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I have a 230V to 6V step-down transformer.

I'm trying to calculate the inductance of the primary side 'roughly' in an easy way.

enter image description here

Here is what I have in my mind: For that I added some illustrations above. I assume when the transformer's secondary is "not loaded" I can model the primary side as in Fig. 1. I also assume the transformer is ideal, so I can model it as a series LR circuit as in Fig. 1. (??)

After these assumptions I do the following steps:

1-) When disconnected from power lines, I measure the resistance of the transformer's primary coil 'R' by a multimeter as 133 Ohm.

2-) Then I turn on the power in no load conditions and measure the rms current through the primary by a multimeter this is called Iz in Fig .1.

3-) Now I know R which is 133 Ohm, Vrms which is 230V and Irms; and of course the mains frequency 50Hz. These all reduces to solving L in an LR circuit.

4-) Then I use the phasor diagrams at Fig. 2, and Fig. 3(rms values as phasors here) for obtaining an equation as follows:

(in the figures: Iz = Irms is the rms current looping the circuit which can be meausred by a multimeter; XL is the inductive reactance, VL is the voltage across it; and VR is the voltage drop accross R)

From the phasor diagrams I write:

$$|Z| = \sqrt{{X_L}^2 + R^2}$$

Vrms = Iz * |Z| so;

$$X_L = \sqrt{\left( \dfrac{Vrms}{Iz}\right) ^2 - R^2} $$

$$L = \frac{X_L}{2 \pi f}$$

$$L = \frac{\sqrt{\left(\frac{Vrms}{Iz}\right)^2 - R^2}}{2 \pi f}$$

I can then plug in Vrms, measured Iz, R and f in the above equation and obtain L.

My questions is:

Assuming the transformer is ideal: Is my equivalent circuit for the primary part in no load condition and the calcuations make sense for a rough calculation?

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  • \$\begingroup\$ For any half-decent transformer, primary inductive impedance >> resistance. Just measure the primary current and assume it's all controlled by the inductance, you'll be right within a few percent. \$\endgroup\$ – Neil_UK Jul 13 '16 at 11:39
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Yes. You want to end up with a model like this:

Ideal transformer

Ignore Xl1 for now and use your numbers to calculate and Xm. R1 should be ~133 ohm.

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  • \$\begingroup\$ Is Xl1 leakage? And Rc is ignored as well? If so my calculation seems ok right? \$\endgroup\$ – user16307 Jul 13 '16 at 10:56
  • \$\begingroup\$ Yes, Xl1 is leakage inductance. It common to sum primary and secondary leakage to one lump on the primary if you only have one secondary winding. Rc is leakage resistance. It should be very high. \$\endgroup\$ – winny Jul 13 '16 at 11:02
  • \$\begingroup\$ thanks. just one more thing. i used unloaded transformer to make the model easier. is the way i calculate L how it is done normally? i mean is this the way in tradition to find out L for an approximate value? \$\endgroup\$ – user16307 Jul 13 '16 at 11:04
  • \$\begingroup\$ I usually just measure it. LCR meter with open secondary = Xm. LCR meter with shorted secondary = Xl1 + (N1/N2)^2*Xl2 \$\endgroup\$ – winny Jul 13 '16 at 11:31

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