Question

How does capacitive interference to Vce leg affects the bjt as a switching circuit? Explain saturation and cut off regions in case of this capacitive interference

Answer 1

Ans : When used as an AC signal amplifier, the Base of the transistor biasing voltage is applied in such a way that it always operates within its “active” region, which is the linear part of the output characteristics curves are used.

However, both the NPN & PNP type bipolar transistors can be made to operate as “ON/OFF” type solid-state switch by biasing the transistors Base terminal differently to that for a signal amplifier.

Solid-state switches are one of the main applications for the use of transistors to switch a DC output “ON” or “OFF”. Some output devices, such as LED’s only require a few milliamps at logic level DC voltages and can, therefore, be driven directly by the output of a logic gate. However, high power devices such as motors, solenoids or lamps, often require more power than that supplied by an ordinary logic gate so transistor switches are used.

If the circuit uses the Bipolar Transistor as a Switch, then the biasing of the transistor, either NPN or PNP is arranged to operate the transistor at both sides of the “ I-V ” characteristics curves we have seen previously.

The areas of operation for a transistor switch are known as the Saturation Region and the Cut-off Region. This means then that we can ignore the operating Q-point biasing and voltage divider circuitry required for amplification, and use the transistor as a switch by driving it back and forth between its “fully-OFF” (cut-off) and “fully-ON” (saturation) regions as shown below.

Cut-off Region:

Here the operating conditions of the transistor are zero input base current ( I_B ), zero output collector current ( I_C ) and maximum collector voltage ( V_CE ) which results in a large depletion layer and no current flowing through the device. Therefore the transistor is switched “Fully-OFF”.