Can a rectified sine wave still induce voltage?

Question

I would like to learn more about the behaviour of a transformer when the primary coil current is not the standard AC sine wave. What happens if the current supplied to the primary coil takes the waveform of a rectified sine, like as shown below?

I feel like there is still a changing level of current in the primary, and therefore a changing magnetic field in the core - am I correct in thinking this? Will the transformer still induce a voltage in the secondary coil? Would the induced voltage be less than with a standard AC sine wave?

Answer 1

Short answer:

• Unipolar current: admissible.
• Voltage with a DC component: bad (the transformer saturates and it won't work that well).

Long answer:

Consider the model of a real transformer.

What happens if the current supplied to the primary coil takes the waveform of a rectified sine, like as shown below?

Applying an unipolar current is admissible for a transformer and it is commonly found on some transformer-based DC-DC converters. You must apply a voltage that will produce that current.

If the secondary is open, then the loop is closed just by the magnetizing inductance \$X_M\$ and \$R_C\$. Assuming \$R_C\$ very high, assuming 0 \$R_P\$ and \$X_p\$, then since \$V_M=L\frac {dI_M}{dt}\$, the voltage at the primary will have to be a discontinuous cosine, with a 0 dc value (voltage and current and time plotted in arbitrary units):

By the way, in the above calculation we assumed that the bandwidth is not limited. A real transformer will have a finite bandwidth and you might not be able to achieve that waveforms.

Instead, applying a voltage with a non-zero DC component will saturate your transformer.

A saturated transformer means a much smaller \$X_M\$. The current at the primary will be very high because of the DC component of the input voltage (which is applied to \$R_P\$, the parasitic resistance of the windings) and because of the very small magnetizing inductance value (which, together with the leakage inductance \$X_P\$ will limit the no-load primary current).

In addition, since the core is saturated, there won't be a good coupling between the two windings, which by the way, will increase also the leakage inductance.

At the end, there will be a extremely poor transmission of the remaining AC component of your rectified sine at the secondary side (and extreme overheat).

Still, if the DC offset is very small (with respect to the AC component), then the parasitic winding resistance will act as a negative feedback. In fact, the drop on the resistor will be larger when the core is saturated, effectively lowering the voltage applied to the magnetizing inductance. However, this will still result in a loss of efficiency.

Answer 2

Yes, a transformer can operate with non-sinusoidal waveforms. The simplest example of this is an audio transformer.

What a transformer can not do is operate at DC. A DC current on the primary winding induces no time varying magnicflux in the core, so nothing is coupled to the secondary. Even more problematic, is that DC creates a static magnetic field in the core which may saturate the core and inhibit AC transmission.

Answer 3

If you couple the transformer through a capacitor to block the DC content it will work fine. The waveform has some high frequency harmonics that may be attenuated depending on the transformer design but generally if the transformer is capable of 5 or 10x the frequency of the sinusoidal waveform before rectification the secondary waveform will look very similar, ignoring the difference in the DC level such that the average DC at the secondary is zero.

If you do not block the DC, the core will likely saturate, greatly distorting the waveform and also the transformer may be destroyed since only the primary resistance will limit the current and resulting \$I^2R\$ heating from the DC component. If you keep the voltage small enough that neither problem arises it will behave similarly to the case with the blocking cap (but it may require dropping the voltage to millivolts). It will also heavily load the source which may affect the results if it does not have negligible source impedance.

The average voltage at the secondary will always be zero, assuming a conventional transformer.

Answer 4

You could find out experimentally. Have you got a signal generator, an oscilloscope, some diodes, and a small transformer, say 1:2 ratio or thereabouts?

Or think about amplifiers which use vacuum tubes: they normally have an output transformer for impedance matching. So, as the sound is not completely ruined, it is reasonable to assume that the waveform output from a transformer is the same as the input waveform.

As the electrons in the primary of the transformer are being wiggled, the electrons in the secondary will be wiggled to match, it's just that the amplitude of the wiggling depends on the ratio of the number of turns.

There can be problems with giving a transformer a DC offset as shown in your diagram - if the transformer core becomes magnetically saturated then there will be distortion in the secondary waveform and heating of the transformer.

The RMS value of a fully-rectified sine wave is the same as a sine wave: \$V_{peak}/\sqrt 2\$, but it has a DC offset of \$V_{peak} \cdot 2/\pi\$ (source: Average and effective values).

Answer 5

COnsider the impedance of diodes when switched and the inductive kick when current is switched off. Next consider the spectrum and bandwidth of the transformer. After this consider the saturation effects of dc on the primary coil

These all reduce the power that you can transfer thru the transformer.

Answer 6

As far as the transformer is concerned, it is equivalent to an alternating current with a DC bias. The DC bias plays no part in the transfer of energy to the secondary winding, although it could produce saturation effects. The output of the secondary will be a non-sinusoidal waveform at twice the frequency of the original (unrectified) waveform; the output waveform will have a component at the original frequency due to the sharp inverse peak (where the original waveform crossed zero).