Chapter 9, Problem 9.81P

In the circuit in Figure P9.81, assume that $Q_1$ and $Q_2$ are identical transistors. If $T=300K$ , show that the output voltage is
$v_O=1.0{\log }_{10}\left(\frac{v_2{R}_1}{v_2{R}_2}\right)$

To determine

To show: The expression for the output voltageis $0.98{\log }_{10}\left(\frac{v_1{R}_2}{v_2{R}_1}\right)$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  

Mark the current directions, voltages and redraw the circuit.

The required diagram is shown in Figure 2

  

The expression for the voltage $v_{22}$ by voltage division rule is given by,

  $v_{22}=\left(\frac{333\text{k}\Omega }{333\text{k}\Omega +20\text{k}\Omega }\right)V_{E1}$

For an ideal op-amp the voltage at both the input terminals are equal and is given by,

  $v_{12}=v_{22}$

Substitute $\left(\frac{333\text{k}\Omega }{333\text{k}\Omega +20\text{k}\Omega }\right)V_{E1}$ for $v_{22}$ in the above equation.

  $v_{12}=\left(\frac{333\text{k}\Omega }{333\text{k}\Omega +20\text{k}\Omega }\right)V_{E1}$

Apply KCL at the inverting terminals of the lower op amp.

  $\begin{array}{l}\frac{V_{E2}-v_{12}}{20\text{k}\Omega }=\frac{v_{12}-v_O}{333\text{k}\Omega }\\ \frac{v_O}{333\text{k}\Omega }=v_{12}\left[\frac{1}{333\text{k}\Omega }+\frac{1}{20\text{k}\Omega }\right]-\frac{V_{E2}}{20}\\ v_O=v_{12}\left[\frac{20\text{k}\Omega +333\text{k}\Omega }{20\text{k}\Omega }\right]-V_{E2}\left(\frac{333\text{k}\Omega }{20\text{k}\Omega }\right)\end{array}$

Substitute $\left(\frac{333\text{k}\Omega }{333\text{k}\Omega +20\text{k}\Omega }\right)V_{E1}$ for $v_{12}$ in the above equation.

  $\begin{array}{l}{v}_O=\left(\frac{333\text{k}\Omega }{333\text{k}\Omega +20\text{k}\Omega }\right)V_{E1}\left[\frac{20\text{k}\Omega +333\text{k}\Omega }{20\text{k}\Omega }\right]-V_{E2}\left(\frac{333\text{k}\Omega }{20\text{k}\Omega }\right)\\ v_O=\left[\frac{333\text{k}\Omega }{20\text{k}\Omega }\right]\left(V_{E1}-V_{E2}\right)\\ V_{E1}-V_{E2}=v_O\left(\frac{20\text{k}\Omega }{333\text{k}\Omega }\right)\end{array}$

The expression for the current $i_2$ in terms of transistor current equation is given by,

  $i_2=I_{S2}\left(e^{\frac{q{v}_{BE2}}{kT}}-1\right)$

The expression for the current through the resistance $R_2$ is given by,

  $i_2=\frac{v_2}{R_2}$

Substitute $I_{S2}\left(e^{\frac{q{v}_{BE2}}{kT}}-1\right)$ for $i_2$ in the above equation.

  $\begin{array}{l}{I}_{S2}\left(e^{\frac{q{v}_{BE2}}{kT}}-1\right)=\frac{v_2}{R_2}\\ e^{\frac{q{v}_{BE2}}{kT}}=\left(\frac{\frac{v_2}{R_2}}{I_{S2}}\right)\\ {\log }_{10}{e}^{\frac{q{v}_{BE2}}{kT}}={\log }_{10}\left(\frac{\frac{v_2}{R_2}}{I_{S2}}\right)\\ {\log }_{10}{I}_{S2}={\log }_{10}\frac{v_2}{R_2}-\frac{q{v}_{BE2}}{kT}{\log }_{10}\left(e\right)\end{array}$

The expression for the current $i_1$ in terms of transistor current is given by,

  $i_1=I_{S1}\left(e^{\frac{q{v}_{BE1}}{kT}}-1\right)$

The expression for the current $i_1$ is given by,

  $i_1=\frac{v_1}{R_1}$

Substitute $I_{S1}\left(e^{\frac{q{v}_{BE1}}{kT}}-1\right)$ for $i_1$ in the above equation.

  $\begin{array}{l}{I}_{S1}\left(e^{\frac{q{v}_{BE1}}{kT}}-1\right)=\frac{v_1}{R_1}\\ e^{\frac{q{v}_{BE1}}{kT}}=\left(\frac{\frac{v_1}{R_1}}{I_{S1}}\right)\\ {\log }_{10}{e}^{\frac{q{v}_{BE1}}{kT}}={\log }_{10}\left(\frac{\frac{v_1}{R_1}}{I_{S1}}\right)\\ {\log }_{10}{I}_{S1}={\log }_{10}\frac{v_1}{R_1}-\frac{q{v}_{BE1}}{kT}{\log }_{10}\left(e\right)\end{array}$

The expression for the transistor current is given by,

  $I_{S1}=I_{S2}$

Substitute ${\log }_{10}\frac{v_1}{R_1}-\frac{q{v}_{BE1}}{kT}{\log }_{10}\left(e\right)$ for ${\log }_{10}{I}_{S1}$ and ${\log }_{10}\frac{v_2}{R_2}-\frac{q{v}_{BE2}}{kT}{\log }_{10}\left(e\right)$ for ${\log }_{10}{I}_{S2}$ in the above equation.

  $\begin{array}{l}{\log }_{10}\frac{v_1}{R_1}-\frac{q{v}_{BE1}}{kT}{\log }_{10}\left(e\right)={\log }_{10}\frac{v_2}{R_2}-\frac{q{v}_{BE2}}{kT}{\log }_{10}\left(e\right)\\ \log \left(\frac{\frac{v_1}{R_1}}{\frac{v_2}{R_2}}\right)=\frac{q}{kT}{\log }_{10}\left(e\right)\left[v_{BE1}-v_{BE2}\right]\end{array}$

Substitute $v_1$ for $v_2$ in the above equation.

  $\begin{array}{l}\log \left(\frac{\frac{v_1}{R_1}}{\frac{v_1}{R_2}}\right)=\frac{q}{kT}{\log }_{10}\left(e\right)\left[v_{BE1}-v_{BE2}\right]\\ \log \left(\frac{R_2}{R_1}\right)=\frac{q}{kT}{\log }_{10}\left(e\right)\left[v_{BE1}-v_{BE2}\right]\end{array}$

Substitute $v_O\left(\frac{20\text{k}\Omega }{333\text{k}\Omega }\right)$ for $v_{BE1}-v_{BE2}$ in the above equation.

  $\begin{array}{l}log\left(\frac{R_2}{R_1}\right)=\frac{q}{kT}{\log }_{10}\left(e\right)v_O\left(\frac{20\text{k}\Omega }{333\text{k}\Omega }\right)\\ v_O=\left(\frac{333}{20}\right)\left(\frac{{\log }_{10}\left(\frac{v_1{R}_2}{v_2{R}_1}\right)}{\frac{q}{kT}{\log }_{10}\left(e\right)}\right)\end{array}$

Substitute $1.38\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{K}}^{-1}$ for $k$ , $300\text{K}$ for $T$ and $1.6\times {10}^{-19}\text{C}$ for $q$ in the above equation.

  $\begin{array}{c}{v}_O=\left(\frac{333}{20}\right)\left(\frac{{\log }_{10}\left(\frac{v_1{R}_2}{v_2{R}_1}\right)}{\frac{1.6\times {10}^{-19}\text{C}}{\left(1.38\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{K}}^{-1}\right)\left(300\text{K}\right)}{\log }_{10}\left(e\right)}\right)\\ =\left(\frac{333}{20}\right)\left(\frac{{\log }_{10}\left(\frac{v_1{R}_2}{v_2{R}_1}\right)}{\frac{1.6\times {10}^{-19}\text{C}}{\left(1.38\times {10}^{-23}{\text{m}}^2\cdot \text{kg}\cdot {\text{K}}^{-1}\right)\left(300\text{K}\right)}\left(0.433429\right)}\right)\\ =0.98{\log }_{10}\left(\frac{v_1{R}_2}{v_2{R}_1}\right)\end{array}$

Therefore, theexpression for the output voltage is $0.98{\log }_{10}\left(\frac{v_1{R}_2}{v_2{R}_1}\right)$ .