4
\$\begingroup\$

In this answer poster uses the rectified secondary's half sines to detect the zero crossings of the mains voltage. Can you do that? I thought there was a phase difference between primary and secondary, depending on the load.

edit
If there is a phase shift, can it be so big that it's unusable for detecting the zero crossing?

\$\endgroup\$
  • \$\begingroup\$ Interesting point about the phase difference. I can't see a reason why there would be a phase difference between two sides of the transformer that would depend on the load and not on the winding of the transformer. \$\endgroup\$ – AndrejaKo Jun 1 '12 at 13:59
  • \$\begingroup\$ @AndrejaKo: that's how I remember it, but I'm not good enough with magnetics that I can explain, therefore the question. \$\endgroup\$ – Federico Russo Jun 1 '12 at 14:11
  • 1
    \$\begingroup\$ With the secondary open circuit, it will have exactly 90 deg phase shift from the current thru the primary. Magnetic field is proportional to current, and open circuit secondary voltage is proportional to change in magnetic field. However, the primary current can be anywhere from 0 to 90 deg out of phase with the voltage, so this gets messy. Optos are better for detecting voltage zero crossings. \$\endgroup\$ – Olin Lathrop Jun 1 '12 at 16:33
  • \$\begingroup\$ @Olin: that confirms what I thought. It would have been a cheaper solution. Thank you. (Why don't you make this an answer? With all due respect, I find it more clear than Mike's) \$\endgroup\$ – Federico Russo Jun 1 '12 at 16:55
  • \$\begingroup\$ @OlinLathrop - True, the phase shift falls from +90 towards zero as the frequency rises. If there is a resistive load, the leakage inductance will eventually 'kick in' and pull the phase shift to -90. Sorry if I wasn't clear Federico. I will try to improve my answer. \$\endgroup\$ – MikeJ-UK Jun 1 '12 at 17:08
2
\$\begingroup\$

EDIT - Answer totally rewritten!

If we consider the equivalent circuit of a transformer below, at low frequencies the response is dominated by the effects of the primary resistance and the primary inductance which form a high-pass filter. The cut-off frequency of this filter is given by \$f_c=\dfrac{R_P}{2\pi.L_P}\$.

Qualitatively, at low frequencies, the primary current is proportional to the input voltage since the primary appears as a resistance. As Olin says in his comment, the secondary voltage is 90° out of phase with the current (it is proportional to the rate of change of current) so we get an overall 90° phase shift.

As the frequency increases, the primary impedance becomes more inductive producing a phase shift in the current. This phase shift subtracts from the original shift and the overall phase shift falls towards zero (but never quite getting there).

If there is a resistive load on the secondary, the leakage inductance eventually comes into effect forming a low-pass filter with the reflected load resistance. This will cause the phase shift to lag further, towards -90°

To take a worked example, I tried to find some specifications for the inductance of mains transformers but manufacturers don't seem to specify this. So I measured a random transformer with a hand-held inductance meter and got 2.5H (hopelessly underestimated since the meter is a cheap audio frequency type). The primary resistance is 47\$\Omega\$. This gives a cutoff frequency of around 3Hz, at which the phase shift will be 45°. At 60Hz, this would have fallen to 2.9°

So that's the "why", but whether this error is significant depends on the application. As others have noted, an opto-isolator may well be a better solution.

Simplified Equivalent Circuit :- Simplified equivalent circuit of a transformer

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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