# 0 current in a circuit using tunnel diode

Suppose we have a diode with negative differential resistance of -40 ohm in the region of 0.3-0.5 V and we put it in parallel with a 40 ohm resistor:

simulate this circuit – Schematic created using CircuitLab

Now Rt = R1*R2/R1+R2 = -1600/0 = infinity. So no current should flow. Is that correct?

Current will flow, but small changes in V1 will not change that current by much.

Differential resistance $$\\ne\$$ resistance.

Edit: See below for a representation of the tunnel diode transfer function (from here):

The negative differential resistance is the negative slope between Ip and Iv. But you may notice that I is always positive. In addition to that positive current through the tunnel diode we also have the 10mA going into the 40$$\\Omega\$$ resistor.

• well then what is the difference between differential resistance and resistance? Apr 24, 2022 at 23:33
• @JunSeo-He Consider an ideal DC-DC converter. Say 10 V to 5 V, 1 A. The output power is 5 W, which is equal to the input power. So the input current is 0.5 A. However, if the input voltage increases, the input current will decrease to maintain constant output power due to the output voltage regulation loop. A negative incremental resistance, but at 10V, the supply still draws 0.5 A, which looks like a 20 ohm resistance. Apr 24, 2022 at 23:46

So no current should flow.Is that correct?

No, this is not correct.
You are over-simplifying things. Take a look at this picture from Wikipedia

You see that from 0 to 13mV the
resistance of this tunnel diode is about (delta 0.013V/0.084A= 1.5 ) +1.5 ohms.
Then, from 13mV up to 54mV it becomes (delta 0.051V/0.0067A= 7.6 ) -7.6 ohms.
Then from 67mV and up it becomes (delta 0.020V/0.0031A= +6.5) +6.5 ohms.

So if you were to put a 7.6 ohms resistance in parallel with the diode you would
end up with a Voltage .vs. Current curve that would look like this.

Explanation: When the tunnel diode negative resistance involves a decrease in current (from 13mV up to 64mV) then an equivalent increase in current from the parallel resistor of +7.6 ohms will compensate and the sum of both device current will create a zero increase current for this region. Beyond 64mV both device exhibit a positive resistance hence the current shows a steady increase.

So, when you mention that "infinite" resistance you are half correct. Only in the region of negative resistance of the tunnel diode will an equivalent impedance equate to infinity. The sum of current could never be zero because some finite quantity of current is necessary for the tunnel diode to enter the negative resistance region, but infinite impedance is possible, only within the tunnel region of the diode.

• Nice work! +1 for the thorough research. Nov 22, 2023 at 22:04

To complement the other answers, and since you ask about it in a comment, let me recall that for a two-terminal, possibly nonlinear resistance the differential resistance is defined as

$$r = \frac{\mathrm{d}v}{\mathrm{d}i}$$

where $$\v\$$ is the voltage across the element and $$\i\$$ is the current crossing it, and the derivative is calculated around the operating point. Likewise, we can define the differential conductance as

$$g = \frac{\mathrm{d}i}{\mathrm{d}v}.$$

For a linear resistor, for which the consitutive relationship is given by $$\v = Ri\$$ or $$\i = Gv\$$, the differential resistance and the differential conductance are constant, equal to $$\R\$$ and $$\G\$$, respectively, independent of the operating point. For a nonlinear resistor, like a tunnel diode, it is $$\i=f(v)\$$ and the differential parameters change, instead, with the operating point, that is, $$\r = r(i)\$$ and $$\g = g(v)\$$.

Now what happens when you connect a nonlinear element such as a tunnel diode in parallel to a linear resistance?

Label as 1 the nonlinear element and as 2 the linear one. Since the two elements are driven by the same voltage, it's better to use conductances. The total current is

$$i = i_1+i_2 = f(v)+Gv.\tag{1}$$

For the tunnel diode $$\f(v)>0\$$ when $$\v>0\$$, and thus the current is never zero for positive $$\v\$$. Now, let's differentiate (1) with respect to the common applied voltage $$\v\$$,

$$\frac{\mathrm{d}i}{\mathrm{d}v} = \frac{\mathrm{d}i_1}{\mathrm{d}v}+\frac{\mathrm{d}i_2}{\mathrm{d}v} = g(v)+G.$$

Thus, if for a certain applied voltage $$\g(v) = -G\$$, we get

$$\frac{\mathrm{d}i}{\mathrm{d}v} = 0,$$

that is, the slope of the tangent line to the $$\i\$$-$$\v\$$ curve of the compound element is horizontal, as shown in the answer by Fred Cailloux. This, however, does not imply a total current of zero, as explained in the above.

# Revealing the secret of negative differential resistance

## Dynamic resistance

The "mystic" phenomenon of negative differential resistance (NDR), particularly the OP's N-shaped NDR, can be easily understood by the extremely simple and intuitive concept of dynamic resistance. It can be easily demonstrated by a simple electrical experiment where a variable voltage source V drives a variable resistor R (rheostat). If you don't mind, let you control the source and I will control the rheostat.

(The picture is from my Wikibooks story about NDR).

## Modified resistance

Now imagine that you are increasing the voltage across the rheostat and, at the same time, I start to increase its resistance. Depending on the rate of change, the current will increase more slowly, will not change or even will decrease. This will create the illusion of increased, infinite or "negative" resistance.

## Negative resistance

So this type of negative resistance is just a vigorously changing dynamic resistance (in the same direction as the input voltage source).

## Neutralized resistance

If we connect such a "super dynamic resistor" in parallel with an ordinary resistor with equivalent "positive" resistance (OP's circuit), we will get a "super dynamic current divider" that maintains a constant (not zero) total current in such a simple way. In terms of resistance, this means the total differential resistance is infinite... but it is hard to imagine.

See more in my Wikibooks story about NDR.

EDIT: I went back a year and a half to supplement my fancy story about this unique phenomenon with CircuitLab experiments. So it became more convincing and attractive.

# CircuitLab experiments

I slightly changed the values of the OP resistors to more convenient ones for the purposes of these concept experiments.

## Experimental setup

According to the explanations above, two resistors are connected in parallel - R1k with a 1 kΩ "positive" resistance, and R-1k with a 1 kΩ "negative" resistance. I have put both names in square quotes because these resistances are neither negative nor positive, but the most common ohmic resistance known since the 19th century. So the negative differential resistance of -1 kΩ is implemented by a variable resistor whose resistance changes from 111 Ω to infinity when Vin varies from 1 V to 10 V.

simulate this circuit – Schematic created using CircuitLab

We need to observe the currents flowing through the resistors. To simplify the schematics, we can combine the resistors with the ammeters into one device ("visualized resistor"). To do this, open the parameters window of each of the two ammeters and set the corresponding internal resistance.

## Step-by-step experiments

To understand the mechanism of this so-called "N-shaped negative differential resistance", let's first examine the circuit at three successive values ​​of the input voltage - 1, 2 and 3 V.

Vin =1 V, R-1k = 100 Ω: At 1 V input voltage, the 1 k positive resistor consumes 1 mA. The negative resistor has a 111 Ω resistance, so 9 mA current flows through it, and the total current consumed is 10 mA.

simulate this circuit

Vin =2 V, R-1k = 222 Ω. When the input voltage increases to 2 V, the negative resistor increases its ohmic resistance to 250 Ω. Now the positive resistor consumes 2 mA but the negative 8 mA, and the total current consumed is again 10 mA (i.e., the left current increases but the right current decreases, and their sum I remains constant).

simulate this circuit

Vin =3 V, R-1k = 375 Ω: Next, the input voltage increases to 3 V, and the negative resistor increases its ohmic resistance to 428 Ω. The positive resistor consumes 3 mA but the negative 7 mA, and the total current consumed is, as usual, 10 mA... and so on...

simulate this circuit

## Automated experiment

To sweep the negative resistance, we can simulate it using a behavioral current source I-1k that produces a current 10 mA - IR.

simulate this circuit

## Graphical representation

From the graphs below, we see that as the input voltage increases, the current through the positive resistor increases and through the negative resistor decreases because its resistance increases. The result is a constant common current which is seen by the input voltage source as an infinite differential resistance.

# Application

simulate this circuit

simulate this circuit

simulate this circuit

simulate this circuit

simulate this circuit

# Conclusions

• The N-shaped negative differential resistor (e.g. a tunnel diode) is a dynamic resistor that increases its static (ohmic) resistance when the voltage across it increases.

• If connected in parallel to an equivalent "positive" resistor, it neutralizes its resistance so that the equivalent resistance is infinite.

• @Root Groves, I updated my answer from a year and a half ago to show the nature of the negative differential resistance of elements with an "N-shaped IV curve" such as the tunnel diode. I hope this helps you to clarify the issue. Nov 24, 2023 at 7:54