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By electrical analogy (force->current) with a mechanical system, I have reached an electrical circuit which is as follows:

It consists of a current source and a (diode + inductance in series) in parallel with this source. The diode is the equivalent I have found for a nonlinear term in my mechanical system, with a behaviour in \$I = \frac{V}{R(1-V/V_0)}\$, so an approximately equivalent the schematic with an ideal diode far from breakdown would be:

schematic

simulate this circuit – Schematic created using CircuitLab

(With an ideal diode, I guess a good approximation would be to have a strong resistor in parallel with the diode, and a weak one in series with it.)

In the mechanical analogy, we plug the left hand side of the circuit with a separate system. One of the ones we use provides for \$t<0\$ a 1 A current and then from \$t=0\$ a lower positive current. We are interested in the electric power/mechanical rate of work that this system provides to the external system, transiently. The voltage is expected to be negative at the top left and go to zero in permanent regime.

I have little background in electricity and would like to know whether this is like any classical circuit? I hadn't intended to make the question so long, but I realized in your answers that I needed to be more precise. Thanks for the answers so far anyway, I gather that this is probably nothing close to a classical circuit.

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  • \$\begingroup\$ If you edit the question and press Ctrl-M there's on on-site schematic editor. I don't know about others but I can't understand that diagram. \$\endgroup\$ – PeterJ Jul 19 '13 at 14:50
  • \$\begingroup\$ What do you mean by 'intensity generator'? \$\endgroup\$ – Adam Lawrence Jul 19 '13 at 14:58
  • \$\begingroup\$ It provides a constant intensity I1 whatever the voltage. Maybe I should have said a current source, sorry, I was no trained in English in this field! \$\endgroup\$ – Joce Jul 19 '13 at 15:11
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    \$\begingroup\$ As drawn, this circuit has a serious issue: the 1A current source produces a CCW current but the diode only allows a CW current. By KCL, your circuit gives the equation \$1 = 0\$. \$\endgroup\$ – Alfred Centauri Jul 19 '13 at 15:40
  • \$\begingroup\$ What's the top left end connected to? or is it just the current source in parallel with the led and inductor \$\endgroup\$ – Iancovici Jul 19 '13 at 15:41
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For a mechanical analogy, we can choose force and velocity as the analogs of voltage and current respectively or the other way around.

Since the (ideal) diode in your circuit allows current in one direction but not the other, the mechanical analog would be a device that allows velocity in only one direction or force in only one direction.


However, the circuit as drawn has a serious issue. If the (ideal) current source produces a current that circulates CW, all is well.

However, since the diode does not allow a CCW current, and since an ideal current source produces a current regardless, the circuit is inconsistent if the current source produces a CCW current.

One fix is to add a resistor in parallel with the current source (making it a non-ideal current source) so that there are two paths for current. When the polarity of the current is such that the diode is "off", the entire current from the source will circulate through the resistive path.

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  • \$\begingroup\$ Yes, this corresponds quite much to what I intended. I didn't mean to put an ideal diode, I used a diode because the Shockley diode equation was a good approximation to a nonlinear mechanical term present in my original model. \$\endgroup\$ – Joce Jul 20 '13 at 20:52
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I'm trying to figure out what you are trying to do and an answer is the only feasible place I have for my "mastications".

The current source, to me, can represent a constant force and if applied to a spring, you would extend the length of the spring giving it some mechanical energy that can be recovered by releasing the force. The spring can be thought of as the inductor.

The diode (again in my brain) is rather like a one-way valve or maybe the rear cog on a bycycle; it allows energy to be put into the wheel to produce forward motion but does not allow the forward motion of the wheel to drive the pedals.

So, given that your circuit makes no practical sense electrically (the diirection of the current will not flow through the inductor because of the diode direction), what may you be trying to achieve mechanically if I swap the electrical components about?

Well, if you reversed the direction of the diode "force" could be exerted on the "spring" and energy would be stored. If you then turned off the "current source" the spring would, (because of the diode) recoil rapidly generating the equivalent of a peak in voltage that is theoretically infinite.

And, if you added a diode and capacitor (correctly), the "recoil" of spring energy could be captured and transferred to the capacitor thus preventing to infinite voltage scenario.

The capacitor would be like a flywheel being accelerated by the energy from the spring. Energy in spring(coil) is transferred to flywheel(capacitor).

But you are two components short and a diode direction correction from achieving this - what does it do mechanically - dunno? What does it do electrically - it's a voltage boost circuit as used in boost regulators.

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  • \$\begingroup\$ Ah, I see now. Sorry, should have read more carefully. \$\endgroup\$ – Phil Frost Jul 19 '13 at 20:03
  • \$\begingroup\$ OK, now that I have added the resistors, I believe the circuit can have some practical sense: the current source will act on the spring/inductor through the 1000 Ohm resistor until an equilibrium is reached. \$\endgroup\$ – Joce Jul 21 '13 at 9:40
  • \$\begingroup\$ @Joce this circuit makes no sense - the diode is always reverse-biased and plays no-part in operations. What are you trying to prove - do you have some mechanical analogy in mind? \$\endgroup\$ – Andy aka Jul 21 '13 at 10:53

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