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• 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

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Question

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）：$$I_{B}=I_{E}-I_{C}=(1-\alpha)I_{E}=(1-\alpha)I_{S}e^{\frac{V_{BE}}{V_{T}}}$$

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