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I am studying the working principle of tunnel diode and I am reading for a voltage range it creates a negative differential resistance.

If we have a current source supplying the circuit with some current, if the current is inside the negative differential resistance region, which will be the voltage of the diode, the green voltage , the orange voltage or the blue voltage?

enter image description here

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    \$\begingroup\$ There are three possible voltage values for the current value shown. \$\endgroup\$ Mar 14 at 22:49
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    \$\begingroup\$ Yes you are correct. \$\endgroup\$
    – Miss Mulan
    Mar 14 at 22:49
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    \$\begingroup\$ The title implies that you are looking for unique ("magical") uses of tunnel diodes. However, the body doesn't say that. Am I missing something? \$\endgroup\$ Mar 15 at 11:35
  • \$\begingroup\$ @MikeWaters you don't miss anything.I think this behavior of the tunnel diode is magical. \$\endgroup\$
    – Miss Mulan
    Mar 15 at 17:06
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    \$\begingroup\$ The "magical" behavior of the tunnel diode in the middle section of its IV curve is that it abruptly and uncontrollably changes its resistance from minimum to maximum or vice versa. As a result, there is a voltage jump at a constant current. \$\endgroup\$ Mar 16 at 7:41
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It depends on how you changed the current before:

  • If the current had not exceed the top (Ipe), then it is the "orange voltage" staying at the first part of the diode IV curve having a positive resistance.

  • If the current had exceed the top (Ipe) and then had not fallen under the bottom (Iv), then it is another (eg, "blue":) voltage that would stay at the third part of the diode IV curve having a positive resistance.

  • You will never see the "green voltage" since it is an extremely rapidly changing voltage (in the middle negative resistance region the operating point is unstable and it "jumps" in a horizontal direction).

The "magical use" of this bistable mode of operation is to make the tunnel diode memorize (act as a latch). For this purpose, the tunnel diode is initially biased by a current Iv < Ibias < Ipe. Then, it is toggled by adding or subtracting a sufficient current so that I > Ipe or I < Iv.

See more about the bistable mode of the N-shaped negative differential resistance and particularly, how the diode "jumps", in the related Wikibooks story.

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  • \$\begingroup\$ So in case we have a switch and we close the switch since the current through the diode had been 0 before we closed the switch then the voltage drop will be equal to the orange voltage? \$\endgroup\$
    – Miss Mulan
    Mar 15 at 1:11
  • \$\begingroup\$ "...under the bottom peak...".Did you mean trough? \$\endgroup\$
    – Miss Mulan
    Mar 15 at 1:18
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    \$\begingroup\$ @Miss Mulan, I have edited it. Maybe I had not said it clearly enough... but I think you understand me. If you want, I can explain here what the clever trick of the tunnel diode is. As a student many years ago, I really wanted to know how it "jumps" and what the trajectory of its jump is... but I managed to figure it out much later. Then I wrote this Wikibooks story... \$\endgroup\$ Mar 15 at 3:42
  • \$\begingroup\$ so it depends from the previous 'state' of the tunnel diode(e.g how much current has passed through the diode) before we measure the current state? \$\endgroup\$
    – Miss Mulan
    Mar 15 at 17:02
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    \$\begingroup\$ @Circuitfantasist Yes. I guess I was speaking generally - to mean that the "green voltage" is not generally inaccessible, although that may be true in OP's specific case. \$\endgroup\$
    – J...
    Mar 16 at 13:15
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If you use a current source, then you can't get a reasonable result.

I built a current/voltage tracer a while back that uses a current source - it sets the current then measures the resulting voltage across the diode.

About a month ago, I got hold of some GI103A tunnel diodes and tried to replicate the typical negative resistance curve trace of a tunnel diode.

This is the result:

enter image description here

It just skips right over the negative resistance range.

My software automatically joins consecutive measurements, so there's a straight line across the negative resistance range. In reality, it just skips right over the whole negative resistance area. The trace should just skip from about 20 millivolts to about 420 millivolts.

To get the negative resistance trace, you have to use a voltage source and measure the current. That's how the traces in the datasheets were produced.

Even Tektronix IV tracers produced IV traces that skip the negative resistance range:

enter image description here

Figure 11A from that Tektronix document shows the IV traces of two tunnel diodes made with a Tektronix 575 curve tracer. Both traces skip the negative resistance range.

This document ftom RCA describes how the negative resistance range can be traced.

enter image description here

The short form is that it takes an IV tracer with a low internal impedance - basically, a voltage source rather than a current source. The tracer "forces" a voltage on the diode and measures the resulting current.

I recommend reading both of the linked documents (Tektronix and RCA.) There's a wealth of information on theory and practicalities of tunnel diodes in both of them.


That "skipping the negative resistance range" is what makes tunnel diodes switch so fast.

As long as the driving source "looks like" a current source (the source impedance is higher than the absolute value of the negative resistance) the voltage range of negative resistance is "forbidden" - from the current, it could have three voltage values. That's impossible, so that range is "skipped," leading to an immediate change in voltage to something outside the forbidden zone.

When the source "looks like" a voltage source to the diode (source impedance is lower than the absolute value of the diode negative resistance) then there's only one current for that voltage - there's no forbidden zone, so it just passes whatever current is required.

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    \$\begingroup\$ (+1) Thanks for posting the link to the RCA Tunnel Diode Manual: I still have my hardcopy, but now I have backup! \$\endgroup\$
    – Ed V
    Mar 15 at 0:05
  • \$\begingroup\$ A Schmitt trigger and latch are reasonable tunnel diode applications where it is driven by current:) \$\endgroup\$ Mar 15 at 11:22
  • \$\begingroup\$ @JRE what do you means the negative differential resistance region is forbidden?Do you mean that it cannot be the green voltage? \$\endgroup\$
    – Miss Mulan
    Mar 15 at 16:56
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    \$\begingroup\$ @MissMulan: Right. When using a current source with a tunnel diode, the green voltage is not possible. No voltage in the range of negative resistance is possible with a current source. \$\endgroup\$
    – JRE
    Mar 15 at 17:01
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That's impossible to tell. It depends in which region the diode operates, exactly as the figure you show informs you!

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It can be either. Because the IV characteristics aren't a one-to-one function only knowing the current isn't enough to determine voltage.

As for which voltage it will be, I'm not sure. Based on how the current source works you may be able to force operation on either the green or orange line.

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