500W inverter circuit not working properly

[bountyhunter]

i dont know please stop the quiz program and help me

Audoguru is not running a quiz program, he is doing what I would have done if I cared... which is try to determine what this thing is supposed to be powering since that defines what the best approach for the design is.

He obviously knows that most equipment designed to run off 240VAC (small coolers, drills, motors, etc) won't work very well with a square wave input. In fact, anything with a basic power transformer in it designed to run off the 240 VAC line which is 50/60 Hz will have a problem with square waves since they have high frequency harmonic content.

Most power inverters have circuits that generate a quasi sine wave voltage ouput.

FYI:

https://www.smps.us/power-inverter.html

There are three basic types of dc-ac converters depending on their output waveform: square wave, modified sinewave, and pure sine wave . The square wave is the simplest and cheapest type, but nowadays it is practically not used commercially because of low power quality. The modified sine wave topologies (which are actually modified squares) provide square pulses with some dead spots between positive and negative half-cycles. They are suitable for many electronic loads, although their THD is about 25%. Priced in the range of $.05-$0.10 per watt, models that employ such a technique are the most popular low-cost inverters on the consumer market today

A true sinewave inverter produces output with the lowest total harmonic distortion (normally below 3%). It is the most expensive type of AC source, which is used when there is a need for clean sinusoidal output for some sensitive devices such as medical equipment, laser printers, stereos, etc

 

[audioguru]

This extremely simple inverter circuit is not even used in a purchased inverter today. Its waveform is a square-wave, its voltage is not regulated and is too low anyway. It does not disconnect the battery when the battery voltage drops too low so the battery will be destroyed. It does not disconnect the load if it is overloaded so it and/or the battery will be destroyed.

 

[aamir1]

what r your comments about this circuit friends is this circuit better than the previous one View attachment 65312

 

[colin55]

The circuit above is very poorly designed. It has the same fault with the driver transitors as the original design. They have no drive-capability. The collectors should be taken to the supply-rail.

 

[alec_t]

The collectors should be taken to the supply-rail.

They are....via the primary winding halves (I think that's what the OP intends, but the schematic has problems showing it) . Or did you mean directly (which would short-circuit the supply)

 

[audioguru]

The circuit above is very poorly designed. It has the same fault with the driver transitors as the original design. They have no drive-capability. The collectors should be taken to the supply-rail.

No.
The transistors are in an ordinary DARLINGTON arrangement which is fine. If the driver collectors go to the positive supply then there will be a huge waste of power in additional series base resistors for the output transistors.

The original circuit uses a TIP122 darlington power transistor driving a TIP35 25A power transistor in a darlington arrangement.

 

[colin55]

The output is very poorly designed.
To get current into the base of the lowest transistor in the staircase, each of the transistors above the lowest must see a supply. This means the collector must be about 1.1v higher than the emitter as this is the collector-emitter minimum voltage. The first transistor is a Darlington and its collector-emitter saturation voltage is 2v. This means the lowest transistor cannot turn on any more than to provide 3.1v for the other two transistors.

 

[Nigel Goodwin]

No.
The transistors are in an ordinary DARLINGTON arrangement which is fine. If the driver collectors go to the positive supply then there will be a huge waste of power in additional series base resistors for the output transistors.

No it's not fine, it makes a massive loss of voltage and has poor drive to the output transistors, meaning they run massively hotter and will provide even less power.

The drivers should be connected directly to the 12V supply (and not the transformer) with resistors feeding the base of the output transistors - I'd also prefer to see PNP drivers to allow more drive as well.

 

[colin55]

For the 2N3055 to deliver current into the base of the 2N6277, it must have a collector voltage that is higher than the emitter.
And for the TIP122 Darlington transistor, it must have a collector voltage that is higher than its emitter.
The minimum collector-emitter voltage for a Darlington transistor is 2v.
The base-emitter voltage for a 2N6277 is about 1.8v to 3.5v (use 2.1v) and for a 2N3055 it is about 0.7v.
This means the TIP122 can only turn on when the collector voltage is 0.7v + 2.1v + 2v = 4.8v.
This means the collector of the 2N6277 cannot be less than 4.8v.
This faulty design can be fixed by taking each of the transistors to the supply-rail via a suitable resistor.
The collector-emitter saturation voltage for the 2N6277 is between 1v - 3v.
This means the transformer sees a higher voltage.

 

[audioguru]

You are saying that a darlington transistor does not work properly?

 

[Nigel Goodwin]

You are saying that a darlington transistor does not work properly?

Depends what you mean by 'properly', it has severe limitations, hence the use of the superior Sziklai pairs to replace them.

Essentially the collector of the driver conected to the colloctor of the output means there's little drive available for the output, and thus a high voltage loss collector to emitter.

 

[colin55]

I am saying that this method of driving the output is a total disaster. Even the output transistor has an enormous collector-emitter voltage drop at 20 amps and the coil sees very little of the 12v supply.