How to plot the I-V curve of a tunnel diode?
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?5 days agotlfong01 14diodetunnelCC BY-SA 4.0 3d ago by Olin Lathrop Copy Link History Edit Close Delete Flag
Question cleaned up, and comments that became a discussion deleted. tlfong01, don’t get used to others cleaning up your questions for you. This site is for questions and answers, not long winded side stories or discussions. Let this be an example of how you should have asked. Others, if you have relevant information for the OP, put it in your own answers. Comments are not for content. — Olin Lathrop 3 days ago
@Olin Lathrop， Thank your very much for your advice. Cheers. — tlfong01 3 days ago · edit · delete
Much better now. Talking shops are discouraged. BUT, the data sheet for the device is still needed. Please supply a link to it. — Andy aka 3 days agoAdd a comment
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.5 days agoOlin Lathrop 1853CC BY-SA 4.0 Copy Link History Suggest edit Delete Flag
@Olin Lathrop, many thanks for your advice. (1) About this TLDR question – I agree that my question is too long. One way to shorten it is to summarize all the long paragraphs into a 2,000 words article, with perhaps 10 pictures and move it from Q&A section to the post/paper section, IF I finally arrive at something useful, either success or failure. Or I can move everything to a GitHub site and write up a summary here. / to continue, … — tlfong01 5 days ago · edit · delete
(2) About using Op Amp for stability and low internal source, resolution, current limiting, etc – I am glad to let you know that I agree with everything you said. This test setup design is a long story, so let me give more details in the next comments. / to continue, … — tlfong01 5 days ago · edit · delete
@tlfong01 – this (until someone tells me differently) is a question and answer site and not a blog and whether you are successful or fail or, whether you post something to github or not, that is of no interest to this site (and anyone please correct me if I’ve misjudged what this site is about). If you want to continue with something (a la “so let me give more details in the next comments. / to continue, … “), please do it elsewhere is my advice. Again, a reminder: this is a Q and A site. — Andy aka 1 day agoAdd a comment+3−0
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.3 days agocoquelicot 297CC BY-SA 4.0 2d ago Copy Link History Suggest edit Delete Flag
Simple and clear explanation… Now only the milliammeter resistance can create a problem. If only we can hide it somehow… — Circuit fantasist 3 days ago
A good answer to my comment above… and demonstration of creative thinking… but there is another problem. If “the voltage is measured at the terminals of the diode with the multimeter”, and the voltmeter is not perfect, it will divert a part of the current through the diode… and an error will appear. It would be more significant if the element under test has high resistance. This is another challenge to your ingenuity:) — Circuit fantasist 3 days ago
@Circuit fantasist. The most banal 3$ digital multimeters like the ones in the OP question have at least 1M ohm input impedance, more for more expensive voltmeters. So, this is pointless. — coquelicot 3 days ago
Good circuit, +1. However, 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. With 100 kOhm fixed and a 10 kOhm pot, you get about the same resolution from a 1-turn pot than you get with the 10-turn full-range pot. — Olin Lathrop 3 days ago
Another suggestion – put the ammeter into the feedback loop (between the op-amp output and the common node where the inverting input and diode are connected). There are two benefits of this trick: first, the op-amp will raise its output voltage to compensate the ammeter resistance; second, both the diode and voltmeter (ADC) will be grounded. — Circuit fantasist 3 days agoShow 3 more comments · Add a comment+2−0
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.
4 days agoTonyStewart 49CC BY-SA 4.0 4d ago Copy Link History Suggest edit Delete Flag
Attractive experiment… If the load line could be shown… — Circuit fantasist 4 days ago
@TonyStewart, Many thanks for you answer, which inspires me to learn more things. I have never heard of Falstad Simulator. So I will google and wiki. And as I mentioned in the beginning sections of my question, I confessed that I don’t clearly know what actually is a load line. I never designed a load line. I only read that for tunnel diode, you can have 3 load lines, one for monostable, one for bistable, one for astable. So perhaps I should design and “implement” them on my little bread board. — tlfong01 4 days ago · edit · delete
I randomly browsed your other answers for more interesting new ideas. One thing that caught my eyes is your answer about using MOSFET as a voltage follower, to replace op amps. I am now thinking of using an op amp in my tunnel diode testing circuit, but I have very little experience and I only know that op amp circuit is difficult. So using mosfet to replace op amp is too good to be true. — tlfong01 4 days ago · edit · delete
Inappropriate comments deleted. Comments are not for content, and certainly not for chatting between two users who didn’t even write the answer the comments are under. Knock it off already! If you want to explain to someone what a “load line” is, or anything else for that matter, put that in your own answer. This space is reserved for feedback to Tony on his answer, nothing else. — Olin Lathrop 2 days agoAdd a comment
BodyWe support Markdown for posts: **bold**, *italics*,
`code`, two newlines for paragraphsLicensesite default: CC BY-SA 4.0 · category default: CC BY-SA 4.0CC BY-SA 4.0 CC BY 4.0 CC BY-NC 4.0 CC BY-NC-SA 4.0 CC0
Leave a Reply