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12 V DC to a 220V AC Inverter AMplfier Design


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i just wan to know the use which pin to connect to the pin Q and inverted Q of cd 4047?

Sorry, I had the pin number of the clock for the CD4047 wrong. I fixed it.
Connect the pin 13 and pin 10 to one gate of a CD4011 which drives the 1st Mosfet. Connect the pin 13 and pin 11 to another gate of the CD4011 and it drives the 2nd Mosfet.

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.
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dear audioguru,

GUru @ master hehe ;D ;D then do u have any suggestion? and i saw u posting in the arron cake forum too....i can see u help lots ppl there... ;D ;D ;D good jobs haha...i will try to generate my circuit and post it soon....any defects there pls help me to detect... ;D ;D thanks

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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.

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Hi Kachew,
It is hard to see your schematic. Make it on the computer with Microsoft Paint or another program.

I didn't check your CD4047 circuit. It should be the same as the 500W square-wave inverter.

You show two CD4001 ICs when only one is needed. A single CD4001 has 4 gates inside.
One gate of the CD4001 needs one input from the pin 13 oscillator pin of the CD4047 and its other input from pin 10. The 2nd CD4001 gate also needs one input from the pin 13 oscillator pin of the CD4047 and its other input from pin 11. The two unused gates in the CD4001 IC should have their inputs grounded.

You have an LM393 dual comparator instead of an opamp. A comparator is not supposed to have negative feedback. It oscillates at a very high frequency when it has negative feedback like yours has. An opamp is designed to have negative feedback.
A comparator has a single transistor at its output with an "open collector" so it doesn't have anything to pull the signal positive.

You have 100k resistors at the outputs of the comparators. A resistor isn't needed.

You have 33k resistors feeding the gates of power Mosfets. Power Mosfets have an extremely high gate capacitance (12nF) which causes them to turn on in 33k x 12n= 396us plus ramping caused by Miller capacitance, which is slow enough to cause the Mosfets to overheat.

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

The circuit needs a single +12V power supply and ground. Yours shows a +12V and a -12V which is 24V.

If you select power Mosfets with a low turn on voltage then a single pair can drive 500W or more.

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dear audioguru.,
thanks for checkin the circuit for me... ;D ;D and so sorry for the unclear circuit coz i draw and use digicam to take the circuit i drawn...audioguru do u hav any recommendation for the software to draw the circuit...coz i dont have any software for that...and for the comparator part shall i change to the opamp?? which opamp is more suit for the inverter? for the power mosfet i will try to find a suitable 1 and redesign the circuit as u guide me...thanks

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dear audioguru,

is it the 396us is the operation cycle for the mosfet?? so its take to 396us to amplified the voltage in the 1st mosfet in the positive cycle and the same time to amplified the negative cycle in the 2nd mosfet...so is it u mean that the resistor in the gate mosfet the larger the better to provide sufficient time to have to cool down time for the 1st mosfet? or the smaller value for the resistor for fast switching??

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do u hav any recommendation for the software to draw the circuit...coz i dont have any software for that.

I use Microsoft Paint program. I make straight lines with the Shift key down and I copy and paste parts and pieces of other schematics and from datasheets.

for the comparator part shall i change to the opamp?? which opamp is more suit for the inverter?

Yes, use an opamp. The LM358 has outputs that go to ground.

is it the 396us is the operation cycle for the mosfet??

396us is the time for the very high value of the 33k resistor to charge or discharge the very high 12nF input capacitance of the Mosfet. You want the Mosfet to switch as fast as possible so it doesn't heat. A low resistance will charge the capacitance quicker, use 1k. 
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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
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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.
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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.
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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...

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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.

post-1706-14279143042194_thumb.png

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