Current components of a PNP Transistor

Advertisements

Advertisements3

Advertisements4

Advertisements6

Advertisements66

The various Current components which flow across a PNP Transistor are as shown in the figure.

For Normal operation

Emitter Junction $J_{E}$ is Forward Biased.
collector Junction $J_{C}$ is Reverse Biased.

The current flows into Emitter is Emitter current $I_{E}$ , $I_{E} = I_{hE}+I_{eE}$ .

This current consists of two components

$I_{hE}$ or $I_{pE}$ – Current due to majority carriers(holes).
$I_{eE}$ or $I_{nE}$ – Current due to minority carriers(electrons).

since $I_{eE}$ is very small $I_{E} \simeq I_{hE}-----------Equation(1)$

All the holes crossing the Emitter junction $J_{E}$ do not reach the Collector junction because some of them combine with the electrons in the N-type Base.

$I_{hC}$ – is the hole current in the Collector.

∴ Base current = Total hole current in Emitter – hole current in Collector.

i.e, $I_{B} = I_{hE}-I_{hC}----------------Equation(2)$ .

If emitter were open circuited $I_{E} = 0$ Amperes which implies $I_{E} = I_{hE}$ from Equation(1) $I_{hE}\approx 0$ Amperes.

Under these conditions, Base-Collector junction acts as Reverse-Biased Diode and gives rise to a small reverse-Saturation current known as $I_{CO}$ .

when $I_{E} \neq 0$ , Total Collector current $I_{C}$ is the sum of current due to holes in the Collector and Reverse Saturation current $I_{CO}$ .

i.e, $I_{C} = I_{hC}+I_{CO}$ .

i.e, In a PNP Transistor $I_{CO}$ consists of holes moving across $J_{C}$ (from Base to Collector) that is $I_{hCO}$ and electrons crossing the junction $J_{C}$ (from Collector to Base) constitutes $I_{eCO}$ .