Question

Under high electric fields, in a semiconductor with an increasing electric field,

• A. the mobility of the charge carriers decreases and the velocity of the charge carriers saturates.
• B. the mobility of the carriers and the velocity of the charge carriers, both increase.
• C. the mobility of the charge carriers decreases and the velocity of the charge carriers becomes zero.
• D. the mobility of the carriers increases and the velocity of the charge carriers becomes zero.

Hint:

Think of drift velocity in a semiconductor like a car's speed. At low speeds, pressing the accelerator more (higher E-field) gives a proportional increase in speed (higher \(v_d\)). At very high speeds, air resistance (scattering) becomes so great that even pressing the accelerator to the floor (very high E-field) won't make the car go much faster (velocity saturation).

Answer 1

The Correct Option is A

Step 1: Recall the low-field relationship. At low electric fields (\(E\)), the drift velocity (\(v_d\)) of charge carriers is directly proportional to the field strength. The constant of proportionality is the mobility (\(\mu\)): \[ v_d = \mu E \] In this region, mobility is approximately constant.
Step 2: Analyze the high-field behavior. As the electric field becomes very strong, carriers are accelerated to high energies between collisions with the crystal lattice. At these high energies, new scattering mechanisms (like optical phonon emission) become dominant and very effective at removing energy from the carriers. This increased scattering rate limits the average velocity that the carriers can attain.
Step 3: Describe velocity saturation. Due to the increased scattering at high fields, the drift velocity no longer increases linearly with the field. Instead, it approaches a maximum constant value called the saturation velocity (\(v_{sat}\)). The velocity stops increasing even if the electric field increases further.
Step 4: Infer the effect on mobility. Mobility is defined as \(\mu = v_d / E\). In the high-field region, as \(E\) increases, \(v_d\) approaches a constant value \(v_{sat}\). For the equation to hold, the mobility \(\mu\) must decrease. \[ \mu(E) = \frac{v_{sat}}{E} \] As E increases, \(\mu\) decreases. Therefore, under high electric fields, mobility decreases and velocity saturates.