DC to AC inverters. How are they boosting voltage?

[fastline]

Looking more at green energy inverters but got me curious as to how they are typically getting voltage up at a 10:1 ratio or more? Is there some digital transformer configuration I am not aware of? I would think they would have a transformer on the front end to boost voltage?

 

[MrAl]

Hi,

There are different ways to do it, but i think the majority of the low cost inverters used a DC to DC converter front end followed by an H bridge back end for converting the DC to AC. The DC to DC converter is a boost converter of course, boosting the voltage up to 145 volts DC. The H bridge then produces plus and minus pulses with a pulse width of approximately 62 percent of the total half cycle period which produces the required 120vrms to the output. Thus the output they usually call a "modified sine" is really plus and minus 145v pulses that are not on for the two entire half cycle periods but are off for roughly 38 percent of the time.

 

[Reloadron]

Transformers are your best friend. On a small and simple scale what if I took 12 VDC and ran an oscillator? Then I use that oscillator to drive a step up transformer? I suggest a simple Google of Inverter Circuits and see where that gets you. There are many designs but in most cases we convert the DC level to an AC signal and drive a transformer with it. I kept it simple as it can get.

<EDIT> MrAl beat me. He even used more words. </EDIT>

Ron

 

[RCinFLA]

They are push pull, PWM regulated switcher. Typically running at 25 kHz to 40 kHz.

Attached is a schematic for an older 300 watt AC inverter.

 

[MrAl]

Hi again,


Ron yeah i must have posted just seconds before you did. I made a slight error in that post though as i remembered the duty cycle to be 62 percent but it is really 68.5 percent.

Just to clear this up...

The goal for the inverter is to provide an output pulse who's RMS value is 120v just like a regular power line in a home. The RMS value for a pulse is:
Vrms=sqrt((1/Tp)*Integral[t1 to t2] Vpk^2 dt)

[LATEX]V_{rms}=\sqrt{\frac{1}{Tp}\int_{t1}^{t2} V_{pk}^2\,dt}[/LATEX]

where t1 is the start time of the pulse and t2 is the end time of the pulse and Tp is the total half cycle period.
Since we can choose t1 to be 0 we have:

[LATEX]V_{rms}=\sqrt{\frac{1}{Tp}\int_{0}^{T} V_{pk}^2\,dt}[/LATEX]

so the integral of Vpk^2 from 0 to the end of the pulse T is:
Vpk^2*T

and divided by the total period Tp we get:
Vpk^2*T/Tp

and finally taking the square root of that we get:
Vrms=Vpk*sqrt(T/Tp)

and normalizing the time period Tp to 1 we get:
Vrms=Vpk*sqrt(T)

and setting Vrms to 120 we have:
120=Vpk*sqrt(T)

and solving for T we get:
T=576/841

which is approximately equal to 0.6848989 which is very close to 68.5 percent. So the 'on' time of the pulse would be 68.5 percent of the total half cycle period.

For a 12vdc to 120vac inverter they start with a boost converter probably using a small transformer to get from 12vdc to 145vdc. This works out better than using a transformer on the output because a transformer on the output has to be much bigger having to handle the low frequency of 50 or 60Hz. On the input side the frequency can be as high as 100kHz or possibly even more, which greatly reduces the required size of the transformer. All that's left to do then is convert it to AC using an H bridge.

It's a little interesting that 145v is exactly half way between 120v and 170v and 170v is the normal peak value of a true sine wave of 120vac.

The way we did it when i worked in the industry is we used different pulse patterns to generate a pseudo sine along with some good filtering to produce outputs with total harmonic distortion under 1 percent even with load. This was possible using dynamic feedback rather than the usual averaging feedback method.