Suppose we have this circuit:


simulate this circuit – Schematic created using CircuitLab

And replace the diode with a tunnel diode. If the voltage stored on the capacitor is close to the voltage in which the tunnel diode has negative resistance , can't this help oscillators run longer than they do or even never stop?


3 Answers 3


A tunnel diode does not have negative resistance -- it has negative differential resistance. A negative resistance is a power source, much like a positive resistance is a power sink, and that energy has to come from somewhere. When using a tunnel diode as an oscillator, you have to bias it to the negative differential resistance point of the IV curve. It draws power from the source that's biasing it while driving oscillations in the tank circuit.

  • \$\begingroup\$ Since the "negative resistance is a power source", what then is the "negative differential resistance"? \$\endgroup\$ Jan 4, 2023 at 6:59
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    \$\begingroup\$ The strongest thing I can think of is a power converter or just a kind of nonlinearity. You can think of it a bit like an escapement in a watch. You have a constant pull from a weight or spring (voltage) that drives a device that snaps between two states and drives the pendulum (driving the tank circuit.) You have a power source corresponding to the potential energy in the spring or weight that drives oscillatory motion using a nonlinearity. \$\endgroup\$
    – Evan
    Jan 4, 2023 at 17:14
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    \$\begingroup\$ Crucially the nonlinearity uses the spring tension or weight tension to push the pendulum towards larger amplitudes. The tunnel diode in an LC circuit does much the same, which can be seen if you trace out the position in the IV curve the tunnel diode during a cycle. However notice that during a cycle, you're always in the upper right quadrant of the IV space, meaning power is in total being consumed from a source. \$\endgroup\$
    – Evan
    Jan 4, 2023 at 17:17
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    \$\begingroup\$ Upper right and lower left correspond to power consumed by load, and the remainder correspond to power delivered by load. Notice a linear negative IV characteristic would go upper left to lower right! \$\endgroup\$
    – Evan
    Jan 4, 2023 at 17:23
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    \$\begingroup\$ I think the views are equivalent. I personally don't like to think in terms of a dynamic resistance -- it's just some thing that has a particular IV curve. I'll think in terms of dynamic resistance when the thing is used as a tunable resistance in a circuit. A dynamic resistance that drives a circuit is too weird for me, especially because the device is pumping oscillations in the circuit, and when I think resistance, I think dissipation. \$\endgroup\$
    – Evan
    Jan 6, 2023 at 16:39

The tunnel diode has an N-shaped IV curve. This means that when the voltage drop across it increases, its resistance increases as well. So, it can compensate "positive" resistance when connected in parallel to it.

The LC tank shown (C1 in parallel to L1) has some internal "positive" resistance. It should be neutralized by equivalent N-shaped negative resistance (tunnel diode) in parallel or by S-shaped negative resistance in series.


It is assumed that this circuit is treated as a series negative resistance oscillator. When your negative resistance oscillator starts, assuming the correct bias to the negative resistance element with an S-shaped IV curve, oscillations build up in amplitude or,

Case 1: when

$$ |R_{negative}| \gt \frac{L_1}{R_{coil + bias} C_1 } = R_{losses} $$

Case 2: when

$$ |R_{negative}| = R_{losses} $$

the oscillations will be steady-state.

Case 3: when

$$ |R_{negative}| \lt R_{losses} $$

oscillations decay to zero. The term \$ { } R_{coil + bias} { }\$ refers to the lumped resistance of the coil and bias resistors and also includes the losses in the diode.

When the power to the circuit is removed, case 3 occurs and oscillations will die down in amplitude to zero and the voltage across the capacitor goes to zero.

In order for this circuit to oscillate, energy must be applied to the diode with a biasing scheme or nothing happens. You might include a resistor in series with your circuit to represent bias resistors and other losses inherent in the other components.

Circuit operation is as follows. The bias on the diode sets the capacitor voltage so that when the diode is on (conducting), the capacitor is discharging through the diode and coil. When the capacitor has discharged long enough, the diode switches off and quits conducting. The voltage across the capacitor builds up from the bias and switches the diode on again. The next cycle starts and oscillation ensues. The voltage across the diode (the breakover voltage) sets the point to where the capacitor quits charging. Once the breakover voltage condition is met the diode goes into the negative resistance region. The above 3 cases still apply.

Also the capacitor \$ C_1 \$ and the inductor \$ L_1 \$ must be in series resonance i.e. their reactance must be approximately equal:

$$ X_{coil} \approx X_{capacitor} $$

The reason that this is approximate is that one of the reactance terms includes information about the amplitude of oscillations. However the condition is close enough to be an approximation. or:

$$ f_{oscillations} \approx \frac{1}{2 \pi \sqrt (L_1C_1)} $$

The term

$$ \frac {L_1}{R_{coil + bias}C_1} $$

was adapted from "Electronics Engineering Handbook" Third edition by D.G. Fink and D. Christiansen page 13-46.

  • \$\begingroup\$ I might add that this is a linear model. It does not include any nonlinear elements. An example of this is a Van Der Pol oscillator which uses a tunnel diode as the active device. \$\endgroup\$
    – user277251
    Jul 28, 2023 at 23:10

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