Does Weak Doping Affect Bipolar Transistor Breakdown Voltage?

[antonantal]

Why is the breakdown voltage higher if the transistor is weakly doped? My intuition says that if the base and collector are weakly doped, the width of the depletion region will be bigger, so the number of extracted electrons/holes will increase, and so will the probability of ionisation by impact that leads to the avalanche multiplication, meaning that the breakdown voltage will be smaller. What is wrong in this?

 

[es]

As you mentioned, a diode with higher doping levels has a larger built-in electric field therefore less external voltage is needed to achive the critical breakdown field. Thus the higher the doping the smaller the datasheet breakdown voltage is.

 

[es]

oh ya, and vice versa