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Which one of the following graphs shows the correct variation of vo with vi? Here, vd is the voltage drop across the diode and the Op-Amp is assumed to be ideal.

Op-amp with input vi at the non-inverting input; its output drives a diode whose cathode is the output vo across load RL to ground, and vo is fed back to the inverting input

  • A.Graph of vo against vi: zero for negative vi, then a straight line of slope one rising from the origin
  • B.Graph of vo against vi: zero until vi equals vd, then a straight line rising
  • C.Graph of vo against vi: nearly zero until vd, then a steep exponential rise
  • D.Graph of vo against vi: a straight line through the origin

Answer: A

The correct graph is (A): \(v_{o} = 0\) for negative \(v_{i}\) and \(v_{o} = v_{i}\) for positive \(v_{i}\).

The circuit is a precision (super) diode. For \(v_{i} \gt 0\) the op-amp output rises until the diode conducts; the feedback taken from \(v_{o}\) then forces \(v_{o} = v_{i}\), and the op-amp output simply sits \(v_{d}\) higher to cover the diode drop. For \(v_{i} \lt 0\) the op-amp output swings negative, the diode is reverse biased, and \(R_{L}\) pulls \(v_{o}\) to 0.

Graph (B), with a threshold at \(v_{d}\), and graph (C), with the exponential diode curve, describe an ordinary diode without the op-amp; the feedback removes the diode drop entirely. Graph (D) is a plain follower with no rectification.

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