A transistor has how many doped areas

The transistor - a jack of all trades






The transistor is probably the most important and diverse component in electronics. Almost no electronic device today would function without this component. Whether as a single element, as it is used here, or in the millions, as it can be found in computers and other electronic circuits. The possibilities of the transistor are so diverse that it is not even possible to list all applications here.


Various transistors


A transistor always has 3 connections. These are denoted by the terms base (B), collector (C) and emitter (E). There are basically 2 different types. Once the NPN and once the PNP types. This name comes from the internal structure. NPN types consist of 2 negatively doped semiconductor layers between which there is a positively doped layer. With PNP this is exactly the opposite.

The type we are using here is the BC548C. And as an equivalent complementary typeBC558C. Complementary type means that both types have the same electrical properties. Only the one, in this case the BC548C, is an NPN type and the other, on the contrary, is a PNP. If you want to know the exact data, you are welcome to take a look at the data sheet.


The European transistor supply is divided into different families, which are determined by the first two letters. The first letter indicates what material the transistor is made of. Are there


A = germanium

B = silicon

C = selenium


Nowadays you can almost only find types from the B group. The A group used to be made a lot. There were almost never any transistors made from selenium. Therefore there are almost no types in this area. The second letter indicates the area of ​​use. This is broken down as follows:


C = General and Small Signal Applications

D = power transistor and darlington types

F = high frequency and field effect transistors

L = high-frequency power transistors

S = low power switching types

U = power switching transistors


A letter or number is often added after the type designation. This indicates which amplification factor the corresponding type has. The BC548C is used here. This means that this type has an amplification factor of approx. 500. A BC548A would have a value of approx. 200 and a BC548B approx. 300. However, these factors do not have to apply to every transistor type. The exact values ​​can be found in the corresponding data sheet.


If there is a number after the type designation, this number indicates the factor * 10. A BC338-40 has a gain factor of approx. 400. This information, whether with a letter or as a value, is only average. They can vary a lot. In order to dimension circuits, however, one always calculates with these mean values. Where the factor plays a major role, trimming potentiometers are used for adjustment.


For the experiments here, you should use the type recommended by me. Other types may lead to different results. Furthermore, the transistors also have many different housings. But even the same housing does not mean that this type can simply be used. It is quite possible that the pin assignment is different. More details can always be found in the corresponding data sheet.



The transistor as a switch



If you want to switch a consumer with the transistor, you can insert it in front of or behind the load. It should be noted, however, that it requires a certain polarity. The current always flows into the collector, this is the connection without the arrow, and the current leaves the transistor again at the emitter, where the arrow is located. If the transistor is to serve as a switch, as here, the consumers are connected to the collector.

If the battery is now connected, nothing happens. This is also correct. You first have to tell the transistor that it should switch. This happens through the third connection, the base.



Here the base of the transistor was supplied with power via a resistor and the button. If the button is now closed, a current flows into the base of the transistor, it releases the collector-emitter path (CE path) and the light-emitting diode lights up. If you let go of the button again, the LED goes out.

But why a transistor? We could switch the light-emitting diode directly via the button and a 470 Ohm resistor.

With this circuit, yes. But it only becomes interesting when we swap the series resistor R1 for a resistor with a significantly higher value. Let's put a resistor with 47 kOhm in the circuit.


If the button is now pressed again, the light emitting diode lights up as if only the resistance of 470 ohms were inside. Anyone who takes the trouble to calculate the current through the base series resistor with the help of Ohm's law, with the assumption that the full operating voltage is applied to it, will quickly find out that this current is not sufficient to make the LED shine.

This shows the strength of the transistor. It amplifies the current that flows into the base and allows a considerably larger current to pass. This factor is called the current amplification factor. With the BC548C this is between 350-600.


In a small graphic you can see the different currents in and out of the transistor. The base current Ib is amplified by the transistor and then allows a current of a certain level to pass (Ic). both currents together then leave the transistor again via the emitter.

In order for the transistor to do its job, it also needs certain voltages. The two most important voltages are the base-emitter voltage (Ube) and the collector-emitter voltage (Uce). The BE voltage is always approx. 0.7 V when the transistor is turned on. If this value is not reached, the transistor makes 'tight'. The BE voltage is set automatically as soon as a base current flows.

The second important voltage is the CE voltage. This is particularly important if you want to determine the power loss of the transistor. The higher the collector current, the higher this voltage. The CE voltage and the collector current should always be kept in mind when developing a circuit. Otherwise it can quickly happen that the transistor is overloaded.

Assuming a base current of 177 µA for our circuit, the transistor would allow a collector current between 62 and 106 mA. Since the light-emitting diode only needs a current of approx. 15 mA, the transistor is overdriven. It is also said that it is full. In switching mode, this effect is even desired in order to be able to guarantee that the transistor actually controls. However, if the transistor is to work as a signal amplifier, saturation should be strictly avoided in order to avoid distortion of the output signal. How this is done will be explained later.


For the sake of completeness, the current and voltage ratios of the PNP transistor should also be briefly shown here. There is not much to say about it. In principle everything is just the other way around as with an NPN. If you look at the drawing, you will see that everything flows upside down.



Perhaps one or the other accidentally touched the base of the transistor with a finger and was allowed to observe an effect that one does not necessarily want here. The LED lights up when you touch the base. This is because the transistor amplifies even the smallest amounts of energy. The current that we absorb through the environment (radio waves, radio signals, electrical radiation from power cables) is sufficient to let the transistor turn on and thus to make the LED shine. But how can you prevent something like that?

You just have to make sure that this energy is dissipated and that the transistor only turns on when a stronger current flows into the base. This is achieved here by using an additional resistor that goes from the base to the minus pole. If you now touch the base, D1 remains dark. It only lights up when the button is pressed.

The additional resistor R2 ensures that the 'wild energy' is dissipated. Due to this additional resistance, the base also achieves a slightly lower current. However, this loss is usually negligible. Instead of 177 µA, the base now gets 145 µA, which is still enough to achieve multiple saturation of the transistor in our circuit.


As shown in a graphic above, the transistor itself needs some voltages in order to be able to work. One of the voltages is the one across the collector-emitter path (CE voltage). If we measure here, we set a value of approx. 0.15 V. With higher loads, this value also increases a little. Together with the collector current, the power dissipation then results for the most part. You should always keep an eye on this, otherwise the transistor could overheat and be destroyed accordingly.


The second important voltage is the base emitter voltage. If the transistor is to be fully controlled, this must reach approx. 0.7 V. If the base is supplied with a series resistor, as in our example, this voltage is set automatically. A measurement proves this.



If you fall below this 0.7 V, the transistor blocks very quickly. In order to test this once, this circuit is set up. If we turn the trimmer, we can see that the voltage remains at approx. 0.7 V for a very long time and then drops very quickly and thus the light-emitting diode goes out quickly.



Twice as much better - the Darlington stage



In a previous experiment, the base of a transistor was touched and the connected light-emitting diode lit up. This effect is undesirable in most cases. In other circuits, however, it is necessary that the really smallest currents are amplified. There it can happen that a single transistor is not sufficient for the amplification. In this case, several transistors are connected in series. This arrangement is called the Darlington pair.

As you can see in this schematic, the emitter of T1 is connected directly to the base of T2. The collectors are grouped together. If the button is pressed here, the LED lights up. So far nothing special. But if we now exchange the resistors R1 and R2 for values ​​that are many times higher, e.g. R1 = 1 MOhm, R2 = 470 kOhm, and press the button again, D1 lights up as well. If these resistors were used in a single transistor, the light-emitting diode would only glow weakly. What happened here?

By connecting the two transistors in series, we have ensured that the gain factors are multiplied. The base current of T1 is amplified. This resulting current is in turn amplified by T2. This creates a transistor circuit that can amplify even the smallest currents. In our example, a base series resistor of 1 MOhm would result in a collector current of at least 0.4 A. This transistor stage is saturated many times over with our LED. What about the tension here?


Since the CE voltage is the most important for most applications, we measure it first. We have to note that the voltage here is a lot higher than with a single transistor.

The base-emitter path of T2 plays a major role here. Since the BE voltage always adjusts to 0.7 V, this voltage is also measured in the total voltage from the collector-emitter.

Since at the Darling level, two BE lines are connected one behind the other, is this tension higher?


If we measure this voltage across the resistor R2, we find here that it is approx. 1.4 V. So we really measure both BE lines one after the other. This fact should be taken into account if you want to use the Darlington stage as an amplifier.


For a better understanding of the current conditions, let's take a look at a graphic. In this example we are supplying the Darlington stage with a base current of 1 µA. Let's just assume a gain factor of 300 for the individual transistors.

The first transistor amplifies the 1 µA to 300 µA. Since the currents from the emitter add up, a current of 301 µA flows there. The base of the second transistor is now supplied with this current and amplifies it again 300 times and then allows a current flow of 90.3 mA.

Since the two collectors of the transistors are connected together, the currents add up here too and a total current of 90.6 mA results if this is not limited by a connected load.

But you don't always have to build a Darlington stage with individual transistors. The industry offers ready-made transistors in one package. These include theBC517 as NPN type or theBC516 for the PNP variant.



The better Darlington stage - The complementary Darlington circuit



In the last chapter the Darlington pair was introduced. With this it is possible to amplify even the smallest currents very strongly. But this circuit has a very big disadvantage. The necessary base-emitter voltage is also twice as high as with a single transistor. With some circuits this is a very big disadvantage.

Another circuit can be used here: the complementary Darlington circuit. This circuit, also known as the Sziklai circuit or Szilkai pair, combines the large current gain and the low base-emitter voltage.

If the adjacent circuit is put into operation, you can quickly see with the multimeter that our typical base voltage of 0.6-0.7V is present again when the button is pressed. The current gain, however, is almost as high as with the conventional Darlington pair.


Understanding how this circuit works is not really difficult. Let us assume that a current of 1 µA flows to the input and we have a current gain of 300 again here, the NPN transistor can allow 300 µA to flow. Since the second transistor is a PNP, the current flows from the emiter to the collector. In order for the transistor to turn on, you have to enable it to let a current flow out of the base.

This is made possible by the first transistor and so the PNP is controlled with 300 µA. This current is again amplified 300 times and a current of 90 mA comes from the collector.


There is also a corresponding complementary substitute type for the PNP Darlington pair. Basically only the two types are swapped here.



In the PNP circuit, the PNP transistor acts as an input stage. If, for example, a current of 1 µA flows from the base, the EC path of the PNP transistor will let 300 µA through and reach the NPN. The NPN transistor amplifies this 300 µA in turn 300 times and so a current of 90.3 mA leaves the NPN transistor, composed of the amplified current of 90 mA and the base current of 300 µA.



The transistor becomes a signal amplifier



In the last tests, the transistor was only used in switching mode. But for other applications, such as hi-fi technology, the transistor is required as a signal amplifier. The output signal must therefore follow with the input.

Let's take our circuit from the beginning of this course and add a trimming potentiometer to it. If the circuit is put into operation and you turn the trimmer, the LED lights up brightly for a long time and then goes out relatively quickly.

Due to the high gain of the transistor, even a small base current is sufficient for it to be fully controlled. Only when the base current falls below the required current does the LED get darker. This effect is not suitable for amplifier applications. Here the output voltage should decrease immediately if the input voltage also decreases. To do this, you only have to rebuild the circuit a little.



If you now put this circuit into operation, the light-emitting diode can be regulated very well. But what is the difference compared to the previous version?



Let's take a look at the voltage ratios of this transistor circuit first.We apply a certain voltage through the trimming potentiometer, which is referred to here as Uin. This voltage is now divided into Urb, Ube and Ulast. As we know, there is always around 0.7 V at Ube. A certain current (Ib) flows into the base of the transistor. This is amplified and ensures a current flow from the emitter, which consists of the collector current Ic and the base current Ib. This current now also flows through the load. What is left of tension can be measured on Urb.

If the voltage Uin is now increased, the voltage not only increases at the load, but also at the base resistance. And so the two voltages regulate each other in such a way that the output voltage follows the input voltage. If the load resistance is large enough, you can even do without a basic series resistor. The base current is also limited here by the load resistance.



Operating point setting of the transistor


It is often necessary to amplify signals that consist of an alternating voltage. There are two options for transistor switching so that they can do it. One possibility would be to supply the amplifier circuit with a symmetrical operating voltage and to let the supply of the circuit only run via the positive or negative pole. However, this is not very useful for small amplifiers. There you resort to another option. You simply raise the working range of the transistor and thus give the transistor the option of steering upwards or downwards.



The diagram opposite shows how this is achieved. When this circuit is put into operation, the light emitting diode lights up. But for a long time not the full luminosity that we are used to from an LED. There are now only approx. 3.8 V at the light emitting diode and the resistor. As a result, less current flows through D1 and it glows darker.


A measurement on D1 / R3 confirms this statement. But why is it only 3.8 V. As some may have suspected, shouldn't 4.5 V be measured here?

The voltage divider, consisting of resistors R1 and R2, feeds a voltage of approx. 4.5 V to the base. The transistor itself needs the 0.7 V at the base-emitter path. The rest of the voltage is now on the load. In our case the LED and the series resistor.


You can see the whole thing here again in a graphic. The two resistors R1 and R2 divide the operating voltage (Ub1 and Ub2) and feed it to the transistor. Together with the load, the transistor again acts as a voltage divider. Here the voltage for the transistor is fixed from the start (Ube) and the rest can then be measured at the load (ULload).

Now you have the possibility to have the transistor controlled even further upwards as well as further downwards. So we have set the operating point of the transistor here.



In order to test the circuit as an amplifier, we also add a trimming potentiometer, which supplies the input of the transistor stage with a controllable voltage.

If you turn the trimmer here, you can control the light-emitting diode in any desired brightness level.


If we supplement the circuit with a capacitor at the input or output, we have a single-stage audio amplifier. The capacitors ensure that the AC voltage is matched to the operating DC voltage of the circuit.



Advantages and disadvantages of the basic circuits


If any load is to be switched with the transistor, the so-called emitter circuit is used. Here the emitter is connected to the zero potential of the operating voltage.

This circuit is characterized by the following properties:

- High current gain

- High input resistance

- Low load on the transistor

- High voltage on the load


This circuit is less suitable for amplifying signals, as the transistor quickly saturates here.


Anyone who has to amplify LF signals is better off with the collector circuit or the emitter follower, as this circuit is also called.

This circuit is characterized by the following properties.

- High current gain

- High input resistance

- voltage gain <= 1

- Amplification of alternating voltages possible

- Lower voltage on the load


There is a third circuit among the circuit types for transistors. The basic circuit. This is mainly used in HF technology and is less suitable for LF and switching applications.



PNP transistor - the not entirely forgotten type


Even if the NPN transistor covers most of the requirements in today's electronics, the PNP is still needed from time to time.


Because with the PNP the emitter switches / controls the positive line, the transistor is usually drawn as 'tilted up'.

With the PNP, all voltage ratios are reversed, which means that the basic circuit of the transistor has to be set up completely 'incorrectly'.



If you want transistor T1 to turn on, you have to allow a small current to flow from the emitter to the base and then to the negative pole via the base series resistor. Only then does the transistor also allow a current to flow from the emitter to the collector via the load.

2 things are immediately noticeable here. The base current has to flow out of the transistor and not into it, as it was previously with the NPN and the transistor controls through from the emitter to the collector. So the current flows into the emitter, so we are clamped to the positive pole of the voltage source, and out of the collector again.