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audioguru
Nobel Prize Winner
Canada
4218 Posts |
 Posted - Nov 04 2009 : 9:25:30 PM
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An inverter that puts out low amps is useless.
I think it wastes most of its battery power.
It is easy to modofy the circuit so that the transistors do not have avalanche breakdown. But then the base current for the transistors will still be too low. |
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Cokaric
New Member
4 Posts |
Posted - Dec 25 2009 : 06:39:21 AM
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I am building ur Inverter base on this...
http://electronics-diy.com/electronic_schematic.php?id=605
I have Googled for last hour or two and finally found ur original schematic... My English lng is poor so I mostly cant understand English electronic terminology.
What I want >>
I want to build ur Inverter, I need to power my computer with it when there is no power, my professor says, with 100 Amph Accumulator it will power my computer for about 8 h, in theory 40 h, but then after I can trow battery.
What battery should I use, what transformer should I use. My grid is ~230V/50Hz. How to modify this schematic to build a 1 kW Inverter. I have 500 W power supply in my computer, what would happen if I connect my computer on 1kW Inverter...
Please can u give me ur email or some way I can contact u... What lng except English do u speak ?
Best regards,
Cokaric |
The more I C, the less I see |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Dec 25 2009 : 10:36:21 AM
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My inverter circuit was originally designed in the Orient for 3rd world countries. It didn't work so I fixed it. It is very simple so it produces a square-wave output that many electronic products do not use, and it does not have voltage regulation so its output voltage is too high with a light load or fresh battery charge and its output voltage is too low with a full load and when the battery has run down.
My 500W computer does not draw 500W all the time, only when it is crunching many numbers. Usually its power is about only 50W.
I speak only English. |
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weqwedas
Apprentece
8 Posts |
Posted - Dec 26 2009 : 01:27:32 AM
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Well seeing as how you're still answering questions, I might as well ask a few of mine. I have been attempting to build an inverter to understand the concept for quite awhile now. I attempted to use both a dual-transistor flip-flop as well as the CD4047 circuit (looks almost exactly like the new light duty inverter circuit you have listed here, but with no amplifiers and a 5.1 volt Zener diode across the CD4047). Both times, however resulted in the flip-flop freezing up and dumping as much current as the 9 volt battery could through one transformer as well as the transistor that was in series with it. It was only after using two 9 volt batteries (one to power the CD4047, one to power the transformers) that I was finally able to get it to work. I noticed that because all of the current was going through the transformer, the CD4047 wasn't able to function correctly (at least that is what I can observe). Finally to the questions:
1. Could this possibly be due to the fact that I am using a 9 volt battery which cannot provide nearly enough current as a car battery?
2. Would adding the two amplifiers make the circuit more efficient or solve the previous problem? What are they there for exactly in your circuit?
3. Is it possible to run it off of a single voltage source without having the CD4047 deprived of amperage?
That should be all for now. If you need clarification on anything, I'll read it later on today. |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Dec 26 2009 : 11:18:38 AM
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quote:
Originally posted by weqwedas
I attempted to use both a dual-transistor flip-flop as well as the CD4047 circuit (looks almost exactly like the new light duty inverter circuit you have listed here, but with no amplifiers and a 5.1 volt Zener diode across the CD4047).
Both times, however resulted in the flip-flop freezing up and dumping as much current as the 9 volt battery could through one transformer as well as the transistor that was in series with it.
Post your schematic here so I don't need to guess about it.
An inverter output is hundreds of Watts. A little 9V battery can supply only 2.2W for about 10 minutes.
Download Attachment:  9V alkaline, 250mA load.PNG
8.28 KB
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weqwedas
Apprentece
8 Posts |
Posted - Dec 26 2009 : 3:23:20 PM
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This one was the original which constantly locked up.
This one is the one that works. The concept was not really meant to power anything, just charge a capacitor. The 45 ohm resistors could probably be moved to the transformer battery side without any complications. It was modeled after the disposable camera circuits in an attempt to charge the capacitor faster.
(Edit) Hmm I just noticed that the zener there is not actually doing anything in that circuit. There used to be a 100 ohm resistor in between B2 and D1. I'll have to add that back in. |
Edited by - weqwedas on Dec 26 2009 5:07:40 PM |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Dec 26 2009 : 8:08:28 PM
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It is not a mains inverter. It makes a high voltage for a zapper or something. the transistors are not flip-flops. The first little ones are emitter-followers driving the larger ones that are common-emitter switches.
The circuit with only one battery has the battery being killed by the high current in the TIP31 transistors that stops the CD4047 oscillator.
The circuit does not need the zener diode since the max allowed supply for the CD4047 is 18V. The CD4047 has a typical output current of only 4ma when its supply is 5.1V. Its output current is 22mA when its supply is 12V.
Since a little 9V battery is very weak then the supply should have a parallel supply bypass capacitor of about 100uF. |
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Cokaric
New Member
4 Posts |
Posted - Dec 26 2009 : 8:21:51 PM
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@audioguru
I was reading from page 1 so I dint saw schematic on page 54, http://www.aaroncake.net/forum/uploaded/Audioguru/20097292438_500W%20inverter%20final.PNG
Can u tell me what is F and what is SW1, I really don't know, I suppose that SW1 is switch and F relay but I am not sure. SW1 is switch does it need to be 60 A. Or maybe is SW1 relay and F fuse? What should be cross section of conductors. What should be cross section on PCB? What battery would u advise me to use?
Did u applied any other changes in ur inverter, did u improved anything and can u post here, please, if u did?
What do u think about this inverter, is MOS-Fet better or urs?
http://electronics-diy.com/electronic_schematic.php?id=609
My professor is sure that urs will work but for this one he said that no1 in our school build it before so he doesn't know if it would work.
I will continue posting in this thread thoug next few weeks, hope I wont disturb u much... Thank u for ur support =)
Best Regards,
Cokaric |
The more I C, the less I see |
Edited by - Cokaric on Dec 26 2009 8:24:56 PM |
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weqwedas
Apprentece
8 Posts |
Posted - Dec 26 2009 : 8:42:26 PM
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quote:
Originally posted by audioguru
It is not a mains inverter. It makes a high voltage for a zapper or something. the transistors are not flip-flops. The first little ones are emitter-followers driving the larger ones that are common-emitter switches.
The circuit with only one battery has the battery being killed by the high current in the TIP31 transistors that stops the CD4047 oscillator.
The circuit does not need the zener diode since the max allowed supply for the CD4047 is 18V. The CD4047 has a typical output current of only 4ma when its supply is 5.1V. Its output current is 22mA when its supply is 12V.
Since a little 9V battery is very weak then the supply should have a parallel supply bypass capacitor of about 100uF.
The flip-flop circuit I was talking about was superseded by the CD4047 chip. What I'm asking is what makes your circuit work where as my circuit freezes? Does the amount of current a car battery can provide mean the difference between oscillating and locking up? |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Dec 27 2009 : 12:59:25 AM
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I have never played with the transformers from a camera's flash so I don't know how much current the transistors will use driving them.
I do know that the current from a little 9V battery is low so that each time the transistors conduct the battery voltage will drop maybe low enough to stop the CD4047 oscillator.
Maybe the frequency of the CD4047 is too low for the little transformers.
A camera's flash unit doesn't use a weak little 9V battery but uses two powerful AA cells instead. Also, the flash unit's transformer is designed for a 3V power supply, not 9V that will increase the current that is used.
In my 500W inverter, SW1 is the on-off switch and F is a 60A fuse to stop a fire if something shorts.
The car battery can supply 500A before its voltage drops and look at the difference in its size and weight compared to a little 9V battery.
Gary's inverter also has a simple square-wave output but it is too complicated. I would use a CD4047 to drive the Mosfets. |
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wasssup1990
Nobel Prize Winner
A Land Down Under
2261 Posts |
Posted - Dec 27 2009 : 02:08:45 AM
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I remember playing with High Voltage siren flashes and camera flashes when I was just a few years old. Going by the sounds they made to charge up those capacitors I would have to say that the camera flash transformer spent most of its time above 2KHz and the siren flasher about 10KHz. I can understand why the siren flasher operated at a higher frequency. The flash interval was short and the transformer was small meaning that the energizing period would have to be shorter since the small transformer coil couldn't store too much energy in a single cycle from its drive circuitry. If the transformer coil was larger (could store more energy) then the drive circuitry could operate at a lower frequency. If the frequency is to low it will be audibly annoying, if it is too high you will need more expensive transistors that have shorter rise and fall times to be efficient but you would be able to flash the Xenon gas tube more frequently.
I hope this becomes useful to someone. |
When one person suffers from a delusion it is called insanity.
When many people suffer from a delusion it is called religion. |
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weqwedas
Apprentece
8 Posts |
Posted - Dec 28 2009 : 10:29:58 PM
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(Borrowed from http://www.repairfaq.org/sam)
This inverter, based off of the disposable flash camera, worked incredibly well, charging the capacitor to 300 volts in only a few seconds. It used the same transformer as my last circuit and only one 9 volt battery with the only different component being the 2sd2504 transistor used for its high gain. Only problem with this circuit in particular is it's DC only, which means you'd have to have a DC-AC circuit in order to use it for AC purposes. |
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wasssup1990
Nobel Prize Winner
A Land Down Under
2261 Posts |
Posted - Dec 28 2009 : 11:55:02 PM
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Yep I did a very similar experiment many years ago. I was able to use a camera flash circuit to power some small home appliances off a rechargeable cell phone battery. I enclosed it all in a self made wooden box and used it for experimentation. It has the potential to stop your heart and cause severe burns if you are caught between the capacitor terminals which is what you should be aware of. You should keep it away from people who are ignorant in this field. They could pick up your circuit and wonder "oh what is this" and then BANG. I have been shocked many times by all kinds of power supplies and I'm still here, but just be carefull.  |
When one person suffers from a delusion it is called insanity.
When many people suffer from a delusion it is called religion. |
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Ibrahim
New Member
Nigeria
2 Posts |
Posted - Jan 04 2010 : 03:55:23 AM
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| please i need sinewave inverter schematic circuit daigram |
ibra |
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audioguru
Nobel Prize Winner
Canada
4218 Posts |
Posted - Jan 04 2010 : 11:45:40 AM
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| A sinewave inverter is fairly complicated. You probably cannot obtain the modern parts it needs in Africa. |
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12/120V inverter again
