Gunn diode oscillator: Given electron drift velocity v_d = 10^7 cm/s and active region length L = 10 × 10^-4 cm, estimate the natural oscillation frequency based on transit time.

Difficulty: Medium

Correct Answer: 10 GHz

Explanation:


Introduction:
Transferred-electron (Gunn) devices oscillate due to the formation and transit of high-field domains across the active region. A first-order estimate of the oscillation frequency can be made from the transit time, which depends on the electron drift velocity and the device length under operating bias.


Given Data / Assumptions:

  • Drift velocity v_d = 10^7 cm/s.
  • Active region length L = 10 × 10^-4 cm = 1 × 10^-3 cm.
  • First-order model: f ≈ v_d / L.
  • Neglect circuit parasitics and harmonics for this estimate.


Concept / Approach:

In the simple transit-time approximation, one oscillation period corresponds to the time a domain takes to cross the active length. Thus T ≈ L / v_d and f ≈ 1 / T = v_d / L. This quick estimate often aligns with the X-band operation of many practical Gunn oscillators using sub-millimeter active lengths.


Step-by-Step Solution:

Convert L: 10 × 10^-4 cm = 1 × 10^-3 cm.Compute transit time T = L / v_d = (1 × 10^-3 cm) / (1 × 10^7 cm/s) = 1 × 10^-10 s.Frequency f = 1 / T = 1 × 10^10 Hz = 10 GHz.Select the closest option: 10 GHz.


Verification / Alternative check:

Typical Gunn oscillators can operate across 5–100 GHz depending on material, field, and length; the calculation is consistent with common X-band designs.


Why Other Options Are Wrong:

  • 1 MHz, 10 MHz, 1 GHz: orders of magnitude too low for the given v_d and L.
  • 100 GHz: an order higher than the transit-time estimate here.


Common Pitfalls:

Using f ≈ v_d / (2L) universally; while subharmonic operation can occur in some circuits, the basic domain transit estimate uses v_d / L.


Final Answer:

10 GHz

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