0
\$\begingroup\$

I am trying to understand how impedence matching works, but I realized I had some issues with basic transformers as well:

  1. If the voltage and current is merely dependent to the input voltage, current and number of turn ratios, I say that the resistance of the secondary part of the transformer is not important? Then why don’t add infinite resistance(even open circuit) to get infinite power?

    In a different way: what happens when there is a resistor at the secondary part of transformer? Voltage and current is only dependent on turn ratios and voltage and current values of the primary part, which are fixed. Then it can disobey ohm’s law?

\$\endgroup\$

2 Answers 2

3
\$\begingroup\$

It appears that you misunderstand how a transformer works. It is true that the secondary voltage and current are a function of the primary voltage and current. However, the primary current depends on the primary voltage and the load that the transformer represents. In an ideal transformer, the secondary load reflects back to the primary divided by the square of the turns ratio. If you place an infinite resistance across the secondary (i.e. an open circuit), then the primary voltage will also see an open circuit (infinity divided by \$N^2\$). Therefore there will be no primary current and hence, no secondary current. Hence there is no power transferred by the transformer.

If you place a finite resistance across the secondary, say R, then that will be reflected back to the primary as \$ \frac {R}{N^2} \$ which is finite. The primary current, per Ohm's law, will be the primary voltage divided by this resistance. The secondary current will be the primary current divided by N, and the secondary voltage will be the primary voltage multiplied by \$ N \$. All of the power from the primary voltage will be transferred to the secondary since an ideal transformer is 100% efficient. However Ohm's is satisfied by both the primary and secondary circuits.

\$\endgroup\$
1
  • \$\begingroup\$ Then can I say that for a transformer the impedence of the inductor used for building transformer is not dependent on the inductor but dependent to the turn ratio and the load inpedence at the output part? Then current flowing in the input part is dependent on the “fake impedence” of the first inductor, therefore dependent on the turn ratio and load impedence. And the voltage is independent from all these, only dependent to the supplier? \$\endgroup\$ Dec 6, 2017 at 6:35
0
\$\begingroup\$

Pin=Vin * Iin
Pout=Vout * Iout
Vout/Vin=n turns ratio= n2/n1
Zin = Zload/n² // Zi where Zi=ωLp

I call Zi the input equivalent impedance with no secondary load for primary excitation current I= Vin/ωLp due to primary inductance.

Losses occur from ideal case above include Mutual coupling ratio <1, I²R for n1 and n2, core eddy current losses and if L saturates from Remanence initial condition or overvoltage then excitation losses rise sharply.

\$\endgroup\$

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.