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Katarina [22]
3 years ago
8

How the LED (light emitting diode) works?​

Engineering
2 answers:
Dennis_Churaev [7]3 years ago
8 0

Answer:

<h3>LEDs (Light Emitting Diodes) convert electrical energy directly into light, delivering efficient light generation with little-wasted electricity.</h3>

igor_vitrenko [27]3 years ago
8 0
Https://www.explainthatstuff.com/diodes.html
This will give you your answer :)
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Which claim does president Kennedy make in speech university rice ?
mafiozo [28]

Answer:  The United States must lead the space race to prevent future wars.

Explanation: Hope this helps

4 0
3 years ago
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The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

Explanation:

From the question we are told that:

Temperature of silicon T=300k

Electron concentration n(x)=10^{16}\exp (\frac{-x}{18})

                                        \frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})

Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

Electron current density Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2

Generally the equation for the semiconductor is mathematically given by

Jn=qb_n\frac{dn}{dx}+nq \mu E

Therefore

-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E

E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

7 0
3 years ago
The "Big Dig" was the nickname of the civil engineering project that redesigned the highway Infrastructure for the city of
zheka24 [161]
Geotechnical since it’s geologicaly based
4 0
3 years ago
Bind hole, 38 diameter, .50 deep
agasfer [191]

Answer:

59.69021

Explanation:

38/.5 x 3.14159

4 0
2 years ago
Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been d
navik [9.2K]

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

7 0
3 years ago
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