Question

9. An n- type germanium semiconductor sample is brought into contact with a p - type silicon sample. The germanium sample has a carrier concentra- tion of 4.5 x 1016cm-3 and the silicon sample has a carrier concentration of 1.0 × 1016cm-3. At 300K the intrinsic carrier concentration of germanium is 2.4 × 1013cm-3 and its band gap is 0.66 eV. At 300K the intrinsic carrier concentration of silicon is 1.45 × 1010cm-3 and its band gap is 1.12 eV. (a) Describe the differences between an n - type semiconductor and a p - type semiconductor. (b) Calculate the energy difference between the centre of the band gap and the Fermi level in both cases. (c) Draw a labeled diagram showing the edges of the bands of the semicon- ductors and the Fermi level(s) when the semiconductors are brought into contact. Show the region near the junction on both sides of the Junction. (d) Name the type of junction and briefly discuss one application 10. Select one of the following devices: Solar-cell, photodetector, avalanche photodiode, LED (light emitting diode) OLED (Or- ganic LED), semiconductor laser diode, VCSEL (vertical cavity surface emitting laser), LCD (liquid crystal display) Explicitly state the device you have chosen For your chosen device, answer the following (i) What is the main purpose of the device? (ii) State two to three applications of the device, including details of the use. (iii) Sketch (including labels) the device. Include all elements necessary for the device to function correctly. (iv) Explain the principles of how the device operates. Refer to your diagram fronm part (c) (iii). Include details of the standard operating conditions (applied cur- rent, direction of current, voltage, temperature etc.). Include band diagrams where appropriate (v) Discuss in detail the basic physical principles underlying the device (vi) Explain the fabrication methods and required processing steps to fab- (vii) Discuss applications of the device. operation ricate this device

Answer 1

Note: I am allowed to answer only 1 question and four of it's sub-parts- Sorry about that

(a)

n- type : It is a semi-conductor that contains predominantly electrons as charge carriers

p -type :It is a semi-conductor that contains predominantly holes as charge carriers

(b)

------- for n-type

= 8.6 X 10^(-5) ev/K X 300 X ln [4.5 X 10^(16)/2.4 x 10^(13)]

= 0.194 eV

For p-type

Ei-Ef = kT ln(p/ni) = 8.6 X 10^(-5) ev/K X 300 X ln(10^16/1.45 X 10^(10)) = 0.347 eV

(c)

geo Po >> >no Neuke PERE ん Nusc Apo no E, le:

(d) This junction is called p-n junction and can be used as a rectifier. Suppose you put an AC voltage across this junction when +ve voltage is on p-type and -ve voltage is n-type ,then the barrier level goes down and and current flows across barrier in form of hole and electrons. However,in the other half of AC power cycle the P-type is negatively charged and n -type is ositively charged ,the potential barrier at junction increases and hence no current flows and thus the output you get is rectified power allowed to flow in one-half of AC cycle.