Question

Which one of the following graphs shows the correct variation of $v_o$ with $v_i$? Here, $v_d$ is the voltage drop across the diode and the Op-Amp is assumed to be ideal. 

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Hint:

A diode after an Op-Amp acts like an ideal level shifter: the output follows the input only after the diode drop is exceeded.

Answer 1

The Correct Option is A

Step 1: Understand the circuit.
The circuit contains an ideal Op-Amp followed by a diode in series with the output. The diode ensures that the output appears only when the Op-Amp forward-biases it. Since the Op-Amp is ideal, it will produce whatever output is needed to make the diode conduct when possible.

Step 2: Output behaviour for $v_i < v_d$.
When $v_i$ is too small, the Op-Amp must generate $v_o + v_d$ internally. However, the diode blocks conduction until the voltage at the diode input exceeds $v_d$. Thus for $v_i < v_d$, the external output $v_o = 0$.

Step 3: Output behaviour for $v_i > v_d$.
Once $v_i$ exceeds $v_d$, the diode forward-biases. The Op-Amp output is now able to appear across the load resistor. Thus $v_o = v_i - v_d$, a straight-line graph with slope 1 starting at $v_i = v_d$.

Step 4: Conclusion.
The graph must be a straight line shifted right by $v_d$, which corresponds to option (B).