Definition of Bipolar Junction Transistor (BJT)

The Bipolar Junction Transistor (BJT) is an active device. In simple terms, it is a current controlled valve. The base current (IB) controls the collector current (Ic).

Regions of BJT Operation

Cut-off region: -

The transistor is off. There is no conduction between the collector and the emitter.(IB=0 therefore Ic=0)

Active region: -

The transistor is on. The collector current is proportional to and controlled by the base current (Ic=ßIB) and relatively insensitive to VCE. In this region the transistor can be an amplifier.

Saturation region: -

The transistor is on. The collector current varies very little with a change in the fill fully base current in the saturation region. The VCE is small, a few tenths of volt. The collector current is strongly dependent on VCE unlike in the active region. It is desirable to operate transistor switches in or near the saturation region when in their on state.

Rules for Bipolar Junction Transistors (BJTS)

1.For an npn transistor, the voltage at the collector VC must be greater than the voltage at the emitter VE by at least a few tenths of a volt; otherwise, current will not flow through the collector-emitter junction, no matter what the applied voltage at the base. For pnp transistors, the emitter voltage must be greater than the collector voltage by a similar amount.

2. For the npn transistor, there is a voltage drop from the base to the emitter of 0.6 V. For a pnp transistor, there is also a 0.6 V rise from the base to the emitter. In terms of operation, this means that the base voltage VB of an npn transistor must be at least 0.6 V greater that the emitter voltage VE; otherwise, the transistor will not pass emitter-to-collector current. For a pnp transistor, VB must be at least 0.6 V less than VE; otherwise, it will not pass collector-to-emitter current.

Basic Equations For The BJT

For npn: VB > VE + 0.6 V

For pnp: VBE – 0.6 V

For both npn and pnp anytime

IE = IC + IB

For both npn and pnp only in the active region

IC = h FE IB = ß IB

IE = IC+ IB = (ß+1) IB ~ ß IB