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# 12 V DC to a 220V AC Inverter AMplfier Design

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150W output plus 38W of heat in the inverter circuit = 188W total.
188W/12V= 15.7A from the 12V battery, into the low voltage winding of the transformer and is the average current in the Mosfets.
15.7A is the average continuous current, the peak momentary current is higher.

If you overload the transformer or the Mosfets then they will overheat and fail.

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well i think it isnt possible or if it is possible then the sytem will be unstable

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well i think it isnt possible or if it is possible then the sytem will be unstable

These are switching inverters that do not have a linear amplifier. They don't have negative feedback. They switch 12VAC at a very high current into the low voltage winding of a transformer and the transformer steps-up the voltage.

They work fine and are not unstable.
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hi Au,

Quote

The Mosfets are turned off for half of the time so their current must be doubled for the inverter to produce the same average output power as a square-wave inverter.

the IRF 540 mosfet only can withstand 23A only, then i need 3 mosfet on each side to withstand that large current or just simply 1 mosfet at each side cos 150W divided 12=15.7A???
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Hi Kachew,
If your inverter will have a max load of 150W then its total power including heating is 188W. The average current from the 12V battery is 188/12= 15.7A but the peak current is 2-root-2 times= 44.4A because it has a modified sine-wave with pauses.

The IRF540 Mosfet is pretty old and has a fairly low max current rating. It is 23A continuous if its case is at 100 degrees C and is a little more if it is cooler. Two of them on each side would be very close to their limit so use 3 on each side.

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hi Au,

can u recommen any better mosfet? thanks

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hi Au,

can u recommen any better mosfet? thanks

I don't use high current Mosfets. The online catalogs of Digikey and Newark have hundreds of them. Some can conduct over 100A. I don't know what you can buy.
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hi au,

can u pls gimme the link? i go and find myself :P thanks cos i need to create the inverter as small as i can

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www.digikey.com They are in some countries.
www.farnell.com and click on the flag of your country.

I look at Canadian and United States pages.

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dear au,

i think i can use this mosfet rite?

IRFZ48NPbF
i will attach the datasheet for your reference....thanks

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Use two of those Mosfets on each side for a 150W modified sine-wave inverter.

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dear Au,

on what gound u state that i shold use the 8.5-0-8.5 to 220V to get the Vpeak for the output of the transformer? as i know output of the transformer is Vrms?
u stated that in order to get Vpeak at the output , the value for the inverter the input of the transformer winding is calculated as 12/8.5 which is nearly square root 2...do u have any site or article to prove this?
the vpeak at the output transfomer become Vpeak as sine wave if i use the 8.5-0-8.5V to 220V transformer? do u have any nice article about transformer?

If your inverter will have a max load of 150W then its total power including heating is 188W. The average current from the 12V battery is 188/12= 15.7A but the peak current is 2-root-2 times= 44.4A because it has a modified sine-wave with pauses.

do u have any site or article on this? so i can understand more...because i dont know which formula you are using to calculate the current in mosfet....which is peak current is 2-root-2 times 15.7A=44.4A because there is pauses between the mosfet .  as i know Ipeak =square root 2 multiply by Vrms.....by there is extra 2 in front?

my theory on inverter is very limited hope you can giv me more information about it thanks

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dear Au ,

i think i get what u mean if this is correct,

the input and output of the transfomer is calculated using Vrms or so call average value?
so in order to get the Vrms in the transfomer we need to 12V/square root 2 = 8.5V in the primary winding of the transfomer? so we can get the output of the transformer Vpeak value as 220Vac peak straight? or the output of the transformer still need to multiply by square root 2 since the output of the transformer is just the Vrms value not Vpeak?

for the current part in the mosfet.....what u means is the mosfet in the inverter on each side is 15.7A drain from the battery so with the modified sine wave because of the mosfet turned off half of the time we need to double the current in the mosfet in order to get the average output power which is 150w continuos
so the formula given to me which is 2 root 2 times 15.7A where the 2 in front is to double the current in the mosfet in order to get same avg output power....is this correct?

but how we know by doing this we can get the same average output? can u explain more and also  why their current need to be double to produce same avg output power for the turn  off for half of  time in the mosfet ....

but i also need some article for my above question which i need to prove it in my thesis.....so my thesis is more solid...
thanks Au

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Hi Kachew,
1) A square-wave inverter has the 12V from the battery switching back and forth in the primary of the transformer. The battery current is nearly continuous and is steady. Its output voltage has a peak value the same as its average value which is the same as the RMS voltage if it is a sine-wave.
2) A sine-wave inverter has the 12V from the battery stepped up to an RMS voltage. Its peak voltage is 1.414 times higher (root 2).
3) A modified sine-wave inverter switches the primary of the transformer on for one side then off, then switches on the other side of the transformer then off. Its current is turned off for half the time so its peak current must be doubled to achieve its rated RMS power output.
Its peak voltage must be about 1.414 times the RMS voltage.

The sine-wave inverter and modified sine-wave inverter must use a transformer with a higher step-up ratio than a square-wav inverter. The ratio of 12V to 8.5V= 1.412 which is close enough to the desired 1.414 (root 2).

I saw articles in Google for modified sine-wave inverters. I didn't save the articles but instead I saved this pic of the waveforms in 120VAC inverters:

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• 2 weeks later...

hi au,

i wan to ask u something....do the copper in the pcb board must have a thicker width? since the mosfet drain a quite high current... is about 16 amps....or double for modified sine wave? if it have thinner width will it affect the efficiency of the inverter?
thanks again

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If the copper traces on your pcb are too narrow for the high current then they will burn out like a fuse.

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dear Au,

i have contruct the circuit and the waveform the oscilloscope is not the expected output i wanted. the output is not a modified square wave instead of square wave only. i attach the waveform for u to view....
can u pls give some opinion on this?? thanks

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dear au,

1 more question....if the dc power supply used to power the inverter circuit which have maximum 3amperes....can it drive the transformer which have the rating of 8.5-0-8.5V?
thanks

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Also look what I described in this post
http://www.electronics-lab.com/forum/index.php?topic=10858.0

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hi an920,

i dont understand the post ... ??? ???

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Hello AN920. Your square wave inverter circuit produces a clock and a clock not? Why does it have so many gates and how does the flip flop help. A flip flop has an extra gate of time delay, which would conflict with that parallel gate.

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hi an920,

i dont understand the post ... ??? ???

It shows that you can have much better control over the output that a fixed duty cycle drive.
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Hello AN920. Your square wave inverter circuit produces a clock and a clock not? Why does it have so many gates and how does the flip flop help. A flip flop has an extra gate of time delay, which would conflict with that parallel gate.

I think you are missing the whole point.
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Hi Kachew,
I looked back at 6 pages and I couldn't find the final schematic for your modified sine-wave inverter.The last one had the Mosfets incorrectly connected as source followers.

The CD4047 has a 100Hz clock output, a non-inverted 50Hz output and an inverted 50Hz output. When they are gated and feed the Mosfets then one Mosfet conducts then turns off for a pause, then the other Mosfet conducts then turns off for a pause.

A 3A transformer would allow only 24W of output which is not even enough for it to warm up a little.

Your 'scope shows that the CD4047 is not gated correctly.

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I just noticed an Edit that occured a while ago.
Another argument about an ordinary low power Cmos oscillator.

Texas Instruments says that the minimum resistor value should be 10k, not 1k.
The capacitor has AC across it so it should be non-polarized.

1k with 4.7uF is wrong.
47k with 0.1uF is correct for 50Hz and 39k with 0.1uF for 60Hz.

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