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ester
Apprentece

Italy
18 Posts
Posted - Feb 26 2011 :  04:08:56 AM  Show Profile  Reply with Quote
hi.I want to build a circuit that inverter 9 v dc to 2000 v ac with 2wat power and frequency 1500 hz with IC 555

audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Feb 26 2011 :  10:28:32 AM  Show Profile  Reply with Quote
The max output current of a 555 is 200mA peak. With a 9V suppy its output voltage is about 3.5V peak.
Then its square-wave output power is only 200mA x 3.5V= 0.7W. So it must use a power amplifier to drive your stepup transformer. Then yje output power is 2 Watts.

The circuit is very simple.
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ester
Apprentece

Italy
18 Posts
Posted - Feb 27 2011 :  02:47:02 AM  Show Profile  Reply with Quote
tanks alot for message.but can you send me circuit ?
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Feb 27 2011 :  08:49:55 AM  Show Profile  Reply with Quote
You are the person who needs a circuit, not me. So buy one, find one or design the circuit yourself. I will help you if you have trouble with your design. The biggest problem is finding a suitable transformer.

What are you powering with 2000VAC at 2W?
What waveform? (Square-wave or sine-wave)
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ester
Apprentece

Italy
18 Posts
Posted - Feb 27 2011 :  09:09:45 AM  Show Profile  Reply with Quote
i have the total form of circuit but bigest problem is when i draw circuit in pspice softwave and change the value of eleman but give not correct response. i send circuit can u help me & say me what is the problem?

Download Attachment: DC-to-AC-with-IC.gif
8.17 KB
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ester
Apprentece

Italy
18 Posts
Posted - Feb 27 2011 :  09:12:14 AM  Show Profile  Reply with Quote
i forget ,the waveform is not important.i should build the suitable transformer.
Edited by - ester on Feb 27 2011 09:14:42 AM
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Feb 27 2011 :  11:36:45 AM  Show Profile  Reply with Quote
The circuit you found is garbage. It has too much voltage loss.
With a 9V supply, the output level of the 555 is about 3V peak and the output of the emitter-followers is about 2V peak. The inductor L1 feeding the transformer reduces the output level to 0.5V peak or less.

Why use a little 9V battery? The circuit will get a little warm so for an output of 2W the battery must supply about 2.2W if your transformer is perfect. Then the current from the battery is 2.2W/9V= 244mA and a little 9V battery voltage will drop to 7V in about 20 minutes.

Download Attachment: 9V alkaline, 250mA load.PNG
8.29 KB
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ester
Apprentece

Italy
18 Posts
Posted - Feb 27 2011 :  4:36:14 PM  Show Profile  Reply with Quote
why garbage???????????????????????i use a little 9V battery because it is my project that my teacher determined.
Edited by - ester on Feb 27 2011 4:44:32 PM
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Feb 27 2011 :  5:48:23 PM  Show Profile  Reply with Quote
quote:
Originally posted by ester

why garbage???????????????????????i use a little 9V battery because it is my project that my teacher determined.

It is a garbage project because it has a high voltage loss as I explained.
I also showed that a little 9V battery is 9V only for a couple of minutes then its voltage quickly drops.
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ester
Apprentece

Italy
18 Posts
Posted - Mar 04 2011 :  09:02:13 AM  Show Profile  Reply with Quote
can u say me how i can eleminate high voltage loss ?
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audioguru
Nobel Prize Winner

Canada
4218 Posts
Posted - Mar 04 2011 :  12:34:46 PM  Show Profile  Reply with Quote
quote:
Originally posted by ester

can u say me how i can eleminate high voltage loss ?

Replace the 555 with a CD4047 oscillator IC that has no voltage loss.
Replace the transistors with Mosfets that have a very low voltage loss.
Use a center-tapped transformer driven from a push-pull circuit.
Eliminate the coupling capacitor C4 and inductor L1.

Download Attachment: Mosfets square-wave inverter.PNG
87.47 KB
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ester
Apprentece

Italy
18 Posts
Posted - Mar 05 2011 :  11:03:08 AM  Show Profile  Reply with Quote
tanks alot.
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