let's visualize a transformer. Transformers consist of 2 coils, primary and secondary coils and both these coils possess some parasitic inductance. When we try to do some impedance matching, we look at the effective resistance seen by the voltage source to which primary coil is connected. And we must take primary coil's parasitic inductance into account for better results but we dont do the same for parasitic inductance of secondary coil. I dont understand why. Can you explain?

  • \$\begingroup\$ what do you mean by 'impedance matching'? Are we talking mains frequencies or RF, power transfer or S-parameters? \$\endgroup\$
    – Neil_UK
    Commented Apr 5, 2018 at 11:24

1 Answer 1


The parasitic inductance of the primary coil can be said to be formed by those turns that don't effectively couple their magnetic field to the secondary therefore those few turns form what is called a leakage inductance and is regarded as equivalent to an external inductor in series with the "ideal" primary.

However, that leakage inductance (as a single component) represents both the primary turns and the secondary turns that don't couple magnetically. In other words, the primary parasitic inductance value that you might read in a data sheet represents (in one component) the leakages of both primary and secondary.

Some data sheets will specify both leakages but they can be grouped mathematically (and quite easily) into one component at the primary.

  • \$\begingroup\$ should not leakage inductor be parallel to primary \$\endgroup\$
    – OnurTR
    Commented Apr 5, 2018 at 13:56
  • \$\begingroup\$ No, leakage inductance is a series "volt-dropping" component - it cannot be in parallel with the main body of the primary inductance (magnetization inductance). Google transformer equivalent circuit. \$\endgroup\$
    – Andy aka
    Commented Apr 5, 2018 at 14:20

Your Answer

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

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