# Current limiting ballast for a transformer

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.

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

I can't figure it out what's going wrong

• 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. Commented Sep 7, 2018 at 18:00
• Why do you hope we can't explain it to you? Commented Sep 7, 2018 at 18:25
• L=XL/(2*π*f) XL=√Z2-R2 Z=230/2=150ohm R=0.5ohm XL=114.9ohm L=0.36H
– Nano
Commented Sep 7, 2018 at 22:11
• That sounds correct. Don’t overlook the incandescent light bulb solution with. Commented Sep 10, 2018 at 22:22
• 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. Commented Sep 30, 2018 at 21:03