Help with transformer inductivity calcs needed :)

[markusw]

In LTspice the transformer models I use have the following parameters as variables:

primary I
primary serial R
secondary I
secondary serial R


In the datasheet of the LT-44 transformer http://www.maplin.co.uk/Media/product_pdfs/hx82.pdf inductivity unfortunately is not indicated.
Are there any easy calculations (approximations) to calculate inductivity from the datasheet?

values given for

impedance, open impedance, freq response, DC resistance, and ?turn number as well as wire diameter?

It would be great just to have an approximation.

Thanks a lot in advance.

Markus

[R.G.]

There are some crude approximations. 

The data sheet tells you that the matched primary impedance is 20K ohms on the primary. It also mentions that the frequency response is no worse than -3db down at the lowest frequency of specification, 300Hz. The only way those can all be true is that the primary inductance is no less than that inductance which will cause an impedance equal to the reflected impedance (20K) at 300Hz.

The inductance must then be Xl <= 2*pi*F*L, so L>= 20K/(2*pi*300) = 10.6 Hy

The datasheet then goes ahead to tell you that the impedance with the secondaries open is greater than 45K at 1kHz. That is, L > 45k/(2*pi*1000) > 7.16 Hy, which would be supported by the 10.6Hy from the first calculation.

The secondary inductance is a bit more slippery. As an approximation, we know that the 3000T on the primary gives you, say, 10Hy. The secondary will have a proportional inductance based on the turns, which are 320x2 or 640. Inductance is related by the number of turns squared, so the secondary inductance is going to be Ls >=(640/3000)**2*10Hy = 0.482Hy if the primary is open.

However, I don't know if LTspice tries to calculate the mutual inductance from those two numbers. It may even try to back calculate the turns ratio from the two inductances, in which case you HAVE to do it that way.

It's an approximation. Try it and see if the answers make sense.

[markusw]

Thanks  lot

Your explanations are exactly what I was hoping for.

Forgot to mention before: LTSpice also wants the coupling factor (<1). Is it possible to give an estimation for this??

Markus

[R.G.]

 

LTSpice also wants the coupling factor (<1). Is it possible to give an estimation for this??

The coupling coefficient "k" for transformers measures how much flux from the primary couples to the secondary as a ratio. If the transformer were perfect, it would be 1.000. That's why LTspice wants it less than one. If the coupling is much less than one, the transformer's not much good, as little of the primary flux gets to the secondary.

So - sure, guess. It's probably more than 0.90, maybe more than 0.95. Start with 0.95 and see if you get answers that make sense. Then try a few runs with different values of k. If it makes little difference in your answers, great, you're done. If it makes a big difference, you'll have to figure out how to measure it.

[markusw]

Thanks a lot again!!

The calculation for Lp seems to be clear. However,initially I had trouble repeating the calculation for Ls

so the secondary inductance is going to be Ls >=(640/3000)**2*10Hy = 0.482Hy if the primary is open.

(640/3000)*2*10Hy = 4.266 Hy ?? Ah got it: its (640/3000)^2*10.6 = 0.482 Hy

When I used 10 Hy for Lp and 1 H for Ls in the LTSpice sims the signal after the passive ringmodulator setup of the Ringstinger increases. So I consider this as an improvement 

Have'nt tested influence of K yet.

Markus


EDIT: Perhaps I should read the posts more carefully:

Inductance is related by the number of turns squared

[markusw]

OK. K definitely has an influence in LTSpice although the results surprise me: while with K=0.8 output voltage after the passive ring modulator is roughly 2.2V pp, it's 1.2V at K=0.9 and 0.1V at K=0.99. Waveshape on the other hand is very symmetric at K=0.99 and shifts to assymetric with lower K values.

Markus