OpAmp - EECE 251 - Saturated Question Final Exam 2010

Initially, consider the OpAmps working in the linear region. - calculate Vo1 (by superposition or KCL). Using KCL: - Negative Feedback V1 = V2 - Node 1 (12 - V1)/4k = 0 + (V1 - Vo1)/40k - Node 2 (10 - V2)/4k = 0 + V2/4k Solving the equations, Vo1 = -65V. This value exceeds the rail limit (12V), then, the opamp is in saturation. The second opamp is in the non-inverting configuration and the resistor R8 does not affect the circuit because there is no current on the input. The output Vo2 = 8V, exceeding the rail limit (6V). Knowing that both opamps are in the saturation region it's necessary to know if they are in negative or positive saturation. Initially, we will assume the values of Vo1 = -12, and, Vo2 = 6. Substituting this values on equation 1 and 2, V1 = 9.81V and, V2 = 5V. - node 3 (6 - V3)/3k = 0 + V3/3k V3 = 3V. As V1 is greater than V2, Vo1 is in negative saturation. As 4V is greater than V3 (V-), the opamp is in positive saturation. Based on: https://youtu.be/KTB-wT8EG0A https://youtu.be/WT_o_OQ7foc