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Inverter Voltage Stepping Up Problem

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ARK5101

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HI,
I designed simple Sine Wave inverter, using using 12DC to AC conversion and stepping up to 220v at the end..I did DC to AC conversion using H Bridge..Bridge works fine at 12 V but when I try to step up that 12v H Bridge output the results are little disturbing..Bridge output drops to 6~7v and at secondary side of transformer just 130`140V not 220v..

I have checked Transformer it steps down fine and is of 500VA but my purpose is not driving 500W load.

Can anybody help me sorting out this problem.
Thanks in advance
 
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KlausST

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Hi,

12V DC with a full bridge means you get max +/- 12 peak output.
This makes max. 8.48V RMS. Usually there is a filter between bridge and transformer that drops the voltage. Maybe you get 8V RMS at transformer input.
To get 230V output the transformer need an 8V : 230V ratio under full load.
A 500W transformer maybe has a no_lod to full_load factor of 1.05.

If you want a standard transformer with 230V input to be operated in reverse, then you have to take the 1.05 twice into account.

--> You need a transformer with 8V / 1.05 / 1.05 = 7.25V nominal output voltage (at full load and 230V rated input voltage)

If your transformer has more than 7.25V rated output, then you will not be able to produce 230V at full load.


****

Next point is that if you use a standard transformer in reverse then (caused by the 1.05 factor) the core is more saturated.
Maybe this causes increased input current.

****

Special care must be taken with your bridge to avoid DC component at the output. Even a small amout of less than 1% of RMS voltage may lead to transformer core saturation.
Additionally with remance in the core may cause problems when switching on bridge.
Try to ramp up voltage. Start with 50% RMS (this should never give a problem) and ramp it up to 100% softly during the next 20 fullwaves.

Hope this helps

Klaus
 

Vbase

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If your inverter is producing square wave then ignore my post.
If your inverter produces sine wave then your bridge is driven by frequency of 15 to 25 KHz PWM, in such inverter your transformer has to be for 15 to 25 KHz, you cannot use 50Hz transformer.
It is possible but low probability, that the 50Hz transformer you use is made of low quality steel and you are induction heating it with the high frequency.
 

ARK5101

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Hi,

12V DC with a full bridge means you get max +/- 12 peak output.
This makes max. 8.48V RMS. Usually there is a filter between bridge and transformer that drops the voltage. Maybe you get 8V RMS at transformer input.
To get 230V output the transformer need an 8V : 230V ratio under full load.
A 500W transformer maybe has a no_lod to full_load factor of 1.05.

If you want a standard transformer with 230V input to be operated in reverse, then you have to take the 1.05 twice into account.

--> You need a transformer with 8V / 1.05 / 1.05 = 7.25V nominal output voltage (at full load and 230V rated input voltage)

If your transformer has more than 7.25V rated output, then you will not be able to produce 230V at full load.


****

Next point is that if you use a standard transformer in reverse then (caused by the 1.05 factor) the core is more saturated.
Maybe this causes increased input current.

****

Special care must be taken with your bridge to avoid DC component at the output. Even a small amout of less than 1% of RMS voltage may lead to transformer core saturation.
Additionally with remance in the core may cause problems when switching on bridge.
Try to ramp up voltage. Start with 50% RMS (this should never give a problem) and ramp it up to 100% softly during the next 20 fullwaves.

Hope this helps

Klaus


Thanks Klaus for your reply.
You are right there is a filter in between Hbridge and Transformer and i noted that point too that filter might be the reason of voltage drop but i was also thinking that high side MOSFETs are not working properly but i checked every thing HBridge is fine..

Try to ramp up voltage. Start with 50% RMS (this should never give a problem) and ramp it up to 100% softly during the next 20 fullwaves.

How should i do that can you give a hint what approach should i go for because i am supplying voltage with a battery..Potentiometer??

- - - Updated - - -

If your inverter is producing square wave then ignore my post.
If your inverter produces sine wave then your bridge is driven by frequency of 15 to 25 KHz PWM, in such inverter your transformer has to be for 15 to 25 KHz, you cannot use 50Hz transformer.
It is possible but low probability, that the 50Hz transformer you use is made of low quality steel and you are induction heating it with the high frequency.

Vbase..No its Pure Sine Wave and Hbridge is driven by 5Khz,filtered to produce 50 Hz and then comes transformer.
 
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KlausST

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Hi,

Ramp up:

Can't you do this with ramping up the pwm? That's how i'd try it.

Good luck

Klaus
 

ARK5101

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Hi,

Ramp up:

Can't you do this with ramping up the pwm? That's how i'd try it.

Good luck

Klaus

That means doing alteration in code !! Can you be little more specific :|
 
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BradtheRad

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I have checked Transformer it steps down fine

This raises a question. Are you using a power transformer which is normally intended to step down? Whereas you have switched around the primary and secondary so that it steps up?

If this is the case then consider that the primary was designed to be connected to the electric company (which is very low resistance). To prevent frying, the primary needs to present a high impedance. It consists of many turns of thin wire.

The primary is not designed to be used as a secondary. That is, it is not designed to have a load in series with it.

So it seems to me that when you attach your load, you are combining two high impedances in series: (1) Your load, and (2) the primary which you are using for a secondary.

I believe your output winding needs to be thicker wire (possibly a different amount of turns), to reduce its resistance, so it can power your intended load.

I could be wrong.
 

ARK5101

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This raises a question. Are you using a power transformer which is normally intended to step down? Whereas you have switched around the primary and secondary so that it steps up?

If this is the case then consider that the primary was designed to be connected to the electric company (which is very low resistance). To prevent frying, the primary needs to present a high impedance. It consists of many turns of thin wire.

The primary is not designed to be used as a secondary. That is, it is not designed to have a load in series with it.

So it seems to me that when you attach your load, you are combining two high impedances in series: (1) Your load, and (2) the primary which you are using for a secondary.

I believe your output winding needs to be thicker wire (possibly a different amount of turns), to reduce its resistance, so it can power your intended load.

I could be wrong.

You could be right but i bought used transformer and I asked the guy if it could be used for inverter project he said it would be fine...

I have a question Silver Core or Copper Core which one is more reliable and has low power loss? Is it possible my Silver Core transformer is not stepping up properly.
 

BradtheRad

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You could be right but i bought used transformer and I asked the guy if it could be used for inverter project he said it would be fine...

One crucial factor is dc resistance of the high voltage winding. It can be measured directly with a meter. The amount is important because it's a clue to how much power you can get from it.
 

babatundeawe

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That means doing alteration in code !! Can you be little more specific :|

Did you try going by the advice KlauST gave to you? I think your success lies in it. Reduce the primary(previously Secondary) windings of your xformer or you get a transformer with 230ACV to 8ACV but i will advice you increase you voltage to 24 volts instead of 12v and preferably get a 230ACv to 17ACv. connect your bridge output to the 17ACV output as you have done before in your previous setup.

Secondly. If you re using SPWM derived from a sine look-up table, then things might be easier. have ur sine value arrays multiplied by 50% or (0.5) and store it up in another table that you can call. or better you do it on the fly.

this is how. Have a variable probably called ScaleFactor here u put the factor you want to use to multiply any of the value you call from your default sine look-up table.

At first initially the variable ScaleFactor to 0.5 (50%) whenever u call any of the values in your look up table you multiply it by Scalefactor (CAVEATE!! THIS MIGHT TAKE A LITTLE OF YOUR PROCESSOR SPEED.) Then pass the result to the pwm register.

e.g
sinval = Sinetable[index] x scalefactor

pwmRegister = sinval

Klaust said ramp it up to 100 % in 20 AC cycles i hope i am right. so you have another variable as counter maybe CycleCount that will count the number of cycles. so for every increase in cycle count, you increase ScaleFactor by maybe 0.25 depending until you get ScaleFator to equal to 1(one), which will now multiply your default sinvalues and then it will be 100%.

i think this can also be termed as soft starting. and if u are smart with you can use to implement an output voltage regulation on your inverter.

Hope it helps and i diddnt get to confuse you the more.

Tunde
 

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