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 :
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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