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Inverter design basics

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samEEEf

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Typically two stages are considered in SMPS inverter design so far I know which sees to be a good design.

Stage 1: Battery (DC) to high voltage DC. This is done using high frequency converter with a ferrite core transformer. Then LC filter
Stage 2: High volt DC to AC. This is actually the inverter where SPWM is used. Then LC filter.

Example: 12v DC----->400v DC----->230v AC (rms)

My question, what if I use only one stage? i.e, 12v dc to 230v AC (rms) just applying SPWM technique and a ferrite core transformer.
 

You can do this BUT the transformer must be capable of handling the output power at 50 HZ. With the conventional way the 50 HZ comes from a pure DC source and is generated via PWM, so there is no low frequency limit, i.e. it will produce a DC output. It is doubtful that a ferrite cored transformer will pass the 50 HZ to the output only its harmonics.
Frank
 
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    mali00

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@chuckey:

I want to explain a bit more -

I want to make a inverter using a SPWM at 50kHz frequency. 12V dc to around 12v SPWM (50kHz modulation, 50Hz sine) will be generated. Using a ferrite core transformer, it will be stepped up to get 230v RMS (325 peak). So, ferrite core will be handled at 50kHz. Is there any gap in my understanding?
 

What you have described so far will generate a 50 kHz output from the ferrite transformer. How will you get the desired 50Hz output from that?
 

I'm thinking in this way -

1107InterfaceConnectionFig4.jpg
 

I'm thinking in this way
The transformer voltage in your drawing contains a low frequency (50 Hz) component. A 50 kHz transformer can't handle it without core saturation. That's the "gap" in your considerations.

To be converted by the transformer, the input voltage must be DC-free, there can't be a slowly varying duty cycle as sketched.
A symmetrical three-level PWM could be upconverted by a transformer. But you need a commutator circuit to make 50 Hz AC.
 

The transformer voltage in your drawing contains a low frequency (50 Hz) component. A 50 kHz transformer can't handle it without core saturation. That's the "gap" in your considerations.

To be converted by the transformer, the input voltage must be DC-free, there can't be a slowly varying duty cycle as sketched.
A symmetrical three-level PWM could be upconverted by a transformer. But you need a commutator circuit to make 50 Hz AC.

Thanks, understood the gap of having 50Hz component with 50kHz through the transformer. :smile:

What is commutator circuit?
 

If you have a square wave, its has a harmonic contents of F,3F,5F,7F... or in your case 50,150,250 KHZ. When you change the mark space ratio, side bands appear each side of these frequencies, INCLUDING 0KHZ. Passing the PWM signal through a low pass filter, removes all components except the sidebands associated with 0 KHZ(DC), i.e. the base band. This is how a conventional SMPS works. This principle is also used in a class D amplifier. This is why your transformer must be able to handle the low frequency components. If you use a ferrite transformer, the low frequency component will not get through, so you get a distorted version of your 50KHZ pulses.
A commutation circuit swops over the polarity.
Frank
 

Not clear about the commutator circuit. :sad:
 

You can see it as controlled rectifier in this case, the transformer can only output 50 kHz AC with variable fundamental wave amplitude (using three-level PWM). So it's output must be rectified to get a low frequency (50 Hz) wave and the polarity must be switched after each halfwave.

It's not more than a principle design idea and not actually better than a regular HV output inverter, I think.
 

Okay, understood....

So, I've to follow the following design -

12v DC----->400v DC----->230v AC (rms)

Stage1
12v DC----->400v DC

Stage2:
400v DC----->230v AC (rms)

In this case, In stage2, no transformer is required. Transformer is only required in stage1. MOSFET or IGBT has to be used in Full bridge configuration in stage2 and output will be taken directly from MOSFET/IGBT.
 

Stage2 circuit will be as below -
H-bridge_inverter_cjc.png
Here DC volt will be 400v.

But problem is - AC output will be floating, ground cannot be common; it is a problem.
 

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