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9 Posts

Posted - Sep 27 2004 :  7:21:45 PM  Show Profile  Reply with Quote
It was my understanding they were using a 12 Vac, center tapped primary. This would then be a 6-0-6. It sounds to me like this is where somes problems were by only obtaining 65 Vac. For a 6-0-6, you would need the 240 Vac secondary to get 120 Vac output.

The terms primary and secondary are reversed here so as not to confuse anyone. The actual transformer will have a 12-0-12 (24 Vac, C.T.) secondary and a 120 Vac primary.

Will

quote:

Hi Will,
If the inverter works properly (I don't know which circuit that you are talking about), the 24V winding's center tap is connected to +12V, so when a transistor grounds one side then the other side will swing to +24V. Select a 24V center tapped transformer that has a high voltage winding that is rated for whatever output voltage that you need, not double what you need.
To get 120VAC output, choose a 120V to 24V ceter-tapped transformer, which is also written as 120V to 12-0-12.





-=Will Matney=-



Edited by - Will on Sep 27 2004 7:36:45 PM

Last time I'll edit, I just cant spell this evening!

Edited by - Will on Sep 27 2004 7:38:18 PM

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Edited by - Aaron Cake on Nov 25 2005 09:12:35 AM
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 27 2004 :  9:26:55 PM  Show Profile  Reply with Quote
Hi Will,
Didn't you read my analysis of this website's 12V to 120V Inverter project that I posted here in May?
Are you talking about people getting an output of only 65VAC into a 75W load?
This simple circuit cannot give much output because the transistors do not have enough current gain and are being forced into avalanch breakdown of their reverse-biased base-emitter junctions.

To provide 75 Watts, the 13.5V battery must supply a current of 5.6A. Each transistor in the inverter must switch this 5.6A through its side of the transformer winding. At 5.6A of collector current, a 2N3055 transistor has a typical current gain of 25, so its base current will be 224mA. But the 180 ohm base drive resistors can supply only 68mA!
With 68mA of base current, the collector current of a typical 2N3055 transistor is only about 3A. So the typical output power is only 40.5W, much of which is used to heat the capacitors and to cause avalanch breakdown of the transistors.
This circuit doesn't work!

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n/a
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9 Posts

Posted - Sep 27 2004 :  10:15:23 PM  Show Profile  Reply with Quote
I hadn't went to look and see what the gain was on the 2N3055. No, I didn't read your post as I just seen the one having problems with a 65 vac output.

If 224 mA is needed then change the resistors to 47 ohms and raise the capacitance to 220 uF. That will give 255 mA with a 67 Hz frequency. Don't see why it won't work if the base current is large enough and as long as the frequency is 60 Hz +. If the voltage rating on the capacitors is large enough, they shouldn't fail providing they're connected with the right polarity. The max on one should be 12 volts with an allowable 24 Vdc peak which is held by the zeners I mentioned. The series resistor-diodes are not needed then.

Will

quote:

Hi Will,
Didn't you read my analysis of this website's 12V to 120V Inverter project that I posted here in May?
Are you talking about people getting an output of only 65VAC into a 75W load?
This simple circuit cannot give much output because the transistors do not have enough current gain and are being forced into avalanch breakdown of their reverse-biased base-emitter junctions.

To provide 75 Watts, the 13.5V battery must supply a current of 5.6A. Each transistor in the inverter must switch this 5.6A through its side of the transformer winding. At 5.6A of collector current, a 2N3055 transistor has a typical current gain of 25, so its base current will be 224mA. But the 180 ohm base drive resistors can supply only 68mA!
With 68mA of base current, the collector current of a typical 2N3055 transistor is only about 3A. So the typical output power is only 40.5W, much of which is used to heat the capacitors and to cause avalanch breakdown of the transistors.
This circuit doesn't work!





-=Will Matney=-

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Edited by - Aaron Cake on Nov 25 2005 09:12:44 AM
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 28 2004 :  04:23:26 AM  Show Profile  Reply with Quote
Hi Will,
That is a good idea. If you use 47 ohm base resistors then this inverter that uses typical 2N3055 transistors should have an output power of a whopping 75 Watts!
You can expand your theory further if you use only 6 ohms for the resistors so that the base current is 2A. Now there will be enough base current so that even 2N3055 transistors that have their minimum guaranteed gain of only 5 will have a collector current of 10A, and the inverter's output will be 120W.
Now that we are talking about 10A collector current, we must not forget that the transistors are part of the oscillator, and have capacitors that must charge and discharge into the transistors' bases. Using 6 ohm resistors, those capacitors will be huge. When the 1st transistor conducts, it turns off the 2nd transistor by its collector cap. When the 2nd transistor turns off, its collector voltage rises to about 27V by center-tapped transformer action. The cap that is connected from the 2nd transistor's collector to the 1st transistor's base must charge without any current-limiting and its current will exceed the 7A maximum base current rating of the 1st transistor for a moment. You could add 3.9 ohm resistors in series with the caps to limit the momentary base current to about 6.7A.

Do you understand about avalanch breakdown of a silicon transistor's reverse-biased base-emitter junction? All silicon transistors have a maximum reverse base-emitter voltage rating of about 7V, whtch if exceeded causes the junction to avalanch like a zener diode and conduct massive current. This current creates hot-spots on that junction that is not cooled well by the transistor's case since the other junction's collector is bonded to the case.
That is bad news for the transistor since a massive current flowing through an uncooled base-emitter junction that has about 7V across it creates a very high temperature at that junction. Poof!

What about the ripple-current rating of the capacitors? That is what caused them to blow-up (even with the correct polarity) in the original circuit. Using 6 ohm resistors, the charge and discharge currents of the caps will be huge. Poof!

So most of the power from this inverter will be used to destroy its transistors and capacitors, and very little power will remain for its load.
It might work a lot better and with improved reliability if the transistors are replaced by Mosfets.

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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 28 2004 :  05:45:04 AM  Show Profile  Reply with Quote
Hi again Will,
I forgot to explain how avalanch breakdown occurs in this circuit.
When the 1st transistor conducts, the capacitor at its collector has been charged to 27V, and the cap's end that is connected to the 2nd transistor's base will attempt to drive the base to about -26V. The cap is very quickly discharged with a very high current because it has a high-current collector of the 1st transistor driving its other end, and the avalanching base-emitter junction of the 2nd transistor is limiting its negative end. So lots of current through the cap and lots through the junction. Poof! Poof!
The other cap and junction have the same problem. Poof! Poof!
This simple inverter is just a poof-maker and is good only for self-destruction!

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n/a
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9 Posts

Posted - Sep 28 2004 :  10:01:29 AM  Show Profile  Reply with Quote
quote:

Hi Will,
That is a good idea. If you use 47 ohm base resistors then this inverter that uses typical 2N3055 transistors should have an output power of a whopping 75 Watts!
You can expand your theory further if you use only 6 ohms for the resistors so that the base current is 2A. Now there will be enough base current so that even 2N3055 transistors that have their minimum guaranteed gain of only 5 will have a collector current of 10A, and the inverter's output will be 120W.

>> I don't think a 2N3055 will do 10 amperes at 12 Vdc. Without looking at the spec sheet, this min. current rating starts to drop off at 5 Vdc. The most I've ever seen them used at was 5 amps IC for 12 Vdc circuits.

Now that we are talking about 10A collector current, we must not forget that the transistors are part of the oscillator, and have capacitors that must charge and discharge into the transistors' bases. Using 6 ohm resistors, those capacitors will be huge.

>> Yes, your correct. The 10 amp rating for one 2N3055 I don't think will work. That Will in fact go up in smoke. Personally, I've never seen one used in any power circuits with a collector current of over 5 amperes or so. That's why I was holding it back. If I were to want 10 amperes IC then I'd use two in parallel.

When the 1st transistor conducts, it turns off the 2nd transistor by its collector cap. When the 2nd transistor turns off, its collector voltage rises to about 27V by center-tapped transformer action.

>> It will actually be higher than this at turn on because of the spike created from the sudden shift in the windings. Thus, the zener diode clamps across the collectors and emitters will throw this to ground and hold it at a reasonable amount. They would have to be at least 24 V + 1 or 2 volts.

The cap that is connected from the 2nd transistor's collector to the 1st transistor's base must charge without any current-limiting and its current will exceed the 7A maximum base current rating of the 1st transistor for a moment. You could add 3.9 ohm resistors in series with the caps to limit the momentary base current to about 6.7A.

>> Exactly. If the base current is not limited, it will run away and destroy the transistor. I'd prefer to use two 2N3055's at each side for this and limit each to 1/2 the current. Here would be a good place to implement that kick start thing I was mentioning. Use a 33 ohm on one base and say a 39 ohm on the other. The slight difference will cause the circuit to always oscillate and not latch up.

Do you understand about avalanch breakdown of a silicon transistor's reverse-biased base-emitter junction?

>>I reckon I do, if not, I flushed two years of school and 23 years experience down the toilet.

All silicon transistors have a maximum reverse base-emitter voltage rating of about 7V, whtch if exceeded causes the junction to avalanch like a zener diode and conduct massive current.

>>Yes it does, this has to be limited in the circuit.

This current creates hot-spots on that junction that is not cooled well by the transistor's case since the other junction's collector is bonded to the case.
That is bad news for the transistor since a massive current flowing through an uncooled base-emitter junction that has about 7V across it creates a very high temperature at that junction. Poof!

What about the ripple-current rating of the capacitors? That is what caused them to blow-up (even with the correct polarity) in the original circuit. Using 6 ohm resistors, the charge and discharge currents of the caps will be huge. Poof!

>>Use a large enough, and correct type of capacitor. When the voltage rating of an electrolytic is increased, so is it's current capacity. This is done by increasing the plates size and the coating thickness on the one plate (anode). Let's say for practical purposes, use a 220 uF @ 450 Vdc cap instead of a 220 uF @ 35 Vdc. The difference in physical size is huge. The larger cap will dissipate heat way better and would probably hold up I would think. I forget what the current carrying capacity is for that cap but it's a good amount. This type I'm speaking of is the radial screw top, computer types.

So most of the power from this inverter will be used to destroy its transistors and capacitors, and very little power will remain for its load.
It might work a lot better and with improved reliability if the transistors are replaced by Mosfets.

>> The Mosfets is a good idea because they're voltage dependant, not current dependant. The capacitance polarity and spikes still have to be corrected though as above.




Will

-=Will Matney=-

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Edited by - Aaron Cake on Nov 25 2005 09:12:55 AM
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n/a
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9 Posts

Posted - Sep 28 2004 :  10:09:55 AM  Show Profile  Reply with Quote
See the explanation in the first post.

The thing is, between you and I, we've about re-designed the entire circuit for the user which I didn't intend to do. I would rather them learn from it and we give help as needed. I'm afraid by about re-designing the whole thing here, some will not learn that much and just build what we tell them to.

There was one person on here who mentioned paralleling a bunch of capacitors in place of one. He was on the right track because the current rating was going up. However, I hope he knew this by learning from the experience and not us telling it.

Will

quote:

Hi again Will,
I forgot to explain how avalanch breakdown occurs in this circuit.
When the 1st transistor conducts, the capacitor at its collector has been charged to 27V, and the cap's end that is connected to the 2nd transistor's base will attempt to drive the base to about -26V. The cap is very quickly discharged with a very high current because it has a high-current collector of the 1st transistor driving its other end, and the avalanching base-emitter junction of the 2nd transistor is limiting its negative end. So lots of current through the cap and lots through the junction. Poof! Poof!
The other cap and junction have the same problem. Poof! Poof!
This simple inverter is just a poof-maker and is good only for self-destruction!





-=Will Matney=-

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Edited by - Aaron Cake on Nov 25 2005 09:12:57 AM
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 28 2004 :  11:12:32 AM  Show Profile  Reply with Quote
Hi Will,
Yeah, I have built unbalanced multivibrators too, to guarantee start-up.
But when used with a supply voltage more than 5V, I always add diodes in series with the emitters to prevent avalanching.
I don't think that limiting the current during avalanching is good enough, after all, the manufacturers just say, "Max reverse base-emitter voltage is 7V. Don't exceed it". In my book that means "Dont do dat!"

The users won't learn much when authors post projects like this one, then say to use more powerful trasistors and transformer to get more output. Only the guy who tried 300V/30A transistors and a rewound transformer from a microwave oven learned that it didn't make any difference.

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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Sep 28 2004 :  11:45:03 AM  Show Profile  Reply with Quote
BTW, Will,
The 500W inverter project that I fixed was reported to deliver 720W. So its output transistors must have had an emitter current of almost 15A.Since its driver transistos were connected in a darlinton arrangement, they probably had a collector current that exceeded that of the output transistors, and all currents went into the transformer. I would never push transistors so hard, therefore rated the circuit at only 500W.
The link to that schematic is posted again here:
http://www.electronics-lab.com/forum/attachments/500Watts_Inverter.gif

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9 Posts

Posted - Sep 28 2004 :  12:09:55 PM  Show Profile  Reply with Quote
quote:

Hi Will,
Yeah, I have built unbalanced multivibrators too, to guarantee start-up.
But when used with a supply voltage more than 5V, I always add diodes in series with the emitters to prevent avalanching.
I don't think that limiting the current during avalanching is good enough, after all, the manufacturers just say, "Max reverse base-emitter voltage is 7V. Don't exceed it". In my book that means "Dont do dat!"

>> Actually, the diodes off the emitters is a good idea and I forgot about that myself. Anyhting to keep the transistor from breaking down.

The users won't learn much when authors post projects like this one, then say to use more powerful trasistors and transformer to get more output. Only the guy who tried 300V/30A transistors and a rewound transformer from a microwave oven learned that it didn't make any difference.

>> The only way to get more output voltage wise is to increase the turns in the winding, i.e., a higher voltage secondary. However, the current will drop accordingly. Wattage going in has to equal wattage going out minus about 5% due to losses in the transformer itself. So if say 300 watts were created by the primary, It takes 315 watts from the secondary. The old saying goes here, you cant get something from nothing.

>>To me, it's best to have more than enough switching transistors/fet's to do the job and keep them running cool instead of pushing their limits. I think the FET idea is the best yet using healthy capacitors. However the current draw wont be there on the capacitors like on the bipolars.

>> Really, the only advantage to this circuit is to set an oscillating frequency. However, in a lot of invertors used to control a DC circuit on the other end, the frequency is raised. The higher the frequency, the more efficient the transformer is. Actually, the amount of iron needed drops as the freq. rises. For portable TV's, etc which use DC circuits anyways, it will work on higher frequencies. But, motors, etc. won't.

Will





-=Will Matney=-

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Edited by - Aaron Cake on Nov 25 2005 09:12:58 AM
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phoenix
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Canada
54 Posts

Posted - Oct 08 2004 :  10:43:00 PM  Show Profile  Visit phoenix's Homepage  Send phoenix a Yahoo! Message  Reply with Quote
Pardon me ...People, Which of the circuits really work?
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audioguru
Nobel Prize Winner

Canada
4214 Posts

Posted - Oct 08 2004 :  10:54:36 PM  Show Profile  Reply with Quote
quote:

The 500W inverter project that I fixed was reported to deliver 720W.
I would never push transistors so hard, therefore rated the circuit at only 500W.
The link to that schematic is posted again here:
http://www.electronics-lab.com/forum/attachments/500Watts_Inverter.gif





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2 Posts

Posted - Oct 12 2004 :  02:29:17 AM  Show Profile  Reply with Quote
Sorry for this silly question, but can anybody tell me how to obtain 230VAC from the 500W 220VAC circuit? Would simply changing the transformer help?
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Posted - Oct 12 2004 :  07:35:47 AM  Show Profile  Reply with Quote
My goodness, just doing a quick search for some inverter designs and I come across this... firstly a multivibrator running a transformer and then pushpull designs... Firstly, inefficient. Secondly, deadly! Where is your isolation?! Should there be a fault with your equipment or your transformer, you have high voltage AC on your 12V side. Anyone who touches your car's chassis will be electrocuted. Far out. Posting things like this should be illegal. You're letting novices try these prototypes and run the risk of getting them, and even yourself, killed. A good inverter has opto-isolation/feedback. You need this feedback to control the PWM incase of variations in the voltage. If anyone is attempting to build these, do not use them for permanent fixtures, and be very careful.

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audioguru
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Posted - Oct 12 2004 :  09:03:27 AM  Show Profile  Reply with Quote
Hi Pprab,
Why do you care about a voltage change of less than 5%? Your mains voltage at home or work changes much more than that.
The 500W inverter was designed to power flourescent lights, electrical tools and international TVs. The lights and tools don't care much about voltage and the TVs run on anything from about 95VAC to 260VAC.
Like most inverters, the 500W one is not regulated, so its output voltage changes a little with the amount of load and the amount of charge in its battery power source.

Hi Pas,
Children should be stopped by their parents from being near dangerous things.
How could anyone be electrocuted if the inverter's output shorted to the 12V side or chassis? Only if they were also holding the other wire of its output!
Maybe a toaster is powered from an inverter that has one of its output wires shorted to a car's chassis and a person was touching the chassis (after removing the paint) and put their hand in the toaster. Ouch! They would get electrocuted, fried and toasted!

Most inverters are cheap and work just fine. You are talking about using opto-feedback and PWM. Are you rich?


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