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Please attach the datasheet for your transformer. Maybe its "0V" terminals are already connected together as a center-tap. Maybe the "0V" terminals are connected together and are connected to the transformer's frame.

You could easily measure with a multimeter how the transformer is made:
1) Disconnect the transformer from the circuit.
2) Use the ohm-meter to see if the "0V" terminals are connected together or to the frame.
3) Use AC volts and plug in the transformer to see if it has 24V with a center-tap.

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Hi, thank you....
the transformer i am using is a 0-15V(multi-tapped)@2A 30VA...According to the specification, it is a type 2156 transofrmer. I can't find the data sheet...But I'll copy the specification from the supplyer here:


0 - 15V (Multi-Tapped) @ 2A 30VA - Type 2156 Transformer


Specifications:
Primary voltage: 240VAC
Secondary Voltage: 6, 9, 12, 15V
Magnetising current: ~65mA
Total Power rating: 30VA
Temp rise above ambient at rated output: ~60

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A Jaycar 2156 transformer is a dinky little thing with a 30VA max rating.
It has a 240V winding and a single multi-tapped low voltage winding.

30VA is far too low for an inverter. 100VA is about the minimum. That is the max power of its load.
The low voltage winding is not center-tapped nor is it two windings like I thought.

You need a transformer with a 10V-0V-10V winding for a modified sine-wave output inverter, or 12V-0-12V for a square-wave output inverter. Some transformers have two identical 10V or 12V windings that could also be used.

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i have some theories about the shematic
in this shematic two op-amps have been used to amplify outputs.but we can use buffer gates insted of them to drive transistors/////// is that right?its so simpler

The buffers in a CD4050 will drive transistors as well as opamps, but is in a 16 pins case. Dual opamps cost about the same and are in an 8 pins case.
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oh yeah
tanx
and these are some other questions i could make
      -why are those 2 transistors(2sc1061and 2n3055) are used to drive 2n3055 trs- enough buffers with a resistor can directly drive 2n3055s.isnt that true?
        -whats the role of those 2A diodes in the circuit?are those anti shoke diods like the ones we put on relays to prevent the relays inductor shock to relay driver transistor?but we put those diods after the relay driver transistor to prevent the shoke to collector but at this inverter they are before 2n3055s trs????
so can we use two diods before the transformators inductor in the wrong posithion to prevent trans inductors to 2n3055 transistors?
          -whats the role of those 8 ./1ohm resistors between emitters?
         

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-why are those 2 transistors(2sc1061and 2n3055) are used to drive 2n3055 trs- enough buffers with a resistor can directly drive 2n3055s.isnt that true?

A 500W inverter uses 600W from 12V which is a current of 50A. The 4 output transistors on each side divide the 50A into 12.5A each. A 2N3055 has a minimum current gain of only 5 at a collector current of 10A and isn't even spec'd at 12.5A but then its minimum gain is probably only 4. Therefore each side's output transistors have a total max base current of 12.5A which is supplied by the single 2N3055 driver transistors on each side. Their max base current of 3.13A is supplied by the 2SC1061 transistors and their base current is supplied by the opamps.

-whats the role of those 2A diodes in the circuit?are those anti shoke diods like the ones we put on relays to prevent the relays inductor shock to relay driver transistor?but we put those diods after the relay driver transistor to prevent the shoke to collector but at this inverter they are before 2n3055s trs????
so can we use two diods before the transformators inductor in the wrong posithion to prevent trans inductors to 2n3055 transistors?

When the output transistors on one side turn off, the transformer makes a positive spike on this side. Since the transformer has a center-tap then the other side makes a negative spike. The negative spike causes the Collector-Base junctions of those output transistors to conduct and then the diode on that side from the bases to ground also conducts. Therefore the spike's voltage is clamped to only -2V on this side and to +2V on the 1st side.

-whats the role of those 8 ./1ohm resistors between emitters?

Each output transistor has a different amount of current gain. The transistor with the most current gain would hog all the current (50A) and therefore would burn out. The emitter resistors create negative feedback so that the transistors have nearly the same amount of current gain and therefore share the current equally.
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Hi....Now I know why the inverter didn't work....Thank you...

Could you suggest any specific transformer that I could use? I am afraid I would pick the wrong one again....And my project is close to the due day....:(

I am in New Zealand, and the two large supply are either Jaycar, or RS component Ltd....Could you please pick up the one needed from the websites for me?
The websites are:
www.jaycar.co.nz

or www.rsnewzealand.com

Cheers.....

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Could you suggest any specific transformer that I could use?

Sorry. I looked at Jaycar but their search engine is stupid. Maybe they don't have a 600VA transformer that is powerful enough. I could spend days asking for a 500VA one, then a 400VA one etc.

The 500W square-wave inverter needs a 600VA 10V-0-10V also called 20V center-tapped, or 12V-0-12V also called 24V center-tapped transformer. It will be huge and heavy and will cost an arm and a leg.
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oh....really....

I looked at some online electronics stores and they have transformers which are 100VA and have 2 0-12V at secondary...I only need the inverter to power a single incandescent lamp so i think 100VA should be enough for me...

How do I know which type of transformer with two 12V output can be implemented as a 24V center tapped? Will that be mentioned in the specifications? What is the difference between those ones and the one I picked before at Jaycar?

Cheers....

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The inverter doesn't need 8 output transistors for only an 80W load (100VA transformer). 4 output transistors will be fine, 2 on each side. It also doesn't need the 2N3055 driver transistors since the 2SC1061 transistors can directly drive the output transistors. The fuse will be 10A.

A transformer that has two 10V or two 12V windings can have them connected in series to make it 20V center-tapped or 24V center-tapped.

The transformer that you tried before has only a single low voltage winding but it has a few voltage taps on it. It doesn't have a center-tap.

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Hi Hericlj,
Good, you built the fixed circuit.
The transformer doesn't have 0V terminals. Its primary winding has a center-tap that is connected to the fused and switched positive 12V, and it has two terminals for the collectors of each side's output transistors.
The emitter resistors of the transistors are connected to 0V and everything with the ground triangle is connected to 0V. The negative terminal of the 12V car battery is also connected to 0V in the circuit.

The circuit doesn't need to be earthed. We say its negative wire is its "ground".



Hi, thank you so much for your help....For me, to  make the inverter seems like an endless process...

Just reviewed the above post you replyed a few days ago...I just had the weird feeling maybe I did something wrong on the circuitry as well...

You said that the circuit doesn't need to be earthed...So if I powered the circuit with a step down transformer(240V-12V) and a bridge rectifier from the main supply, should I connect all the nodes with triangle (including the emitterors of the transistors) to the negative node of the rectifier? And the centre tap of the transformer is connected to the positive node (12VDC) of the rectifier?The collectors to the remaining terminals of the transformer? 

Cheers....
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[i powered the circuit with a step down transformer(240V-12V) and a bridge rectifier from the main supply

12VAC when rectified with a bridge rectifier makes a peak voltage of 15V and needs a huge capacitor to filter it into 15VDC. It is fairly high for the circiuit but is safe for the parts.

should I connect all the nodes with triangle (including the emitterors of the transistors) to the negative node of the rectifier?

The triangles on the schematic are connections to the negative terminal of the power supply (battery). The emitters of the transistors don't connect directly to it but to resistors that connect to it.

And the centre tap of the transformer is connected to the positive node (12VDC) of the rectifier?

No. It connects to the switched +12V side of the on-off switch.

The collectors to the remaining terminals of the transformer?

To the low voltage terminals, not to the high voltage terminals. The high voltage terminals are the output.
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And the centre tap of the transformer is connected to the positive node (12VDC) of the rectifier?

No. It connects to the switched +12V side of the on-off switch.


...So what I did was to connect the center tap of the transformer to the positive node of the rectifier...How wrong was that? Is that the problem to my inverter?

Could you tell me more about the switched +12V side of the on-off switch? Isn't that the DC supply to the IC? I thought it was a power switch, so I didn't make that part...

Cheers...
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A stepdown transformer and a bridge rectifier makes pulsing DC that won't power an electronic circuit. A huge filter capacitor (20,000uF or more) must be used to smooth it into usable DC.
The center-tap of the transformer in the inverter circuit, pin 8 of the LM358 dual opamp and pin 14 of the CD4047 all must be connected to the positive of the power supply.

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  • 2 weeks later...

Good day Audioguru. I've read all the topics posted and they are all helpfull.
Im a mechanical engineer by profession and I like to learn some electronics as well.
  My question is how can I make the 500W inverter to use a 24V dc power input?

I will Patiently waiting for reply
   
thanks in advance

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My question is how can I make the 500W inverter to use a 24V dc power input?

Change the transformer's voltage.
Change all the transistors to higher voltage ones but use less.
Add a voltage regulator for the oscillator and add gain to the opamps.
It is like a complete re-design.

You are a mechanical engineer? Then how can I cut the weight of my car to half so it can go nearly twice as fast?
Similar. ;D
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Hi Xerex,
I use www.datasheetarchive.com to get datasheets but 1st I need to know the part's number.

The max supply voltage for a CD4047 is 18V. Use a 12V zener diode.
The max supply voltage for an LM358 is 36V so it is fine with 24V. It needs to have a gain of 2 so its 12V input from the oscillator is amplified to 24V for the transistors.

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