Topic: Analog Electronics
Transistors $Q_1$ and $Q_2$ are identical and $\beta>>1$ in the circuit shown in the figure below. The output voltage is $(V_t=0.026 \text{ V})$.
- A. $2 \log_{10}(\frac{v_2}{v_1}\frac{R_1}{R_2})$
- B. $\log_{10}(\frac{v_2}{v_1}\frac{R_1}{R_2})$
- C. $2.3 \log_{10}(\frac{v_2}{v_1}\frac{R_1}{R_2})$
- D. $4.6 \log_{10}(\frac{v_2}{v_1}\frac{R_1}{R_2})$
Answer: B. $\log_{10}(\frac{v_2}{v_1}\frac{R_1}{R_2})$
Explanation: $\frac{V_1}{R_1}=I_{C1}$ and $\frac{V_2}{R_2}=I_{C2}$
$I_{C1} = I_S e^{V_{BE1}/V_T} \Rightarrow V_{BE1} = V_T \ln(\frac{I_{C1}}{I_S})$
$I_{C2} = I_S e^{V_{BE2}/V_T} \Rightarrow V_{BE2} = V_T \ln(\frac{I_{C2}}{I_S})$
$V_0 = \frac{333}{20}(-V_{BE1}+V_{BE2}) = \frac{333}{20}V_T \ln\left(\frac{V_2}{R_2}\times\frac{R_1}{V_1}\right)$
$= \frac{333}{20} \times 0.026 \times 2.303 \log_{10}\left(\frac{v_2}{v_1}\frac{R_1}{R_2}\right) \approx \log_{10}\left(\frac{v_2}{v_1}\frac{R_1}{R_2}\right)$.