All about Sim

[GibbsHelmholtz]

Hi Gibbs, The bug can be related to any thing, but likely to the 555 model. Generally, only a nominal functioning is modelized. A 555 is specified for a power voltage range 4.5v - 16v. If you power it with 1.5v, the model doesn't apply. Another example is transistor 2N2222A. We can use it as a negative resistance by connecting its emitter and collector in a reverse way. But this is not simulable with an ordinary 2N2222A model because it's out of a normal functioning.

The simulations are restricted by Kirchoff's equations saying that 1-1=0. If you observe "1", you can be sure there is a bug or the model is not used in its domain of validity.

I know you are convinced, otherwise you would have already built the setup with a real 555, isn't it?   

A negative resistance must be considered as a power supply. It really provides energy. Spice allows for negative resistance values. If you put one in a circuit, you get a "perpetual motion".
But in the real world, a negative resistance is negative only around an operating point obtained by biasing the dipole with a DC source, and it is this DC source that provides energy.

I'm now revisiting the Kapanadze's device with this idea that a negative resistance (may be in a spark gap?) could be the cause of energy gain in his coupled circuits. I'm not yet convinced if it is a scam or if he has really something.

His version is just raw.  You can narrow down what the problem is by modify the circuit.  It can be use at higher voltage and different transformer.  I see it as an opportunity to understand more about sim that could simulate OU.  Your information about 2N2222A is interesting because Lasersaber just build a joule ringer in a weird way that may related to your statement.  Of course if we model his circuit on sim that doesn't incorporate this mode of function, the sim wouldn't run. 

I don't understand why Kirchoff's logic cannot allow for OU.  If t1 = 1-1 = 0, and t2 = 2-2=0 .  It could means an oscillation is building up.  Kirchoff's still hold but it is OU . 

You are right about a source must be introduce to the sim.  I think the problem is the sim does not account for ambient condition and we must manually add them in to simulate OU.  For example, we want to model transmission line.  If we put 2 wire parallel, the sim does not know what the capacitance between the wire is.  The wire could be subjecting to an environment around a tesla coil, but the sim wouldn't know. 

The negative resistance in the real world use a DC source bias, but at this point of operation, the DC source would provide less compare to a non negative resistance.  It is still OU in a relative way.  I think sim should incorporate negative resistance modeling.  It's more accurate with the real world and could simulate OU. 

OU to me isn't something that excite me anymore.  It is just a normal thing like... matter of fact.  lol

[exnihiloest]

...
I don't understand why Kirchoff's logic cannot allow for OU.  If t1 = 1-1 = 0, and t2 = 2-2=0 .  It could means an oscillation is building up.  Kirchoff's still hold but it is OU.

The sum of currents at a node is zero and the sum of voltages around a loop is zero (the sign being taken into account), therefore no extra-energy can appears from Kirchoffs laws, provided that only models of real components are used.
Kirchoff's laws can show only an apparent OU, when a hidden extra-energy source is included and not clearly identified. But when you include it also in the analysis, OU disappears. For example if you have a neg resistance, and you look at the current, you see that it flows in a reversed direction compared to a positive resistance, meaning that the neg resistance has a role of a generator. The neg resistance acting as a voltage source, it is a generator for the Kirchoff's viewpoint, and so there is not OU.

You are right about a source must be introduce to the sim.  I think the problem is the sim does not account for ambient condition and we must manually add them in to simulate OU.
...

Exactly! For example, if I put a pure negative resistance, it is the same as using an extra-energy source. The sim will show OU coming from the negative resistance, but a pure negative resistance that we can use in Spice is not a real component. I can modelize it with an op-amp of gain 2 and with its input+ connected to the output with a positive resistance: I simulate this way a negative resistance, I get a circuit that can be really built, but the sim will not show OU because in this case the extra-energy is now provided by the power supply of the op-amp, which is included in the sim.

In fact we can simulate any OU device because we can include neg resistance or any other component models specially designed to bring a gain from nothing, but if we don't know where the extra-energy really comes from (and how), we have no guarantee that the sim modelizes the real circuit.

[GibbsHelmholtz]

The sum of currents at a node is zero and the sum of voltages around a loop is zero (the sign being taken into account), therefore no extra-energy can appears from Kirchoffs laws, provided that only models of real components are used.
Kirchoff's laws can show only an apparent OU, when a hidden extra-energy source is included and not clearly identified. But when you include it also in the analysis, OU disappears. For example if you have a neg resistance, and you look at the current, you see that it flows in a reversed direction compared to a positive resistance, meaning that the neg resistance has a role of a generator. The neg resistance acting as a voltage source, it is a generator for the Kirchoff's viewpoint, and so there is not OU.

Exactly! For example, if I put a pure negative resistance, it is the same as using an extra-energy source. The sim will show OU coming from the negative resistance, but a pure negative resistance that we can use in Spice is not a real component. I can modelize it with an op-amp of gain 2 and with its input+ connected to the output with a positive resistance: I simulate this way a negative resistance, I get a circuit that can be really built, but the sim will not show OU because in this case the extra-energy is now provided by the power supply of the op-amp, which is included in the sim.

In fact we can simulate any OU device because we can include neg resistance or any other component models specially designed to bring a gain from nothing, but if we don't know where the extra-energy really comes from (and how), we have no guarantee that the sim modelizes the real circuit.

Of course I mean we should add a generator for sim to go OU.  This generator is the negative resistor.  However, I see a problem in our thinking.  You seems to say the input to the negative resistor is the same as negative resistor output.  If so, there is no OU.  I also do not see what's the point of negative resistor in this. 

I see many definition for negative resistor online. Very much they say the slope is negative.  I think this is an inaccurate definition and description of what a negative resistor is.  Slope being negative in different quadrant could means different things. I have a way to define this negative resistance thing.

If I^2R/ IV is not 1  or IR/V is not 1.  This is the only way to differentiate if the resistor obeys Ohm's law or not.  If it does not, then we could  immediately recognize potential for free energy.

[exnihiloest]

Of course I mean we should add a generator for sim to go OU.  This generator is the negative resistor.  However, I see a problem in our thinking.  You seems to say the input to the negative resistor is the same as negative resistor output.  If so, there is no OU.  I also do not see what's the point of negative resistor in this.
...

I don't see where is the problem. It is the same thing with a generator. OU is just a mind view of an inexhaustible extra-energy source. If it is coming from a neg resistance or from a perpetual voltage generator, it doesn't change anything since the moment you include it in a sim.

So I don't understand what you mean. A negative resistance is a dipole where current and voltage variations are proportional but of opposite directions (one increases when the other decreases). In real neg resistances, this applies only in a voltage and current range around a particular point of polarisation. The required energy to produce the effect is provided by the polarization source. Outside of the range, the dipole becomes non linear and loses its property of negative resistance.
In an ideal neg resistance, the effect doesn't depend on a biasing voltage point, it is available for all currents and voltages, including variations around V=0.
We can see real neg resistances where a limited variation of voltage/current around a functioning point is concerned, or ideal ones, as following:

The reverse sign of the power due to the reverse sign of the current compared to the voltage, indicates that the energy is provided or consumed. The relative direction of current and voltage in a neg resistance shows that it acts as a generator.

Now, what is OU? "OU" is only an idea of the consequence of an inexhaustible source of energy that we could tap. There are two ways to see it, depending on what type of "perpetual motion" we invoke. If we speak of type 1, then OU in a closed system is extra-energy coming from nothing. It is likely that none of us think it is possible. We preferably refer to perpetual motion of type 2. In this frame, OU is extra-energy due to an unknown source and coming from outside of our (apparently) closed system into it, proving that our system was not really closed.
A sim can't show OU:
- a sim follows the equations of academic science, implying energy conservation in a closed system. A sim can't guess nor detect that there would be a source of extra-energy adding up in the system if it is not talked that there is one.
- and if we include the extra-energy source, then this source becomes conventional from the viewpoint of the simulation, and so the sim will not show OU, the energy is still conserved because our system includes now the energy source, so it is really closed.

When we include a neg resistance in a sim, the signal in the sim can diverge but there is still an energy balance between production and consumption. It is the same thing when we connect a (positive) resistance to a voltage generator in a sim: the energy becomes infinite with time, but this doesn't prove OU.
Only the inextinguishable character of the energy source can prove "OU": this is always presumed in a sim when using a generator or a neg resistance, therefore OU can be shown and proved only ouside the frame of a simulation.

[GibbsHelmholtz]

Exn.,

I think you still believe that energy comes from nothing.  When we put a generator in a sim, it means that generator is a free energy source.  Where do negative resistance gets its energy from?  Somewhere, but we don't care.  We just tell the sim to include it in and stop questioning its programmer.

The definition you provide for negative resistance is limited.  We do not need the current to flow in the opposite to generate OU.  If you put in 10 and the neg. resistance give out 15, it's good enough.  Again, where do the neg. resistance gets its extra 5 units from... we don't care, as long as it does in reality.

[exnihiloest]

...
I think you still believe that energy comes from nothing.

You are wrong. I know that energy must come from somewhere. Sorry to quote myself: "If we speak of type 1, then OU in a closed system is extra-energy coming from nothing. It is likely that none of us think it is possible. We preferably refer to perpetual motion of type 2."
I'm among "us". Otherwise I wouldn't have used "us" nor "we".

When we put a generator in a sim, it means that generator is a free energy source.  Where do negative resistance gets its energy from?  Somewhere, but we don't care.  We just tell the sim to include it in and stop questioning its programmer.

That's what I said too.

The definition you provide for negative resistance is limited.  We do not need the current to flow in the opposite to generate OU.  If you put in 10 and the neg. resistance give out 15, it's good enough.  Again, where do the neg. resistance gets its extra 5 units from... we don't care, as long as it does in reality.

1) "My" definition is not mine, it is the definition that you can find anywhere. It is the same definition as a positive resistance! R=U/I. When U/I is positive, the resistance is positive. When U/I is negative, the resistance is negative. If the negative effect occurs only in a range around a particular polarisation point U and I, then the negative resistance is given by R=dU/dI (<0) although U/I>0.
2) "the current to flow in the opposite to generate OU" is not in the definition of a negative resistance, it is the consequence of the definition. R=U/I, therefore if I and U are of same sign, R is positive, if U and I are of opposite sign, meaning that the current flows in a reverse direction compared to the positive resistance, R is negative. So simple...

We don't care to know the real energy source in a sim. We can have a voltage generator, a current generator, a neg resistance or whatever else to generate energy.
But for a sim we need to know exactly how it provides voltage and current i.e. the U - I relation.
A voltage generator provides a voltage U at its terminals whatever the load and the current. A current generator provides a current I whatever the load and the voltage. A neg resistance provides a voltage U=-RI. You just need to put "U=-R*I" between the nodes of the neg resistance to modelize it by Kirchoff's laws: spice does it so.
It's really as basic as a positive resistance.

[GibbsHelmholtz]

You are wrong. I know that energy must come from somewhere. Sorry to quote myself: "If we speak of type 1, then OU in a closed system is extra-energy coming from nothing. It is likely that none of us think it is possible. We preferably refer to perpetual motion of type 2."
I'm among "us". Otherwise I wouldn't have used "us" nor "we".

That's what I said too.

1) "My" definition is not mine, it is the definition that you can find anywhere. It is the same definition as a positive resistance! R=U/I. When U/I is positive, the resistance is positive. When U/I is negative, the resistance is negative. If the negative effect occurs only in a range around a particular polarisation point U and I, then the negative resistance is given by R=dU/dI (<0) although U/I>0.
2) "the current to flow in the opposite to generate OU" is not in the definition of a negative resistance, it is the consequence of the definition. R=U/I, therefore if I and U are of same sign, R is positive, if U and I are of opposite sign, meaning that the current flows in a reverse direction compared to the positive resistance, R is negative. So simple...

We don't care to know the real energy source in a sim. We can have a voltage generator, a current generator, a neg resistance or whatever else to generate energy.
But for a sim we need to know exactly how it provides voltage and current i.e. the U - I relation.
A voltage generator provides a voltage U at its terminals whatever the load and the current. A current generator provides a current I whatever the load and the voltage. A neg resistance provides a voltage U=-RI. You just need to put "U=-R*I" between the nodes of the neg resistance to modelize it by Kirchoff's laws: spice does it so.
It's really as basic as a positive resistance.

Hehe,

It's just a reverse psychology we have sometimes.  I used to believe energy can comes from nothing.  It doesn't bother me until I see cases like money triangle and system where the big guy eat little guy and little guys eat even more little guys under the belief that something comes from nothing.  We're better than that. 

I know what you mean with the negative, but what if the case is like this:  We have a voltage source of 1 V and a 1 Ohm resistance.  When connected, we get 1 amp flow.  Nothing strange, positive resistance at play.  Now what if we connect and the voltage still reads 1 V but 2 amps flow.  R*I still positive, but there is something wrong with the resistance.  It gives more energy dissipation than the input.  That's what I mean.

[GibbsHelmholtz]

I've been thinking a little bit about negative resistor in reference to breaking down dielectric.  Is it the same as stripping electrons?  I have a vision that free electrons actually gives you more energy than the input into removing it.  

Let's say to remove the electron, the energy needed is Escape energy defines as minimum energy requires to separate electron from proton from infinitely close.  However, since electron in orbit, it already has some energy as a fraction of the Escape energy.  The input require to remove the electron in orbit is:

E(input) + E(orbit) = E(escape)

or, Velocity(input) + Velocity(orbit) = Velocity(escape)

Please excuse my model, I see it better with orbital mechanics.  

After the electron is free from the atom, we can convert its potential energy to kinetic energy.  We then have energy equals to the Escape Energy.  However, the input require to do this is less than the Escape Energy.

Is it this simple? Hm...

[Peterae]

SM describes the ability to run with gain and it appears to get this he requires fast send off, he relates this to the way a rectifier valve works which in my mind would equate to electron emission, so to run with gain we need to emit electrons, do something special so they pick up free energy on the way or gain energy and then collect the new total energy. simple really