8 00 ! Required information Consider the circuit given below. 0/2 points awarded 3 ΚΩ www t=0 6kM Scored R 1.5i Vc 1 μF 10 V If R = 5.00 kQ, determine vao+). The value of va(0) is 1.4545 V.

I want to know what does it look in a breadboard circuit, because I want to created it but I not sure it is build properly, can you give me an illustuation base on this image, it do need to real, something like virutal example

Charge neutrality Since doped semiconductor remains electroneutral, the concentration of negative charges equals the concentration of positive charges. n+ Na,ionized p+Nd,ionized np = n; 2 2 N-Na N N d d р + 2 2 n = Nd-Na 2 + Na - 2 Na +n₁ 2 71/2 1/2 2 2 +n Concentration of electrons and holes 1. Calculate concentrations of electrons and holes at room temperature in Si and Ge with donor concentration of 1.5x10¹7 cm³ and acceptor concentration of 8x1016 cm-3. 2. Will these concentrations change much with the temperature increase to 100°C?

Answer the questions on the end of the image please

Answer these two questions on the end of the image, please    1.Calculate intrinsic carrier concentration for Si, Ge and GaAs at temperatures -20°C, 20°C (room temperature) and 120°C     2.Compare the obtained data with n and p shown on previous slide 25

Can you help me achieve the requirements using Arduino? I have encountered some issues with these requirements. Q.2: Suppose you have two push buttons connected to ports (0 & 1) and four LED's connected to ports (6-9). Write a program to flash ON the odd LED's if we press the switch 0 for 4s, flash ON the even LED's if we press the switch 1 for 5s and flash ON all the LED's otherwise for 6s.

Charge carrier concentration in doped semiconductor: compensation n = Na - Na Na - Na >> ni n-type p = n₁²/n 2 if N₂ >> N₁, n = N₁_ and _p=n² / Na d p = Na-Nd p-type Na-Na >> n₁ d 2 n = n₁₂²/p 2 if N₁ >> N₁, p = N₁ and n = n² / Na a n-type Dopant compensation: Examples d n = Na-N₁ = 4×10¹ cm¯ -3 ++++++ n = 4×1016 cm-³ N=6×1016 cm-3 p=n/n=1020/4×1016 = 2.5×10³ cm p-type -3 p=Na-N₁ =8×10 −6×1016 = 2×10¹6 cm³ n=n²/p=1020/2×101 =5×10³ cm‍³ N2×1016 cm³ ++++++ N=6x1016 cm-3 N = 8×1016 cm-3 p=2×1016 cm‍³ The resulting charge carrier concentration in compensated semiconductor approximately equals the difference between the donor and acceptor concentrations. Charge carrier concentration in n-type and p-type semiconductors 1. Calculate concentrations of electrons and holes at room temperature in Si containing 2x1017 cm³ of donors and 8x1016 -3 cm³ of acceptors. Assume that Na, Nd >> n;. αν 2. Calculate concentrations of electrons and holes at room temperature in Ge containing 2x10¹7 cm³ of…

lonization energy of dopants in semiconductors lonization energy of shallow donors and acceptors can be evaluated using hydrogenic model: lonization energy E Hion and orbital radius a, of hydrogen atom Hydrogen Atom moe4 EHion = 13.6 eV a = 8ε²h² Απερη mee² = 5.2918 x 10-11 m lonization energy Eion and orbital radius D,A of donors and acceptors electron m* e4 Eion = ~50 meV 8K² &²h² 4πεερη2 "D,A 1 nm m*e² Orbit of an electron bound to a donor in a semiconductor crystal. Energy levels of donors and acceptors Conduction Band ↓ Ec -Ed Donor Level Donor ionization energy Acceptor ionization energy Acceptor Level Εα Ev Valence Band Ionization energy of selected donors and acceptors in silicon Donors Acceptors Dopant Sb P As B Al In Ionization energy, Ec-Ed or Ea-E, (meV) 39 44 54 45 57 160 Hydrogenic model of donors and acceptors Calculate the ionization energies and orbit radii of donors and acceptors in Si and Ge. Dielectric constant of silicon is k = 11.7. Dielectric constant of…

I need help in construct a method in matlab to find the voltage of VR1 to VR4, rhe current, and the power base on that circuit Nominal or Theortical: E1 = 3V , E2 = 9V, E3 = 1.5V R1 =10Kohm, R2 =2Kohm, R3 =1Kohm, R4 =16Kohm