Question

A semiconductor Hall device at T=300 K has the following geometry: d=10-3 cm, W=10-2 cm, and L=10-1 cm. The following parameters are measured: Ix=0.5 mA, Vx=15 V, VH = -5.2 mV, and Bz=0.1 tesla. Determine the (a) conductivity type; (b) majority carrier concentration; and (c) majority carrier mobility.

Answer 1

For an type semiconductor , the hall voltage is given by: …… (1) Here, is the current in direction, is the magnetic field in direction , is the electron concentration, is the thickness and is the electron charge. For type semiconductor , the hall voltage is given by: …… (2) step: 2 of 4 (a) Provided in the question, hall voltage is negative. From the equation (1) when hall voltage is negative then conductivity type is . Hence, it is n-type semiconductor. step: 3 of 4 (b) Calculate majority carrier concentration. Rearrange the equation (1) for majority carrier concentration. Substitute for , for , for , for and for . Hence, the majority carrier concentration is . step: 4 of 4 (c) Write the expression for the majority carrier mobility. Here, is the length of the conductor, is the voltage in the x-direction and is the space charge width of the conductor. Substitute, for , for , for , for , for , for and for . Hence, the majority carrier mobility is .