If you want to increase the reverse bias capacitance of a P+/N junction diode at a given reverse bias voltage, which of the following action (assume the diode remains to be P+/N diode after the change) is the most effective method?

Question

BlackTom AI · Accepted Answer

Question restatement: If you want to increase the reverse bias capacitance of a P+/N junction diode at a given reverse bias voltage (with the diode still being P+/N after the change), which action is most effective? Options: increase P+ doping; reduce P+ doping; increase N doping; reduce N doping.

Analysis of how depletion capacitance behaves: For a PN junction in reverse bias, the depletion width w is approximately w ≈ sqrt[ (2ε (V_bi + V_R)) / (q (N_A + N_D)) ], where ε is the permittivity, V_bi is the built-in potential, V_R is the reverse-bias voltage, q is the elementary charge, and N_A and N_D are the acceptor and donor dopant concentrations on the P and N sides, respectively. The depletion capacitance per area is C ≈ ε / w, which leads to C ∝ sqrt( N_A + N_D ) for a given reverse bias. In short, increasing the total dopant concentration on either side generally increases the capacitance, but the detailed effect in a P+/N structure depends on which side is doped more heavily and how the depletion region shifts.

Option 1: Increase the P+ doping. Rationale: Increasing N_A raises the sum N_A + N_D, which tends to increase C according to the relation above. However, because the P+ side is already heavily doped (P+), further increasing P+ doping yields diminishing returns in reducing the depletion width on that side and may cause only a modest change in C compared with altering the lighter-doped side. In practice, the impact is present but not as strong as changing the less-doped side in an asymmetric junction.

Option 2: Reduce the P+ doping. Rationale: Lowering N_A decreases the total N_A + N_D, which tends to decrease C (w increases, so C decreases). This action would reduce the reverse capacitance, not increase it.

Option 3: Increase the N doping. Rationale: Increasing N_D raises the total N_A + N_D, which tends to shorten the depletion width and raise the depletion capacitance. In a P+/N junction where the N side can still be effectively doped, boosting N doping directly increases C more markedly, especially if the N region was relatively lightly doped before. This is the most effective lever among the given options for increasing C under a fixed reverse bias.

Option 4: Reduce the N doping. Rationale: Decreasing N_D lowers N_A + N_D, broadens the depletion region, and reduces C. This exactly works against the goal of increasing reverse capacitance.

Putting it together: While multiple changes influence capacitance via the depletion width, increasing the N doping on the N side directly increases the sum N_A + N_D and tends to produce a more pronounced rise in reverse bias capacitance for a P+/N junction under the stated condition. In contrast, reducing either dopant or increasing the dopant on the already heavily doped P+ side either lowers capacitance or provides a smaller effect. Therefore, the most effective method among the options is to increase the N doping.

If you want to increase the reverse bias capacitance of a P+/N junction diode at a given reverse bias voltage, which of the following action (assume the diode remains to be P+/N diode after the change) is the most effective method?单项选择题

类似问题

If you want to increase the reverse bias capacitance of a P+/N junction diode at a given reverse bias voltage, which of the following action (assume the diode remains to be P+/N diode after the change) is the most effective method?

A project has a required return of 12.6 percent, an initial cash outflow of $42,100, and cash inflows of $16,500 in Year 1, $11,700 in Year 2, and $10,400 in Year 4. What is the net present value?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单