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Note: I'm not an electrical engineer, I'm a materials engineer. So maybe this question has an obvious answer, but I couldn't find one with a Google search. I also have no idea if I tagged this properly, sorry!

I was reading through this open access paper on evolvable circuit components and was pretty much following it, right up until the demonstration section at the end. In the caption to the following figure, the authors note:

Measuring the current at the terminal where VIn1 is applied, as shown in panel c, causes the current from both inputs to have the same sign.

Why should the ammeter have that effect when it's moved from the position in a to the position in c?

Schematic and measurements

(I'm also a bit confused by the final graph in b; does the negative current reading here imply that current is flowing back from ground when V2 is applied after training?)

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    \$\begingroup\$ The resistor has been moved as well. \$\endgroup\$ – Ben Voigt Sep 26 at 18:02
  • \$\begingroup\$ @BenVoigt Sorry, it's been like 5 years since I've studied circuits, i didn't realize those would be electrically different \$\endgroup\$ – realityChemist Sep 26 at 18:40
  • \$\begingroup\$ V1,V2 are bipolar voltages and currents. But the training correction seems poor \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 26 at 20:39
  • \$\begingroup\$ @SunnyskyguyEE75 By that you mean they have different signs? I know, but what would they then have the same sign after the change in c? \$\endgroup\$ – realityChemist Sep 26 at 20:47
  • \$\begingroup\$ I assume the voltages were low impedance and constant voltage \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 27 at 1:52

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