6
\$\begingroup\$

I'm trying to simulate a transformer by using a circuit simulation software. In my requirements sheet, three specifications are given for the transformer:

  1. Maximum power (\$P_m\$) of the transformer will be 4VA.
  2. Magnetizing inductance (\$L_?\$) will be 30\$\mu\$H.
  3. \$\frac{N_1}{N_2}\$ ratio will be 4.

(I don't know what "magnetizing inductance" is. Is there a term like that? Can it be mutual inductance (\$L_m\$)? Are \$L_m\$ and \$L_?\$ the same parameters?)

The turns ration must be a function of these parameters, isn't it?

Like:

\$\frac{N_1}{N_2}=f(P_m, L_1, L_2, L_m)\$

Where;
\$L_1\$: Inductance of the primary winding.
\$L_2\$: Inductance of the secondary winding.

My simulation software does not allow me to directly set \$\frac{N_1}{N_2}\$ ratio. Instead, it allows me to set the \$L_1\$, \$L_2\$, \$L_m\$ parameters.

So, my question is, how do I indirectly set the turns ration (\$\frac{N_1}{N_2}\$) by choosing \$L_1\$, \$L_2\$ and \$L_m\$ parameters?

\$\endgroup\$

2 Answers 2

Reset to default
11
\$\begingroup\$

It sounds like you're dealing with a SPICE simulator.

In SPICE, you define a transformer as a coupled inductor, with the following ratio:

\$\dfrac{N_1}{N_2} = \sqrt{\dfrac{L_1}{L_2}}\$

There's also a coupling factor K that comes into play.

The magnetizing inductance of a transformer is the inductance you measure across the primary winding with the secondaries open-circuit (floating). It's a function of the core material and geometry, air gap and number of turns.

Your \$L_1\$ and \$L_m\$ should be one and the same.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Some of your statements are incorrect since you seem to be ignoring leakage inductance. \$\endgroup\$
    – Olin Lathrop
    Jan 10, 2012 at 20:35
  • 1
    \$\begingroup\$ I used your formula with K=1 to see that it is working very good in my circuit simulator. Next, I gave different values to K to find this more general formula: N1/N2 = K * sqrt(L1/L2). \$\endgroup\$
    – hkBattousai
    Jan 10, 2012 at 20:59
  • 2
    \$\begingroup\$ Depending on the degree of accuracy that you want to achieve in your simulation, and depending on the type of transformer, the leakage inductance can often be ignored. The coupling factor, K, can be used to take care of the leakage inductance in case you want to stick with the ideal two-winding representation instead of the T-model (leakage inductances in series with ideal transformer's windings). K=1 is often good enough for a simulation; "good" real tranformers often have a K very close to 1 (e.g. K = 0,995). For power converters (e.g. flyback), however, it's good to use leakage inductances. \$\endgroup\$
    – zebonaut
    Jan 11, 2012 at 7:41
  • \$\begingroup\$ @AdamLawrence "Your L1 and Lm should be one and the same. ". Could you please elaborate , why L1=Lm ? \$\endgroup\$
    – Israr
    Jul 3, 2021 at 13:00
4
\$\begingroup\$

Magnetizing inductance is the total inductance seen into one of the windings with all other windings open, minus the leakage inductance. The magnetizing inductance of each winding is the inductance of that winding that is shared or coupled with other windings. The total inductance seen from outside (with all other windings open so that they don't matter) is that plus the leakage inductance.

The leakage inductance is the inductance "private" to a winding. You can model the whole transformer as a ideal transformer with separate non-coupled inductors in series with each winding. These non-coupled inductors represent the leakage inductance.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.