Use for the Tesla Switch

[nilrehob]

However, what I wrote about the charge Q is correct, as far as I can tell, and *that* is what really matters.

I just did a very simple 2-stage-test (several times for each stage and with the very same components).
The switch is manually set to on, then to off again (coil saturated, spike goes to caps).
* In the parallel test the voltage on the caps went from zero to 13.7V.
* In the serial test the voltage on the caps went from zero to 32.4V (16.2V each).

Energy(serial) / Energy(parallel) = 16.2^2 / 13.7^2 = 1.4

Its definitely more efficient with the caps in series.
But why?
Is it because of Your proposed explanation with the equations, or is there other explanation?

Comments anyone?

[nvisser]

I think
In parralel , the capacitance will double and will take much longer to charge to a certain voltage
In series your capacitance will be less and it will charge up much faster.
It is what you get when you pulse a coil with rotating magnets
Of cause if you work out a resonant parralel cap for your coil( if you know it`s mH) at a certain switching frequency you will get a much higher voltage after the diode on your capacitor. Parralel resonance. It will also draw much less current from your source

[nilrehob]

I think
In parralel , the capacitance will double and will take much longer to charge to a certain voltage
In series your capacitance will be less and it will charge up much faster.
It is what you get when you pulse a coil with rotating magnets
Of cause if you work out a resonant parralel cap for your coil( if you know it`s mH) at a certain switching frequency you will get a much higher voltage after the diode on your capacitor. Parralel resonance. It will also draw much less current from your source

Yes, resonance, thanks for reminding me, it's been on my list for a long time, I better look into it.

E=1/2 * C*V^2 and Q=C*V so E=(Q^2)/(2*C) means less capacitance gives more energy.
I did a video: YouTube - Charge two capacitors

[nvisser]

Resonance: F=0.159/square Root(lc)
F In Hz, L In Hendrys, C In Farads
C= (0.159/f)^2/l

[lamare]

Its definitely more efficient with the caps in series.
But why?
Is it because of Your proposed explanation with the equations, or is there other explanation?

This is quite an interesting result, because it cannot be explained by conventional theory. Conventional theory says: you put a certain amount of energy in the coil (magnetic energy), which is converted back to electrical energy the moment you open the switch. According to this theory, you should end up with the same amount of energy in the caps, regardless of wether or not you charge them in series or in parallel.

The component conventional theory does not consider, is that what happens outside the circuit. However, the moment you open the switch, you get a sudden increase in voltage of the coil terminal, a sharp rising gradient. This creates an electro-static (as "static" as an "atom" really is a-tomair, inseperable) shockwave just outside the circuit, a desruption of the environmental electric field itself, propagating at the speed of light.

This appears to be an effect that is completely separate from the normal, much slower, matter (electron) bound current *inside* the surface of a wire.

Such an electro-"static" shockwave, is apparantly guided by matter, it apparantly tends to follow your wires, it appears to be guidable with wires in the direction of a lower potential. Or at least: there appears to be some phenomenon that remains in the direct vicinity of matter, of wires, that can be guided.

Since this phenomenon apparantly is completely different from the the normal current flow, it appears to be independently powered. It does not appear to consume any "traditional" energy. However, it is capable of producing a certain amount of charge in a capacitor. Hence, this phenomena is capable of converting a certain amount of some form of energy we did not put in the system ourselves into useable energy in the form a charged capacitor.

So the charge that ends up in the capacitors is the summation of the effects caused by two separate energy sources: the collapsing magnetic field in the coil and the energy source responsible for the rapid expansion of the electric field itself, whatever that may be.


Since the shockwave guided by our wires, powered by an unidentified energy source, travels *outside* the wires and components, it appears to see no resistance, it's strength appears to be unaffected by the resistance of the matter, the wires, the caps, it is guided by.

That means it acts like a *current source* rather then a *voltage source* and that is very interesting.

Because if you put two caps in series, you get a higher voltage as a result of the collapsing magnetic field, because of the smaller capacitance.

And, because of that, you get a sharper pulse, a faster change of the electric field surrounding the components and the wires.

In other words: we create a stronger "radiant" event outside our wires *and* we get to use the same "radiant" juice twice!!!!


So, charged in series, we get more "radiant" energy for free out of the vacuum, while paying the same "price" in terms of energy we have to put into the system ourselves.

[lanenal]

energy wasted on resistance

I'd adventure that the energy wasted on resistance (e.g. the coil and the diode) is smaller in the serial test (the time for the current to drop to zero is about half).

I just did a very simple 2-stage-test (several times for each stage and with the very same components).
The switch is manually set to on, then to off again (coil saturated, spike goes to caps).
* In the parallel test the voltage on the caps went from zero to 13.7V.
* In the serial test the voltage on the caps went from zero to 32.4V (16.2V each).

Energy(serial) / Energy(parallel) = 16.2^2 / 13.7^2 = 1.4

Its definitely more efficient with the caps in series.
But why?
Is it because of Your proposed explanation with the equations, or is there other explanation?

Comments anyone?

If you can provide me with the source voltage, coil inductance, coil resistance and your diode type (voltage drop), as well as the capacitance, I can probably find the result according to traditional theory.

[nilrehob]

If you can provide me with the source voltage, coil inductance, coil resistance and your diode type (voltage drop), as well as the capacitance, I can probably find the result according to traditional theory.

I know, there is nothing strange about this, if you think about it it's part of a backward-Tesla-switch.
If you do some traditional theory calculations on a regular Tesla-switch you will see that you always loose energy for each switch without getting anything for it at all, because you charge from serial and give to parallel.
If you could do the opposite, make a backward-Tesla-switch, charge from parallel and give to serial, wouldn't you gain energy?
But how could you overcome the potential and charge from parallel to serial?
Maybe a Bedini-circuit could do the potential leap for you?

[nilrehob]

If you can provide me with the source voltage, coil inductance, coil resistance and your diode type (voltage drop), as well as the capacitance, I can probably find the result according to traditional theory.

Oops, forgot:
source voltage = about 13V
coil inductance = 177 mH
coil resistance = 24.3 ohm
diode type = 1N4007
capacitance = 47 uF for each cap

[lanenal]

Thanks for the numbers. Here are the estimation (Ignore the Diode):

Saturation Current: I = 13/24.3 = 0.535 A
Energy in the inductor: L*I*I/2 = 0.02533 J

Ref: Resistors with Inductors and Capacitors

tau = 2L/R = 2 * 0.177 / 24.3 = 0.014568
alpha = 1/tau = 68.644

parallel case: Cp = 47 uF * 2 = 94 uF
omega_o = 1 / sqrt( L Cp ) = 245.16
omega_r = sqrt( omega_o * omega_o - alpha * alpha)
= 235.35
frequency = omega_r/(2 pi) = 37.46 Hz
Time for current drop to zero (0.25 cycle):
t = 0.25/37.46 = 0.006674 s

Energy wasted on resistor: (This is an approx)
Er = t*(R*I*I/2) = 0.02321 J

Energy stored in capacitor =
Energy in the inductor - Energy wasted on resistor
= 0.02533 - 0.02321 = 0.00212 J
Final voltage on the capacitor = sqrt(2 * Energy / Cp)
= 6.716 V (oops, not very close: over-estimated Er)


serial case: Cs = 47 uF / 2 = 23.5 uF

You may try this by your self.

Oops, forgot:
source voltage = about 13V
coil inductance = 177 mH
coil resistance = 24.3 ohm
diode type = 1N4007
capacitance = 47 uF for each cap

[nilrehob]

Thanks for the reply.

Time for current drop to zero (0.25 cycle):
t = 0.25/37.46 = 0.006674 s

Shouldn't it be 0.5 cycle?

Energy wasted on resistor: (This is an approx)
Er = t*(R*I*I/2) = 0.02321 J

I don't think you can use I as is here as I is not constant over time.

I'm thinking more in terms of charge (as in Coulomb).
The energy in the saturated coil must have some relationship with charge Q.
If so and if Q sent to n caps in parallel, Q is divided between the caps giving each cap Q/n.
But if Q is sent to n caps in series, each cap will have charge Q.
Ignoring losses here of course.

[nilrehob]

Taking many and small steps towards a Tesla-switch I have made a video of a new simple one-battery-variant at youtube:
YouTube - Tesla-switch part 6

Enjoy!

[nilrehob]

And here is a video showing the benefits of driving in parallel and charging in series, the Tesla-switch in reverse:
YouTube - Tesla-switch part 7

Enjoy!

[Matthew Jones]

Beneficial Change

@Hob

If its possible you should change the configuration of your wheel. Somehow suspend the coils above the disc. And add North and South magnets so the coil drives from both ends.

See attached picture.

You'll see a better recovery effect.

Matt

[nilrehob]

@Hob

If its possible you should change the configuration of your wheel. Somehow suspend the coils above the disc. And add North and South magnets so the coil drives from both ends.

See attached picture.

You'll see a better recovery effect.

Matt

Hi Matt, Yes I'm planning to do that, but in my next motor using a HD-drive. Thanks!

[nvisser]

Found this file on the net last night
Answered a lot of my questions .
I`m sure this resonance effects can be used on the Tesla switch
See page 11
http://www.byronwine.com/files/plans.pdf