Negative resistance, impact on voltage source?

Question

According to this, if \$R_1=R_2=R\$ then input resistance is \$R_{in}=-R_3\$, and if we assume that op amp is ideal. So negative resistance is in parallel with voltage source \$V_s\$.

• Am I understand this well, in this case ideal voltage source is acting as load and input resistance \$R_{in}=-R_3\$ is acting as generator?

• So how is it possible that ideal voltage source \$V_s\$ which internal resistance is 0, acting as load?

• If \$V_s\$ is a real voltage source with finite internal resistance, what impact on the voltage source, i.e. battery, will be in this case? Can we somehow destroy that battery with excessive current?

Answer 1

Yes, your opamp circuit presents resistance of -R3 to ground at the positive opamp input. I think your question is more about what will a voltage source do when presented with a negative resistance.

The current will flow backwards thru the voltage source. That means the negative resistance is producing power and the voltage source dissipating it. That's OK. In fact negative resistances always produce power unless the voltage across them is zero, just like positive resistances always absorb power unless the voltage across them is zero. This is one reason why we don't have negative resistors like we do positive resistors. The negative resistors would have to produce power.

To see who was paying attention when he was talking about negative resistances, my circuits professor in college ended by saying "... and I have a jar of them in my office. Anyone that wants to see one can come by later." and then looked around to see who laughed. I was surprised how many were staring blankly wondering why a few of us were chuckling.

I think the confusion comes from the fact that we rarely run into negative resistances, and that you are used to thinking of a voltage source as producing power, not absorbing it. If so, you need to broaden your outlook. The only thing an ideal voltage source is guaranteed to do is to hold the voltage across it constant. That is true whether is has to source or sink current to do it. There are instances in regular circuits where we have voltage sources that are intended to work by sinking current. That is basically what a shunt regulator is. A Zener diode is a passive component that does this.

You have to realize that a voltage source, particularly an ideal voltage source used for theoretical analysis as above, is not the same as a power supply. Power supplies may strive to emulate an ideal voltage source, at least for some limits within first quadrant operation, but a true ideal voltage source works for all currents from -∞ to +∞.

If the voltage source had a positive resistance, it would be the same as having an ideal voltage source with a positive resistance in series with it. Series resistances add, whether they are negative or positive. The effect from the voltage source point of view is simply the sum of the two resistances.

Note that if the voltage source resistance is too high, then the circuit becomes unstable. Consider a open-circuit negative resistance. It would be stable at exactly 0 volts, but as soon as there was a little voltage on it, it would quickly run away to either positive or negative infinite voltage, depending on the sign of the starting voltage.

Obviously real circuits can't go to infinite voltage. In the case of your opamp circuit, the opamp output can only to its supply rails. Once that happens, it will fail to operate as a negative resistance.

Answer 2

Let's analyze what's going on in this circuit:

Let's assume that the voltage at each input terminal of the opamp are equal.

That means through R1 we have the following current:

\begin{equation} I_1 = \frac{Vs}{R1} \end{equation}

This current can't come from the opamp input terminal, so it must come through R2.

\begin{equation} I_1 = I_2 = \frac{Vo - Vs}{R2} \end{equation}

Let's solve for the output voltage.

\begin{equation} Vs \cdot R2 = R1 \cdot (Vo - Vs)\\ Vo = \frac{Vs (R2 + R1)}{R1} \end{equation}

And finally, there may be some current flow through R3.

\begin{equation} I_3 = \frac{Vs - Vo}{R3}\\ I_3 = \frac{Vs - \frac{Vs (R2 + R1)}{R1}}{R3}\\ I_3 = Vs \frac{-R2}{R1 \cdot R3}\\ I_3 = Vs \frac{-1}{R3} \end{equation}

So our initial assumption of current flowing out of the positive Vs terminal was wrong, it's actually flowing in. So you are right that this circuit acts similar to a generator/charger for the voltage source since it's giving power back to the source rather than taking it out.

There are a few key take-aways:

1. The source voltage is lower than the opamp output voltage. How much lower is dictated by the R3 resistor.

2. In practice, the opamp is not ideal. It has a limited voltage supply limit, and can only output a maximum amount of current. Most jelly-bean type signal opamps can only source/sink a certain amount of current, say on the order of 40mA, so it may happen that with a particular setup the source will never reverse bias and it will drain as normal.

So what if the source isn't ideal and has some internal resistance? The equations will change slightly (you can analyze it yourself), but the end result is you'll get slightly less current flowing into voltage source.

Will this damage/destroy a battery source? Well, it highly depends on the battery you have, what actual resistors you have, and how close your opamp is to ideal. Certain batteries you should not try to charge (generic Alkaline batteries have a higher tendency to leak if you try to charge them).

Answer 3

As far as I understood, the circuit is just helping the generator \$V_s\$ to behave as its internal resistence is 0 by pumping into the load ( that has to be connected to \$V_s\$ ) an extra current. So there is no difference if the interlan resistance o the generator is already 0. In any case there is a limit to how much extra current can be pumped, depending on both Vs and the maximum output voltage of the amplifier.

Answer 4

There is no real mystery, the "negative resistance" term makes it sound like it is some magical device but it isn't.

First imagine a much simpler scenario. A battery H connected to another battery L of lower voltage via a resistor R. Simply, H will start "charging" L, which basically means that current will flow into it. How much current flows depends mostly on H's voltage and R.

Now, what if you control the value of H's voltage so that the amount of current that ends up flowing into L depended on L's voltage? Specifically, so that if L's voltage increases, the current flowing into it increases, and if L's voltage decreases then the current flowing into it decreases (proportionally). It would be mathematically equivalent to having a negative resistor instead. This is exactly what the op-amp circuit does. Our R is R3 in the circuit, our battery L is the Vs voltage source, and our special H battery that changes voltage according to L's voltage is the op-amp circuit, adjusting its output voltage so that our special condition is met.

This will work until the op-amp's output gets close to its supply voltage, lifting the curtain and exposing our trick.

Answer 5

My answer

It is simple and short: There is no use of this arrangement (negative "resistor" connected to a voltage source). In order to be useful, there must be a load connected in parallel. Then the voltage source will supply the load and the negative resistor will "help" it by passing an additional current through the load. Depending on the degree of the "help", various results can be obtained.

Example 1 : Load cancelling

As an illustration of this extravagant circuit technique, I will show in four steps how the resistive load of a voltage divider can be cancelled by an equivalent negative resistance. As a result, the voltage divider has the "feeling" that there is no load connected (open circuit).

1. Voltage divider unloaded. This humble circuit of two resistors in series (or a potentiometer) supplied by a perfect voltage source (with zero output resistance) behaves as a "bad" voltage source (with some output resistance). But when it is not loaded - Fig. 1, its "badness" does not manifest itself and its output voltage is exactly as it should be.

Fig. 1. Voltage divider unloaded

2. Voltage divider loaded. The problem of the high output resistance is that when a load RL is connected to the voltage divider - Fig. 2a, it "sucks" a current IL and the output voltage VL drops.

Fig. 2a. Voltage divider loaded by a resistor RL

This phenomenon can be seen everywhere in our life - if living beings are loaded, e.g. by a heavy load (Fig. 2b), they bend.

Fig. 2b. Gravitational analogy of a loading

The classic solution of this problem is to connect a voltage follower before the load to provide the load current instead of the voltage divider. This means to isolate the load from the voltage divider… to increase the load resistance "seen" from the voltage divider output.

3. Voltage divider "helped" - concept. But there is another more extravagant idea that can be borrowed from life - to "help" the voltage divider in parallel by an additional current source providing all the load current needed. This should be not the ordinary constant current source but a "proportional to voltage current source" that produces a reverse current I = -VL/RL.

We can implement this powerful idea by connecting in series a variable voltage source BH producing two times higher voltage VH= 2VL and another resistor R = RL acting as a voltage-to-current converter - Fig. 3a. This network acts as a negative resistor with resistance -RL that neutralizes the equivalent positive resistance RL and the result is infinite resistance (open circuit).

Fig. 3a. Voltage divider "helped" by an additional current source - concept

As an example of life, if someone has to raise a heavy weght - Fig. 3b, we can help it by an equivalent "anti-weght" (a powerful idea of mechanics that is widely used in lift systems, cranes etc.)

Fig. 3b. Gravitational analogy of an anti-weight

The brake booster in cars is another well-known non-electrical implementation of this idea. When the driver presses the brake pedal, the booster engages "in parallel" and assists the driver.

4. Voltage divider "helped" by INIC. In the end, all that remains is to implement this powerful idea through an op amp circuit - Fig. 4. It is simply a non-inverting amplifier (acting as the variable voltage source BH above) and a resistor R = RL in series to the op-amp output. This circuit is known as a "current-inversion negative impedance converter" (INIC).

Fig. 4. Voltage divider "helped" by an INIC

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The trick of load cancelling is clever but it implies the load is a steady resistor; the input voltage and the potentiometer can vary.

Example 2 : Howland current source

Another well-known application of this powerful idea is the legendary Howland current source aka Howland current pump - Fig. 5.

Fig. 5. Howland current source - classic circuit diagram

We can see in the conceptual circuit diagram (Fig. 6), that really the idea is the same - an imperfect current source (on the left) is "helped" by an additional current source (on the right) so that the load current stays constant when the load varies.

Fig. 6. The idea behind the Howland current source

The "helping" current source acts as a negative resistor with resistance -R that neutralizes the positive resistance R of the imperfect current source; so the resulting "internal" source resistance is infinite.

The operation of the Howland current source is visualized in Fig. 7 by voltage bars and current loops.

Fig. 7. The operation of the Howland current source visualized

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In contrast to the load canceller above, in the Howland circuit solution the load can vary. The negative "resistor" -R (INIC) neutralizes the steady "internal" resistance R of the imperfect current source.

Answer 6

It's really difficult to understand your questions (you might try editing your question to use shorter sentences), but I'll take a stab at the last one.

Yes, if the source resistance of V_S is greater than R3, the opamp will have net positive feedback and it will "run away". An ideal opamp will supply infinite voltage, but any practical opamp will stop at its supply rails, and at that point the current will be limited to the supply voltage divided by the (positive) resistance of R3.