For a Si, one sided abrupt junction with NA = 2 × 1019/cm3 & ND = 8 × 1015/cm3 & ni = 9.65 × 109/cm3 The junction capacitance at zero bias is ______(in mF/cm2). Given,ϵr = 11.7a)0.68b)0.72Correct answer is between '0.68,0.72'. Can you explain this answer?

Question

Answer

And ND = 0, the depletion width can be calculated using the following formula:

W = sqrt((2 * ε * Vbi) / ((1 / NA) + (1 / ND)))

Where:

- W is the depletion width
- ε is the permittivity of Si (11.8 * 10^-12 F/cm)
- Vbi is the built-in potential, which can be calculated using the following equation:

Vbi = (kT / q) * ln(NA * ND / ni^2)

Where:

- k is the Boltzmann constant (1.38 * 10^-23 J/K)
- T is the temperature in Kelvin
- q is the electron charge (1.6 * 10^-19 C)
- ni is the intrinsic carrier concentration of Si at the given temperature, which can be calculated using the following equation:

ni^2 = Nc * Nv * exp(-Eg / (kT))

Where:

- Nc is the effective density of states in the conduction band (2.8 * 10^19 cm^-3 at 300K)
- Nv is the effective density of states in the valence band (1.04 * 10^19 cm^-3 at 300K)
- Eg is the energy gap of Si (1.12 eV at room temperature)

Assuming a temperature of 300K, we can calculate:

- ni^2 = 1.5 * 10^10 cm^-6
- Vbi = (0.026 / 1.6) * ln(2 * 10^19 * 1.5 * 10^10 / (1.5 * 10^10)^2) = 0.821 V

Plugging these values into the depletion width formula, we get:

W = sqrt((2 * 11.8 * 10^-12 * 0.821) / ((1 / 2 * 10^19) + (1 / 0))) = 0.634 μm

Therefore, the depletion width for a Si one sided abrupt junction with NA = 2 × 10^19/cm^3 and ND = 0 is 0.634 μm.

Similar GATE Doubts

A one-sided abrupt pn junction diode has a depletion capacitance CDof 50 pF at a reverse bias of 0.2 V. The plot ofversus the applied voltage I/for this diode is a straight line as shown in the figure below. The slope of the plot is _______x1020 F-2 V-1.

The built-in potential of an abrupt p-n junction is 0.75V. If its junction capacitance (Cj) ata reverse bias ( VR) of 1.25V is 5pF, the value of Cj in pF when VR = 7.25V is ________________. Correct answer is '2.5'. Can you explain this answer?

The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.

In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side is

For an abrupt p-n junction diode, the doping concentrations on p-side is 9 × 1016 cm3 and an n-side is 5 × 1016/cm3. The diode is in reverse bias operating mode with total depletion width of 4 μm. Then the depletion width on n-side is __________ μm.

Top Courses for GATE

Suggested Free Tests

Related Content

Mock Test Series

Subject-wise Courses

Other Engg Branches

Related Exams

Company