For the typical flow direction in a PNP transistor:
For the typical flow direction in an NPN transistor:
The IV input characteristic of a Si-NPN transistor looks exactly as the IV characteristic of a Si-diode. This is not surprising, because the base-emitter is a PN junction and it is also a PN junction we have in a diode from the Anode to the Cathode.
For the diode and for the transistor the IV characteristic shows the current through the PN junction as a function of the voltage across the PN junction. In the transistor input characteristic, the current through the PN junction is considered to be the current IB into the base which continues through the PN junction to the emitter.
Figure 62.3.1 shows an example of IV output characteristics of a transistor. What makes this figure particularly interesting is that it actually shows many IV output characteristics in the same graph. For each specific value of the base current IB, we have a different output characteristic.
Similarly, if we look at the case where IB = 40 µA, we see the same pattern but with different values: The collector current rises quickly until it reaches 4 mA, then remains constant as VCE continues to increase.
The transistor essentially behaves like a current-controlled switch. It can turn on and off a current in the main flow direction that is much larger than the control current IB. This is where the amplification property of transistors comes from.
Also sometimes denoted as hFE, the current gain β varies by transistor type:
Figure 62.4.1 shows an IV output characteristic of a transistor together with an indication of the three different states. Understanding these states is crucial for proper transistor operation.
The cutoff state represents the "OFF" condition of the transistor. In this state:
When VBE is smaller than VBE(on) (the base-emitter voltage needed to activate the PN junction), the "BE diode" is off. This means IB = 0, which consequently leads to IC = 0.
The saturation state represents the "fully ON" condition. Here's what happens:
When VBE equals VBE(on), the "BE diode" turns on, allowing current flow. For example, with IB = 60 µA and β = 100:
\[ \beta I_B = 100 \cdot 60 \text{ μA} = 6 \text{ mA} \]In the saturated state:
\[ 0 < I_C < \beta I_B \]The active state is where the transistor operates as an amplifier. Following our previous example with IB = 60 µA:
In this state, IC remains constant at β·IB while VCE can vary. This is the region where the transistor is most commonly used for amplification.
For IB = 0 we have VCE = VCC. For increasing IB through the active region, we have an increasing IC = βIB and following a decreasing potential at the collector. At point S, the collector current IC cannot grow further and remains constant at higher IB values.