In order to know in which mode the transistor will operate, or to *force* the transistor in a certain operating mode, the principle of the load line could help. Without knowing anything yet about the transistor, one knows that in the shown circuit the voltage across the transistor can never exceed the power supply voltage V_{CC} (the open circuit voltage), and that the current through the transistor can not be larger than V_{CC} / R_{C} (the short circuit current). The operating point defined by the voltage V_{CE} across the BJT and the current I_{C} through the BJT will always be on the line connecting the open circuit point to the short circuit point.

Figure 3.15 (left) The intersection of the transistor characteristic with the load line (in blue) yields the operating point. (right) The base current determines the collector current, and that will determine the voltage across the transistor.

The equation of the load line follows from Kirchhoff’s voltage law:

V_{CE} + I_{C} . R_{C} = V_{CC}

The operating point corresponds to the intersection of the load line with the transistor characteristic. It turns out that the base current determines in which area the operating point will be:

- If I
_{B}= 0 A, then the transistor is in cut-off mode. The collector current is 0 A and the operating point (Q_{0}) corresponds to the open circuit point. This is logical since the transistor behaves as an open switch. - If I
_{B}increases, then initially I_{C}increases. The voltage across the resistor then also increases, and as a result the voltage across the transistor decreases. The operating point (Q_{1}and Q_{2}) shifts along the load line from the lower right to the upper left. - At some point an increase in I
_{B}has barely any effect. The transistor is driven into saturation. The voltage across the transistor is nearly 0 V and the power supply voltage is applied across the resistor. The operating point (Q_{3}and Q_{4}) nearly corresponds to the short circuit point. Further increase of the transistor current or a further decrease of the transistor voltage is very hard and is bounded by limits. The (theoretical) limit for the current is V_{CC}/ R_{C}(the short circuit current).

So, there are two ways of forcing the transistor into saturation: increasing the base current and increasing the collector resistor. In the latter case the short circuit current will decrease, resulting in a less steep load line.

When analyzing a circuit, one can not always easily predict in which area the transistor will operate. One way of finding out is to start from an assumption and to check afterwards whether or not it was valid.

## Transistors - Level 1

Time to start practicing. In the following exercises the circuit at the base of the transistor is not drawn, but an indication on the value of the base current is given. Your task is to predict what will happen if a certain parameter changes.

Part of my answer to question 6 (raise the resistor value to bring the transistor into saturation), is said to be incorrect. Could this be a mistake? (same goes for question 7, but vice versa)

The site is really helpful, by the way!

You’re completely right.

Thanks for mentioning this.

It’ corrected now.

And congratulations for achieving this level!

It’s not a really big remark but answer 6 to both questions 2 and 3 make my OCD senses tingling.