DETERMINATION OF A POINT OF OPERATION

The characteristics of the transistors are used to determine the operating conditions of these transistors.

One determines thus the value of the electric quantities intervening in this operation.

The most used characteristic is that which relates to the output circuit.

Let us consider the network of characteristics of figure 11.

That is to say a tension VCE equalizes with 10 volts and a current IB equal to 70 µA.

To determine current IC correspondent with these two parameters given, it is enough to trace a vertical starting from the tension VCE = 10 V ; this vertical cuts the characteristic relative to IB = 70 µA to point A. From this point A, it remains to trace a horizontal line which cuts the vertical axis of the Cartesian reference mark. In this case, one finds IC = 20 mA.

Point A is called not operation. Indeed, this point A makes it possible to know the three parameters relating to the operation of the transistor.

You notice that it is enough to know two parameters among three to determine this point of operation.

Let us suppose for example that one knows IC = 14 mA and IB = 50 µA.

It is enough to trace a horizontal line corresponding to IC = 14 mA ; this one meets the characteristic IB = 50 µA at the point B. One then traces the vertical resulting from the point B, which makes it possible to determine a tension VCE equal to 9 volts.

We now will examine the operation of the assembly located on figure 12.

The common transmitting assembly is fed by a pile of 24 volts. The variable resistor RB makes it possible to vary basic current IB.

Resistance RC is the resistance of load located in the collector of the transistor.

The terminal voltage of the resistance of load is VR. The sum of tensions VR and VCE is equal to VCC is 24 volts.

We wish to know the report/ratio which exists between three sizes VCE, IC and IB. We have for example the network of characteristics of exit with parameter IB.

It is enough to choose a certain number of values for current IC, then to calculate tension VCE for each one of these values. That will enable us to as many place points on the network of characteristics.

For IC = 5 mA,

VR = RC x IC = 800 x 5 x 10^-3 = 4 volts, is VCE = VCC - VR

from where VCE = 24 - 4 = 20 volts.

That gives us the co-ordinates of a first point which we name A on figure 13.

In the same way, for IC = 10 mA, VR = 8 volts and VCE = 16 volts (not B).

For IC = 15 mA, VCE = 12 volts (point C)

For IC = 20 mA, VCE = 8 volts (point D)

For IC = 25 mA, VCE = 4 volts (point E)

You notice that these five points are aligned ; also, one can make pass a line by these points which one calls right-hand side of load.

This line of load represents the whole of the points of operation for the transistor. I.e. for each point of the right-hand side, one can determine three sizes VCE, IC and corresponding IB.

This line meets the two axes of the reference mark at the points P and Q. the point Q corresponds to a tension VCE of 24 volts and to a current IC no one. The point P corresponds to IC = 30 mA and VCE = 0 volt ; in this case, tension VR is equal to tension VCC.

As you know it, it is enough to know two points to determine a line. However, in this case, it is possible to know these two points to plot the straight line of load.

We will take an example. That is to say a resistance of load RC equalizes to 1,3 KW and a voltage supply VCC equal to 16 volts.

When IC are null, we know that VR = 0 volt and VCE = VCC = 16 volts. We have thus to determine the Q' point on the horizontal axis (figure 13). To determine the second point, we poseVCE = 0 volt. In this case VR = 16 volts and IC = VR / RC = 16 / 1,3 x 103 = 12,3 mA.

The second P' point thus corresponds to IC = 12,3 mA. (Application of the law of Ohm).

It remains to connect these two P' points and Q' and thus, one plotted the straight line of load relative to RC = 1,3 KW and VCC = 16 volts.

Its position is definitely different from that examined before.

These two examples thus show that the position of the right-hand side of load is related to VCC and of RC.

It is not always possible to determine the two points located out of the two axes of the reference mark.

Let us take the case of the network located on the figure 14.

That is to say VCC = 24 volts and RC = 0,3 kW.

When IC = 0 mA, VCE = VCC = 24 volts. We thus know the point Q located on the horizontal axis.

For VCE = 0 volt, we have IC = VR / RC = VCC / RC is IC = 24 / 300 = 80 mA.

However, the point corresponding to IC = 80 mA is located apart from the graph, therefore it is advisable to seek another point.

One thus takes IC = 50 mA.

One deduces VR = RC x IC = 300 x 50 x 10^-3 = 15 volts

and VCE = VCC - VR = 24 - 15 = 9 volts.

Thus, the second point S corresponds to VCE = 9 volts and to IC = 50 mA.

The line of load is thus determined by the points Q and S.

Consequently, it is easy to determine the values of sizes VCE and IC if current IB. is fixed

For example, the characteristic corresponding to IB = 110 µA meets the straight lines of load at the point T (figure 14).

This point makes it possible to know VCE = 14 volts and IC = 33,5 mA.