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So, guys, I hope you can explain to me how to calculate the necessary parameters for a ballast to limit the current on the primary side of a high voltage transformer.It is for a tesla coil project. To stop excessive current being drawn through the transformer's winding when the secondary is short-circuited.

Transformer specs.
Pri: 230 V/50 Hz, 2.09 A Resistance: 0.5ohm Sec: 5 kV, 80 mA

So how much inductance I need to limit the current to 2A?

Solution. The total resistance Z of a choke coil consists partially of the actual coil resistance R and partially of the inductive resistance XL in a linear relationship of the frequency f and the inductivity L as XL = 2 πf*L and Z = √ R2+XL2 The resistance of the coil R can be measured directly over the coil with an Ohm-meter but usually R hardly plays a role compared to the inductive resistance. We want to reduce the current flowing through our primary of the HV transformer to 2 A maximally at 230 VAC. This means that the choke coil must have a total resistance Z of 230/2 = 115 Ω. We measure a resistance R over the coil of 0.5 Ω than the inductive resistance must be √Z2-R2 or √((115Ω2)-(0.5Ω2)) = 114.9 Ω. The next step is to calculate the inductivity by arranging our formula into L = XL/(2*π*f)
or L = 114.9/(2*π*50) = 365 mH.

Problem. So I build a choke of the calculated value and test it out. enter image description here

But the ballast is not limiting the current of the calculated target Its 2 times higher

enter image description here I can't figure it out what's going wrong

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    \$\begingroup\$ Why are you wanting to limit the current? Anything wrong with a fuse or using a more powerful transformer? In other words your question is made unclear because your stated requirements are unclear. \$\endgroup\$ – Andy aka Sep 7 '18 at 18:00
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    \$\begingroup\$ Why do you hope we can't explain it to you? \$\endgroup\$ – Transistor Sep 7 '18 at 18:25
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    \$\begingroup\$ L=XL/(2*π*f) XL=√Z2-R2 Z=230/2=150ohm R=0.5ohm XL=114.9ohm L=0.36H \$\endgroup\$ – Nano Sep 7 '18 at 22:11
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    \$\begingroup\$ That sounds correct. Don’t overlook the incandescent light bulb solution with. \$\endgroup\$ – winny Sep 10 '18 at 22:22
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    \$\begingroup\$ You can do that but no need since you already know your inductance (366.8 mH) and number of turns (hopefully). Cross section area with a ruler. Now you can solve IL=NAB for B. \$\endgroup\$ – winny Sep 30 '18 at 21:03
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You actually need a device with a non-linear response. Typically this would be a 100 watt incandescent light bulb. This would be in series with the 230 VAC power to the transformer.

If there is no arcing the current is low and the bulb is dim or not lit at all. When arcing occurs the bulb flickers and heats up, increasing resistance and limiting the current to the transformer to about 1/2 amp (with a 250 volt 100 watt bulb).

For more current simply put more lamps in parallel until you reach the maximum current you want. 2 bulbs would give you 1 amp maximum current. 4 bulbs would give you 2 amps maximum current.

You could use 200 watt bulbs but they do cost more and tend to have a short life. It depends on how many hours of arcing you want.

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  • \$\begingroup\$ Thank you very much for your answer,I em familiar with that option. But i em simply interested with the idea of inductive ballast,and i want to learn the math behind it. \$\endgroup\$ – Nano Sep 23 '18 at 11:25

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