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Vinvika [58]
3 years ago
14

For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active

mode. If IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA, calculate the base transport factor, emitter injection efficiency, common-base current gain, common-emitter current gain, and ICBO. If the minority stored base charge is 4.9 * 10-11 C, calculate the base transit time and lifetime.

Engineering
1 answer:
Sonja [21]3 years ago
6 0

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 =  0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 =  0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97  = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

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OlgaM077 [116]

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

0+F(t_{2} -t_{1} )=0.2v_{2}

if F=2 kN and t2-t1=2x10^-3 s. Replacing

0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

T_{2} =\frac{1}{2} mv_{2}^{2}  \\T_{3} =0\\U_{2---3} =-F_{res} x

Replacing:

\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N

3 0
3 years ago
The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:
Serggg [28]

Answer:

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b) 1.38233

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Explanation:

Find the void ratio using the formula:

y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1

Here;

G_{s} is specific gravity of soil solids

y_{w} is unit weight of water = 998 kg/m^3

w is the moisture content = 0.17

e is the void ratio

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Saturation Ratio Formula:

w*G_{s} = S*e  ..... Eq2

S is saturation rate

Substitute Eq 2 into Eq 1

y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w}  }{1+e}

14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233

Specific gravity of soil solids

G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765

Saturated Unit Weight

y_{s} = \frac{(G_{s} + e)*y_{w}  }{1+e} \\=\frac{(4.878811765 + 1.38233)*998  }{1+1.38233}\\\\= 2622.902571 N/m^3

7 0
3 years ago
A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as
Ulleksa [173]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

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v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 =  101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ =   254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP  = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP  = 9196.41/(0.00055501 -  0.000055501) = 18410899.5 kPa

4 0
3 years ago
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Makovka662 [10]

Answer:

The correct option is;

A. be in compliance with school attendance requirements

Explanation:

The requirements to acquire a special restricted driver's license for driver's license for under under 18 that have had a beginner's for up to 180 days include a minimum of 40 hours driving practice 10 of which should be in the dark. The application for the conditional drivers license is to be signed by the parent or guardian and the application is to be accompanied with proof of acceptable school attendance

At 17, after holding the special restricted drivers licence for a year without issues you can obtain the full drivers license.

7 0
3 years ago
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Sin∝ = 1 / 2

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∝ = Sin⁻¹ 0.5

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3 0
3 years ago
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