Question 14.3: Designing an offset-compensating network The noninverting am...

Since the offset due to the biasing current I_{ B } is minimized by the resistance R_{ x } , the output offset voltage will be contributed mostly by V_{ io } \text {. For } V_{ io }=6 mV \text { and } V_{ nt }=V_{ CC }=15 V , Eq. (14.38) gives 6 mV =15 R_{ c } / R_{ b } .

V_{ io } \approx \frac{R_{ c } V_{ nt }}{R_{ b }}=\frac{R_{ c } V_{ CC }\left(=V_{ EE }\right)}{R_{ b }}                                                 (14.38)

Letting    R_{ c }=10 \Omega, we get

R_{ b }=\frac{15 R_{ c }}{6 mV }=25 k \Omega

Letting R_{ b }=10 R_{ nt ( max )}=10(R / 4) , we get

R=\frac{4 R_{ b }}{10}=4 \times \frac{25 k \Omega}{10}=10 k \Omega                               (potentiometer)

The network will change the voltage gain from

1+\frac{R_{ F }}{R_{1}}=1+\frac{100 k \Omega}{10 k \Omega}=11

to      1+\frac{R_{ F }}{R_{1}+R_{ c }}=1+\frac{100 k \Omega}{10,010 \Omega}=10.99

causing a 0.09% error.