Creating a circuit with negative differential resistance

Question

Is it possible to construct a circuit with negative differential resistance by only using components that have no negative differential resistance (e.g., resistors, diodes, BJTs, etc)? If possible, what are the examples. If not, is there any theorem suggesting so?

By negative differential resistance, I mean a circuit where the current decreases when the voltage increases. Note that components with negative differential resistance (such as tunnel diode) can’t be used to construct the circuit in question. Power source can’t also be used.

Answer 1

Why does voltage decrease when current increases?

By negative differential resistance, I mean a circuit where the current decreases when the voltage increases.

Well, true... But what is the physical meaning of this statement? How does this magic work? Is it something useful and therefore desirable? If so, what can we use it for? Let's understand it thinking as humans and explain it without using meaningless verbal clichés.

Basic idea

The purpose of a negative resistor is to do the opposite of what a "positive" resistor does, i.e. to add rather than dissipate power. True negative resistors do this by adding voltage in series or current in parallel from their own built-in sources. But how is the humble resistor to do so when it does not have that? The only thing it has is resistance and that is something that dissipates power.

But wait, it can reduce its resistance and thus increase the power added by an external source!

So the idea in general, as we know it from life, is to initially do something bad and then start reducing it, creating the illusion that we are doing good. In this particular case, the negative differential resistor (NDR) has some initial resistance which creates a voltage drop ("bad"). But then it begins to decrease its resistance as the current increases, thus creating the illusion that it is adding power ("good"). Simply put, NDR acts as a dynamic resistor.

NDR acting this way (including OP's one) are called "S-shaped" or "current-driven" while the opposite NDR (tunnel diode, lambda diode in the Heart's answer, etc.) are called "N-shaped" or "voltage-driven".

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Let's now explore this phenomenon in detail through step-by-step CircuitLab experiments, first with a "positive" resistor and then with an S-shaped NDR.

Positive resistance

The behavior of a 1 kΩ positive resistor matches our intuition confirmed scientifically by Ohm's law.

I = 1 mA, V = 1 V, R = 1 kΩ

^{simulate this circuit – Schematic created using CircuitLab}

I = 1.5 mA, V = 1.5 V, R = 1 kΩ

^{simulate this circuit}

I = 2 mA, V = 2 V, R = 1 kΩ

^{simulate this circuit}

We also see graphically that when we increase the current through the resistor, the voltage across it increases proportionally, and the resistance does not change (R is static). The resistor IV curve does not move.

Note that on the graph, the voltage is plotted (as is generally accepted) along the abscissa and the current along the ordinate.

Negative differential resistance

Conceptual circuit

To understand exactly how NDR does this magic, let's simulate it with a variable resistor R.

I = 1 mA, V = 2 V, R = 2 kΩ: Its initial (static) resistance is 2 kΩ but in this static state we cannot say how much NDR is.

^{simulate this circuit}

I = 1.5 mA, V = 1.5 V, R = 1 kΩ: When the current rises to 500mA, the trick is that we reduce R by 1 kΩ, and the voltage drop instead of rising, drops.

^{simulate this circuit}

I = 2 mA, V = 1 V, R = 500 Ω: Then the current increases by another 500 mA, but we reduce R by 500 Ω, and the voltage drops even more.

^{simulate this circuit}

In the graphical representation, when the current increases, the resistor IV curve rotates counterclockwise, and the voltage decreases. The intersection points lie on the -1 kΩ NDR IV curve (in green).

Why voltage decreases when current increases

The explanation of this mysterious phenomenon (never done in textbooks) turned out to be very simple:

Basically, Ohm's law is a function of one variable (input quantity) - Vout = Iin.R. So as we increase the input current Iin, the output voltage Vout increases proportionally with the factor R (constant). If we begin to vary the resistance at the same time as the current, Ohm's law becomes a function of two variables (input quantities) - Vout = Iin.Rin. Depending on the rate of change of R, we have the following cases:

• First, if we decrease the resistance at a smaller rate of change than that of the current increase, the voltage will increase but less, and we will have the illusion of lower resistance. An example is a poor quality Zener diode.

• Then, if we decrease the resistance at the same rate of change as the current, the voltage will not change at all, and we will have the illusion of zero resistance. An example is a perfect Zener diode.

• Finally, when we decrease the resistance at a higher rate than that of the current, the voltage starts to change in the opposite direction, and we think of it as a negative resistance. An example is a neon lamp.

Implementation

The NDR "helps" the input source by changing its resistance in the appropriate direction. This behavior can be implemented by a 2-terminal transistor circuit with positive feedback as the dashed circuit below.

^{simulate this circuit}

The resistors set the region of negative resistance and the slope of the curve there through the loop gain. If it is too high, the circuit becomes a Schmitt trigger.

Note that on the graph, the current is plotted along the abscissa and the voltage along the ordinate because CircuitLab puts the input quantity on the abscissa.

Simplified version: The circuit can be further simplified by omitting the base-emitter resistors...

^{simulate this circuit}

... but the negative resistance range becomes very small.

Why positive feedback?

Now I will try to answer this "never-answered textbook question" as well.

Negative feedback can help us to decrease the resistance in transistor circuits with low gain... or almost to zero it in op-amp circuits with huge gain (the so-called virtual ground). But more than that (to go below zero) it cannot; there is no way that the gain can become more than infinity:-) Let's see it in the 2-transistor circuit above after we remove one or the other transistor.

Without Q1

Thus, Q2 acts as a "transistor diode" with negative feedback between collector and base. As we know, such a configuration acts as a voltage stabilizer whose internal resistance is close to zero.

^{simulate this circuit}

As you can see in the graph below (in the same region as above), the circuit IV curve represents small but positive resistance.

Without Q2

Then let's remove Q2. Now Q1 acts as a "transistor diode" whose internal resistance is close to zero.

^{simulate this circuit}

The result is the same - the circuit IV curve represents the same small positive resistance.

Positive feedback comes to our aid, which makes resistance go beyond zero. To introduce it, we connect the two transistors in this interleaved (collector-base) fashion forcing them to drive each other. If the loop gain is more than one, the circuit becomes a latch.

Op-amp analogy. For the same reason, in a VNIC (current-controlled true negative resistor), in addition to negative, there is also positive feedback, and the voltage of the virtual ground drops below zero. Let's then investigate this dual negative resistance phenomenon.

True negative resistance

I will briefly consider here only the so-called voltage-inversion negative impedance converter (VNIC) because of three reasons:

• It corresponds to the transistor circuit above showing negative differential resistance.

• It is another example of using positive feedback in negative resistance circuits.

• The other answers show schematics of only the dual current-inversion negative impedance converter (INIC); so it would be interesting to see the dual version.

It does the same as the above circuits do, but not through dynamic resistance but through dynamic voltage. So the op-amp's role is to add a voltage proportional to the current that virtually reduces the resistance.

Virtually decreased resistance

If in a transimpedance converter we greatly reduce the gain (10), the voltage at the "virtual ground" will not be zero but will have a significant positive value (91 mV). The 10 kΩ‎ resistance will be virtually decreased about 10 times and the input current source will "see" 914 Ω‎.

^{simulate this circuit}

Virtually zeroed resistance

If we enormously increase the amp gain (100000), the op-amp will completely compensate for the voltage drop across R, and the virtual ground voltage will be almost zero (only 10 uV). As a result, the 10 kΩ‎ resistance will be virtually decreased to only 0.1 Ω‎.

^{simulate this circuit}

But even if we keep increasing the gain, we will not get the voltage (resistance) below zero.

Negative resistance

We know the remedy - positive feedback. To introduce it, we connect the R1-R2 voltage divider between the op-amp output and non-inverting input. As a result, the voltage at the inverting input changes its polarity; hence the name "voltage inversion". The resistance "seen" by the input current source is R = -V/I = -10 kΩ‎.

^{simulate this circuit}

Conclusions

Negative vs positive resistance

• In both, there is a linear relationship between current and voltage (Ohm's law), but negative resistance adds power while positive resistance dissipates power.

• So, when added to equivalent positive resistance, the negative resistance neutralizes it and the result is (virtual) zero.

Differential vs absolute negative resistance

• Both are dynamic but negative differential resistor is actually a dynamic positive resistor while the absolute negative resistor is a dynamic source (not a resistor).

• The combination of a negative differential resistor and source constitutes a true negative resistor.

S-shaped vs N-shaped negative differential resistance

• In S-shaped (current-controlled) negative differential resistor, the resistance vigorously decreases when the current increases; so it behaves as an "overacting voltage stabilizer".

• In N-shaped (voltage-controlled) negative differential resistor, the resistance vigorously increases when the voltage increases; so it behaves as an "overacting current stabilizer".

Positive vs negative feedback

• To make the stabilizers above overact, it is necessary to apply moderate positive feedback...

• ... or to combine it with negative feedback.

• To ensure stability, the negative feedback must dominate over the positive one.

See also another related answer of mine.

Answer 2

Absolute negative resistances without as external energy source cannot exist, as they would violate the laws of thermodynamics. A normal (positive) resistor puts out heat to the surroundings – voltage times current gives us the power dissipated While a positive resistance consumes power from current passing through it, a negative resistance produces power (or amplify).

From Frequency of Self-Oscillations, by Janusz Groszkowski:

In a positive resistor, electric energy is absorbed under the action of the flowing current and transferred into heat or other forms of energy. A negative resistor, on the contrary, has the ability to deliver electric energy which can be used outside it. Thus, the negative resistor is like arrangement containing a source of electric energy.The negative resistors which may be realized with modern electronics are, as a rule, non-linear.

From Optoisolation Circuits: Nonlinear Applications in Engineering, by Ofer Aluf:

Is it more correct to say that a circuit element has a negative differential resistance region than to say that it exhibits negative resistance because even in this region, static resistance of the circuit element is positive, while it is the the slope of the resistance curve which is negative.

In other words, the components listed below, use some specific region of I-V, curve which shows a differential negative resistance.

From Wiki:

List of negative resistance devices

• Tunnel diode, resonant tunneling diode and other semiconductor diodes using the tunneling mechanism.
• Gunn diode and other diodes using the transferred electron mechanism IMPATT diode,TRAPATT diode and other diodes using the impact ionization mechanism.
• Some NPN transistors with E-C reverse biased, known as negistor.
• Unijunction transistor (UJT) thyristors.
• Triode and tetrode vacuum tubes operating in the dynatron mode.
• Some magnetron tubes and other microwave vacuum tubes.

With operational amplifiers is possible to make a NIC (Negative Impedance Convertert), which utilizes a combination of active and passive components to generate a negative input-output impedance relationship.

Below an image from Wiki:

As you've asked about implementations with BJT, a NIC can be built using transistor based circuits, through feedback and biasing techniques. There are current and voltage versions of NIC (INIC, VNIC). An example of INIC's design using BJTs appears on A thesis on the design and application of negative impedance converter devices in network synthesis. There, other configurations of NIC are presented.

Answer 3

Edit: I misread the question, which referred to differential resistance, presumably meaning an V-I curve with a region of negative gradient. In that context my answer below is not a good one, except in the sense that the designs have exclusively negative gradient. I also use an explicitly prohibited additional power supply. Sorry about that. Circuit Fantasist's answer nails it.

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Since a negative resistance must necessarily be a source of energy, Your insistence that a power source cannot be used is absurd.

In the circuits shown above, it's clear that current is flowing inside the voltage source from higher potential end to lower. Current is positive from the perspective of the voltage source. That means, by passive sign convention, power \$P\$ in the voltage source is:

$$ P = I \times V = +2mA \times +2V $$

Positive power indicates that the voltage source is receiving energy, and that energy is coming from the -1kΩ resistor, which a real-life resistor obviously cannot do.

The negative impedance converter comes close to achieving what you ask, by employing an op-amp that can derive the required energy from its own power supply:

^{simulate this circuit – Schematic created using CircuitLab}

In that design, one side of source V1 must be grounded, and current \$I\$ will be:

$$ I = -\frac{V_1}{R_{X}} $$

It's the equivalent of the lower circuit, with an actual negative resistance (which is only possible in a simulator, of course), with the caveat that one side of the source must be grounded.

The inconvenience of having to ground one side of the source can be overcome by combining two negative impedance converters, as show here:

^{simulate this circuit}

This design permits you to apply arbitrary potentials to nodes A and B, and have that source of potential difference both sink and source current regardless of their common-mode value.

The trick is to have their outputs share the common-mode potential, finding an equilibrium where the junction between R2 and R3 (node C) has a potential \$V_C\$ exactly half way between the input potentials \$V_A\$ and \$V_B\$:

$$ V_C = \frac{V_A + V_B}{2} $$

That's shown on voltmeter VM1. This common-mode potential is a surrogate "ground" for each negative impedance converter, about which the pair can operate. You can replace R2 and R3 with a single resistance of 10kΩ. The only reason I used two resistors was to illustrate the principle of common-mode at work here.

Consequently, \$V_A\$ is half of \$V_1\$ above \$V_C\$, and \$V_B\$ is half of \$V_1\$ below it, and you can see intuitively that one op-amp will source current \$I\$ and the other will sink that same current.

This design requires that all resistances be extremely precise. Any imbalance in the symmetry of this circuit will result in an imbalance of these currents, which need to be identical.

Also you must ensure that \$V_A\$ and \$V_B\$ always lie within the op-amps' acceptable input voltage range, which will be constrained (as usual) by the op-amps' own supply potentials.

Answer 4

Here's a very simple way to do it, a circuit known as a "lambda diode":

^{simulate this circuit – Schematic created using CircuitLab}

When the voltage source is close to zero volts, the JFETs behave like a resistor, with a linear I-V curve. But as the voltage increases (in the positive direction), the JFETs begin to turn off, causing the current to reach a maximum and then fall exponentially until the gate junction breaks down.

This only works in one direction; the lambda diode has a diode-like I-V curve in the negative direction, as the gate junctions of the two JFETs are forward biased.

Obviously, this would also work with depletion-mode MOSFETs, but JFETs are generally easier to get ahold of.

Answer 5

Is it possible to construct a circuit with negative differential resistance by only using components that have no negative differential resistance (e.g., resistors, diodes, BJTs, etc)? [...] Note that components with negative differential resistance (such as tunnel diode) can’t be used to construct the circuit in question.

You said tunnel diodes are excluded because they're already negative-resistance devices, which would make the solution too easy - but BJTs are okay. Fortunately for me, your question still contains a loophole that permits a trivial answer: if you abuse BJTs carefully, BJTs are negative-resistance devices too under the right conditions. Although these modes of operations are not normally used, they have legitimate niche applications.

There are two ways of achieving that.

Base-Emitter Junction Reverse Breakdown

The simplest way to create negative resistance from a BJT is to intentionally induce a base-emitter junction breakdown. Negative resistance is exhibited during the breakdown, demonstrated by the fact that it's possible to make oscillators similar to tunnel diode circuits.

Source: BJT negistor applications, by sv3ora

This configuration is known among hobbyists as the "negistor". Most generic small-signal transistors should work, including the 2N2222. Only a small voltage is needed, around 12 V or so. But beware that reverse-biasing a base-emitter junction can cause gradual device degradation, so it's likely not stable enough for long-term use.

Source: BJT negistor applications, by sv3ora

Collector-Emitter Junction Reverse Breakdown

The negistor oscillator works, but the base-emitter breakdown is a slow process as it's relatively "gentle". On the other hand, a collector-emitter junction breakdown is a more violent process due to the "avalanche breakdown" effect. In this process, a local hotspot increases the mobility of more charge carriers, which in turn creates more hotspots. This semiconductor phenomenon is similar to the Townsend discharge process found in nature, in which one electron knocks out more electrons away from their gas molecules, causing an exponential growth of free electrons, which is responsible for creating electric arcs in the air.

In the following circuit, we charge a 20 pF capacitor from a 150 V power supply via a 220 kΩ resistor. Once the voltage across the capacitor is greater than the collector-emitter breakdown voltage, an avalanche breakdown suddenly occurs, discharging a sharp pulse into the 50 Ω resistor.

In power electronics, secondary breakdown effect is normally destructive to the transistor and is a common source of failure. But here, the total pulse energy is limited by a 20 pF capacitor, so the transistor will continue to work, forming a relaxation oscillator with periodic discharges.

As you can see, this technique is useful for creating fast pulse generators. With even a common transistor like a 2N3904, I personally measured a signal rise time of 970 picoseconds, well above the bandwidth of basic oscilloscopes. Here, the waveform is measured with a 1 GHz oscilloscope. It's amazing to see that such a fast pulses can be created from one simple component, especially in the 1980s.

(Warning: Basic oscilloscopes have 1 MΩ inputs so the output is safe to probe with a 1:10 probe. But, if you want to repeat this experiment with a high-speed oscilloscope like this one, you must attach a 30 dB attenuator to protect your 50 Ω input stage from destruction by 150 V!)

Other transistors are capable of generating even faster pulses. According to a widely-read article by Jim Williams,

82% of samples of the 15V high-speed switch 2N2369, manufactured over a 12-year period, were capable of generating avalanche breakdown pulses with rise time of 350 ps or less.

There are also transistors specifically designed and optimized for use in this breakdown mode, known as avalanche transistors.

The disadvantage of all avalanche transistor circuits is that a high voltage is needed to induce collector-emitter breakdown, but it's a worthwhile sacrifice. A very weak low-current voltage source to gradually charge a capacitor is already enough.

Reverse Breakdown is not the same as Negative Resistance

A common follow-up question is, "A Zener diode also breaks down, why can't it be used as a pulse generator, then?"

It's worth mentioning that although both methods make use of reverse breakdown, but not all reverse breakdowns imply negative resistance. Oscillators work because of the negative resistance exhibited by some particular kinds of reverse breakdown. During the breakdown, its IV curve must have a discontinuous "snapback". This phenomenon is found in spark gaps, gas discharge tubes, thyristors (SCRs), Shockley 4-layer diodes, and BJTs in "negistor" and "avalanche transistor" configurations. This is what makes a relaxation oscillator possible: If you build a capacitor discharger, the device will keep being ON for a while even after the capacitor voltage drops below the initial breakdown threshold, allowing the discharge to complete.

Ordinary Zener diodes also experience reverse breakdown, but their IV curve doesn't have a "snapback", so they don't have negative resistance. If one attempts to build a relaxation oscillator, capacitor voltage would stay at a point of equilibrium, rather than generating periodic discharges.

Further Readings

• Wikipedia Article: Avalanche transistor

• Linear Technology Application Note 47: High Speed Amplifier Techniques, Appendix D: Measuring probe-oscilloscope response

• Blog: Avalanche Pulse Generator Build Using 2N3904

Answer 6

By negative differential resistance, I mean a circuit where the current decreases when the voltage increases.

Here is an example of a simple circuit that has this feature. When the applied voltage [Vcc] is between 4V & 10V the input current decreases as the applied voltage (Vcc) increases.

However, outside of that voltage range, the current does not follow this rule. The simulation result shows how the current in D1 responds as the voltage at [Vcc] starts at 0Vdc, then increases linearly up to 30Vdc.

Circuit Performance
The graph below plots I in the diode vs voltage applied to node [Vcc].
At Vcc= 4.0V, I=5.14mA.
At Vcc=10.0V, I = 0.48mA.
Differential resistance:
Change in voltage = 10 - 4 = 6.0V.
Change in current = 0.48mA - 5.14mA = -4.66mA.
Resistance = 6.0V / -4.66mA = -1.29kΩ

Note that the current is a minimum when the applied voltage is 10V; as the applied voltage increases above 10V, the current starts to increase slowly, but at 21V the rate of increase sharply accelerates - this is undesirable, and hopefully will be addressed in a future refinement of the circuit.

Circuit Description & Operation.
Q1 is a constant current sink, but needs the applied voltage [Vcc] to reach about 3.5V before its current becomes "constant". Q2 then reduces the voltage at the base of Q1 in response to increasing applied voltage at Vcc, which in turn reduces the current in Q1 collector.

D1 is not really required; it was included to prevent accidental reverse conduction in a practical circuit and to provide a convenient means of showing the current flow in the simulation.

Image below: Schematic

Voltage sources V3 and V4 are only used to bias the transistors, they do not contribute to the current flowing in the input terminal. They are set to 2.8V and 1.4V, both of which are multiples of 0.7V, which is the forward voltage of a diode - which is a hint as to how to implement these in a real circuit (hint: use multiple forward-biased diodes). Alternatively use a circuit known as an "amplified diode" (often seen biasing the output stage of a class AB audio amplifier), in which case the voltages used for biasing can be adjusted to almost any value by changing the resistors, refer image below:-

EDIT 2024-MAR-27: Improved Circuit
The circuit below is an improvement on the previous ckt posted above. It is still only made up of passive parts: resistors and BJTs. The performance is enhanced as follows:

1. It enters the negative resistance region at a lower voltage (better);
2. Once the current reaches its minimum value, it remains at that value for voltages up to 30V (rather than increase when the input voltage exceeds 20V, as the previous circuit did).

Circuit Performance
The graph below plots I in the diode vs voltage applied to node [Vcc]. The cursors shown are at the start and end of the negative resistance region, as follows:
At V_Stim= 2.00V, I = 5.09mA.
At V_Stim=10.0V, I = 1.00mA.
Differential resistance:
Change in voltage = 10 - 2 = 8.0V.
Change in current = 1.00mA - 5.09mA = -4.09mA.
Resistance = 8.0V / -4.09mA = -1.96kΩ

The image below shows the same chart, but now the cursors have been moved to the end of the negative resistance region, up to the maximum permitted applied voltage of 30V. These cursors show that the current remains quite constant below 150uA for this voltage range.

Circuit Description & Operation
Q1 is a constant current sink, but needs the applied voltage [V_stim] to reach about 1.9V before its current becomes "constant". Q2 then reduces the voltage at the base of Q1 in response to increasing applied voltage at Vcc, which in turn reduces the current in Q1 collector.

The circuit is quite similar to the previous circuit; the main changes are:

1. The constant voltage sources have been implemented using the "amplified diode" as suggested previously.
2. The voltage sensing circuit has been improved, Q5 & Q2 form a voltage follower with low offset voltage, so that the current in R3 is proportional to input voltage; this current is subtracted from the bias voltage at Q1 base such that current in R1 (hence current in Q1 collector) is reduced linearly as input voltage increases.

Image below: Schematic

Current sources I1 & I2 can be implemented in various ways using only resistors and BJTs. If a constant power supply is available, eg: +12V, then I1 & I2 can be replaced with just one resistor each to give equivalent current; in this case remove R7 and connect current sources to node [Vcc]. Performance is almost identical provided Vcc > 8VDC.

Other improvements:
Circuit performance could be greatly improved with the addition of an error amplifier to correct the error in R1 current.

To Adjust the Transfer Function:
The transfer function has three main parameters:
Starting current: the current at the start of the negative resistance region.

Starting voltage: the input voltage at the start of the negative resistance region.

Slope: The rate at which current reduces for each volt increase at [V_stim].

1. Starting current:
   Adjust R1. This current is inversely related to the value of R1.

2. Starting voltage:
   Adjust R9. Increasing R9 will increase the starting voltage.

3. Slope:
   Adjust voltage divider R4 & R5 to set the slope. Increasing R5 will increase the slope (making the plot of input current vs voltage steeper). This will also set the voltage at which the input current hits zero.

Answer 7

Any sane switching mode power supply, when powering some load, will behave as negative diferential resistance over their rated input voltage range.

This is because their efficiency does not change much with the change of the input voltage so they draw almost constant power, determined by the load demand.

E.g. a 100W, 95% efficient PSU will draw 1.05A at 100V input, but less than 0.5A at 240V input.