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Question

A TRAPATT diode has the doping concentration NA = 2 x 1015/cm3 and a current density of 20 kA/cm2. The value of avalanche zone velocity is given by :

This question was previously asked in
UGC NET 2016 Paper 3 Electronic Science Question Paper (10-Jul-2016)
The correct answer is

6.25 x 107 cm/s

To find the value of the avalanche zone velocity for the given TRAPATT diode, we need to use the relationship between current density (\( J \)), doping concentration (\( N_A \)), charge of an electron (\( q \)), and the avalanche zone velocity (\( v \)). The formula that relates these variables is:

\(J = q \cdot N_A \cdot v\)

Where:

  • \(J\) is the current density.
  • \(q \approx 1.6 \times 10^{-19} \text{ C}\\) is the charge of an electron.
  • \(N_A\) is the doping concentration.
  • \(v\) is the avalanche zone velocity.

Given:

  • \(N_A = 2 \times 10^{15} \, \text{cm}^{-3}\)
  • \(J = 20 \, \text{kA/cm}^2 = 20 \times 10^3 \, \text{A/cm}^2\)

Plug the known values into the equation:

\(20 \times 10^3 = 1.6 \times 10^{-19} \times 2 \times 10^{15} \times v\)

Simplifying the equation to solve for \(v\):

\(v = \frac{20 \times 10^3}{1.6 \times 10^{-19} \times 2 \times 10^{15}}\)

Calculating the right side:

\(v = \frac{20 \times 10^3}{3.2 \times 10^{-4}}\)

\(v \approx 6.25 \times 10^7 \, \text{cm/s}\)

Thus, the avalanche zone velocity of the TRAPATT diode is \(6.25 \times 10^7 \, \text{cm/s}\).

This corresponds to the correct answer: 6.25 x 107 cm/s.

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