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4hv.org :: Forums :: Tesla Coils
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TC equivalent circuit

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Crunchy Frog
Thu Feb 03 2011, 08:22PM Print
Crunchy Frog Registered Member #2422 Joined: Tue Oct 06 2009, 02:41AM
Location:
Posts: 85
I'm trying to construct a mathematical equivalent circuit of a tesla coil with a non-resonant primary (L only). However, different sources have some disagreement about how. On wikipedia ( Link2 ), the leakage inductance is to the left of the magnetizing inductance, and the transformed impedance of things connected to the secondary is in series with the secondary impedance.

However on Link2 , he puts the leakage inductance on the right of the "primary" inductance, and the secondary inductance and capacitance in series.

The other thing he says is that the voltage ratio in a loosely coupled transformer is closer to the square root of the ratio of the inductances. If impedance transformation ratio is the square of the ratio of voltages, then it is just the ratio of the inductances. But if the secondary inductance is transformed by the ratio of inductances, it would always appear as the primary inductance. Which doesn't make sense.

So can anyone explain this?
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Wolfram
Thu Feb 03 2011, 08:27PM
Wolfram Registered Member #33 Joined: Sat Feb 04 2006, 01:31PM
Location: Norway
Posts: 971
Have you seen this Link2 page? There's some information relevant to your questions there, especially on page two.
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Frosty90
Fri Feb 04 2011, 12:31AM
Frosty90 Registered Member #1617 Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139
If you ingore the C and R, the 'usual' transformer model is a T like in the wiki link, the I bit of the T is the magnetising (mutual) inductance, and the bits across the top are the leakage inductances. You can transform this T circuit to an equivalent pi circuit (a star/delta transformation really), which would look like the second link.

If you make the assumption that Lm is very small, then when you do the transformation, the left had vertical leg of the pi is approximately equal to Lp (the primary leakage) and the right had leg is approximately equal to Ls, if you assume that Lm is very small, which it is when you have only loose coupling.
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