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I am now analyzing the following Chua's circuit.

enter image description here

The triangle at the bottom denotes zero electric potential (I struggle to find the right symbol). \$N_R\$ is the Chua's diode. So the circuit is non-linear.

This version of the circuit can be modeled by the following system of ODEs, which can be found on Wikipedia.

$$ \frac {dx}{dt}=\alpha [y-x-f(x)],$$ $$ RC_{2}\frac {dy}{dt}=x-y+Rz, $$ $$ \frac {dz}{dt}=-\beta y. $$

x(t), y(t), and z(t) represent the voltages across the capacitors C1 and C2 and the electric current in the inductor L1 respectively.

My question is, how can this circuit work if it does not have a power supply? The only component supplied by the battery is the opamps inside the Chau's diode. Such power supplies are not part of the main circuit, so how can the main circuit gain power? The initial conditions of the differential equations are $$x=y=z=0, t=0$$ (since all electric potentials are zero before we finish connecting the circuit). Those initial conditions will generate the solution \$x=y=z=0\$ for all time.
If that is the case, then how can we observe the double scroll pattern? Everything should be constantly zero.

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  • \$\begingroup\$ The circuit you show is fully linear. Thus no bifurcations possible. \$\endgroup\$ – analogsystemsrf Jun 25 at 2:51
  • \$\begingroup\$ @analogsystemsrf R2 is the Chua's diode, which is non-linear \$\endgroup\$ – Jethro Jun 25 at 2:53
  • \$\begingroup\$ @analogsystemsrf Now I have changed it to make it more clear \$\endgroup\$ – Jethro Jun 25 at 2:55
  • \$\begingroup\$ You lost your ground symbol in the edit trail. \$\endgroup\$ – Scott Seidman Jun 25 at 13:14
  • \$\begingroup\$ Doesn't a chua diode always have an op-amp in it that is powered? \$\endgroup\$ – Keeta Jun 25 at 13:49
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The power comes from Chua's Diode.
It is not a passive device. It actively provides power as you can see quickly if you look at its i-v-curve:

i-v curve of Chua's Diode (image from Wikipedia)
i-v curve of Chua's Diode (image from Wikipedia)

The i-v-curve of Chua's diode occupies the 2nd and 4th quadrant. A passive device would occupy only 1st and 3rd Quadrant:

i-v-characteristics of power sources and passive devices (Image from Wikipedia)

See section "Types of I–V curves" in Wikipedia article Current–voltage characteristic.

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IT MUST SUPPLY POWER.

Follow this reasoning:

  • A transistor is a negative resistance (NR) device only when it is biased with some power source, so it is not a negative resistance thru the origin of V vs I.

  • The same is true for gas tubes and other passive NR devices.

  • A Power supply with a transistor with gain is a negative resistance combination.

  • If we say a load draws positive power then a generator provides negative power.

  • The part shown has negative resistance thru the origin, therefore, it is not a passive device.

  • thus the Op Amp example shown is an active realization of this transfer function
  • The left side uses a negative impedance bridge called a gyrator to a negative resistance thru the origin with a cap to simulate an inductive filter.

enter image description here

CHUA Simulator with pots

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  • \$\begingroup\$ What about the initial conditions? They make all voltage and current zero. What is wrong? \$\endgroup\$ – Jethro Jun 25 at 5:33
  • \$\begingroup\$ And you say the word transistor again and again, but there are no transistors on your diagram. I am really confused. \$\endgroup\$ – Jethro Jun 25 at 6:27
  • \$\begingroup\$ Op Amps are made from transistors . Got it? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 25 at 6:28
  • \$\begingroup\$ Oh I get it. But what about the differential equations in the question? They should not give zero solutions... What's wrong? \$\endgroup\$ – Jethro Jun 25 at 6:29
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    \$\begingroup\$ Oh hey, neat to see the chua example circuit I contributed being useful! \$\endgroup\$ – Hearth Jul 13 at 16:09
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Without knowing what a "Chua diode" is, one would get the impression from this question that this circuit is supposed to be a perpetual-motion or free-energy device.

The first thing I would look at in that context is the inductor. Unless you place the device in an RF-free zone such as a Faraday cage, the inductor will couple with the magnetic element of ambient EM radiation, near the resonant frequency defined by the RC parallel tank formed with C2, and thereby absorb some of that energy into the circuit. This is exactly how certain kinds of radio antenna work, and Nikola Tesla actually proposed a global electrical-energy distribution system on this principle.

In practice, a "Chua diode" is a theoretical construct, typically realised in practice by building it from op-amps. These op-amps require a external power supply which is, conventionally, not usually shown in the diagram. A typical op-amp for this purpose is the TL074, a quad package requiring only a single power supply for all four op-amps required for making a "Chua diode".

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Even if all currents and voltages being 0 is an equilibrium of your system of differential equations, if this system is chaotic, it must be an unstable equilibrium. As soon as even the tiniest current or voltage enters the system (as a result of picking up radio waves, or from static discharge from you touching the circuit while building it, or from moving the wires through the magnetic field of the Earth, or from noise in the amplifier in the diode, or any number of other sources), it will start kicking itself around the state space chaotically and not return to the equilibrium point.

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  • \$\begingroup\$ That's exactly what I want! Thanks! \$\endgroup\$ – Jethro Jun 25 at 21:48

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