12 V DC to a 220V AC Inverter AMplfier Design

[audioguru2]

Very good trick AU. You are genious. From where u learned it?

I read the instructions. ;D

 

[kachew]

dear audiouru,

Sketch the circuit of a modified sine-wave inverter and I will check it for you. Don't forget the transformer must have an output voltage about 1.414 times higher than the transformer in a square-wave inverter, to create the peak voltage of the sine-wave. Therefore instead of a 230V to 24V center-tapped (12V-0-12V) transformer you need a 230V to 17V center-tapped (8.5V-0-8.5V) transformer

may i know wat is the power rating for the transfomer???is it usin the equation p=vi? is so in order to get 150w at output 220v then we need 0.68A and at the input then we must have at least 12.5v?

how about i add LM358 amplifier to increase the output current from the CD4047 in order to increase the efficiency?
At only 50Hz or 60Hz, I don't think the switching speed makes any difference. If it was 50kHz or 500kHz then a lot of current is needed to charge and discharge the high gate capacitance of the Mosfets for them to switch quickly so they don't get hot. An LM358 is about the slowest opamp that is made, even slower than an old 741 opamp. It would be lousy for driving Mosfets.

tought u said this to me before??? why call me to use the LM358 again??? or do you have any other suggestion???

The 100k resistors in series with the capacitors at the transformer have way too high a value to do anything.

Au arent the resistor used for the current equalising and as pull down resistor?? how about compare it with sasi inverter using the mosfet??
thanks again...

for the mosfet...the low turn on voltage means that the resistance inside the mosfet must be low enough to turn on the mosfet ??? by equation V=IR???
thanks for your reply.... ;D ;D

 

[audioguru2]

may i know wat is the power rating for the transfomer???is it usin the equation p=vi? is so in order to get 150w at output 220v then we need 0.68A and at the input then we must have at least 12.5v?

A 150W inverter needs a 150W or higher transformer. The input must be calculated from the battery voltage (13.2V?) minus the loss from the Mosfets then multiplied by the root of two to get the peak voltage of the modified sine-wave for each half of the winding.

why call me to use the LM358 again??? or do you have any other suggestion???

The LM358 is slow but is much quicker than a 33k resistor trying to charge and discharge the high gate capacitance of the Mosfets. I think the outputs of the CD4047 can directly drive the gates directly much quicker. The outputs of the CD4047 are like a series 1k resistor.

how about compare it with sasi inverter using the mosfet??

Sasi's inverter uses a Mosfet driver IC that supplies very high currents to charge and discharge the high gate capacitance of the Mosfets quickly, through 47 ohm resistors.

for the mosfet...the low turn on voltage means that the resistance inside the mosfet must be low enough to turn on the mosfet ??? by equation V=IR???

I mean and should have said to select Mosfets with a very low on-resistance. Then they would have a very low voltage drop and therefore won't heat much.

 

[kachew]

dear audioguru,

1.can i still use the transformer u recommended earlier ???(8.5-0-8.5)
2. so for my circuit do i need to add the LM358 or straight use the CD4047??
3. in font of the power mosfet i still need to put 1k resistor? if i use low on resistance???

 

[kachew]

dear audioguru,

as u can see in the sasi inverter circuit it also uses CD 4047 but there sasi uses 4.7k not 47ohms
and may i know what is the use of the capacitor and resistor near the transformer ?? thanks

View attachment 39562

 

[audioguru2]

1.can i still use the transformer u recommended earlier ???(8.5-0-8.5)

Are you making a square-wave inverter or a modified square-wave inverter?
Its voltage also depends on how much is the voltage loss of your Mosfets and what transformer voltage is available.
For a modified sine-wave inverter, 8V-0-8V to 10V-0-10V will be fine.
For a square-wave inverter, maybe 10V-0-10V to 12V-0-12V will be fine.
These simple inverters don't have a regulated output voltage.

2. so for my circuit do i need to add the LM358 or straight use the CD4047??

I don't think the LM358 is required. Connect the outputs of the CD4047 directly to the Mosfets.

3. in front of the power mosfet i still need to put 1k resistor? if i use low on resistance???

Use 1k or connect directly. It doesn't make much difference.

as u can see in the sasi inverter circuit it also uses CD 4047 but there sasi uses 4.7k not 47ohms

His circuit has 4.7k resistors feeding an LM358. The resistors and the opamps are not required.

what is the use of the capacitor and resistor near the transformer ??

They absorb voltage spikes.

 

[kachew]

dear audioguru,

if i wanted to make and modified sinewave inverter with a single mosfet as you recommended which power mosfet can u recommen me? cos i weak in looking the datasheet...so audioguru can u teach me how to calculate the loss in the mosfet so i would know the power rating used for this modified sinewave ....thanks again...

 

[kachew]

this the circuit ...thanks for your every help audioguru ....really appreciate it

 

[kachew]

this the circuit ...thanks for your every help audioguru ....really appreciate it

View attachment 39567

 

[audioguru2]

Hi Kachew,
Your circuit has the outputs of the unused gates shorted to ground. You have the inputs grounded which is good, but then their outputs will try to go high. Leave the unused outputs disconnected.

With the 6 Mosfets, the max continuous output power will be about 600W or more. The voltage will be a little high with a light load and a little low with full load.

View attachment 39568

 

[kachew]

dear audioguru,
so the rest of the part is ok already? for the ouput in order to get lesser like 200W i shall reduce the mosfet or change the transformer value???

 

[audioguru2]

Use 2 Mosfets and a 200W transformer for a 200W inverter.
The IRF540 Mosfet is cheap and common. It can conduct a peak current of 33A. Therefore its RMS max current is 23A. It has a max on-resistance of 44 milliohms which produces a voltage drop of 1.0V. The 10V-0-10V transformer gets a loss of 1V in the Mosfets so the circuit's voltage is 9V RMS times the 23A= 207W.
Modern Mosfets with a lower on-resistance and a more powerful transformer can make a more powerful inverter.

 

[kachew]

dear audioguru,

thanks again for the help and giving so much information to me.. ;D ;D

 

[Morro]

hi Audioguru and kachew,

is the inverter circuit works?? have anyone tried before?? what is the usage of the 0.01/5w parallel resistor used for?? and the diode 1N5408??? thanks for reply

 

[audioguru2]

Hi Morro,
I don't think the 0.01 ohm resistors are required.
The diodes are also not needed because power Mosfets have the diode built-in them.

 

[Morro]

hi audioguru,

the diodes and the resistor not used to absorb voltage spikes?
thanks for reply.. ;D

 

[audioguru2]

Hi Morro,
Four 0.01 ohm resistors in parallel like that is 0.0025 ohms, which is the same as a short piece of wire. It does nothing in the circuit.
All power Mosfets have the diode inside as a part of how they are made. In fact, it is a high-power zener diode which protects the Mosfet from d-s voltage spikes.

View attachment 39573

 

[Morro]

dear audioguru,

the oscillator CD 4047 and the CD4001 really can make it to become modified sine wave? i think in order to make modified sine wave i think we should change from NOR and inside the cd 4001 to XOR gate?
thanks for your reply...

 

[audioguru2]

the oscillator CD 4047 and the CD4001 really can make it to become modified sine wave? i think in order to make modified sine wave i think we should change from NOR and inside the cd 4001 to XOR gate?

I think NAND gates or NOR gates will work. I didn't look at XOR.
Here are waveforms with NAND gates. NOR would make similar waveforms except the inputs go low to produce a high output.View attachment 39597

 

[Morro]

hi audioguru,

how about the efficiency of thr circuit? it seems having efficiency only 50%, if so is there any method to improve the output so that having more steps? thanks for reply... ;D ;D