Talk:Negative impedance converter

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you can easily explain this by drawing 2 voltage sources, each at Vs, but otherwise unconnected. from here the amplifier makes one of the Av*Vs and the other is just Vs. Av*Vs - Vs is applied across the resistor. this is the explaination for what it does, and you can easily show that the two sources could be merged into a single source. Using a single source makes the circuit "simpler" but adds more confusion as it seems there is an additional feedback path. only when the reader realizes that there is no feedback because there is no series resistance on the ideal voltage source does this become clear.

This also allows the application section to make sense. with an added series impedance, another feedback path is formed if there is only one source. this prevents the negative resistance from canceling out the series resistance, and can cause the circuit to latch in cases where an increase in output voltage would lead to a larger differential voltage.

likewise, the analysis's statment of the summing point constraint is not terribly valid. there are plenty of opamp circuits that fail this, furthermore, this circuit is very close to being one of such circuits. —Preceding unsigned comment added by 63.229.181.173 (talk) 04:04, 1 May 2008 (UTC)

Lengthy derivation

Is the lengthy input impedance derivation in the "Basic circuit and analysis" section really necessary? I'm inclined to just state the final result and link to the input impedance article or something. -Roger (talk) 17:52, 23 February 2009 (UTC)

Negative impedance is a novel result, but it's true that it seems like something trivial that follows from the circuit. Nevertheless, I think most references at least have a brief commentary on circuit operation. I think it would be fine to reduce the section to a paragraph, but I don't think it should be removed entirely. An alternative is to take the approach used on, say, Bernoulli principle where derivations are hidden by a collapsed table that can be expanded with a click. —TedPavlic (talk) 14:00, 24 February 2009 (UTC)

Moved content

Circuit dreamer's discussion has been moved to user space. Hellbus (talk) 17:47, 4 February 2012 (UTC)

Questions on this circuit description

I think a circuit analysis (math) is essential, especially with a speculative circuit like this. These NIC circuits all seem to ignore some basic properties of opamps and rules about circuit analysis. For example there is an element that is both a current source and a voltage source but there is nothing that says what the relationship is between the two. There is an independent I and V so I suspect the analysis is incorrect. The reference cited on the negative resistance page has a date 1953. Are there any more recent references or any evidence someone has actually built this circuit and demonstrated the claimed negative resistance effect?Zen-in (talk) 07:06, 14 December 2013 (UTC)

I believe there is an error in the transfer function for this circuit going back to the first editor of this page. He used the wrong method to derive the transfer function by starting with Va = Vb instead of using the equation Vo = (Va - Vb)α, where α = the open loop gain. That method is ok for some opamp circuits but is incorrect in this case. The correct transfer function I believe is:
$R_{\text{in}} \triangleq \frac{V_s}{I_s} = -R_3 \frac{R_1}{2R_1 + R_2}.$
I'm not convinced the circuit would really work as stated. An increase in Vs might just cause it to rail.Zen-in (talk) 08:42, 14 December 2013 (UTC)
There is nothing speculative in this circuit. It is based exactly on the "basic properties of opamps and rules about circuit analysis". To show it, here is another ("functional") analysis based on the properties of the classic non-inverting amplifier and the two versions of the Ohm's law (I = V/R and V = I.R). I (a still banned Wikipedian) give it to you, Wikipedians editing this talk page, as a gift (Spinningspark, sorry; I hope you will forgive me this little "trespass":)
Current-inversion NIC (INIC)
The op-amp keeps up the voltage drop VR1 across R1 equal to the input voltage VS (the op-amp acts as a voltage-to-voltage converter or voltage follower) by passing a current I2 = VR1/R1 = VS/R1 through the resistor R2 (so R1 acts as a voltage-to-current converter). The current I2 creates a voltage drop VR2 = I2.R2 across R2 (so R2 acts as a current-to-voltage converter). The op-amp keeps up the voltage drop across R3 equal to the voltage drop across R2 (the op-amp acts as another voltage-to-voltage converter or voltage follower) by passing a current IS = VR3/R3 = VR2/R3 = (I2.R2)/R3 = ((VS/R1).R2)/R3 through the input source. So, the input resistance is really -R3.R1/R2.
If R2 = R3 (the usual case), the circuit injects the same current IS = -I2 that would be drawn by the resistor R1 if it was connected directly to the input source. So, it behaves as a "negative resistor" R1 having the same voltage as the positive R1 but with an inverted current; thus the name of the circuit - "negative impedance converter with current inversion" (INIC). The circuit "inverts" every positive/negative element (resistor, capacitor or inductor) connected in the place of the resistor R1 to the "opposite" negative/positive element with equivalent impedance; it is just a "current inverter" (actually, the very INIC consists of the two resistors R2 and R3, and the op-amp).
"The circuit would really work as stated" (would be stable) if the internal resistance Ri of the input source is low enough; in the case of an "ideal" voltage source (Ri = 0), it is unconditionally stable. More precisely, the inequality Ri/(Ri + R3) < R1/(R1 + R2) must be satisfied so that the negative feedback dominates over the positive one. Circuit dreamer (talk, contribs, email) 23:50, 14 December 2013 (UTC)
Your first statement I2 = VR1/R1 = VS/R1 is incorrect. VR1 = VO - VS. You cannot ignore VO in deriving the transfer function. And just stating a few "Ohms Law" equalities and then pulling a transfer function out of the air is not algebra. Zen-in (talk) 00:09, 15 December 2013 (UTC)
I checked my derivation and found the mistake. The transfer function as stated in the page is correct. I copied the derivation to user space, where most of the other discussion is. Zen-in (talk) 02:32, 15 December 2013 (UTC)