Bipolar transistor breakdown

  • Thread starter antonantal
  • Start date
  • #1
218
0
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?
 

Answers and Replies

  • #2
es
70
0
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.
 
  • #3
es
70
0
oh ya, and vice versa
 

Related Threads on Bipolar transistor breakdown

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
778
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
11
Views
2K
Replies
4
Views
14K
Replies
1
Views
2K
Replies
12
Views
4K
Replies
6
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
11
Views
8K
Top