Finding the effects of thermal drift on the output of an inverting op-amp circuit The inverting amplifier in Fig. 14.5(a) has R_{1}=10 k \Omega, R_{ F }=100 k \Omega \text {, and } R_{ X }=R_{ F } \| R_{1}=9.091 k \Omega . The op-amp parameters are V_{ io }=\pm 6 mV , I_{ B }=500 nA , I_{ io }=\pm 200 nA \text {, and PSRR }=150 \mu V / V . The thermal drifts are D_{ v }= 15 \mu V /{ }^{\circ} C , D_{ i }=0.5 nA /^{\circ} C \text {, and } D_{ b }=0.5 nA /{ }^{\circ} C \text { at } 25^{\circ} C . The temperature is 55°C. The DC supply voltages change from V_{ CC }=15 V \text { to } 12 V \text { and }-V_{ EE }=-15 V \text { to }-12 V . The input voltage is v_{ S }=100 mV \text { (DC) } . Determine the output voltage v_{ O } \text { if (a) } R_{ X }=R_{ F } \| R_{1}=9.091 k \Omega \text { and (b) } R_{ x }=0 .
Question 14.2: Finding the effects of thermal drift on the output of an inv...
\begin{aligned}\Delta T &=55-25=30^{\circ} C \\\Delta V_{ DC } &=(15+15)-(12+12)=6 V \\\Delta V_{ io } &=D_{ v } \Delta T+ PSRR \Delta V_{ DC }=15 \times 10^{-6} \times 30+150 \times 10^{-6} \times 6=1.35 mV \\\Delta I_{ io } &=D_{ i } \Delta T=0.5 \times 10^{-9} \times 30=15 nA \\\Delta I_{ B } &=D_{ b } \Delta T=0.5 \times 10^{-9} \times 30=15 nA\end{aligned}
(a) With offset-minimizing resistance R_{ X } , the total output voltage of the inverting amplifier is given by
v_{ O }=-\frac{R_{ F }}{R_{1}} v_{ S } \pm\left(1+\frac{R_{ F }}{R_{1}}\right)\left(V_{ io }+\Delta V_{ io }\right) \pm R_{ F }\left(I_{ io }+\Delta I_{ io }\right) (14.25)
\begin{aligned}=&-\left(\frac{100 k \Omega}{10 k \Omega}\right) \times 100 \times 10^{-3} \pm\left(1+\frac{100 k \Omega}{10 k \Omega}\right) \times(6+1.35) \times 10^{-3} \\& \pm 100 \times 10^{3} \times(200+15) \times 10^{-9} \\=&-1000 mV \pm 80.85 mV \pm 21.5 mV \\=&-1102.35 mV ( min ) \text { or }-897.65 mV ( max )\end{aligned}
(b) With R_{ x }=0 , the total output voltage of the inverting amplifier is given by
v_{ O }=-\frac{R_{ F }}{R_{1}} v_{ S } \pm\left(1+\frac{R_{ F }}{R_{1}}\right)\left(V_{ io }+\Delta V_{ io }\right)+R_{ F }\left(I_{ B }+\Delta I_{ B }\right) (14.26)
\begin{aligned}=&-\left(\frac{100 k \Omega}{10 k \Omega}\right) \times 100 \times 10^{-3} \pm\left(1+\frac{100 k \Omega}{10 k \Omega}\right) \times(6+1.35) \times 10^{-3} \\&+100 \times 10^{3} \times(500+15) \times 10^{-9} \\=&-1000 mV \pm 80.85 mV +51.5 mV \\=&-1029.4 mV ( min ) \text { or }-867.65 mV ( max )\end{aligned}