In a Schottky barrier contact between a metal and an n-silicon. If the doping concentration of the n region is halved, which of the following statement is/are correct?

Question

BlackTom AI · Accepted Answer

When the n-region doping concentration Nd is halved in a Schottky barrier, several interrelated quantities change.
Option 1: The depletion width increases
- The depletion width W in a Schottky contact at zero or small bias is proportional to sqrt(εs Vb / q Nd), where Vb is the built-in potential. Reducing Nd tends to widen the depletion region because fewer dopants mean the depletion region must extend further into the semiconductor to balance charges.
- Additionally, halving Nd reduces Vb only modestly (since Vb ∝ ln(Nd) for a Schottky contact with metal work function fixed), but the dominant Nd in the denominator makes W larger overall. Therefore, this option tends to be correct: the depletion width increases.

Option 2: The depletion width decreases
- This is inconsistent with the reasoning above. Since Nd decreases, W does not decrease; it tends to increase. Hence this option is incorrect.

Option 3: The built-in potential increases
- Vb for a Schottky contact with a fixed metal work function is given by Vb ≈ (kT/q) ln(Nd Nm / ni^2). If Nd is halved, the product Nd Nm decreases, so the logarithm—and thus Vb—decrease slightly, not increase. Therefore the built-in potential does not increase; this option is incorrect.

Option 4: The built-in potential decreases
- As explained, halving Nd reduces the term inside the logarithm, so Vb decreases modestly. This makes this statement correct.

In summary, the statements that correctly describe the situation are: the depletion width increases (Option 1) and the built-in potential decreases (Option 4). The other options describe opposite trends and are not correct.

In a Schottky barrier contact between a metal and an n-silicon. If the doping concentration of the n region is halved, which of the following statement is/are correct?多项选择题

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