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RFID Zapper - some basic questions

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Hello everyone!

I'd like to build a RFID Zapper, really simple device, but still i've got some questions.

The zapper circuit looks like this:
RFID Zapper

I've got a nice 150µF 385V capacitor from a TV:

Where should i connect the negative? To the middle lead (at the center of the capacitor)?

Also can i, instead of the diodes, use a 2A bridge rectifier, or is ~230V too much for that bridge? The leads are close, like 3mm apart.

Can i change the resistor value? (i have some 1MΩ 1Watt resistors i could use instead) Also why is the resistor "before" the bridge? I think it should be put "after" the bridge, connected to the negative lead of the capacitor, coil and, from the other side, to the negative lead of the bridge.

Thanks in advance!
 
Nice, I love exploding wire too !

> Can i change the resistor value?

You could use a different value, but 1Mohm will probably take too long to charge the cap. Do the math !

> Where should i connect the negative

It's probably written on the cap...

> Also can i, instead of the diodes, use a 2A bridge rectifier, or is ~230V too much for that bridge?

If you can't read the voltage rating on this bridge, better buy some diodes rated for the intended voltage (ie, 600V ones).

> Why is the resistor "before" the bridge? I think it should be put "after" the bridge

No difference

PS : when testing any circuit which will create a violent discharge like this, wear safety goggles like the Brainiac dudes. Fasten your coils securely to a non-magnetic material, or they will fly. And it will also probably zap anything including credit cards, clelphones, etc.
 
Thanks for the reply !

It's probably written on the cap...

Well I wouldnt have asked if it'd be so obvious. The only thing are those pluses in circles like (+) on the first image and an arrow pointing down. IMO it means that the middle lead is negative and the outer circle is positive. Still I need and expert opinion, the voltage is too high to experiment with this, i dont want to get the cap explode!

And thanks for the warning, i will be careful :)

BTW
I did some math and, assuming i'd use 1/4MΩ 4Watt resistor (made from 4 * 1MΩ resistors) and loading the capacitor to the 85% voltage, i got 71 seconds.

But.. using the 2.2kΩ resistor it'd be ... 0.6 seconds! which seems impossible!

Is my forumla wrong?
https://www65.wolframalpha.com/input/?i=-+(2.2)+kOhm+*+150uF+*+ln(1-0.85)
 
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Well, I think whoever designed the markings for this capacitor should be flogged, because they're so obscure you got a 50% chance of putting it in the wrong way. Examine the PCB it came from, and if all else fails, connect it to a low voltage (like 5V) supply with a 1K resistor in series and look at the voltage. If it behaves normally, it's plugged the right way, if the voltage on the cap doesn't go above 1 volt, it's in reverse. A big cap like this should withstand a few milliamps in reverse for a few seconds...
 
Ok i will try it, but i also asked somewhere else and maybe i'll get my answer.

I hope a bridge made from 4* 1N007 diodes will be enough (they are rated 1A). Also please look at the results i've got, those for 2.2k resistor seem to be pretty crazy !
 
That's not how you calculate the resistor...

You suppose the worst case, ie the button is always closed, so the resistor should not burn. Suppose you want to dissipate 4 watts max, because you don't want to use a fan, so P = V²/R, this gives you the resistor value, in your case about 15k. Then you calculate the charge time to, say, 80%, which is about 3 seconds. If you need faster, then get a smaller resistor, like 2.2k 25W, which is going to be quite expensive, or do it ghetto style, use a 230V 10W lightbulb as the resistor.
 
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Thanks for the reply again.

The middle lead is positive and all the rest is negative, as the negative part is connected to the casing of the cap. (i got a reply on other forum)

So, i've done it, as the circuit is really simple, and it worked. Cap loaded nicely to ~310V and then it was discharged through the coil. The thing is i cant say it destroyed the rfid card i got, so i need to check it on some reader and then i'll be sure.
Or maybe there is another way? I mean, make another coil and connect the ampere meter, or capacitor with voltage meter. Though, this "emp" impule is so "tiny" i dont know it'd be detectable by such means.
 
That's true but there is a way to find out whether it is zapping things (without putting there working cellphone etc).

I connected a coil to a small 3mm red LED (which runs on about 2.5V). When i zapped it I could see spark in the LED chip, and it was certainly destroyed by HV, as it didnt work after that.

Wondering, as the rfid tags got antennas to get energy, rfid readers got them too. Will this zapper be able to manage to kill rfid reader too?
 
That's true but there is a way to find out whether it is zapping things (without putting there working cellphone etc).

I connected a coil to a small 3mm red LED (which runs on about 2.5V). When i zapped it I could see spark in the LED chip, and it was certainly destroyed by HV, as it didnt work after that.

Wondering, as the rfid tags got antennas to get energy, rfid readers got them too. Will this zapper be able to manage to kill rfid reader too?

I think that to kill the reader it going to be much harder to get to, to even attempt to kill it. It seems pretty safe to say that that you are going to kill pretty much anything electronic with a pulse like that , that has not been specifically designed to ward off several thousand volts...which I dont think the average tag has.
 
Can you please help me understand why on earth you would want to destroy an RFID reader?

But to help you out on the topic, I have done a few projects with RFID readers before, and with the one that I was using it had a max voltage supply of 5.75V DC, when my LDR gave it a 10ms burst of 7V, it rendered the reader inoperable. It shouldn't take too much voltage to destroy it. As long as you can get an arc from the source to the reader, it will almost definitely be destroyed (unless I built the circuit, then it will be just a little harder :))
 
It's not like that, I'm just asking :) IMO it'll be easier to kill some device that got antenna such as the reader. Without it, it's sure harder to induce voltage by the pulse.

Lately I got to new capacitors.
From the equation 1 in this application note (note) it seems that increasing current twice, the magnetic field will be also two times stronger.

To increase the current twice I need to put there another 150uF cap. Is that correct?

Also, could You tell me how can I calculate the thickness of my copper wire. Right now it's 1mm thick, but as I increase the current flowing in the coil I must put there a thicker wire?
 
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