Curve tracer for transistors

In my previous story, I told you how my students and I made simple curve tracers in the laboratory and used them to study the IV curves of various 2-terminal elements (resistors and diodes). Now I will show you how, in the next exercises, we improved our devices to be able to study transistors.

As I have already said, such professional devices exist both for the study of diodes and transistors but they are closed and their internal structure is inaccessible to students. That is why, again I preferred that they design and make curve tracers themselves to understand them. I have been using this approach since 2015, when I had to organize and lead the laboratory exercises on Semiconductor Devices in the ITI department. I have also written this story in Bulgarian.

Transistor IV curves

Transistors are semiconductor elements with three terminals - base, emitter and collector. To get their IV curves, we decompose them into two 2-terminal "elements" - input (base-emitter) and output (collector-emitter), and then examine them in the familiar way. Thus we obtain the input characteristic and the output characteristic of the transistor respectively. This laboratory experiment is devoted to the study of transistor output characteristics. But let's first see what properties they have...

Output characteristic

So, the relationship between the collector current Ic and the collector-emitter voltage Vce of the transistor, represented graphically, is called the output characteristic of the transistor. In fact, it is an IV curve like that of the resistor and the diode, but here it can be controlled by the base current Ib or the base-emitter voltage Vbe. It is accepted that the voltage is applied along X and the current along Y. While the IV curve of a resistor is a straight line passing through the origin of the coordinate system, the output characteristic of a transistor is highly non-linear with a typical horizontal part. This means that the output part of the transistor behaves like a current-stabilizing element (see How does a transistor maintain a constant current?)

Measuring by an "ideal" voltage source...

A possible way to measure the IV curve of a resistor with resistance R was to apply a voltage V across it and read the current I flowing through it according to Ohm's law, written in its "straight" form I = V/R. This method was not suitable for obtaining the vertical part of the diode IV curve (the blue characteristic) because it was difficult to obtain a sufficient number of points... but it is not of interest here. The horizontal part of the output characteristic of the transistor we are interested in does not have this problem.

IV curve I = f(V) of a resistor R and transistor T obtained by a varying "ideal" voltage source.

... by an "ideal" current source...

Conversely, using an "ideal" current source, we could pass a current I through a resistor of resistance R and read the voltage V across it according to Ohm's law written in its dual form V = I.R. This method was suitable for taking the vertical part of the diode IV curve because it was easy to get enough points, but now it does not allow us to get enough points in the horizontal part of the output characteristic of the transistor.

IV curve V = f(I) of a resistor R and transistor T obtained by a varying "ideal" current source.

... by a real voltage source

Above we came to the conclusion that, with the help of an "ideal" voltage source (with a vertical IV curve), we can conveniently measure the horizontal part of the transistor output characteristic... and with an "ideal" current source (with a horizontal IV curve) - the vertical part of the transistor output characteristic.

But, as before, to get the full output characteristic, we can simply use a real voltage source (with a sloped IV curve):

IV curve of a resistor R and transistor T obtained by a varying real voltage source.

Building the device on the blackboard

Here we have to solve an additional task. Since the transistor is a controlled device, we need to accompany the simple curve tracer with a circuit for driving the transistor base. Here is a possible scenario for building the new circuit:

"Inventing" the input part

1. We assemble a variable voltage source using a constant voltage source (AC-DC converter) and a potentiometer P.

Variable voltage source assembled by a constant voltage source and potentiometer P.

Constant voltage source (12V/1A adapter).

As a voltage regulator, we use our favorite linear potentiometer with a resistance of 1 k, not forgetting to twist its terminals 90 degrees so that it contacts reliably in the sockets.

We use our well-known linear potentiometer with a resistance of 1 k.

2. We convert the variable voltage source into a variable current source by adding a 10k resistor Rb in series and "shorting" the output through a milliammeter.

Variable current source made by adding a base resistor

The ammeter with which we will measure the base current is again a digital multimeter DT830B

3. We break the circuit and include the BEJ base-emitter junction. Thus, the input part of the circuit, with the help of which we will adjust the position of the output characteristic on the screen, is ready. Now we need to assemble the output part of the previous lab experiment.

"Inventing" the output part

1. Electrical diagram. It is almost the same as the simpler diode test characterization from last time. Only the diode D is added, which passes only the positive half-waves to the n-p-n transistor T.

The right part of the new circuit diagram repeats the circuit diagram of the simple curve tracer.

AC voltage source (step-down transformer for 17V).

We apply the same trick by choosing the resistance of the resistor R with a value of 1 kom. Thus, the value of the voltage Vy [V] directly corresponds to the value of the current Ic [mA].

2. We decouple the control circuit. At first glance, the resistor R acts as an emitter resistor and the circuit resembles an emitter follower with a series negative feedback that subtracts the output voltage from the input one. But this is undesirable here because we want the input voltage to be applied directly to the base-emitter junction. Therefore, the input part we developed above must be fed with a "floating" transformer (as it actually is). As a result, no base current flows through the resistor R.