# 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 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 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.

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.

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).

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.

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