Question

The break down voltage of a transistor with its base open is BV_CEO and that with emitter open is BV_CBO, then 

(a) BV_CEO = BV_CBO 

(b) BV_CEO > BV_CBO 

(c) BV_CEO < BV_CBO 

(d) BV_CEO is not related to BV_CBO

Answer 1

Correct Option(c) BV_CEO < BV_CBO 

Explanation:

The given voltage ratings are reverse breakdown voltages.

BV_CEO – Voltage between the collector and emitter with base open

BV_CBO – Voltage from collector to base with emitter open

The mechanism involved for such breakdown is due to Avalanche.

The equation relating these breakdown is

BV_CEO= BV_CBO/(β)^1/N

This shows that voltage in open base configuration is smaller by

(β)^1/N