Question 12.10: For the antilog amplifier in Figure 12–31 , find the output ...

First of all, notice that the input voltage in Figure 12–31 is the inverted output voltage of the log amplifier in Example 12–9 . In this case, the antilog amplifier reverses the process and produces an output that is proportional to the antilog of the input. So the output voltage of the antilog amplifier in Figure 12–31 should have the same magnitude as the input voltage of the log amplifier in Example 12–9 because all the constants are the same. Let’s see if it does.

V_{\mathrm{OUT}} \cong-I_{\mathrm{EBO}} R_{\mathrm{F}}\operatorname{antilog}_e\left(\frac{V_{\text {in }}}{25  \mathrm{mV}}\right)=-(40  \mathrm{nA})(68  \mathrm{k} \Omega) \operatorname{antilog}_e\left(\frac{0.175 \mathrm{~V}}{25  \mathrm{mV}}\right)

=-(40  \mathrm{nA})(68  \mathrm{k} \Omega)(1100)=-3 \mathrm{~V}

P R A C T I C E EXERCISE

Determine V_{\text {OUT }} for the amplifier in Figure 12–31 if the feedback resistor is changed to 100 kΩ.