I saw this problem on a sample test and was stumped by the purpose of the diode. Since it's an ideal op amp, the 2 inputs (Vp, Vn) are the same. You know that Vp = 2Vi and so you also know that Vp = Vo = 2Vi. This leads me to think that Vo is totally unaffected by the diode. That was my intuition at least. This led me to believe that this would be the answers to part A and B, but apparently it's this. This obviously means my rationale is wrong that the diode actually plays a role in all of this. Please help me understand where I went wrong. Thanks!
This Non-Inverting amplifier is unity gain when the diode conducts with the Op Amp more negative than the Resistor at Vo and it is open loop when the diode is reverse biased when OpAmp Vout is positive while Vo=0V.
Since we know during linear operation Vin(+) = Vin(-) must be true and the output must be at whatever level it can to achieve this. Thus when V0=0 the Op Amp must be at -0.7 lower to drive the resistor at 0V.
But if Vin(+) goes >0 then the loop opens and Op Amp is now like a comparator and goes as high as possible to V+ while the diode is now reverse biased and the resistor is still at 0V.
That is until the 30V Zener breakdown voltage is reached , when it conducts with 100% feedback current and becomes a unity gain non-inverting amplifier again.
So we have 3 voltage ranges where ;
- Vin(+) < 0 Av=+1 Vo=Vin(+)
- Vin(+) > 0 Av= ∞ Vo=0V
- Vin(+) > Vz Av=1 Vo=Vin(+)
Now you can plot it and figure out the rest.
It does not matter if the diode is silicon or Schottky or an LED, as far as the DC operation now as an ideal diode. It would affect the maximum voltage range but V+/V- limits were not defined.
Normally a small signal silicon diode is most commonly used.
We also know that the Op Amp out is current limited, so it cannot be used for any power rectification.)