2

I am thinking of the circuit pictured below, which, while the op-amp is operating in its linear region, gives rise to the relation

$$ v_0 - v_1 = -i_0 r , $$

where the direction of positive current is indicated on the diagram below. The thing that's really bugging me is that I also have the relation

$$ i_1 = - i_0 ; $$

that is, if \$ i_0 \$ is flowing into the NIC, then \$ i_1 \$ is also flowing into it. This really disturbs me because it means that I can't black-box this as a single-port circuit element --- charge isn't conserved. This is exactly what Wikipedia seems to be doing, however, when discussing Chua's Circuit.

Have I made an algebra error; that is, is \$ i_1 = - i_0 \$ not true? Or is there some method of dealing with a dually-emitting source like this? Again, I'm trying to get at Chua's Circuit, so if you would like to conduct an example analysis for me, that would be a prime target. Alternatively, I think that if you hook this guy to a capacitor then you get an oscillator of some sort, so that would make for a more compact example.

Thanks in advance for any help I can get on this. If anything needs clarifying, I would be more than happy to explain myself to you.

enter image description here

Jordan
  • 151
  • 6

3 Answers3

1

I think, the equations as given by you are correct. There is no contradiction. Both currents (\$I_0\$ and \$I_1\$) are going into the circuit and into the opamp's output. Where is the problem?

EDIT: Here are the formulas, replacing \$r\$ with \$R\$ and \$S\$ with \$R_s\$:

$$I_1 = \frac{V_{-}-V_{1}}{R} = \frac{V_{out}-V_{-}}{R_s}$$

$$I_0 = \frac{V_0 - V_{out}}{R_s} = -\frac{V_{out} - V_+}{R_s} = -\frac{V_{out} - V_-}{R_s}$$

so \$I_0 = -I_1\$, and

$$V_0 - V_1 = I_1R = -I_0R$$

Greg d'Eon
  • 3,807
  • 4
  • 24
  • 50
LvW
  • 24,857
  • 2
  • 23
  • 52
  • I thought the equations will be correct only if \$v_1=0\$. – nidhin Feb 27 '15 at 09:22
  • Why? Something wrong in my equations? – LvW Feb 27 '15 at 12:45
  • Nop. My mistake. Your calculations are correct. – nidhin Feb 27 '15 at 13:44
  • I understand the equations, and I understand that there is no explicit contradiction, but how should I look at this when it's plugged into a larger circuit? For example, a [relaxation oscillator](https://en.wikipedia.org/wiki/Relaxation_oscillator). – Jordan Feb 27 '15 at 14:01
  • Is it your primary goal only to understand the working principle of the relaxation generator - or, in particular, based on the NIC-theory? The most direct and most simple explanation is based on the general feedback theory - without using the NIC approach. – LvW Feb 27 '15 at 14:09
  • My primary goal is to understand the implementation of Chua's Circuit on the Wikipedia page. I think that this would be easiest if I could black-box the NIC, but a feedback explaination would be fine if I can generalize it. Additionally, I am disturbed that people would actually call the NIC a thing and give it it's \$ v = -ir \$ equation if it can't actually be black-boxed, so there's a good deal of desire to learn how to deal with the NIC as an abstract circuit component. – Jordan Feb 27 '15 at 14:27
  • No - it is not an "abstract" component; it can be seen as a voltage controlled current source (a positive resistor is a current sink) - and the current goes through the controlling source. Of course, you can model it as a black box. But note that it works as a linear NIC only in case the negative feedback overrides the positive feedback. – LvW Feb 27 '15 at 14:42
  • So, do you know if you could do an example analysis of a relaxation oscillator, viewing the NIC as a voltage-controlled current source? It's okay if it goes nonlinear for a little bit, as long as it stays black-boxed. – Jordan Feb 27 '15 at 14:49
  • Didn`t you read the working principle description in Wikipedia? Or don`t you understand it? – LvW Feb 27 '15 at 17:22
  • I don't see a section in Wikipedia's article about relaxation oscillators called "Working principle description", but I've read "Comparator-based relaxation oscillator\General Concept" and "Example: Differential Equation Analysis of comparator-based Relaxation Oscillator". I actually do not follow the example they give; if you could walk me through it, though, I would be very grateful. In particular, I'm having difficulty seeing how they get \$ A = V_{out} \$, since \$ V_{out} \$ is time-dependent. – Jordan Feb 27 '15 at 17:52
  • It`s not too complicated. Start with capacitor "empty" (no charge). So we have no negative feedback and the output is at the pos. (or neg.) rail, per accident. This holds as long as the voltage at the pos. input is larger than the voltage at the negative input, which is rising slowly (the cap charges to pos. values.) As soon as the cap voltage exceeds the voltage at the pos. terminal, the output quickly jumps to the neg. max. voltage (supply rail) anfd the whole procedure starts again - now from the neg. side. – LvW Feb 27 '15 at 18:17
  • Okay, I think I understand now! Thank you! My error was that \$ V_{out} \$ actually is a constant (on the time intervals we're looking at). There's one thing still bugging me, though: is the ground in the picture a physical Earth-ground, or is it just a measuring point for 0 V? Both of the grounds have current flowing into them, so if they're not Earth-grounded then Kirchhoff's Current Law doesn't hold. – Jordan Feb 27 '15 at 19:30
  • OH MY GOODNESS look [what I just found](http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=13&ved=0CGQQFjAM&url=http%3A%2F%2Fwww.faculty.umassd.edu%2Fxtras%2Fcatls%2Fresources%2Fbinarydoc%2F3622.doc&ei=etvwVOC4FJGWyASRzIHoDw&usg=AFQjCNFAqtSALZw_xA8tv4-EH13L8GKz7g&sig2=mITQPFu99JDTH0EoWxDYMA)! I think that this clears everything up for me ... I was just always thinking of the power supply circuit as totally separate from the oscillatory circuit. I'll build a relaxation operator tonight and report back if this was the hole in my thinking. – Jordan Feb 27 '15 at 21:19
  • I think that the relaxation oscillator is a bad example -- the "op-amp" isn't actually an op-amp at all; it really is important that it's a comparator. Otherwise, one can't just assume that \$ V_{out} \$ jumps straight to \$ V_{dd}\$. I used a 411 op-amp to build this to test, though, and it worked fine, so I'm starting to doubt that a NIC can be black-boxed at all, and I'm getting confused about op-amp rules . . . so, do you think that a NIC can be black-boxed, even in a four-terminal box? By the way, thank you for your patience with me so far. – Jordan Mar 01 '15 at 00:08
  • The unit called "opamp" is an opamp - independent on its particular application. This device can be operated in linear mode (amplification) or with positive feedback as a comparator (with hysteresis!). – LvW Mar 01 '15 at 09:53
  • How can I tell which it does? I read [this question](http://electronics.stackexchange.com/questions/112472/how-are-positive-and-negative-feedback-of-opamps-so-different-how-to-analyse-a), but I can't find an input voltage on the oscillator, so I can't use nidhin's answer exactly. I tried leaving the op-amp output as \$ A (v_+ - v_-) \$ and taking the limit \$ A \rightarrow \infty \$ once I was done, but I got the same answer as when I assumed that negative feedback dominated (the wrong answer). Also, assuming negative feedback in the NIC is correct, and the oscillator is nearly the same . . . – Jordan Mar 01 '15 at 14:12
  • Don`t be confused - the linked question has nothing to do with your problem (it concerns LINEAR operation only). However, the relaxation oscillator works strongly non-linear - between the output extremes only!. The opamp acts as a switch between +Vs and -Vs. – LvW Mar 01 '15 at 16:53
  • How can I tell when it's linear and nonlinear; for example, in the case of the NIC (linear) versus the case of the oscillator (nonlinear)? – Jordan Mar 01 '15 at 17:28
  • So, if you're still listening: I've been working on the r-oscillator problem, and I think that it only oscillates if the capactitor is connected to a VNIC. Additionally, even if one assumes linear behavior initially from the op-amp, one finds that eventually it experiences hysteresis. I'm sidetracked at the moment trying to figure out what happens if one replaces the capactitor with a capacitor and inductor, but if this gives you any ideas, I think that it's a lead . . . – Jordan Mar 06 '15 at 14:03
1

I'm pretty sure that this cannot be viewed as a single-port element unless \$v_1\$ is grounded (or, at least, fixed). If \$v_1\$ is grounded, then this can indeed be viewed as a single-port element in which the current and voltage at the input are related via the equation \$v_0 = -i_0 r\$, as long as the op-amp is operating in its linear region. This happens as long as $$2 |v_0| \leq |V_{CC}| , $$ where \$V_{CC}\$ is the op-amp's supply voltage, assumed symmetric.

I think that the NIC will behave similarly if \$v_1\$ is not grounded, or if the op-amp's supply voltage is not symmetric, but at the very least the condition for linear operation will change.

Having current flowing out of both of the NIC's terminals is perfectly fine; it's being sourced by the op-amp's supply rails. If you only look at the NIC's input terminal, you won't even notice the effect (i.e., you can't put it just anywhere in a circuit and expect Kirchoff's laws to hold).

The reason that the analysis with the relaxation oscillator wouldn't work is because the NIC displayed above is an INIC, whereas a relaxation oscillator is hooked to a VNIC. These two circuits behave the same in their linear regions, but when their op-amps saturate they behave very differently. In particular, the VNIC can do a hysteresis thing that keeps the op-amp entirely outside of its linear region, neglecting transients, which makes the relaxation oscillator oscillate. The INIC cannot support the correct form of hysteresis and, thus, will not oscillate (in the relaxation oscillator setup).

Jordan
  • 151
  • 6
0

From that wikipedia entry:

A "locally active resistor" is a device that has negative resistance and is active, providing the power to generate the oscillating current.

The opamp will drive current in and out of its output pin in order to equalise its terminal voltage. This current flows in its power rails. So it's not a dual-port element but a four-port element with the other terminals going to the power supply.

pjc50
  • 46,540
  • 4
  • 64
  • 126
  • Do you know if you could do, as an example, an analysis of, say, a [relaxation oscillator](https://en.wikipedia.org/wiki/Relaxation_oscillator) for me? It's when this thing is plugged into a larger circuit that I lose the ability to reason about what happens. – Jordan Feb 27 '15 at 14:04
  • I'm far too lazy to do any kind of analysis! I call your attention to http://crossgroup.caltech.edu/chaos_new/Chua_docs/works.html which says that you should just treat that subelement as a resistor of value -R and substitute it in accordingly. – pjc50 Feb 27 '15 at 15:01
  • ☺ Thank you, but it seems to me that they're wrong. Treating it as a simple negative resistance in the relaxation oscillator leads, I believe, to the conclusion that the voltage across the capacitor increases exponentially into infinity. Obviously, this will be cut short by the maximum voltage output of the op-amp in our non-ideal negative resistor, but point is that this thing can't just be treated like a negative resistance. – Jordan Feb 27 '15 at 15:09