transformer for 555 taser gun

[colin55]

Here are all the 100 - 555 Circuits:

http://www.talkingelectronics.com/projects/50 - 555 Circuits/50 - 555 Circuits.html

 

[DerStrom8]

Here are all the 100 - 555 Circuits:

http://www.talkingelectronics.com/projects/50 - 555 Circuits/50 - 555 Circuits.html

But I still don't see proof that they stole yours. How do we know you didn't steal theirs?

I'm not trying to insinuate that you're a thief. I'm just trying to make a point about proof of ownership.

P.S. I just noticed this at the bottom of the page:

Not copyright 6-7-2011 Colin Mitchell You can copy and use anything.

 

[canadaelk]

Asides from who did what...I still don't get how this transformer works in a flyback mode. Could somebody explain that? Thanks! E

On a personal note, having been involved in lifesafety systems, I find publications of circuits like these irresponsible. Taser the cat before you shove it into the microwave? E

 

[audioguru]

When the darlington transistor turns off then the stored energy in the inductance of the transformer has nowhere to go so it produces a high voltage spike as it tries to keep the current flowing. I think the voltage spike will be clamped to the avalanche VCE breakdown voltage (80V or more) of the darlington.

He, he. Hold both wires of a relay (an inductor) in one hand. Apply a 9V battery. When you disconnect the battery then your hand will feel a high voltage shock!
DO NOT hold the relay wires in each hand or your heart (in between) might stop.

The spark coil for a car engine works like that. When the points disconnect then a high voltage is produced.
The flyback transformer in a CRT TV also works like that to produce the high voltage for the CRT.

But this simple circuit does not produce the 50,000V of a stun gun. It is only the few thousand volts of an piezo lighter.

 

[canadaelk]

I understand the back-EMF part but not what it has to do with a Darlinton Q. Direct me to some reading, if you can (simple stuff!). E

 

[audioguru]

The darlington has a high current gain that is not used in this circuit.
With a 12V supply then the output transistor produces about 11.5V into the 8 ohm winding of the transformer which is a current of 11.5V/8 ohms= 1.44A. An ordinary power transistor needs a base current of 0.144A to saturate well and the output of a 555 has an output current that is up to 0.2A so it will work fine without a darlington. Of course the value of the base resistor must be re-calculated for 0.144A.

The OP will use a 9V battery so the current in the 8 ohm winding is only 1.1A.

 

[colin55]

I think the voltage spike will be clamped to the avalanche VCE breakdown voltage (80V or more) of the darlington.

That comment has absolutley nothing to do with the output of the transformer.

When the darlington transistor turns off then the stored energy in the inductance of the transformer has nowhere to go so it produces a high voltage spike as it tries to keep the current flowing.

That comment has no relevance or bearing on simplifying or explaining how the high voltage is produced.

It's no wonder newcomers have no chance to understand the operation of circuits like this.

 

[DerStrom8]

That comment has absolutley nothing to do with the output of the transformer.



That comment has no relevance or bearing on simplifying or explaining how the high voltage is produced.

It's no wonder newcomers have no chance to understand the operation of circuits like this.

You know Colin, to be perfectly honest, you're not helping one little bit. As usual, you seem to only be arguing with what other people are saying, and how "they're wrong". With all due respect, unless you can produce VALID arguments about this, I think it would be better for you to just leave the people here to their jobs and stop posting replies such as those. You're only making yourself look worse.

 

[canadaelk]

collin55: It's no wonder newcomers have no chance to understand the operation of circuits like this.

May I suggest than that you explain it. As all there is is the schematic without any kind of, well, anything. E

 

[DerStrom8]

Deleted Off Topic.

Moderation.

 

[audioguru]

That comment has absolutley nothing to do with the output of the transformer.

Of course it does. The transistor will have a VCE avalanche breakdown at 80V or more on the low voltage side of tythe transformer. The turns ratio of the transformer procuces an output voltage that is 11.2 times higher than 80V or more. The 4-stage voltage doubler increases it more.

If a transistor with a higher VCE rating is used then the high voltage will be higher.

 

[colin55]

When the Darlington transistor turns on, current flows though the primary of the transformer to produce magnetic flux in the core.
This flux cuts the turns of the secondary and since there are many more turns of the secondary, the voltage produced by the secondary is many times greater than the voltage across the primary.
But in this circuit, the high voltage produced by the secondary at the moment is not as high as will be produced in the next part of the cycle and we will not be concerned with this fairly low voltage.
The output of the 555 is a square-wave and after the transistor has been turned on for a short period of time, it is abruptly turned OFF.
This is when an amazing thing happens.
The magnetic flux produced by the primary abruptly stops.
The energy in the core of the transformer (in the form of magnetic flux) ceases to receive magnetic energy. This magnetic flux is called expanding flux, even though it may not be received at an increasing rate at the time of the stoppage.
As soon as the receiving flux (energy) is turned off, the energy in the core starts to collapse and when it does, the magnetic lines of force reverse direction.
These magnetic lines of force cut all the turns in the transformer (both the primary and secondary) and the collapse is very rapid.
This produces a voltage in the turns that is not only very high but in the OPPOSITE DIRECTION.
The actual speed of collapse is called the “Q-factor” of the inductor (transformer) and this is also known as the “Quality Factor.” The Q-factor can be 10, 100 or even 1,000 and is the ratio of the voltage produced by the output when a sinewave is delivered to the transformer compared to the voltage produced when the input voltage collapses very quickly.
There is no way to determine the “Q” of the transformer from any winding data or size or shape. The transformer must be used in a circuit to determine this.

As a side point. The BD679 transistor has been tested to have a breakdown at about 270v - as proven im my Xenon circuit - so the 80v rating in the data sheet has no bearing.

 

[audioguru]

Colin,
I believe in a manufacturer's maximum allowed voltage ratings, you don't.
The manufacturer has tested millions of their transistors so they know that some will break down a little higher than their maximum voltage rating. How many millions did you test when they had a breakdown voltage 3.4 times higher?

When a transistor breaks down then it is like a short circuit that squashes the flyback voltage spike.

What is the breakdown voltage of the cheap little transformer?

 

[canadaelk]

colin55:

Your penultimate line: There is no way....

Exactly! Give us such transformer to prove you right. E

 

[colin55]

Deleted Abusive.

Moderation.

 

[colin55]

I believe in a manufacturer's maximum allowed voltage ratings, you don't.

Just buy one and try it.

I bought 20,000.

 

[canadaelk]

Deleted Off Topic.

Moderation.

 

[DerStrom8]

Just buy one and try it.

I bought 20,000.

But you can't EXPECT something to operate well beyond its ratings. You're trying to use it in an environment that it wasn't designed for, and that creates a very unstable and poor overall design.

 

[DerStrom8]

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[redpepper007]

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