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"Determine and graph v OUT as a function of v IN for the circuit shown in Figure 16.31. In doing so, model the diode as shown in Figure 16.29, and assume that the Op Amp is ideal. Also, contrast the input-output behavior of the circuit shown in Figure 16.31 with that of the half-wave rectifier studied in the chapter on nonlinear analysis."

I'm confused what happens when the diode is off. Can someone explain? here's the circuit: enter image description here thanks in advance.

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    \$\begingroup\$ Practically, when the diode is "off", the op amp saturates at the negative rail, and that is why this is a bad circuit. For the purpose of the exercise, they probably want you realize that no current is flowing. \$\endgroup\$ – Matt Young Jan 6 '16 at 15:27
  • \$\begingroup\$ @MattYoung Why is that a bad thing. The reverse bias current of the diode should be well bellow the max current of the OpAmp. \$\endgroup\$ – vini_i Jan 6 '16 at 15:32
  • \$\begingroup\$ @vini_i That is probably worth a question by itself. \$\endgroup\$ – Matt Young Jan 6 '16 at 15:36
  • \$\begingroup\$ @Matt: Ideal opamps don't have problems driving to the negative rail. \$\endgroup\$ – Olin Lathrop Jan 6 '16 at 15:36
  • \$\begingroup\$ @OlinLathrop That is why is prefaced the statement with practically. \$\endgroup\$ – Matt Young Jan 6 '16 at 15:37
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When you have a problem analyzing a circuit like this in a particular state, start by noting the voltages and currents you do know on the schematic, then see if you can figure out what others are from that. As you figure out more, it should be possible to figure out even more. Eventually you figure out the whole circuit.

In this example, you want to examine what happens with negative input. Set the voltage source to -1 V. That obviously makes the opamp + input also -1 V. You don't know yet what the opamp output and the diode are doing. Start by assuming the diode is a open circuit, and see where that goes. You should be able to see the voltage on the opamp - input from inspection then. Now you have both opamp inputs, so you should be able to see what the output is doing. Now you know both ends of the diode. Is it really reverse biased as we assumed, or is it forward biased and we therefore need to revise our assumption of the opamp - input voltage? Either way, you get to a stable condition where you know all the voltages and you should be able to see what is going on.

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Since the op-amp is stated to be ideal, for an input below 0V, the output voltage should increase (in a negative direction) until the op-amp is balanced. Since the diode is not stated to be ideal, that will happen when the diode breaks down in reverse (maybe a few hundred volts, or more, depending on the diode), unless the input voltage is low enough that Is satisfies the op-amp's appetite for balance.

In reality, an op-amp cannot supply voltages beyond its (unseen here) supply rails, and the diode would sensibly be rated to break down at a higher voltage, so the op-amp output sits near the negative rail ('saturated') and the diode is reverse biased (aka 'off') with only a bit of leakage current (some nA typically) flowing.

The ugly reality- op-amps can have poor (and poorly characterized) behavior when you allow them to saturate- very long recovery times to balance again (due in part to the internal compensation circuitry IIRC). Also, this circuit allows the differential input voltage of the op-amp to be large for large negative inputs (that will result in conduction between the inputs in many modern bipolar op-amps so it won't work well as a rectifier if the input gets below about -600mV). It also requires a negative supply on the op-amp- it's a pretty horrible circuit in a plethora of different ways.

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