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?(adsbygoogle = window.adsbygoogle || []).push({});

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# Bipolar transistor breakdown

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