Figure 1 shows a simplified structure of an N-channel enhancement mode MOSFET - AQA - A-Level Physics - Question 1 - 2021 - Paper 8

The terminal connected directly to 0 V when the MOSFET is used as a simple switch is the source.

To determine the minimum value of $V_{GS}$ needed for the lamp to operate at full power, we first calculate the current using the formula for power:

$P = I imes R$ Substituting the known values:

$0.65 = I imes 154$

Solving for current ($I$), we find:

I = rac{0.65}{154} ext{ A} \approx 0.00423 ext{ A} \approx 4.23 ext{ mA}

Next, we need to read from the graph (Figure 2), where the drain-source current $I_{DS}$ shows that for a current of approximately 4.23 mA, a gate-source voltage $V_{GS}$ of 3.4 V is necessary.

The relevance of the data provided lies primarily in its implications for energy management and efficiency in mobile devices. For instance, the CPU of a mobile phone utilizes a significant number of transistors (approximately $8.5 imes 10^9$) which indicates a complex operation that relies on effective switching. Given the low power consumption of individual MOSFETs (with a leakage current of around 10 nA), understanding the $T_{loss}$ value becomes crucial for optimizing the overall battery consumption of the device. Moreover, the battery capacity and the ability to remain functional for approximately 12 hours on standby also highlights the importance of creating energy-efficient designs in modern mobile technology.