Circuits II · Two-port networks and interconnection
An ideal operational amplifier is supplied from ± 12.0 V and saturates at those rails
Problem
An ideal operational amplifier is supplied from \(\pm 12.0\,\mathrm{V}\) and saturates at those rails. In the linear region the input currents are zero and \(v_+=v_-\). The independent source \(v_{\mathrm{in}}\) is connected directly to the noninverting input. Resistor \(R_g=5.00\,\mathrm{k}\Omega\) is connected between the inverting input and ground. Resistor \(R_f=20.0\,\mathrm{k}\Omega\) is connected between the output and the inverting input. No separate load is attached. Practical constraint: the linear relation may be used only while \(|v_o|\) is strictly less than \(12.0\,\mathrm{V}\). (a) Determine the closed-loop voltage gain \(v_o/v_{\mathrm{in}}\) that applies in the linear region. (b) Determine the closed interval of \(v_{\mathrm{in}}\) for which the amplifier remains linear. (c) For \(v_{\mathrm{in}}=2.00\,\mathrm{V}\), determine \(v_o\) and \(v_-\), and state whether \(v_+=v_-\). (d) For \(v_{\mathrm{in}}=2.80\,\mathrm{V}\), determine the actual output voltage after the rails are enforced, determine \(v_-\) from the resistor divider attached to that output, and state whether a virtual short exists between the input terminals.
Hint
Noninverting gain is \(1+R_f/R_g\) only while the output is strictly inside the rails.
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