Question 10.12: Calculation of Worst-Case DC Output Voltage A certain direct...

Because the circuit is linear, we use superposition, thereby dividing the problem into several relatively simple problems.

The circuit, including the source resistances, is shown in Figure 10.42(a).
Since the circuit is linear, we can use superposition, considering each source separately. Because the impedances for the two inputs are the same, the effects of the bias currents balance and, therefore, can be ignored.

The offset current flows through the parallel combination of R_{\text{in}} and the sum of the source resistances as shown in Figure 10.42(b). Thus, the differential input voltage arising from the offset current has a maximum value given by

V_{I\text{off}}=\frac{I_{\text{off}}}{2}\frac{R_{\text{in}}(R_{s1}+R_{s2})}{R_{\text{in}}+R_{s1}+R_{s2}}=6.67 \text{ mV}

The circuit with only the offset voltage source activated is shown in Figure 10.42(c). The differential input voltage resulting from the offset voltage is found by noting that part of the input offset source voltage appears across the input terminals and the rest appears across R_{s1}  and R_{s2} . The portion across the input terminals can be computed by use of the voltage divider principle as

V_{V\text{off}}=V_{\text{off}}\frac{R_{\text{in}}}{R_{\text{in}}+R_{s1}+R_{s2}}=4.17 \text{ mV}

Multiplying by the amplifier gain, we find that the maximum output voltage caused by the offset current source is 0.667 V and the maximum output voltage caused by the offset voltage source is 0.417 V. These voltages are maximum values and they can have either polarity, so the total output voltage can range between –1.084 and +1.084 V.