Question

A circuit with a diode is shown. The ratio of the minimum to maximum small-signal voltage gain \[ \frac{\partial V_{out}}{\partial V_{in}} \] is to be found (rounded to two decimals).

Hint:

Small-signal gain changes depending on diode state: OFF → resistive divider; ON → clamped node.

Answer 1

Correct Answer: 0.7

The diode characteristic indicates: - For $V_D<0.7\,$V, the diode is \textit{OFF}. - For $V_D \approx 0.7\,$V, the diode conducts with a very steep slope $\Rightarrow r_d \approx 0$ (ON). --- Case 1: Diode OFF (high resistance)
Circuit becomes purely resistive: \[ V_{out} = V_{in}\left( \frac{2k}{2k + 2k} \right) = \frac{1}{2} V_{in}. \] So: \[ \left( \frac{dV_{out}}{dV_{in}} \right)_{\min} = 0.50. \] --- Case 2: Diode ON (short circuit approximation)
When diode conducts, the node to the right is clamped, and the effective gain becomes: \[ V_{out} = V_{in} \left( \frac{2k}{2k + 0} \right) = 1. \] Thus: \[ \left( \frac{dV_{out}}{dV_{in}} \right)_{\max} = 1. \] --- Final Ratio: \[ \frac{\left( \frac{dV_{out}}{dV_{in}} \right)_{\min}} {\left( \frac{dV_{out}}{dV_{in}} \right)_{\max}} = \frac{0.50}{1} = 0.50. \] Rounded: \[ \boxed{0.50} \]