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enter image description here

I know this circuit is called 'Negative Impedance Converter' and for this case, works as -100 ohm resistance.

What is the voltage of the lowest bifurcation when I apply triangular wave(0 to 0.1 to 0 V, period 0.1 s)?

I thought the current flows from the ground, so the voltage should be negative, but I checked that's not correct with my LTspice simulation.

How can I calculate the voltage properly?

(+) LTspice simulation enter image description here

I apologize for my haste,(I should've fully checked some errors I've mentioned.) but I think the question is being developed by your advice.

Here is my second try. The input voltage has uV scale and position of Vcc, Vee were corrected. There are 2 new questions. enter image description here

Why is the voltage of right next point to the input(plotted over the circuit) not identical with the given input??

And why can't I acquire the voltage of the blue-squared point over R3(by clicking there)?

(+)(might be last edit...) I found the position of resistances in the figure above is wrong, so I've corrected them.

I found that the supply voltage of op amp affects the result(voltage of blue-squared point over R3). I know that 'voltages of +- input voltage be identical when there is feedback resistance' and 'output voltage of op-amp is A(V+-V-)'.

How I understand the output voltage can be nonzero is 'V+ and V- is actually have very small difference and A is sufficiently large'. Is it correct?

jihoon
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  • What do you mean by "voltage of the lowest bifurcation"? "How can I calculate the voltage properly?" - you want to calculate it manually rather than using LTspice? – Bruce Abbott Dec 11 '20 at 08:27
  • Yes. The bifurcation means the point(marked as blue square) over the R1. I want to calculate it manually so that the result match with the simulation. The simulation says that the voltage of the point is about 2/3 of input wave. – jihoon Dec 11 '20 at 09:02
  • What did your simulation result show? Please embed a picture. – Andy aka Dec 11 '20 at 09:19
  • Green one is given and blue one is the result. – jihoon Dec 11 '20 at 10:25
  • In your diagram, the supply voltages are not correct!! I suppose, you have simulated the passive part of the circuit only and the opamp is "dead". – LvW Dec 11 '20 at 11:32
  • I didn't see that. I'll try again with much smaller peak of voltage right now.(Is it right attempt?) And I just have noticed Vcc and Vee are changed each other. – jihoon Dec 11 '20 at 11:46
  • All I have to do is showing NIC works as negative resistance in circuits with providing any example with LTspice. Please check edited version of my question and if you feel the need, give an another example that help me to achieve my goal. – jihoon Dec 11 '20 at 12:09

1 Answers1

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In your first attempt, the supply voltages were not correct. OK - you have corrected this.

At the same time, however, you have also changed the node where the input voltage is applied.

Note that there are two basic options for an NIC:

(1) Short circuit stable (input voltage at the non-inv. input) with a working negative feedback.

(2) Open-circuit stable (current source resp. voltage source with a suffieciently large source resistor at the inverting input) with some positive feedback.

Your second circuit cannot work because the pos. feedback is dominating (negative feedback signal is shortened by the signal source). The output voltage is latched at the supply voltage.

LvW
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