about 0.154k coupling so then Q=1/0.154=~6.4935
As I could understand, ark loading is a few hundreds kOhms then secondary impedance should be "k" times of ark loading so the ark loading is ~50kOhms/0.154k=~324.67kOhms.
But I still can't understand how I can use 324.67kOhm and 6.4935 number in choosing Lsec, Csec, wire length and wire thikness. How Q and ark loading relate to each other and relate to Lsec and Csec?

Many thanks in advice.

Intra

Mon Jun 11 2018, 12:43AM

Registered Member #2694 Joined: Mon Feb 22 2010, 11:52PM
Location: Russia, Volgograd (Stalingrad).
Posts: 97

I made a simulation for 18AWG and 26AWG again but now with the same number of turns and secondaries length for both coils.
simulation of 18AWG made without spacing
simulation of 26AWG (0.405mm) made with spacing ~0.59562mm for acheive the same wiring length and number of turns (~160cm) but with 26AWG instead of 18.
Result is surprised me:
Both secondaries starts to have 43kOhm impedance unloaded.
Q unloaded drops on 52% for 26AWG coil BUT simulation in ScanTesla stay quitely almost the same.
And as on-time stay quitely almost the same then as I could understand if I wire the secondary with 26AWG with that kind of spacing instead of using 18AWG, the 26AWG coil will works almost the same as 18AWG coil could do?

Many thanks in advice

Uspring

Mon Jun 11 2018, 11:32AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

The primary tank is a mmc?

The primary tank is the series circuit of the MMC and primary L.

I have 75pcs of TPC FAG86-9301--C 3.1uF 600Vrms 150Arms in 15S5P configuration which gives 9kV 1.0(3)uF 750Arms and which connected to each other with tin-galvanized 25*6*~128mm copper busses and brass bolts\studs.
How can I calculate or measure primary tank impedance?

Primary impedance is Zpri=1/(2*pi*f*Cpri), where f is the primary resonance frequency.

JavaTC told me use about 0.154k coupling so then Q=1/0.154=~6.4935
As I could understand, ark loading is a few hundreds kOhms then secondary impedance should be "k" times of ark loading so the ark loading is ~50kOhms/0.154k=~324.67kOhms.

IMHO JavaTCs optimal coupling applies only to SGTCs. It certainly doesn't take into account arc loading. The way to proceed in calculating secondary impedance would be to start from an arc load Rarc, then calculate the best secondary impedance by k*Rarc. Rarc changes radically during arc growth and is pretty much an unknown, though. Empirically values of around 50kOhms for the secondary impedance work well. The secondary impedance is given by 2*pi*f*Lsec, where f is the secondary resonance frequency.

I made a simulation for 18AWG and 26AWG again but now with the same number of turns and secondaries length for both coils.
simulation of 18AWG made without spacing
simulation of 26AWG (0.405mm) made with spacing ~0.59562mm for acheive the same wiring length and number of turns (~160cm) but with 26AWG instead of 18.
Result is surprised me:
Both secondaries starts to have 43kOhm impedance unloaded.
Q unloaded drops on 52% for 26AWG coil BUT simulation in ScanTesla stay quitely almost the same.
And as on-time stay quitely almost the same then as I could understand if I wire the secondary with 26AWG with that kind of spacing instead of using 18AWG, the 26AWG coil will works almost the same as 18AWG coil could do?

With the same number of turns and the same length of the secondary you will get a similar inductance, so the impedances are the same. The unloaded Q difference is due to the differing resistances of the wire you use. I'd go with a thicker wire to reduce heating up of the secondary. Maybe add 10-20% more turns to increase the impedance a bit, if possible.

Intra

Tue Jun 12 2018, 02:30AM

Registered Member #2694 Joined: Mon Feb 22 2010, 11:52PM
Location: Russia, Volgograd (Stalingrad).
Posts: 97

Uspring wrote ...

The max current in the primary due to arc loading is dependent on the choice of the primary tank impedance. Typical primary Q due to loading maybe around 5 or 10, so you have to choose the primary L and C accordingly to match the bridge capabilities.

and

Uspring wrote ...

Primary impedance is Zpri=1/(2*pi*f*Cpri), where f is the primary resonance frequency.

How is primary Q relate to Zpri?
For fat long sparks will be better small MMC and lot of primary turns or huge MMC and a few primary turns?

Uspring wrote ...

I'd go with a thicker wire to reduce heating up of the secondary. Maybe add 10-20% more turns to increase the impedance a bit, if possible.

Regarding to my strange traction to use 26AWG which I already have and avoid to buy any other wire while I already have one, will it be acceptable\reasonable to winding two 26AWG wires in parallel?
As I can understand this should:
double thickness to reduce heating up,
double Q unloaded,
makes possible winding wires without spacing,
makes possible adding ~20% number of turns (from 1600 to 1900) for increasing impedance a bit,
and makes possible save same winding length, ~same Fres (38-43kHz), ~same Zsec.
Or not?

Many thanks on advice.

MRMILSTAR

Tue Jun 12 2018, 03:28AM

Registered Member #62119 Joined: Sun Feb 04 2018, 04:59AM
Location: Cedar Rapids, Iowa
Posts: 136

You can model spark loading with JAVATC. In the top load portion of JAVATC, you can supply more than one top load component. In addition to my toroid, I modeled a spark as a cylinder with a diameter of 0.02 inches and whatever length you want to simulate. The cylinder is vertical only so there is that limitation. JAVATC will combine all of the top load components to compute the top load capacitance. Now, I don't know if this will figure in any way into the coupling computed by JAVATC because I don't know the underlying equations used by JAVATC but it it is very useful to predict top load capacitance.

Uspring

Tue Jun 12 2018, 10:30AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

How is primary Q relate to Zpri?

The effective primary Q, which is determined by arc loading, limits the the primary current to which the primary tank will ramp up to. Max primary current is given by Iprimax = Vinp * Q/Zpri.

For fat long sparks will be better small MMC and lot of primary turns or huge MMC and a few primary turns?

A too small MMC together with a large primary L implies a big Zpri (=sqrt(Lpri/Cpri) ). If Zpri is too large, than you will not get enough current into your primary, i.e. less than your bridge can handle. For a large MMC with a low primary L, Zpri will be small and you will have to shorten the burst in order not to overload the IGBTs. So the optimum is somewhere in the middle, where you get currents high enough to provide a lot of input power but not so high as to invoke the OCD. This has been discussed here

and

Regarding to my strange traction to use 26AWG which I already have and avoid to buy any other wire while I already have one, will it be acceptable\reasonable to winding two 26AWG wires in parallel?

It may be difficult to wind to wires in parallel. Also this will double the height of your winding, which will lower the inductance significantly.

MRMILSTAR wrote:

You can model spark loading with JAVATC. In the top load portion of JAVATC, you can supply more than one top load component.

Secondary Q is determined by the resistive i.e. consumptive part of the arc load. Adding top loads only increases the capacitive part.

Intra

Tue Jun 12 2018, 11:52AM

Registered Member #2694 Joined: Mon Feb 22 2010, 11:52PM
Location: Russia, Volgograd (Stalingrad).
Posts: 97

Uspring wrote ...

It may be difficult to wind to wires in parallel. Also this will double the height of your winding, which will lower the inductance significantly.

I mean, when
"simulation of 26AWG (0.405mm) made with spacing ~0.59562mm for acheive the same wiring length and number of turns (~160cm) but with 26AWG instead of 18."
then this gives 160cm wiring length because of spacing and 1600 turns. so inductance here is around 143365uH
but if it be wired with two 26AWG in parallel then for 1600 turns it will be 128cm length and 175850uH
but you said about adding 10-20% turns so when wiring two 26AWG in parallel becomes 1900 turns then wiring length becomes 152cm and inductance becomes 212162uH

So looks like it won't double the height and won't lower the inductance.

Uspring

Tue Jun 12 2018, 03:05PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Ok, but you might have some trouble winding that.

Intra

Tue Jun 19 2018, 01:23AM

Registered Member #2694 Joined: Mon Feb 22 2010, 11:52PM
Location: Russia, Volgograd (Stalingrad).
Posts: 97

what here:

Uspring wrote ...

Although the above equation is exact, changes of Qsec, fsec and f due to arc loading make estimates fragile.

and here:

Uspring wrote ...

The values are similar but not the same. The correct equation is:
Qpri = (Qsec/k^2) * fpri*fsec/f^2*(1 - f^2/fsec^2)^2 + fpri/fsec*1/(k^2 * Qsec).
fpri is the primary resonance f.

means "f"?

fpri as I understand is a 1/2*pi*sqrt(Lpri*Cpri)
fsec as I understand is a 1/2*pi*sqrt(Lsec*Csec)
right?

I found Antonio's post useful and made a spreadsheet with:

Antonio wrote ...

I would avoid considerations of Q, since the system is not a second-order circuit, and consider the impedance that the inverter wants to see, considering the maximum input current and the maximum voltage applied to the primary:
R=(4/pi)*Vmax/Imax
And then design the system as an impedance matching network to match this impedance to the impedance at the secondary. The required calculations are in my site:

The formulas for the doubly terminated case can be worked to produce the following:
Given the secondary inductance and capacitance, Lb and Cb, and the primary capacitance Ca:

Required primary inductance La=Lb*Cb/Ca+Ca*R^2 (note the last term. This is not a SG Tesla coil where La*Ca=Lb*Cb, although the difference is small).
Driving frequency F=1/(2*Pi*sqrt(Lb*Cb)). The secondary resonance.
Voltage gain: n=sqrt(Lb/Ca)/R. The assumed load is then R*n^2.
Bandwidth, if the primary side had R in series:
B=Ca*R*sqrt(2)/(2*Pi*Lb*Cb)
Coupling coefficient: k=sqrt(B^2/(2F^2+B^2))

The number of cycles for stabilization of the output is approximately F/B.
Note that this design is the same obtained for the lossless case for a certain R and a certain mode, since the tuning relations are identical.

These designs are idealized. Apply the usual discount of secondary capacitance considering capacitive streamer loading.

considering my secondary and found a bit inconsistency between his pack of formulas and block of:

Uspring wrote ...

How is primary Q relate to Zpri?

The effective primary Q, which is determined by arc loading, limits the the primary current to which the primary tank will ramp up to. Max primary current is given by Iprimax = Vinp * Q/Zpri.

For fat long sparks will be better small MMC and lot of primary turns or huge MMC and a few primary turns?

A too small MMC together with a large primary L implies a big Zpri (=sqrt(Lpri/Cpri) ).

and

Uspring wrote ...

How can I calculate or measure primary tank impedance?

Primary impedance is Zpri=1/(2*pi*f*Cpri), where f is the primary resonance frequency.

With Vinp*Q/Zpri I got only 78% of Iprimax from Imax which I picked up for R=(4/pi)*Vmax/Imax to calculate Ca for calculation of n, load, B and k for my secondary+topload.
I'm almost sure I shouldn't use Qsec which is 1/k in Iprimax=Vinp*Q/Zpri and should use Qpri=(Qsec/k^2)*fpri*fsec/f^2*(1-f^2/fsec^2) ^2+fpri/fsec*1/(k^2*Qsec), right?

Beside of this as I mention I made a spreadsheet and figured out that I still didn't get how to balance MMC and Lpri according to pack of formulas from Antonio's post.

For a few values of MMC in some range with different Imax I got different values of load, coupling, voltage gain, bandwidth, etc, and methinks I'm completely missing something on this way. Attached a few screenshots.

Will MMC and Lpri will be balanced beetween each other if the load on chosen current will be some kind of 10 of Zsec? Something of 500000kOhms load for 50kOhms Zsec.

How to define high and low impedances and when the impedance is high, low and too high or too low?

Also I found your Vmax/Imax < k/(2*pi*f*C) on Tardief's thread and tries to calc it for my coil so for results I got from Antonio's pack of formulas I realized that in my setup Vmax/Imax=k/(2*pi*f*C) and then if I rebuild spreadsheet for desirable Vmax/Imax=~0.5*k/(2*pi*f*C) then it will mismatch with Antonio's formulas.

Not sure I'm understand how all this works.

Many thanks in advice.

EDIT:
oh, It seems I completely forgot about Ohm's law. =\
with 565V on bus it looks like no 100A or 200A may happen on that config without variable Rpri and no variable Rpri should happens. Even one string from the table is possible - string with current on 565V bus. And it seems I use R=(4/pi)*Vmax/Imax wrong way. =\

Intra

Tue Jun 19 2018, 01:37AM

Registered Member #2694 Joined: Mon Feb 22 2010, 11:52PM
Location: Russia, Volgograd (Stalingrad).
Posts: 97

Forgot add the spreadsheet itself
] balancing_mmc_and_primary_on_chosen_secondary_in_d r.zip[/file]

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