Home
You Are Here: Home » Old Forums Archive » Computers » Tron Box

Tron Box


— Posted by Abe on 10:10 pm on Mar. 11, 2002

I tried to make this shit. Everytime i plug this in my AC source it blows the resistor. I need some tecnical asistence.
The Diagram:
——————R—–F—-
I I I I
I I I I- AC
(C) (C) (C)
I I I I- SOURCE
I I I I
—————————–

(C)=CAPACITOR
F =FUSE
R =RESISTOR
I,- ARE WIRE
PARTS LIST:
(3) ELECTROLYTIC CAPACITORS RATED AT 50V(LOWEST) .47UF
(1) 20-30OHM 1/2 WATT RESISTOR
(1) 120VOLT FUSE (AMP RATING BEST TO USE AT LEAST HALF OF TOTAL
HOUSE CURRENT OR EVEN LESS IT KEEPS YOU FROM BLOWING YOUR
BREAKER JUST IN CASE…)
(1) POWER CORD (CUT UP AN EXTENSTION CORD. NEED PLUG PART AND WIRE)
(1) ELECTRICALLY INSULATED BOX


— Posted by ActionTimedDeath on 7:16 am on Mar. 12, 2002

Why are you doing this?

The capacitors will appear to the AC as essentially a piece of wire – if there’s enough of them they might as well be! And if the fuse is a big bit of wire then you’re connecting a small (power rating) resistor across whatever source you’re using (120V????!!!) – so making resistor toast and a lovely smell. You’ll probably blow up the caps too – great for taking your eyes out.

Stick to lighting fires in fields, not your house!


— Posted by Abe on 9:18 am on Mar. 12, 2002

I thing I’ll need non-polarized capacitors rated for the supply voltage.  Maybe I could use a condensor (like the ones in fluorecent lamps).  I am guessing, if somebody knows more than me, please reply.


— Posted by Zambosan on 10:11 am on Mar. 12, 2002

ActionTimedDeath knows more than you… why don’t you listen???  Nonpolarized caps are still capacitors… and a capacitive reactive load looks very much like a short circuit to an AC signal, especially a large-ish one (and 3 caps in parallel are represented by a lumped parameter model of a cap with a capacitance equal to the sum of all 3 of them).  Plus, as ATD pointed out, a 110v RMS signal is going to dissipate roughly 403 watts across a 30 ohm resistor, while only drawing about 3.6 Amps, which your wall circuit will happily supply, and the fuse will probably happily ignore.  So of *course* you’re going to blow the resistor.

Here’s a tip… the “Tron” box is a fraud.  The concept is that you create a resonant “tank” circuit that feeds back an inverted waveform to cancel out the generated one, so the meter registers no power.  Not possible, unless you’re enjoying a pipe dream.


— Posted by drybilgeltd on 1:41 pm on Mar. 13, 2002

Hey Abe: I do my best to not make negative comments. ActionTimedDeath has made all the negative comments needed for me. My input will be positive information as best as I can. Please accept my apoligies beforehand if my comments hurt. They are not intended to.

Any time a resistance is present there will be an energy loss. Watthour meters are sensitive to the phase relationship between the line voltage and line current. In a purely resistive circuit peak current occures at the same time as peak voltage. In a circle there are 360 degrees. Twice during a complete cycle of AC power there is a peak ( one positive and one negative ). Also twice during a complete cycle of AC power there is a transition through zero. Current through an ideal capacitor across an AC line will lead ( by 90 degrees )  the applied voltage. By using a math term called cosign of phase angle you will have the power facter.  Leading or laging, the power factor of 90 degrees is zero. As soon as a resistance shows up there will be a phase angle less than 90 degrees. When the current lags the applied voltage, it is inductive in nature. Charging a capacitor is like putting air into a tank from an air  compressor. To start, the tank is empty ( no pressure in it ) and maximum air flowing from the compressor. ( The hose connecting the tank to compressor and the volume capacity of the compressor may be considered as resistance ).  As the tank gets filled up, the flow will decrees. When the tank and compressor have equal pressure there is no flow. The flow rate change will follow the curve of a natural logarithm that all things in nature follow just like water dripping out of a leaky buckit. As the water level goes down ( less pressure differance ) the leakage rate goes down.  The reactance ( AC resistance ) of a cap ( in ohms ) is 1/2*pie*freq ( in hz )*C ( capacitance  in farads ). I used the word pie to meen the greek letter that I can not find on my keyboard. For an inductance, the formula is 2*pie*freq*L ( inductance in henries ). The peak voltage on sinewave power is the square root of 2 ( sometimes called root2  or just ROOT ) times the RMS ( heating power equivelant ) voltage. It is the RMS value that lights up a tungstan light bulb. By measuring the heat produced by a tungstan bulb the RMS value can be determaned for any wave form. Us mad scientist types use a device of this type called a thermal converter.  Your cap would need a voltage rating for safety purposes of at least 1.5 times the supply voltage.

If I said any more in this post ever one would turn their computer off.

    I care. Respectfully

  Drybilgeltd


— Posted by Zambosan on 2:37 pm on Mar. 13, 2002

The RMS value for a given waveform is usually computed analytically; as drybilgeltd mentioned, this is 1/sqrt(2) * Vp for a pure sinusoid.  More generally, the calculation is:

sqrt[(1/N) integral|0..N[ f(t) ^ 2 ] ]

Reactive circuit theory aside, the Tron box is still a fabrication.  The circuit you have is a single-order low pass filter.  Given the capacitor values, you have effective values:

R = 30 ohm
Ceff = 1.41 uF

The corner (-3dB) frequency is 1/RC = 23,641 rad/s or 3,764 Hz.  Rolloff is 6 dB/octave for a first-order butterworth filter.  Therefore, since your AC signal is ~60Hz, you’re just going to see an imperceptibly attenuated signal, slightly phase shifted, at the output.  It might as well be a wire.  The reason your resistor keeps frying is that your caps can’t tolerate the voltage, so they short out and the resistor dissipates too much power.


— Posted by drybilgeltd on 7:21 pm on Mar. 13, 2002

Another reason the resister fries would be that the caps ( already identified as being electrolytics ) are most likely the normal polarized types ( a positive terminal and a negative terminal ) rather than AC rated non-polar types even if they had a suitable voltage rating. At 60 hertz,  most non-polar caps would not tolerate their full voltage for long do to the heating from the AC current flowing through them ( motor starting electrolytic caps as compared to motor run oil filled caps ). To get the meter to read zero or even better read backwards the resistive component needs to be zero or negative ( negative resistance ? ),  in other words –  supply power to the power company. Draw a circle with a large cross  ( + ) through it , above the center line is positive resistance  ( dissapates power ), to the right is capacitive reactance, to the left is inductive reactance and below is negative resistance ( supplies power ).


— Posted by Zambosan on 9:37 am on Mar. 14, 2002

Drybilgeltd:he’s *not* going to get the meter to read zero!!! The only way to do *that* is to not consume any power, or like you said, supply power to the power company. That’s not “getting free power”, which is what the lousy text file he’s reading is trying to convince him he can do. If he had a purely reactive circuit, composed of ideal capacitors (which are a convenient model for lumped-element analysys, and don’t actually exist), the power factor would be zero ( cos(pi) ), and he would be consuming no power. Any real circuit, however, is going to have some finite resistance, which translates into a positive value along the Real axis. Regardless of the Imaginary (reactive) value, you’re never going to reach a phase angle of +/- pi, so you’ll always be dissipating power somewhere in the circuit.

He shouldn’t be using an electrolytic in the first place for a purely AC circuit, as the polarity is going to be incorrect exactly 1/2 the time; to use an electrolytic, you need to make sure the DC bias is high enough to prevent signal excursions below zero volts. Anways, an electrolytic cap that gets reverse biased at that high of a margin over its rated voltage will usually “pop” a lot quicker than a resistor will burn, since the dielectric, which is a light oil, boils at a pretty low temperature and the cans are hermetically sealed.

In conclusion, no selection of components is going to do what you want it to do, unless you buy a Honda generator and plug its output into the grid. :biggrin:

Oh yes, and the Real axis for complex frequency domain analysis is drawn along the X axis, not the Y.

(Edited by Zambosan at 9:39 am on Mar. 14, 2002)


— Posted by balor on 2:45 pm on Mar. 14, 2002

On a more practical aspect,I install electricity meters as part of my job. Every meter being produced now( that I work with) has a ratchet mechanism on the aluminium disc that revolves so it doesnt turn backwards and hence reducing the numbers on the meters.The most effective free electricity(called black and blue boxes here) boxes that I have come across are connected across the meter and are just so obvious assuming that the seals arent broken.The probes of the box are shoved up under the cover to touch the terminals in the meter.This leads to very obvious burning marks on the plastic cover and wood backing.Just the most obvious thing over here I see is that both sides of the fuse cutout before the meter is supposed to be sealed at both sides but most usually only seal it on one side so cut the seal at the meter side of the cutout.This allows the fuse to be partially withdrawn and a wire to be inserted and the fuse to be replaced leaving you with a phase.To get your neutral just go off the earth spike or your neutral block on your mcb/fuse board.


— Posted by drybilgeltd on 10:52 am on Mar. 15, 2002



Quote: from Zambosan on 1:37 am on Mar. 14, 2002
Drybilgeltd:he’s *not* going to get the meter to read zero!!! The only way to do *that* is to not consume any power, or like you said, supply power to the power company. That’s not “getting free power”, which is what the lousy text file he’s reading is trying to convince him he can do. If he had a purely reactive circuit, composed of ideal capacitors (which are a convenient model for lumped-element analysys, and don’t actually exist), the power factor would be zero ( cos(pi) ), and he would be consuming no power. Any real circuit, however, is going to have some finite resistance, which translates into a positive value along the Real axis. Regardless of the Imaginary (reactive) value, you’re never going to reach a phase angle of +/- pi, so you’ll always be dissipating power somewhere in the circuit.

He shouldn’t be using an electrolytic in the first place for a purely AC circuit, as the polarity is going to be incorrect exactly 1/2 the time; to use an electrolytic, you need to make sure the DC bias is high enough to prevent signal excursions below zero volts. Anways, an electrolytic cap that gets reverse biased at that high of a margin over its rated voltage will usually “pop” a lot quicker than a resistor will burn, since the dielectric, which is a light oil, boils at a pretty low temperature and the cans are hermetically sealed.

In conclusion, no selection of components is going to do what you want it to do, unless you buy a Honda generator and plug its output into the grid. :biggrin:

Oh yes, and the Real axis for complex frequency domain analysis is drawn along the X axis, not the Y.

(Edited by Zambosan at 9:39 am on Mar. 14, 2002)


 Zambosan: In your post you have made mention of a light oil being used as the dielectric in electrolytic capacitors. Go back to your text books and you will find the dielectric to be aluminum oxide on an aluminum plate ( this plate is one electrode ) and the other electrode is the electrolyte, an second aluminum plate is used simply to make connection to the electrolyte. Most books do not mention what chemical is used for the electrolyte, but after a lot of digging I found it, Aluminum Borate.  If you are familiar with Water Barrel Rectifiers, One plate is Aluminum and the other is Lead,  using a solution of Sodium Tetraborate  ( Borax ) or Sodium Bicarbonate ( Baking Soda ) for the electrolyte. When first hooked up to AC it will conduct in both directions. After a while an oxide film developes on the aluminum and then you have rectifier. I made these when I was a child for a 6 volt car battery charger with a 150 watt Lionel train transformer. The ectrolytic cap you were refering to is a polarized type and you are right that it will die. But the resistor will burn up for a safety fuze effect, many circuits do it this way for a cheep fuze ( in quanity a 1/2 watt resistor cost about 2 cents in American money ).  Non Polar electrolytics have an oxide film on both aluminum plates to work on AC, motor starting caps and speaker crossover networks. Oil filled caps are used for motor run and across the line for power factor correction applications, they will handle their full ac voltage at 60 HZ. In some electronic circuits two polar caps ( of same value ) will be wired back to back in series to handle AC voltages ( but not recomended by me ). There are Tantalum electrolytic caps that use sulfuric acid for the electrolyte but that is not the caps we are dealing with. My comment to dray a circle was just to give a picture. I did not intend to get into Smith charts for incident power verses relected power and transmission line losses, which is way off the thread topic.

Respectfully Drybilgeltd


— Posted by Zambosan on 1:54 pm on Mar. 18, 2002

Correct, the electrolyte itself is not the main dielectric component, although it does possess some dielectric properties.  Older electrolytics used to contain an oil containing poly-chlorinated biphenyls, but those have been retired for environmental reasons, so that part of my post was indeed incorrect.  However, the thrust of my post was that all the electrical theory is probably confusing the crap out of the original poster, and gives the appearance that the circuit can indeed get your meter to read zero, all you have to do is select the right components.  Which isn’t true at all.


— Posted by ActionTimedDeath on 11:38 am on Mar. 19, 2002

The older style meters with a spinning disc (in the UK anyway) can be cheated by turning them on their sides!


— Posted by balor on 3:30 pm on Mar. 19, 2002

Thats right Actiontimedeath.The meter even reads less if it is tilted in any direction at a minimum of 10 degrees.


Leave a Comment

Scroll to top