Hi,

I'm in the very early stages of putting together the design for a DC/AC inverter running 24v dc ->230v ac. One of the major design considerations seems to be how long to stay in dc for. At the extremes, you can either invert 12v and step up with a large transformer, or use a DC/DC converter to get 230v then invert. I can see there must be some compromise between DC/DC converter switching losses that will increase with step up ratio, and losses associated with a large transformer, which will become less as the DC converter output rises, but have not had much luck quantifying this compromise. If anone out there has prior experience/words of wisdom, they would be gratefully recieved.

Cheers,

Ben

 

Well first off, a DC/DC converter is no such thing, it's a DC/AC/DC converter - you would normally use a high frequency AC to transform the inoput voltage upwards (which is why it's converted back to DC afterwards).

Once you have a high voltage DC you then chop it to give a 50Hz AC signal, usually using a stepped waveform which better approximates a sinewave - items requiring a 'proper' sinewave mains input usually work off these supplies.

The older way is to convert from DC to AC at 50Hz, using a square wave, this is far more inefficient and a great deal heavier - plus many items won't work off a square wave.

As I see it, there's no competition, go for the modern high frequency (cheap and efficient) system, that provides a better output waveform.

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Yes - I had been thinking of going for Quasi-sine. not only is it easier and more efficient that full sin, but the stepped waveform should allow deadtime to avoid any risk of shorting out the bridge.

I realise that you can go via AC using a bridge circuit and rectifier, but surely a simple boost converter circuit is just DC/DC...

So the argument for stepping up the DC supply before swapping to AC is that it avoids the use of a large, heavy, and lossy transformer? I had a peek inside a comericail inverter, which appeared to be almost entirely switch mode, with only one tiny coil, which may just have been an inductor for a step up converter. I had a go at quantifying the transformer losses vs. converter switching losses, but got a bit bogged down.

Thanks for the response,

Ben

 

It's unlikely to be a simple DC/DC converter if it carries signficant power. The typical step-up converter doesn't make very good use of the inductance, which is the expensive and hard-to-design part. It's more likely a high frequency switching transformer, much more effective in terms of inductor size and efficiency, particularly when higher buck/boost ratios are involved.

 

No, it converts to AC first, and usually uses a single recifier to convert back to DC. Anything switch-mode is AC.

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