Question

A pn step junction is fabricated in silicon with a donor concentration of 5 x 10¹⁵ cm⁻³ on one side of the junction and an acceptor concentration of 2 x 10¹⁵ cm⁻³ on the other. A bias voltage of 0.3 V is applied to the p-side (relative to the n-side). Find the depletion width W at 300 K. Repeat when the bias voltage is reversed.

Select 2 correct answer(s)
O W = 1.04 μm
O W = 0.567 μm
O W = 0.322 μm
O W = 0.899 μm
O W = 1.323 μm
O W = 1.84 μm when the bias voltage is reversed
O W = 0.929 μm when the bias voltage is reversed
O W = 1.131 μm when the bias voltage is reversed
O W = 0.355 μm when the bias voltage is reversed
O W = 0.527 μm when the bias voltage is reversed

Answer 1

Depletion width W of a pn junction depends on bias voltage; it narrows under forward bias and widens under reverse bias. Numerical calculation of W requires specific formula and semiconductor properties.

Step-by-step explanation:

To find the depletion width W of a silicon pn junction at 300 K with given concentrations of donors and acceptors and a bias voltage, we can use semiconductor physics principles. When forward-biased (positive side connected to p-type), the depletion region narrows, reducing the potential barrier and allowing current to flow more easily. Conversely, reverse biasing (positive side connected to n-type) widens the depletion region, increasing the potential barrier and hindering current flow.

The question involves calculating this depletion width under both forward and reverse biases. However, the student needs to provide the equation used to calculate the depletion width as it is dependent on various factors such as the permittivity of silicon, applied bias, and the concentrations of the dopants. Without the specific equation or more context, providing a numerical answer would be speculative.