Electric Fence Energizer

[WJoubert]

Hi

I want to build a simple (I think) energizer.
Details are as follows=

Input Voltage: 9Vdc to 12Vdc
Current consumption: 50mA @ 9Vdc 80mA @ 12Vdc
Output Open V: 4000V
Output into 500 OHMS: 2800V
Energy over 500 OHMS: 0.9J
Impulse Rate: 1.1 Second
Impulse duration: 250uS

Can anyone please help
Thanx

 

[Russlk]

Your specs sound lethal, I am not getting involved.

 

[TheOne]

I agree, the energy dumped into 500 Ohms in 250uS will be 3.92 J, two and a half time more powerful than a police tazer :!:

 

[WJoubert]

Can you then help me with a design this is specs that I got for a PetFence. That is why it is 0.9J & 2800V.
I would like to build it myself

Thanx

 

[TheOne]

Where does the 500 Ohms come from?

 

[TheOne]

My quick calculation show that if you charge a 0.47uF capacitor to 2800V and discharge it through a 500 Ohm resistor for 250uS you should have dumped about 0.9J of energy into the resistor. Maybe someone can verify this

This was an error: See my later post

 

[Roff]

I'm also curious where the 500 ohms came from. Perhaps the resistance from a cow's wet nose to ground?

In any case, if you consider the equivalent circuit to be a 160nF capacitor charged to 4000 volts, in series with a resistance of ≈214 ohms, dumped through a zero resistance switch, you will get (I think) 0.9 joules delivered to a 500 ohm load, with a pulse width (90% to 10%) of about 250usec.

 

[TheOne]

I also calculated that after the 250uS discharge there is still about 966V over the cap. The 2800V value may be because that voltages lower than 900V were found to be too low to have a desired effect, taking into account skin/fur resistance of the animal?

This was an error: See my later post

 

[Roff]

Here is a schematic and sim results of the model I proposed. Remember that WJoubert said the open circuit voltage is 4000 volts. As I said above, the pulse width is measured 90% to 10%.

 

[TheOne]

Ron

Looking at your graph there is something I can't add up. Does the green curve indicate the voltage over the cap while discharging into the 500 ohms, thus the part responsible for supplying the energy to the load?

The moment you start your discharge you should have enough energy stored to dump into the load. If I read your graph correct at the start of the discharge cycle you will have only [160nF x (2800v)^2]/2 = 0.63 J of energy stored in your cap.

Unless I am reading your graph/sim wrong.

 

[TheOne]

I think my original calculation was also flawed. I overlooked a factor 2 somewhere when doing my integration. The value should be about 1/2 of 0.47uF (0.235uF) ~ closest standard value 0.22uF

This seems to check out on simulation. 0.86 J at the beginning of the discharge, with 0.009 J energy left in the cap after 250uS.

0.235uF gives 0.92 J and 0.013 J after 250uS

The 214 Ohm in Ron's circuit will give the required drop down to 2800V from 4000V the moment the 500 Ohm is connected due to voltage divider action. So the 500 and 214 Ohm resistors will be in series over the capacitor. The capacitor will be charged up to 4000V in the 1 sec allowed between energy pulses. The fence voltage will be taken from the connection between the resistors. So as long as the 500 Ohm load is not connected there will be 4000V at the open end of the 214 Ohm resistor. The moment the 500 load connects, the voltage will be divided down to 2800 and at the same time the capacitor will start to discharge.

The 4000V could be generated using something like a strobe psu and some capacitance voltage multipliers after the transformer from this example (you only need the part up to the transformer) and add the multipliers (caps and diodes) **broken link removed**
onto the secondary winding.
http://www.hut.fi/Misc/Electronics/circuits/strobo_12v.html
Because of the light load the inverter should draw little current and the 80mA from 12V should be possible. (you have 1 sec to charge the cap every time) With a 680k resistor from the 4000V supply to the cap, it will charge up in about 0.8 sec, with a peak initial current at 6mA and tapering off towards zero after 0.8 sec

 

[Roff]

Here is the way I analyzed it:

Rsource=Rload*(Vopen-Vpeak)/Vpeak
Rsource=500*(4000-2800)/2800
Rsource=214.3

Jout=0.9
Jtotal=Jout*(Rsource+Rload)/Rload
Jtotal=(214.3+500)/500=1.2857

The basic equation for energy stored in a capacitor:
J=0.5*C*V^2
Solving for C,
C=2*J/(V^2)
C=2*1.2857/(4000^2)
C=161nF
Tau=(Rsource+Rload)*C
Tau=114.3usec

If we define pulse width from as being from 90% to 10% (arbitrary maybe, but fairly common for an exponentially decaying pulse),

Pulse Width=2.2*Tau=251usec (I bypassed the derivation of the factor 2.2, because it is well known).

Admittedly, 0.9 joules will not be delivered during the 251usec, but I think the manufacturers could have used the same thought process to come up with total energy first, and then derived (or measured) pulse width afterwards.

The difference between TheOne's derivation and mine is that he constrained the 0.9 Joules to be delivered within 250usec (with some energy remaining in the cap), while I calculated the capacitance required to deliver 0.9 Joules, then calculated the resultant pulse width.
This has become just an interesting intellectual exercise, and I don't know which method (if either) is correct, nor do I particularly care.

Edit:
After I posted this, I concluded that Jtotal should be:

Jtotal=Jout*(Rsource^2+Rload^2)/(Rload^2)

If this is correct, this makes

C=150nF (instead of 160nF), and

PulseWidth=232usec

Not quite as close as before, but still not too far off.

 

[TheOne]

You are right Ron, it is impossible to say what the manufacturer meant. In any case I think we have given WJoubert enough to chew on . If he uses either method he will be close to his goal.

 

[WJoubert]

Thanx

Thanx

RON H & The One......

I will try my best.......with all the info given.... :roll:

 

[Goytil]

Hallo!!

I want to build a simple energizer.

Where can I find some electric wire diagrams.

I have bought some but I would like to now how it works. I am an electrician, working whith 230 -400 V, I have not worked with electronics since i was in school.

Can anyone please help

 

[bryan1]

I've made up a simple electric fence controller using coil from a vehicle and also a couple from an electronics kits. Now this subject has been delt with quite a few times so a search of this forum would give all the details you need. Don't ask me to go find them as I'm only suggesting you search before asking and off memory one of the threads did deal with using lethal currents.

Cheers Bryan

 

[smut]

Hello,

Hi!!

Did you get help in designing the energizer, am being asked to design the energiser too, can you help me!!!
reply direct to:s200241010@students.polytechnic.edu.na

Smut

Hi

I want to build a simple (I think) energizer.
Details are as follows=

Input Voltage: 9Vdc to 12Vdc
Current consumption: 50mA @ 9Vdc 80mA @ 12Vdc
Output Open V: 4000V
Output into 500 OHMS: 2800V
Energy over 500 OHMS: 0.9J
Impulse Rate: 1.1 Second
Impulse duration: 250uS

Can anyone please help
Thanx

 

[Spyker]

Hi,

I bought a "taser" at a fleemarket, but the main components are cast in urathane.

We broke it open when it stopped working and found a small transformer with three windings, what appears to be a medium power transistor and two blue capasitors.

We successfully build the car coil with 555 circuit, but there is no comparison between the two.

The car coil circuit is a joke.

We also tried the voltage multiplier circuits, and could only get a small spark.

The fleemarket taser packs one hell of a punch, makes a lot of noice and you can smell the argon in the air.

Does anyone know how to build one of these proper "tasers".

I want to use it for a valid reason and have no intension of injuring anyone.

Thanks.

 

[xinyuanwanli]

Learning some thing!!

 

[Spyker]

xinyuanwanli - You might be learning something, not me?

All I am getting is frustrated.

All that work and all I can get is a small spark.

And then some guy in China can do it with less than ten components.