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# How to plot the I-V curve of a tunnel diode?

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I am trying to understand tunnel diodes by experimenting with them. Research tells me they can have negative resistance, and can be used to build a high frequency oscillator. Tunnel diodes are supposed to have this I/V characteristic:

I don't really understand negative resistance, so I thought plotting the I-V characteristics myself would help me understand. Unfortunately, that didn't work.

What I did:

• Connect PSU in series with a protective, current limiting resistor, and the tunnel diode.
• Increase/decrease voltage level, in steps as fine as mV, and measure corresponding current values, as fine as 100 µA.
• Record each point on a graph of current as a function of voltage.

Here is my test setup:

As I increase voltage from 0 to IpeakV, the current increases as expected. Once approaching the IpeakV, the current suddenly jumped from tens of µA to some 400/500 mA. In other words, I just missed the most important measurements, those of the negative resistance region.

Why can I not measure the current as soon as the gradually increasing voltage V enters the negative region? How can I tell the tunnel diode not to "skip" the tunnel?

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## How to test a Tunnel diode in 25 words or less.

With DC bias = ~490mV with fine tuning and a small signal swing of 60mV you can generate a IV negative slope of -16 Ohms.

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Apparently you want to measure the current/voltage relationship of a tunnel diode. The tricky part is that the voltage isn't unique for currents over parts of the range. From your question:

Note that the current is still unique as a function of voltage. One solution is therefore to sweep the voltage and measure the current.

This needs to be done carefully, since small changes in the voltage can result in large changes in the current. You also need the voltage source to be very stiff (low impedance) so that the wildly changing current drawn by the device doesn't cause control instability.

One way to achieve this is with high open loop gain and negative feedback. If you are only feeding back the output voltage, and the control signal to drive the pass element is reasonably insensitive to the resulting current, then the system will be stable.

I haven't done this, but I'd probably start with an opamp driving an emitter follower, where the emitter voltage is fed back and is therefore ultimately what is regulated. The opamp output drives the base. The base voltage won't need to change much, even with large changes in the load current.

You might also want to add some current limiting so that you don't fry the device under test when the voltage goes a little too high. This limit needs to be above I1 in your diagram.

Another approach is to put a resistor in parallel with the diode under test. Make the resistor small enough so that the combined diode+resistor exhibits positive I/V slope over its whole operating range. You measure the I/V characteristics of the combined device, then do the math on the result to deduce the I/V characteristics of just the diode. This approach will require good accuracy, since the resistor is essentially adding noise to the desired signal. You still need to have sufficiently accuracy left after subtracting off the known "noise" signal.

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Don't put the "protective" series resistor: connect directly your regulated power supply to the diode (with a short wire) and measure the current: the voltage regulator is all what you need to protect the diode and ensures you are not missing the negative resistance part. Then you increase very slowly the voltage until the curve is completed.

I once traced the VI curve of a a Gunn diode that way, with a stupid LM317 regulator. The only difference is that the Gunn diode has a higher voltage graph.

But since you say you have a well regulated power supply that allows fine tuning from 0V to 5V (at least), there should be no problem.

Otherwise, it is also easy to build an adjustable regulator for low voltages by driving a rail to rail opamp follower with a pot. This is sufficient for this task, since only few tens of mA are needed.

Now only the milliammeter resistance can create a problem.

Not a problem at all if your power supply has an integrated ammeter. But even without that, it is unlikely that the few tens of millivolts drop from the milliammeter would cause instability and missing the negative resistance part of the diode. Of course, the voltage should be measured at the terminals of the diode, and not after the milliammeter!

OR, with the pot-follower solution above, here is the schematic:

The voltage is measured at the terminals of the diode with the multimeter.

I'd break the pot up into a fixed resistor for the top part, and a pot to only allow up to a volt or so max for the bottom part. That reduces the settings that might cause damage, in addition to giving you higher adjustment resolution

In other words:

Yes, of course. My circuit was just "quick and dirty", but if it has to be used several times, it should be made safe as suggested by Olin.

Circuit Fantasist has also suggested the following:

This should work as well, despite I somewhat dislike the idea to add inductance inside the control loop (due to the terminals of the ammeter, or worse, due to the ammeter itself if an analog ammeter is used), near a negative resistance.

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## Setup

We can best understand the behavior of the tunnel diode in the region with negative resistance if we imagine it as a self-variable (dynamic) resistor R driven by a variable voltage source V - Fig. 1. If OP has a subtle sense of humor, I suggest we conduct this experiment in the form of a fun (but useful) game where he is the voltage source and I am the "tunnel diode":)

Fig. 1. A setup for measuring the tunnel diode IV curve in the negative resistance region (Wikibooks)

## Graphical representation

The voltage VA across and the current IA through the two elements are the same. So their IV curves can be superimposed on the same coordinate system - Fig. 2. The IV curve of the variable resistor is a straight line (in orange) beginning from the coordinate origin and having a slope depending on the instant (static) resistance R. The IV curve of the input voltage source is a vertical line (in red) shifted to the right from the Y-axis. The intersection point A (aka operating point) represents the instant magnitudes of the current IA and the voltage VA.

Fig. 2. The graphical representation of the circuit operation as two superimposed IV curves (Wikibooks)

## Operation

1. Low positive resistance region. When OP gradually increases the input voltage from zero to the beginning of the negative resistance region, I keep a low constant resistance R... and OP (i.e., the voltage source) imagines it as a steep line (IV curve).

2. Negative resistance region. When reaching the negative resistance region, I decide to play a trick with OP and begin increasing R vigorously while he gradually increases the input voltage. As a result, in Ohm's law I = V/R, both V and R change in opposite directions and with different rates of change. The resistance rate is higher than the voltage rate so the current decreases and OP sees a negative resistance. In Fig. 2, my (R, tunnel diode) IV curve begins rotating clockwise; the operating point A moves down and pictures the negative resistance part of the tunnel diode N-shaped IV curve.

3. High positive resistance region. After the negative resistance region, OP continues increasing the input voltage. My resistor possesses high positive resistance and OP imagines it as a sloping IV curve.

## More considerations

Let's see what is the most important to measure the tunnel diode IV curve. IMO this is not the scope. You have to see that when increasing the voltage across the diode, the current through it decreases. For this purpose, first at all, you have to remove (destroy, neutralize) any resistance in series. Here is my philosophy about how we can do it...

The problem. What is this resistance? First, this is the internal resistance of the voltage source and next, this is the ammeter resistance (as you provably now, VOM measure the current by measuring the voltage drop across a small resistance). How do we do this magic? The usual answer is, "It is very simple, just apply a negative feedback". Yes, but you probably are not satisfied with ready-made formal explanations and you want to understand the idea behind all this. Here is what it is ...

Solution. To zero a resistance actually means to zero the voltage drop across it (V = I.R = 0). The natural way is to replace the existing resistance with a "piece of wire"; then really V = I.0 = 0. Only, we cannot do it in this way. However, since we are inventive enough, we decide to do it in an artificial way - by adding a voltage equal to the voltage drop in a series manner, according to KVL. As a result, the total voltage across this network will be zero... as though the resistance is zero. Wonderful, isn't it?

We can add the compensating voltage in two ways:

1. Increasing VIN with VR. First, we can make the input voltage source to raise its voltage with the value of the voltage drop. We can do it by applying a series negative feedback to an op-amp and putting (hiding) the ammeter resistance into the feedback loop. This idea is implemented in Olin's and coquelicot‭'s answers. A disadvantage of this solution is that the ammeter (ADC) is floating. Also, you can want to test another device by current; then the device will be floating.

2. Adding VR to VIN. The second idea is just amazing - instead to increase the input voltage with VR, we decide to add a compensating voltage VR to the input voltage. This means to connect an additional voltage source in series to the diode so that its voltage adds to the input voltage according to KVL. For this purpose, its voltage must be negative in regards to ground (you can see for yourself if you travel along the loop). The sum of the compensating voltage and the voltage drop across R is zero (VR - VR = 0) and the so-called virtual ground appears.

We can explain this technique in terms of resistances. The op-amp output can be considered as a *negative resistor" with equivalent but negative resistance -R that is added to the positive resistance R. The result is zero resistance.

Implementation. Thus you have two possibilities to artificially zero the undesired resistance - by a non-inverting configuration or by an inverting one. I prefer to use the second; this old setup from 90s is implemented this way (here is a movie showing how a rectifying diode can be investigated by voltage). It is actually an inverting amplifier with buffered input and output. For your purposes, you have to connect the tunnel diode in the place of R1 and the ammeter (VOM or movement) in the place of R2. If you want to measure the current by a grounded ADC or a microcontroller port, then use the op-amp output voltage as a measure of the current. The only problem is that it is negative (I = -Vout/R).