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2 years ago

A diode fed with a constant current I=1mA has a voltage V = 690 mV at 20°C. Find the diode voltage at −20°C and at +85°C.

Engineering

1 answer:

jeyben [28]2 years ago

7 0

<h2>Answer:</h2><h2></h2>

602.23mV at -20°C

852.88mV at +85°C

<h2>Explanation:</h2><h2></h2>

===> First, let's get the saturation current of the diode

The current (I) through a diode is given by;

I = $I_{S}$ x $e^{r}$ ---------------------(i)

Where;

$I_{S}$ = saturation current of the diode

r = $\frac{V}{V_{T} }$

$V_{T}$ = thermal voltage

V = diode voltage

<em>At 20°C, the thermal voltage (</em> $V_{T}$ <em>) of a diode is 25 x 10⁻³V</em>

<em />

<em>From the question, the following are given;</em>

I = 1mA = 1 x 10⁻³A

V = 690mV = 690 x 10⁻³V

<em>Find the value of r by substituting the values of V and </em> $V_{T}$ <em> into the equation;</em>

=> r = $\frac{V}{V_{T} }$

=> r = (690 x 10⁻³) / (25 x 10⁻³)

=> r = 27.6

<em>Substitute the values of r and I into equation (i) to give;</em>

1 x 10⁻³ = $I_{S}$ x e²⁷°⁶

1 x 10⁻³ = $I_{S}$ x 9.69 x 10¹¹

<em>Solve for </em> $I_{S}$ <em />

$I_{S}$ = 1 x 10⁻³ / (9.69 x 10¹¹)

$I_{S}$ = 0.103 x 10⁻¹⁴A

$I_{S}$ = 1.03 x 10⁻¹⁵ A

---------------------------------------------------------------------------------------------

===> Second, let's get the diode voltage at -20°C using the diode voltage formula as follow;

V = $V_{T}$ x ln (I / $I_{S}$ ) --------------------(ii)

Where;

V = diode voltage

$V_{T}$ = thermal voltage = k x T / q

k = Boltzmann's constant = 1.38 x 10⁻²³ Joules/kelvin

T = temperature = -20°C = 273 - 20 = 253K

q = absolute value of electron charge = 1.6 x 10⁻¹⁹C.

I = diode current = 1mA = 1 x 10⁻³A

$I_{S}$ = saturation current (calculated above) = 1.03 x 10⁻¹⁵A

Solve for $V_{T}$ ;

$V_{T}$ = kT/ q

$V_{T}$ = 1.38 x 10⁻²³ x 253 / (1.6 x 10⁻¹⁹)

$V_{T}$ = 218.2 x 10⁻⁴V

Substitute these values into equation (ii)

V = 218.2 x 10⁻⁴ x ln [(1 x 10⁻³) / (1.03 x 10⁻¹⁵)]

V = 218.2 x 10⁻⁴ x ln (0.97 x 10¹²)

V = 218.2 x 10⁻⁴ x 27.6

V = 6022.32 x 10⁻⁴V

V = 602.23 x 10⁻³V

V = 602.23mV

<em>Therefore, the diode voltage at -20°C is 602.23mV</em>

------------------------------------------------------------------------------------------------------

===> Third, let's get the diode voltage at +85°C using the diode voltage formula as follow;

V = $V_{T}$ x ln (I / $I_{S}$ ) --------------------(ii)

Where;

V = diode voltage

$V_{T}$ = thermal voltage = k x T / q

k = Boltzmann's constant = 1.38 x 10⁻²³ Joules/kelvin

T = temperature = 85°C = 273 + 85 = 358K

q = absolute value of electron charge = 1.6 x 10⁻¹⁹C.

I = diode current = 1mA = 1 x 10⁻³A

$I_{S}$ = saturation current (calculated above) = 1.03 x 10⁻¹⁵A

Solve for $V_{T}$ ;

$V_{T}$ = kT/ q

$V_{T}$ = 1.38 x 10⁻²³ x 358 / (1.6 x 10⁻¹⁹)

$V_{T}$ = 308.8 x 10⁻⁴V

Substitute these values into equation (ii)

V = 308.8 x 10⁻⁴ x ln [(1 x 10⁻³) / (1.03 x 10⁻¹⁵)]

V = 308.8 x 10⁻⁴ x ln (0.97 x 10¹²)

V = 308.8 x 10⁻⁴ x 27.6

V = 8522.88 x 10⁻⁴V

V = 852.88 x 10⁻³V

V = 852.88mV

<em>Therefore, the diode voltage at +85°C is 852.88mV</em>

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Laminar flow normally persists on a smooth flat plate until a critical Reynolds number value is reached. However, the flow can b

grandymaker [24]

Answer:

At L = 0.1 m

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975K W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5 W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5 dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2 dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5 = 3.48K(0.1)^-0.5 W/m^1.5

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5 = 3.48K(1)^-0.5 W/m^1.5

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K W/m^1.5

h⁻_turb = 7.8848K W/m^1.8

At L = 1 m

h⁻_lam = 3.48K W/m^1.5

h⁻_turb = 4.975K W/m^1.8

3 0

2 years ago

A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensil

Gemiola [76]

The radius of the specimen is 60 mm

<u>Explanation:</u>

Given-

Length, L = 60 mm

Elongated length, l = 10.8 mm

Load, F = 50,000 N

radius, r = ?

We are supposed to calculate the radius of a cylindrical brass specimen in order to produce an elongation of 10.8 mm when a load of 50,000 N is applied. It is necessary to compute the strain corresponding to this

elongation using Equation:

ε = Δl / l₀

ε = 10.8 / 60

ε = 0.18

We know,

σ = F / A

Where A = πr²

According to the stress-strain curve of brass alloy,

σ = 440 MPa

Thus,

$sigma = 50,000 / \pi (r)^2\\\\440 X 10^6 = \frac{50,000}{3.14 X (r)^2}\\\\r = 0.06m\\r = 60mm\\\\\\$

Therefore, the radius of the specimen is 60 mm

3 0

2 years ago

The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use

Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$

7 0

2 years ago

A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses

yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature $T_1$ = 25° C

$T{\infty} =600^0C$

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}$

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}$

$Bi = \dfrac{2.5}{231}$

Bi = 0.0108

The time constant value $\tau_t$ is :

$\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}$

$\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}$

$\tau_t = \dfrac{2702* 0.025*1033}{100}$

$\tau_t = 697.79$

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]$

where;

$Q = -\Delta E _{st}$ which correlates with the change in the internal energy of the solid.

So;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}$

The maximum value for the change in the internal energy of the solid is :

$(pVc)\theta_1 = -\Delta E _{st}max$

By equating the two previous equation together ; we have:

$\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}$

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

$0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}$

So;

$0.75= [1-e^{\dfrac {-t}{ 697.79}}]}$

$1-0.75= [e^{\dfrac {-t}{ 697.79}}]}$

$0.25 = e^{\dfrac {-t}{ 697.79}}$

$In(0.25) = {\dfrac {-t}{ 697.79}}$

$-1.386294361= \dfrac{-t}{697.79}$

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

$\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}$

$\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}$

$\dfrac{T - 600}{25-600}= 0.25$

$\dfrac{T - 600}{-575}= 0.25$

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0

2 years ago

Liquid oxygen is stored in a thin-walled, spherical container 0.75 m in diameter, which is enclosed within a second thin-walled,

nadezda [96]

Answer:

Explanation:

the solution is well stated

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2 years ago