Topic: Analog Circuits
In this circuit, the relationship between $V_{i}$ and $V_{0}$ is
- A. $V_{o}=V_{i}e^{V_{o}/V_{\tau}}$
- B. $V_{0}=V_{T}ln\frac{V_{i}}{R_{s}I_{S}}$
- C. $V_{o}=V_{i}-I_{C}R_{S}$
- D. None of above
Answer: D. None of above
Explanation: $I_{1}=\frac{V_{i}-0}{R_{s}}=I_{c}$. $I_{C}=I_{s}e^{\frac{V_{BE}}{V_{T}}}$. $V_{BE}=V_{T}ln\frac{I_{C}}{I_{S}}$. Also $V_{o}=-V_{BE} = -V_{T}ln\frac{I_{1}}{I_{S}} = -V_{T}ln\frac{V_{i}}{I_{S}R_{s}}$.