# An inverter without a transformer?

I am trying to build an inverter without a transformer. I want to use a 12 V battery connected to a boost converter which will boost the 12 V to 120 V, then simply attach an H-bridge to the 120 V boost converter output to create an AC voltage. Is this going to work?

• It'll create an AC voltage. Commented Apr 7, 2021 at 13:10
• I have a grid tie solar inverter that does just that. Solar cells into a boost converter then a h bridge to generate the ac. Commented Apr 7, 2021 at 16:28
• Yes and no. You can't boost 10:1 without a transformer. But it doesn't need to be a 60 Hz mains transformer. It can be a much smaller switch-mode transformer or auto-transformer. Also if you want sine wave output your high voltage DC bus needs to be around sqrt(2) * 120V. So that is more like 170VDC. Commented Apr 7, 2021 at 18:29
• How much power do you need to convert? If you buy an equivalent inverter from Pep Boys and open it up, you will see something of how complex and non-trivial this is. And they will still have a transformer of sorts in the form of a serious high frequency inductor, often a big toroid with multiple windings on it. You may want to start your journey in this arena by starting with something a little simpler like a joule thief, but if you simply must, search Digikey for a reference design that does what you want. Best of luck to you. Commented Apr 8, 2021 at 0:20

boost converter which will boost the 12V to 120V,

Transformer will be much more efficient and cheaper than boost converter for that high output/input voltage ratio.

Say you need 120V 10A, if it is 100% efficient you will input 12V 100A.

If you use a transformer, then you can find low-RdsON low voltage FETs that will handle that current, chop it into a high frequency square wave, and feed it to a small transformer designed for that frequency. You can get high voltage boost ratios efficiently. You also get isolation for free.

But in a boost converter, you need the FET to handle the full input current, and also the output voltage. Also it will get Vds voltage swings equal to the output voltage. So you need an impossible combination of low RdsON, high voltage, and low capacitance, because high voltage FETs have higher RdsON, and low RdsON FETs have higher capacitance.

It's okay if you need high voltage at very low current, but at high power it is quite unforgiving.

So-called modified sine wave inverters use a DC rail that is approximately $$\\sqrt{2}\$$ times the RMS voltage of the sine that is being simulated, so about 170V for 120VAC, and a duty cycle that causes the RMS output to match the sine wave (image from here):

Duty cycle should be 50%, so the RMS value and the peak value match that of a sine wave of 120V RMS.

• @user263983 RMS value is proportional to the square root of the duty cycle. Peak value is sqrt(2) higher than desired RMS value. So 50% is correct. Commented Apr 7, 2021 at 18:03
• @user263983 So what is the RMS value of 120 * sqrt(2) * sqrt(0.5)? Commented Apr 7, 2021 at 18:56
• You right. My mistake. Commented Apr 7, 2021 at 18:58

Yes, some very cheap inverters do exactly that: make a high-voltage square wave.

That said, to do this directly you need to create a power supply that is more than twice 120V. This is because the AC waveform is actually about 170V peak, and 340V peak-to-peak.

You also need to modify the duty cycle of the wave to achieve the same RMS value as a sine: make it only 70.7% of the half-wave period.

Further, one of your output lines needs to be a neutral. So no cheating: you need to swing the hot leg 170V above and 170V below that neutral. That means a + and - 170V power supply, or use a single-ended 340V modified sine and large capacitor to couple it. (Yes, I'm aware of floating neutral generators. In many situations they pose a severe safety hazard.)

This is not impossible, mind you. A car stereo amp does exactly that (and more) to create window-shattering levels of power. I suppose you could even drive one with a 60Hz sine and get AC power off of that.

But guess what? All this gets a lot easier (and cheaper) if you use a transformer. Then you get additional safety in the bargain. Your proposed inverter uses a transformer anyway, why not combine the functions into one?

• It is possible to use a single high voltage supply, but you get a "hot neutral" as a result. To do this, you take the "Hot" wire positive for one half-cycle, then take the "Neutral" positive for the next half-cycle (or part thereof). This is done in small inverters and some small inverter-type generators, like the Honda EU2000i. Commented Apr 7, 2021 at 18:47
• And that would be a hazard if any connection to it isn’t double-insulated. I’d add an isolation transformer. Commented Apr 7, 2021 at 19:13
• I don't think you need more than twice 120. You do need the 170V for peak, but because the inverter stacks in input voltage in opposite directions off ground/neutral, you get the 340V peak to peak output with only isolated 170V input available. Peak to peak of the output is more than twice 120, but without a second opposing phase, 340V will never be present across any two conductors.
– K H
Commented Apr 8, 2021 at 2:27
• OP is attempting to do this without a transformer. I’m stipulating no voltage swing on neutral (ideally it should be bonded to earth.) That rules out a push-pull approach. So you need -/-170V to make a single-ended hot. Commented Apr 8, 2021 at 2:35

No.

120 V AC is the RMS voltage. Peak voltage is $$\ 120 \sqrt 2 \ \text V \$$.

Tip: 'V' for volt.

• For a square wave? I think it is safe to assume that his H-bridge is four saturating switches. Commented Apr 7, 2021 at 13:58
• @AnalogKid, no, for the AC waveform OP is trying to generate. That peak voltage will be the H-bridge top supply rail. Commented Apr 7, 2021 at 15:36
• H-bridge - duh - never mind. With a 120 V source and a squarewave output, the output will be 240 Vp-p, or 120 Vrms. Commented Apr 7, 2021 at 18:09