Question 6A171

The diode detector operates based on the non-linear current-voltage (I-V) characteristic of a diode. For **weak signals**, the diode's exponential I-V curve can be approximated by a Taylor series. When only a small AC voltage is applied, the output current contains terms proportional to the input voltage (V) and the square of the input voltage (V²). The V² term is responsible for detection, producing an output proportional to the square of the input signal amplitude. This is known as a **square-law detector**. For **strong signals**, the diode is driven well into its conducting region during the positive half-cycle and deeply into cutoff during the negative half-cycle. In this regime, it acts more like a rectifier, passing the positive peaks and blocking the negative troughs. The average output voltage becomes nearly directly proportional to the peak amplitude of the input signal, exhibiting a **practically linear** response. Therefore, the diode detector is square-law on weak signals and practically linear on strong signals, making both A and B correct. C is incorrect because it is not square law on strong signals.