Well, Vasile it is true... but note that this "resistor type" arrangement is just as accurate as your "constant-current source" setup... Why? How do we explain this paradox? And what is the input and what the output quantity here - the voltage across the diode... or the current through the diode... or the voltage across the resistor... or the total input voltage?

I have used this picture only to illustrate the measuring setup (the trick with the floating scope)...

I think it is the Voltage. The voltage V is independent quantity (normally on the plot's horizontal axis) and the current I is dependant quantity on vertical axis of the plot. If you have some modern power supply with precise voltage and current measurements, you can record the plot directly with this device. Of course, before the measurement, set the device current limit to maximum rate of the diode. In the case of power diode, also the cooling should be provided. You can provide the measurement manually, but many modern power supply allows to do it by computer control via dedicated interface (RS232, HPIB, etc..).

Jan, the only problem is that there are three voltages here... and each of them can act as an input voltage...

 Cyril, sorry, I do not understand. Where are three voltages? There is only one voltage across the diode and there is one current flowing through diode. Nothing else. Both are precisely measured by any serious power supply. Now many devices can provide this. And some of them pretty precisely and fast (you should make the measurement of full V-I characteristic at constant temperature). E.g., we are using this one:  http://www.keithley.com/products/dcac/highspeedpower/2280S/?mn=2280S-60-3

Well, Jan... I will try to explain what I mean when saying that there are three voltages and each of them can act as an input quantity...

First, in all cases, we (somehow) change the voltage produced by the potentiometer; so the overall voltage is the first input voltage.

Then, we can adjust it in such a way that to set desired values of the voltage drop across the diode (e.g., 500 mV... 550 mV... 600mV... 650 mV...) So the voltage drop across the diode acts as an input voltage.

Finally, we can adjust the potentiometer voltage in a manner that to set desired values of the voltage drop (e.g., 1V... 2 V... 3 V...) across the resistor having some constant resistance (e.g., 1 kohm) to set a desired current (1 mA... 2 mA... 3 mA...) through the diode. Now the voltage drop across the resistor acts as an input voltage.

In this way, we always change the overall input voltage with the purpose to set the desired value of the voltage across either the diode or resistor. In this way, we act as a manual negative feedback system (servo) that keeps the desired voltage across the corresponding element.

Here are two "op-amp analogies" equal to the two situations above. First imagine a non-inverting amplifier where we replace the resistor R1 (between the inverting input and ground) with a diode so that the op-amp to keep the desired voltage (equal to some input voltage) across the diode. Then we replace the resistor R2 with the diode so that the op-amp to keep the desired voltage (equal to the input voltage) across the resistor R1.

Vasile, you had a very good look at my problems in the academic laboratory taking into account the specific financial constraints...

I have a nice functional generator producing the needed triangle wave... but still, during the first lab, I preffered to use small transformers producing AC 24 V peak value. They look more reliable, clear and intelligible for these junior students in this humble arrangement.

Regarding the reverse biasing I think that the diode will not die:) if we connect a current limiting resistor in series... and this resistor will not affect the IV curve. The problem is the very high reverse breakdown voltage of the ordinary diodes... I will try to investigate base-emitter junctions of bipolar transistors or LEDs instead ordinary diodes to see the breakdown.

Dear Cyril,

Dear Colleagues,

I would like this time to give a direct answer to the question of cyril. The input to the measuring circuit of the diode is the power supply voltage which is applied to the potentiometer to divide it and apply it on a series combination of the diode and series resistance as a current limiting and sensing device. Varying the potentiometer setting will change the voltage across the diode and the current through it. 

It is the connection of the current measuring instrument and the voltage measuring instrument. In case of forward bias the connection shown , for example, AC IV set up is suitable. However, for reverse bias one has to connect the diode voltage measuring instrument across the diode and the series resistance while keeping the voltage measuring instrument across the series resistance unchanged. This is to avoid the falsification of measuring the  diode current in the revere bias case by the current passing through the voltmeter as the voltmeter may have a smaller  internal resistance 

than the diode.

Well... as "a picture is worth a thousand words", then let's illustrate the written above...

1. Setting the voltage VF across the diode.

2. Setting the voltage VR across (current I through) the resistor.

3. Setting the voltage VP produced by the potentiometer.

Dear Cyril,

It looks like that you are not convinced with the above answers. 

In the three configurations you have the same circuit. It is the measuring circuit of the I-V characteristics of the diode. After seeing your very clear drawings, i can say all designations are possible and results in the same I-V characteristics and the same measuring errors.

On the other side, generally in circuit notation, if the circuit have ports then one can speak from input ports and output ports.

Have a nice time.

Dear Abdelhalim,

Your explanations are excellent... but still I continue thinking about the functional relationships in this arrangement...

Thanks for the support,

Cyril

Here are more viewpoints at this humble circuit of two elements in series...

1. Conventional voltage divider with grounded output.

Here we change the total voltage as an input and take the voltage drop across E2 as an output.

2. Rarely used voltage divider with floating output.

Now we change the total voltage as an input and take the voltage drop across E1 as an output.

3. Voltage-to-voltage converter with negative feedback (or current-to-voltage converter) - version 1.

In this case, we change the total voltage so that to set the voltage drop across E2 as an input and take the voltage drop across E1 as an output. Actually, we set the E2 current as an input; so this arrangement acts as a current-to-voltage converter with negative feedback.

4. Voltage-to-voltage converter with negative feedback (or current-to-voltage converter) - version 2.

Finally, we change the total voltage so that to set the voltage drop across E1 as an input and take the voltage drop across E2 as an output. As above, we actually set the E1 current as an input; so this arrangement acts as a current-to-voltage converter with negative feedback.

From this perspective, in the humble voltage divider, the voltage drop across each of the two resistors represents the current through the other resistor... and can be used to plot its IV curve...

Dear Cyril,

All configurations are possible. It is the matter of the tranfer characteristic you seek.

The first picture shows the definition of input and output as per natural voltage divider. If we assume that the reference point is at the negative of the power supply voltage, figure 2 will represent the tranfer characteristics taking the output from  the higher potential side of the potential divider. It complements the previous tranfer characteristics, in the sense that T2=1-T1 where T2 is the transfer ratio in the second case and T1 is that of the case1. In case 3 the tranfer ratio T3=E1/ E2 and the slope of this straight line will be R1/R2. In the fourth case  T4= E2/E1 which is the inverse of the previews case and its slope will be R2/R1. 

Nice analysis, Abdelhalim!

It is also interesting to find some relationship with the transistor amplifying stage...

Thanks for the support,

Cyril

Yes Cyril, Not at all!

The circuits are classified into two terminal networks, four terminal or two port networks and multi port networks. Filters network are basically two port networks. Amplifiers also are two port networks. Mixers are three port networks. Multiplexers are many to one port network and the demultiplexer is the opposite. Ports are related to the signal flow schemes in the circuit.

Concerning basic amplifiers i would like to take this opportunity to pay your attention to my notes on this topic in the Link:https://www.researchgate.net/publication/256346392_The_bipolar_transistors_theory_and_basic_applications

My best wishes

Book The bipolar transistors, theory and basic applications

Remarkable work, Abdelhalim! The experimental arrangements are particularly interesting for me. Now I see that you have entered very deeply into this matter while I have only some particular insights.

Regards,

Cyril

Abdelhalim

Thank you for the Bipolar Transistor Tutorial. 

After 40 years, this was interesting to read, and clearly written.

Glen

You may consider the supply current since in plays an important role for I-V characteristics.