Circuit plans for a DC AC Inverter? Output 10 KVolts

[Reloadron]

It's hardly a pure sine wave. It's also really inefficient. Plus it's nowhere near 10KV. Why are you building it?

I think to get the idea.

 

[mneary]

With RA = 0, the DIS pin of the 555 is trying to discharge the entire battery. This will take the entire lifetime of the 555 (which won't be very long). You need a different circuit to get 50% duty cycle from a 555.

When you step up voltage in an ordinary transformer you step the current down. Your 12V 300mA produces 230V at only 16mA. If you hook a 12V 300mA light to it, 16mA is not enough to light it.

If your transformer is over unity then maybe it produces 230V 300mA from 12V 300mA. But I think you need a special meter.

 

[Sparky_s]

It's hardly a pure sine wave. It's also really inefficient. Plus it's nowhere near 10KV. Why are you building it?

I'm just trying to prove some tests I have seen in the net. Since I'm very skeptic about strange experiments, I have seen one that 4 or 5 guys have succesfully replicated. So I want to see what happens with that circuit. It's very simple.

The reason I couldn't open it at work was because of where it was hosted. Our Internet is a partially secured network. Anyway, no problem.

OK in your diagram I see a BD138 transistor configured as a NPN while in fact the BD138 data sheet shows me a PNP transistor as seen here. If you had a NPN as drawn it would work. If I have some time this weekend I may try to breadboard the circuit.

Ron

Thanks very much Ron. A friend told me that I needed a NPN transistor, and as you say, the bd138 is the one he told me that I needed to use, and the bd138 is a PNP. I think it could be other of the problems. What transistor I should use?

I won't ask again to that friend. I have learnt the lection.

With RA = 0, the DIS pin of the 555 is trying to discharge the entire battery. This will take the entire lifetime of the 555 (which won't be very long). You need a different circuit to get 50% duty cycle from a 555.

When you step up voltage in an ordinary transformer you step the current down. Your 12V 300mA produces 230V at only 16mA. If you hook a 12V 300mA light to it, 16mA is not enough to light it.

If your transformer is over unity then maybe it produces 230V 300mA from 12V 300mA. But I think you need a special meter.

Thanks, I removed the RA because when I calculated the values in a website program, the program showed a RA = 0 ohms, so I decided to remove the RA. I know it could be other of the reasons, so I need to find other circuit to get 50-60 HZ Square Wave DC.

The first step I want to do is to get the 220VAC of the output of the transformer. Then, I will look for other things. But right now, that is the idea.

 

[mneary]

Not all people who write web pages for 555 understand the chip. They design calculators that happily tell you to build circuits that blow up. The good news for you is that a 45% duty cycle should be good enough for this circuit, and the simple calculators provide a reasonable value for RA.

To test your 230V output, you must have a tester that draws no more than 16mA. It could be a bulb rated at 240V, 3 watts or less. Or a pair of 120V, 1.5W bulbs in series. Try it first with just your voltmeter. (Remember that 240V can harm you.)

 

[audioguru]

The 555 will not oscillate in your circuit without Ra becase pin 7 is open-collector and needs a resistor to pull it high.

Don't use a calculator program that is wrong. Instead, use simple arithmatic to calculate the resistor and capacitor values. the simple formula is on the datasheet and there is a graph that also parts values.

Don't post schematics at ImageShack or Imagen who take all day to wake up and the schematics will soon disappear. Instead, post the schematics here attached to your reply.

 

[Dvinchy]

What about my circuit? is wrong?

It is no idea to connect the capacitors as well as there is AC voltage into output in the scheme. There is DC-pulsed current into the output so the capacitor will probably have been completly charged in a short while after starting it means stop-current through the load.

 

[Dvinchy]

It was response to 26

 

[blueroomelectronics]

I'm just trying to prove some tests I have seen in the net. Since I'm very skeptic about strange experiments, I have seen one that 4 or 5 guys have succesfully replicated. So I want to see what happens with that circuit. It's very simple.

Would be helpful to post links.