Given these conditions:

1. Due to LOG114 symmetry, select either I₁ or I₂ as the signal input pin. Choosing I₁ as the reference makes the resistor network around A₄ simpler. (Note: Current must flow into I₁ and I₂ pins.)
2. Select the magnitude of the reference current. The signal (I₂) spans eight decades, therefore set I₁ to 1μA − four decades above the minimum I₂ value. (Note that the value does not have to be placed in the middle. If I₂ spanned seven decades, I₁ can set three decades above the minimum and four decades below the maximum I₂ value.) This configuration results in more swing amplitude in the negative direction, which provides more sensitivity (ΔV_O4 per ΔI₂) when the current signal decreases.
3. Use Equation 1 to calculate the expected range of log outputs at V_LOGOUT:
   Equation 15. $$\begin{gathered} \mathrm{For} {\mathrm{I}}_2=10\mathrm{m}\mathrm{A}: \\ {\mathrm{V}}_{\mathrm{L}\mathrm{O}\mathrm{G}\mathrm{O}\mathrm{U}\mathrm{T}}=0.375\times \log \left(\frac{1\mathrm{\mu }\mathrm{A}}{10\mathrm{m}\mathrm{A}}\right)=-1.5\mathrm{V} \mathrm{For} {\mathrm{I}}_2=100\mathrm{p}\mathrm{A}: \\ {\mathrm{V}}_{\mathrm{L}\mathrm{O}\mathrm{G}\mathrm{O}\mathrm{U}\mathrm{T}}=0.375\times \log \left(\frac{1\mathrm{\mu }\mathrm{A}}{100\mathrm{p}\mathrm{A}}\right)=1.5\mathrm{V} \end{gathered}$$

   Therefore, the expected voltage range at the output of amplifier A₃ is:

   Equation 16. $-1.5\mathrm{V}\le {\mathrm{V}}_{\mathrm{L}\mathrm{O}\mathrm{G}\mathrm{O}\mathrm{U}\mathrm{T}}\le 1.5\mathrm{V}$
4. The A₄ amplifier scales and offsets the V_LOGOUT signal for use by the ADC using the equation:
   Equation 17. ${\mathrm{V}}_{\mathrm{O}4}=(-{\mathrm{G}}_{{\mathrm{A}}_4}\times {\mathrm{V}}_{\mathrm{L}\mathrm{O}\mathrm{G}\mathrm{O}\mathrm{U}\mathrm{T}})+{\mathrm{V}}_{\mathrm{O}\mathrm{F}\mathrm{F}\mathrm{S}\mathrm{E}\mathrm{T}}$

   The A₄ amplifier is specified with a rated output swing capability from (V−) +0.5V to (V+) − 0.5V.

   Therefore, select the final A₄ output:

   Equation 18. $0\mathrm{V}\le {\mathrm{V}}_{\mathrm{O}4}\le 2.5\mathrm{V}$

   This output results in a 2.5V range for the 3V V_LOGOUT range, therefore a gain of 5/6 is needed for A₄.

5. When I₂ = 10mA, V_LOGOUT = −1.5V. Using Equation 27 in step 4:
   Equation 19. $0\mathrm{V}=\frac{-2.5\mathrm{V}}{3\mathrm{V}}\times (-1.5\mathrm{V})+{\mathrm{V}}_{\mathrm{O}\mathrm{F}\mathrm{F}\mathrm{S}\mathrm{E}\mathrm{T}}$

   Therefore, V_OFFSET = 1.25V

   This makes the A₄ formula:

   Equation 20. ${\mathrm{V}}_{\mathrm{O}4}=\frac{-5}{6}\left({\mathrm{V}}_{\mathrm{L}\mathrm{O}\mathrm{G}\mathrm{O}\mathrm{U}\mathrm{T}}\right)+1.25$

   Replacing V_LOGOUT with Equation 1 gives the system's overall function to be:

   Equation 21. ${\mathrm{V}}_{\mathrm{O}4}=-0.347\times \log \left(\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}\right)+1.25\mathrm{V}$

   The external resistor values for A₄ can be seen in Figure 7-8.

Figure 7-8 Dual Supply Configuration Example for Best Accuracy Over Eight Decades