Optoelectronics and Photonics 2nd edition. Chapter
3.16
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3.16 Heterostructure Consider a Type I heterostructure as shown in Figure 3.27. (a) If Egl
Wdep P-Type N -Type Q1. Consider the PN junction at equilibrium shown in the figure above....
Wdep P-Type N -Type Q1. Consider the PN junction at equilibrium shown in the figure above. Both N-side and P-side has same doping density NA ND 1017 /cm3. Assume both electron and hole mobility to be same, i.e Me - 1000cm2/Vs. a equilibrium energy band diagram. Find (EF Et at(i)x-0. (ii) x »xn (iii) X <K_Xp Find the value of built-in voltage and total depletion width (5+5 points) Find electron and hole densities at (i) x = 0. (ii) x...
Q1. i) Will the current flow from A to C, for the connection shown in Figure....
Q1. i) Will the current flow from A to C, for the connection shown in Figure. 12 Support your answer (11 with proper reasons. Both D1 and D2 are silicon diodes. D1 D2 A_* B Figure 1 ii) Draw the equivalent circuit diagram for Ideal Diode under forward bias and reverse bias. iii) Determine the value of I for the circuit shown in Figure. 2 R - 5V 15.2 V 5 kΩ Ge Figure 2 iv) Compute Vo and Ir...
9. An n- type germanium semiconductor sample is brought into contact with a p - type...
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....
ECE question
P and N type semiconductors are formed with an acceptor and donor concentration of 1×1017 cm-3 and 1×1016 cm-3 , respectively,
intrinsic carrier concentration is 1×1010 cm-3 and relative permittivity (єs
) is 12є0 @ 300K.
Given, permittivity of free space (є0
) 8.85 × 10-12 Farad/meter, KT @ 300K 0.0259 eV, q = 1.602 × 10-19 coulombs
A. Calculate the following quantities @ 300K
1. Potential Drop (in V) and Maximum Electric Field (in V/cm) across PN-junction [2 +...
EENG 245 Physical electronics HW 1 1) The NaCl crystal is cubic, and can be described...
EENG 245 Physical electronics HW 1 1) The NaCl crystal is cubic, and can be described as follows. Na atoms sit at the corners and faces of a cube, and Cl atoms sit in between two Na atoms. This means that a Clatom is found half-way along each of the cube edges, and there is a Cl in the center of the cube. (We could also have described the lattice by interchanging Na and Cl in the description above.) Another...
Wdep P-Type N -Type Q1. Consider the PN junction at equilibrium shown in the figure above. Both N-side and P-side has same doping density NA ND 1017 /cm3. Assume both electron and hole mobility to be same, i.e Me - 1000cm2/Vs. a equilibrium energy band diagram. Find (EF Et at(i)x-0. (ii) x »xn (iii) X <K_Xp Find the value of built-in voltage and total depletion width (5+5 points) Find electron and hole densities at (i) x = 0. (ii) x...
Q1. i) Will the current flow from A to C, for the connection shown in Figure. 12 Support your answer (11 with proper reasons. Both D1 and D2 are silicon diodes. D1 D2 A_* B Figure 1 ii) Draw the equivalent circuit diagram for Ideal Diode under forward bias and reverse bias. iii) Determine the value of I for the circuit shown in Figure. 2 R - 5V 15.2 V 5 kΩ Ge Figure 2 iv) Compute Vo and Ir...
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....
EENG 245 Physical electronics HW 1 1) The NaCl crystal is cubic, and can be described as follows. Na atoms sit at the corners and faces of a cube, and Cl atoms sit in between two Na atoms. This means that a Clatom is found half-way along each of the cube edges, and there is a Cl in the center of the cube. (We could also have described the lattice by interchanging Na and Cl in the description above.) Another...
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