Inverter 12 to 220V 300W transformer Help

I am trying to design an inverter.
Vin 10-14VDC
Output 220VAC (modified sin wave)
50Hz
300W

I don't have any inverter design experience, so I swollowed my pride and tried to reverse engineer a few different automotive inverters. Most ot them where crap, but I did find one that seemed to be a decent design.
Long life electrolytic caps, small transformer, compact design. The design used two TL494's and an op amp for the control of the boost and DC to AC 50Hz.
Boost stage TL494 Osc=50khz
Push pull primary drives each FET at 25KHz

Transformer primary is CT (Vin)
CT is at one end of the transformer not the middle. lower inductance?
Secondary is not CT
Marking is EI40-12-220

Does anyone know how I can find the specs of this transformer?

Using the TL494 to generate the 50Hz seems like a bad idea (min osc=1Khz for TL494), but they seem to make it work.
thanks in advance for any help.


Note: There are other pieces to the design, battery charger, solar input etc. In case you where wondering why I don't just buy an inverter.

What you're looking at is sometimes called a "transformerless" inverter. It does not use a bulky 50Hz power transformer isolation, but instead uses a smaller high frequency transformer in a DC-DC converter (sounds like a push pull converter in your case), which converts the ~12V input to around 400VDC (or whatever is needed to generate the desired AC output). The actual inverter part is usually just a full bridge of MOSFETs with a filter on the output, which inverts the 400VDC into 220VAC. These types of inverters are generally nice, but are much more complicated to design.

Thanks for your response

Push-pull to boost the DC. Then full wave bridge.
Each half of the Push pull boost on the primay is switching at 25KHz.
Primary CT is on one end of the transformer.
No CT on secondary.

Not sure how to go about designing something like this.

EEenergy said:

Thanks for your response

Push-pull to boost the DC. Then full wave bridge.
Each half of the Push pull boost on the primay is switching at 25KHz.
Primary CT is on one end of the transformer.
No CT on secondary.

Click to expand...

Yeah, that sounds like a push pull. Often a push pull will also have CT on the secondary, but not always (it just requires a different rectifier).

Designing, the DC-DC converter stage is significantly more complicated than the circuitry for a normal inverter. The design of an optimal high frequency transformer is especially difficult, and is somewhat of a power electronics black art (though if you don't need very optimal performance then it's not hard). If you want to give it a try, you should start by reading up on SMPS design in general. The reference text Switch Mode Power Supply Handbook by Billings is excellent for this.

Thanks again for your response.

A good thing about working for a small company is you get to design a little bit of everything. A bad thing about working for a small company is you get in over your head sometimes.

Looks like I have my work cut out for me on this one.

Understandable. Fortunately there are lots of people on the forums who have experience in such things, and this sort of design is pretty common, so you have help when you need it. But you should still go out and and read various white papers and reference texts before diving into an actual design.

How this circuit works is that, the 12V DC is first stepped up to high voltage DC. It's converted to high voltage DC because a ferrite transformer can be used if it's driven at high frequency. However, the high frequency output can not be used to power mains devices. So, it's converted to DC and then this DC is converted back to 50Hz using a full-bridge converter stage employing 4 MOSFETs (or other similar "switches", eg IGBT or BJT).

The transformer has a primary winding (centre-tapped) and a secondary. The turns-ratio is proportional to the voltage ratio. So, if the voltage ratio required is 10:300 (10 for low voltage), the turn ratio be a little greater than 10:300. So, you can use a turns ratio of 10:320.

Now, decide how many turns you require for the primary. Consider 2 turns first. Then, calculate if the maximum flux density is acceptable. If not, increase to 3 turns and keep on increasing if required. So, if you use 3 turns (for example), and the turns ratio is 10:320 (which is equal to 3:96), you have to use 3 turns + 3 turns for the primary and 96 turns for the secondary.

You can use this formula to calculate flux density and/or use it to find the required number of turns:



You can find the information about the core area and maximum flux density (denoted Bmax) from the datasheet, or you can use a value, such as 2000G or 2500G or 3000G (these are common values of Bmax).

Hope this helps.
Tahmid.