Using the summer in Fig. 12.64, and the OTAs specified in Example 12.30, produce the following function

Repeat for the function $\begin{aligned}&v_{o}=10 v_{1}+2 v_{2} \\\\&v_{o}=10 v_{1}-2 v_{2}\end{aligned}$

Question

Using the summer in Fig. 12.64, and the OTAs specified in Example 12.30, produce the following function

Repeat for the function [latex]\begin{aligned}&v_{o}=10 v_{1}+2 v_{2} \\&v_{o}=10 v_{1}-2 v_{2}\end{aligned}[/latex]

Accepted Answer

Comparing Eq. (12.79) ([latex]i_{o 1}=g_{m 1} v_{1} \quad i_{o 2}=g_{m 2} v_{2} \qquad i_{o}=i_{o 1}+i_{o 2} \qquad v_{o}=\frac{i_{o}}{g_{m 3}}=\frac{g_{m 1}}{g_{m 3}} v_{1}+\frac{g_{m 2}}{g_{m 3}} v_{2}[/latex]) with the desired function, we see that [latex]g_{m 1} / g_{m 3}=10[/latex] and [latex]g_{m 2} / g_{m 3}=2[/latex]. With only two equations and three unknowns, we must choose one [latex]g_{m}[/latex] value. Arbitrarily selecting [latex]g_{m 3}=0.1 \mathrm{mS}[/latex] yields [latex]g_{m 1}=1 \mathrm{mS}[/latex] and [latex]g_{m 2}=0.2 \mathrm{mS}[/latex]. The corresponding bias currents are [latex]I_{A B C 1}=50 \mu \mathrm{A}, I_{A B C 2}=10 \mu \mathrm{A}[/latex], and [latex]I_{A B C 3}=5 \mu \mathrm{A} .[/latex]

For the second case, we simply invert the sign of [latex]v_{2}[/latex] as shown in Fig. 12.67. This is yet another advantage of OTA versus op-amps. Again choosing [latex]g_{m 3}=0.1 \mathrm{mS}[/latex] yields the same bias current as the first case.

Question 12.32: Using the summer in Fig. 12.64, and the OTAs specified in Ex...

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