simulate this circuit – Schematic created using CircuitLab

Recently I've been trying to improve my understanding of inverter technology. I understand there are three main types: square wave, modified sine and pure sine. The square wave type works by creating a square wave (that oscillates between two voltage levels with the same magnitude and opposite polarity, such as +12V and -12V). This square wave voltage is then stepped up to somewhere around 230V by using a transformer. The modified sine type works roughly the same way, except there is a "pause" between the positive and negative pulses. This is better, because the waveform now contains less of the high-frequency harmonics.

The output after stepped up by the transformer is what I don't understand. With the modified sine and square wave we are giving pulses of essentially DC voltage to the primary of the transformer. When DC voltage is put to a transformer, the current starts to rise. The secondary is outputting voltage only when the current in the primary is changing (because only then the magnetic field is changing in the core, and a changing magnetic field is what we need to produce voltage on the secondary). Therefore what we are getting out of the secondary is just short spikes of voltage at the same moments when the voltage and current changes across the primary. So what really is the advantage of the modified sine wave if we are still just inserting pulses of DC voltage to the primary? I understand that the waveform that goes into the primary now has less harmonics, but what difference does it make to the output from the secondary?

I conducted the following experiment: I built a circuit that outputs a modified sine wave. The circuit has an astable multivibrator, that has each of the two outputs connected to a monostable multivibrator. The astable outputs a square wave at 50Hz. When the voltage goes high, it triggers a monostable through a MOSFET, that outputs a high state for around 5ms. Then the voltage goes down for another 5ms and the same thing happens with the other side of the astable. The output waveform looks good:

enter image description here

The blue waveform is offset down for clarity. The timescale is 8ms.

At the top is a sketch of my set-up, omitting some safety details.

The blue waveform from the previous picture goes into the MOSFET marked with "blue" and the yellow one to the "yellow" MOSFET.

And finally, here is the output from the transformer secondary:

enter image description here

The time scale of the oscilloscope is set to 2ms. The transformer is rated for 50Hz.

As I expected: I'm seeing spikes of high voltage, spaced with rougly 10 ms between them, corrensponding to the times when the voltage from the oscillator transitions to positive or negative. The smaller oscillation in between is another thing I don't quite understand, but I suppose it has something to do with voltage being induced to the transformer from the cutting of the current when the pulse goes down. Here is a close up to the oscillation (left):

enter image description here

So, to conclude this long question: Is this spiked output anywhere close to what is expected from modified sine wave inverters, or is there something fundamentally wrong with my understanding and/or my circuitry? Clearly it does not look right. Quality equipment is not easily found in my native country so I have not been able to purchase a real inverter and examine the output. All the texts I've read about inverters just show a picture of the modified sine waveform like the one I have (just imagine either the blue or the yellow waveform as negative), and just state that this waveform is good because there are less harmonics, but none talk about how the output from the secondary acctually looks. So: what exactly is the use of the modified sine if the acctual output of the inverter is not a modified sine? Both the modified sine and the square types just seem to give voltage spikes. Any insight into this would be much appreciated!

EDIT: This is what the voltage at the center tap on the primary look like. Scope says the frequency is 100Hz, timescale is 8ms. Not what I expected, but I can't really tell what to make of it.. The image is a bit fuzzy but it's basically a square wave with ripples.

enter image description here

  • \$\begingroup\$ Are you using a 9Volt battery? Those little batteries can't put out much current, which might be why you are seeing such short pulses. Might also be the way you are driving the transistors. What does the signal look like at the center tap of the transformer on the low side? And what does it look like on the other two terminals on the low side? Maybe the voltage to the transformer doesn't look like you expect it to. \$\endgroup\$
    – JRE
    May 6, 2017 at 19:32
  • \$\begingroup\$ @JRE I added a picture of the center tap voltage. Yes, I'm using a 9V battery. At the other terminals the voltage looks the same as at the center tap. As for how I'm driving the mosfets, I've simply connected the oscillating waveforms to the gate pins. \$\endgroup\$
    – S. Rotos
    May 6, 2017 at 20:05
  • \$\begingroup\$ Looks like the voltage from the battery is dropping with each pulse - the little battery can't supply the current you are trying to draw through the transistors and the winding. I expect that is a part of the reason your output voltage on the high side looks messed up. You need a better powersupply on the low side. \$\endgroup\$
    – JRE
    May 6, 2017 at 20:32
  • \$\begingroup\$ @JRE Interesting, thank you! I'll try to get a better power supply and try again. I'm curious though: I cannot immediately see how the duration of the pulse is affected. I mean, I would more like expect the magnitude of the output to be less, but I'm still getting hundreds of volts. \$\endgroup\$
    – S. Rotos
    May 6, 2017 at 20:40
  • \$\begingroup\$ What's the spec on your transformer? Are you by any chance saturating it with 45 mVs? \$\endgroup\$
    – winny
    May 6, 2017 at 21:03

1 Answer 1


You have multiple problems and misconceptions.

Your second paragraph starts ...

The output after stepped up by the transformer is what I don't understand. With the modified sine and square wave we are giving pulses of essentially DC voltage to the primary of the transformer. When DC voltage is put to a transformer, the current rises quickly.

When a DC voltage is put to a transformer, the current rises slowly. If your pulse timing, your input voltage, and your core flux carrying capability are properly matched, then the current will continue rising across the width of your input pulse. If your core runs out of flux carrying capability (runs into saturation), then your current will rise quickly. But that's not a good operating region for a transformer.

If you want to drive a transformer with 'underlapping' pulses (pulses that do not overlap), then you must use a full H bridge. You are driving it push pull. When one drive pulse ends, a current is flowing, and as that FET turns of, it has nowhere to flow. Now it wants to change fast, and so requires a very high voltage to do it. That energy is probably being 'caught' by the body diode in the other driver turning on.

If your power supply does not have the current output capability to meet the transformer current demand, then its voltage will collapse.

  • \$\begingroup\$ Yes, I meant that the current rises not instantly but still quite fast (compared to the duration of the pulse). I understand that I need the current to rise somewhat linearly (or rather, rise in the linear-ish region of the exponential curve) when the pulse is high if I want the output to resemble a modified sine wave. I'll get a better power supply and try again. \$\endgroup\$
    – S. Rotos
    May 7, 2017 at 17:38
  • \$\begingroup\$ @S.Rotos NO, it's not fast compared to the duration of the pulse, it rises throughout the pulse. \$\endgroup\$
    – Neil_UK
    May 7, 2017 at 18:48
  • \$\begingroup\$ Hello Neil, I tried to use a more powerful supply (I used a lead acid battery). It still does not work, the output looks the same as before. \$\endgroup\$
    – S. Rotos
    May 10, 2017 at 18:23
  • \$\begingroup\$ I also edited my question regarding the rise of current, you're right, that was a confusing sentence. \$\endgroup\$
    – S. Rotos
    May 10, 2017 at 18:35
  • \$\begingroup\$ Now I got it working better, it seems that I had a second problem: My mosfets were not driven fully into saturation. I now have another problem with my circuit which I will make into another question. By this problem is solved so I will accept your answer. \$\endgroup\$
    – S. Rotos
    May 15, 2017 at 19:35

Your Answer

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

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