Toggle navigation
BuboFlash
Do you want BuboFlash to help you learning these things? Or do you want to add or correct something?
Click here
to log in or create user.
Contents
Flashcard status
Presentation
Question
Answer
Question
双极性结型晶体管的伏安特性：
[...]
Answer
发射极（Emitter）：
\(\begin{align}I_{E}&=I_{S}\left(e^{\frac{qV_{BE}}{KT}}-1\right)\approx I_{S}e^{\frac{qV_{BE}}{KT}}=I_{S}e^{\frac{V_{BE}}{V_{T}}}\\&=I_{C}+I_{B}=(\beta+1)I_{B}\end{align}\)
集极（Collector）：
\(\begin{align}I_{C}&=\alpha I_{E}=\alpha I_{S}e^{\frac{V_{BE}}{V_{T}}}\\&=\beta I_{B}\end{align}\)
基极（Base）：
\(I_{B}=I_{E}-I_{C}=(1-\alpha)I_{E}=(1-\alpha)I_{S}e^{\frac{V_{BE}}{V_{T}}}\)
Question
双极性结型晶体管的伏安特性：
[...]
Answer
?
Question
双极性结型晶体管的伏安特性：
[...]
Answer
发射极（Emitter）：
\(\begin{align}I_{E}&=I_{S}\left(e^{\frac{qV_{BE}}{KT}}-1\right)\approx I_{S}e^{\frac{qV_{BE}}{KT}}=I_{S}e^{\frac{V_{BE}}{V_{T}}}\\&=I_{C}+I_{B}=(\beta+1)I_{B}\end{align}\)
集极（Collector）：
\(\begin{align}I_{C}&=\alpha I_{E}=\alpha I_{S}e^{\frac{V_{BE}}{V_{T}}}\\&=\beta I_{B}\end{align}\)
基极（Base）：
\(I_{B}=I_{E}-I_{C}=(1-\alpha)I_{E}=(1-\alpha)I_{S}e^{\frac{V_{BE}}{V_{T}}}\)
Summary
status
not learned
measured difficulty
37% [default]
last interval [days]
repetition number in this series
0
memorised on
scheduled repetition
scheduled repetition interval
last repetition or drill
Details
No repetitions
Add new event manually
Discussion
Do you want to join discussion?
Click here
to log in or create user.