All categories

3 years ago

A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?

Engineering

1 answer:

Illusion [34]3 years ago

3 0

Answer:

0.0659 A

Explanation:

Given that :

$I_{0} = 6nA$ ( saturation current )

at 25°c = 300 k ( room temperature )

n = 2 for silicon diode

Determine the saturation current at 100 degrees = 373 k

Diode equation at room temperature = I = Io $\frac{V}{e^{0.025*n} }$

next we have to determine the value of V at 373 k

q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1

Given that I is constant

Io = $\frac{e^{0.025*2} }{31.08}$ = 0.0659 A

You might be interested in

An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610

KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + $\frac{q}{h\pi Di}$

Ts = $30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000} }$

$Ts=30+\frac{294}{0.3928} \\$

$Ts =30+748.4\\Ts = 778.4$

The surface temperature is therefore 778.4°C if the cable is bare

6 0

2 years ago

Why was information revolution different or unique <br>

Xelga [282]

Answer:

It made information easily accessible and ensured individuals became more vast in subject topics of interest.

Explanation:

Information revolution is different and unique and it came with the advent of computers and the internet. A lot of information is stored there which is too large and complex for the human brain.This helped people to access information without much stress as informations about almost every subject is on the Internet.

Individuals can check the informations up and become more vast in interested topics.

8 0

3 years ago

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length

masya89 [10]

Answer:

Time period = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration $9.4 \times 10^{-5} kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s$

Time period $= \frac{210}{5.04\times 10^{-3}} = 41654.08 s$

3 0

3 years ago

What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a

Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18 )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0

2 years ago

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

nydimaria [60]

Answer:

The value of Modulus of elasticity E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

Beam deflection is = 0.15 in

Explanation:

Given data

width = 5 in

Length = 60 in

Mass of the person = 125 lb

Load = 125 × 32 = 4000 $\frac{ft lbm}{s^{2} }$

We know that moment of inertia is given as

$I = \frac{bt^{3} }{12}$

$I = \frac{5 (1.5^{3} )}{12}$

I = 1.40625 $in^{4}$

Deflection = 0.15 in

We know that deflection of the beam in this case is given as

Δ = $\frac{PL^{3} }{48EI}$

$0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}$

E = 85.33 × $10^{6}$ $\frac{lbm}{in^{2} }$

This is the value of Modulus of elasticity.

Beam deflection is = 0.15 in

6 0

2 years ago