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Nesterboy [21]
4 years ago
12

1. Think about what you’ve learned about BJTs, both from lecture and from lab. What aspects of the transfer functions of the CE

amplifier and the emitter follower can you sketch on your own (i.e. without having the PSpice simulation or Labview measurement)? What value does PSpice add to your ability to analyze these circuits?

Engineering
1 answer:
sladkih [1.3K]4 years ago
8 0

Answer:

Check all the explanations below. Also check all the attached files for additional information

Explanation:

A Bipolar Junction Transistor is a semiconductor device consisting of two P-N Junctions connecting three terminals called the Base, Emitter and Collector terminals. The current and the gain of theses transistors is determined by the arrangement of these terminals. The BJT uses both electrons and holes as the charge carriers. The BJT can be of two types, the PNP and the NPN transistors.

The common emitter amplifier is a three basic single-stage bipolar junction transistor and is used as a voltage amplifier. In this configuration, the emitter terminal is common to both the collector and the base, the input to the amplifier is taken from the base while the output is taken from the collector.

The common emitter amplifier circuit is attached as a file to this solution.

The current gain of a common emitter amplifier is \beta = \frac{\triangle I_{c} }{\triangle I_{B} }

where I_{c}  and I_{B} are the collector and emitter currents respectively

The voltage gain of the amplifier is given by A_{v} = \beta \frac{R_{c} }{R_{B} }

R_{c} and R_{B} are the collector and base resistance respectively

In an emitter follower configuration, emitter follows the voltage on the base. It is also called the common collector configuration. The collector is common to the emitter and base terminals.

The common collector configuration is shown in the second file attached

The output voltage on a common-collector amplifier will be in phase with the input voltage, making the common-collector a non-inverting amplifier circuit.

The current gain = \beta + 1

Voltage gain = 1 ( Approximately)

Pspice helps in the simulation and analysis of electronic and electrical circuits without going through the rigors of using a pen and paper. It also allows us to test with real data and make proper choice of values.

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A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4
Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0
3 years ago
Urgent!!! <br>List the assumptions that can be taken into account in torsion analysis.​
krok68 [10]

Answer:

Explanation:

In the development of a torsion formula for a circular shaft, the following assumptions are made: Material of the shaft is homogeneous throughout the length of the shaft. Shaft is straight and of uniform circular cross section over its length. Torsion is constant along the length of the shaft.

5 0
3 years ago
At a construction site, there are constant arguments and conflicts amongst workers of different contractors and sub-contractors.
jeyben [28]

Interpersonal skill is required by the construction manager to resolve the arguments and conflicts among workers and between the contractors.

Answer: Option(d)

<u>Explanation:</u>

  • Interpersonal skills are a set of behavioral action that involves the person to interact freely with other peoples effectively.
  • A person before interacting with others should pay attention, listen to their words and give responses to them after they finish with their arguments.
  • In the above answer, the construction manager has to develop interpersonal skills to resolve the issues, conflicts, and arguments raised by his employees.
  • The manager must listen to their needs and demands possessed and should take necessary action in resolving the issues.
  • For that, the manager has to interact freely and effectively with their employees in resolving the issues.
6 0
4 years ago
You have a piece of film paper that is 3 in x 5 in. You fix it inside the back of a pinhole camera with a focal length of 5.5 in
shepuryov [24]

Answer:

In order to take a portrait, the distance of the mascot from the camera should be approximately 7.33 feet

Explanation:

The size of the film paper = 3 in. × 5 in.

The focal length of the camera = 5.5 in.

The height and width of the guinea pig = 4 ft.

The height of the aperture above the ground = 2 ft.

Therefore, we have;

Magnification = Height of image/(Height of object)

Withe the 3 in. wide film, we have;

Magnification = 3 in./(4 ft.) = 3 in./(48 in.) = 0.0625

Magnification = Length of camera/(Distance of object from pin hole)  

∴ Length of camera/(Distance of object from pin hole) = 0.0625

Length of camera = Focal length of the camera = 5.5 in.

Therefore;

5.5 in./(Distance of object from pin hole) = 0.0625

Distance of object from pin hole = 5.5/0.0625 = 88 inches = 7.33 ft

Therefore, the camera should be approximately 7.33 ft. from the mascot to take a portrait.

3 0
4 years ago
A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i
ruslelena [56]

Answer:

point B where Q_B = 101.25 \ in^3  has the largest Q value at section a–a

Explanation:

The missing diagram that is suppose to be attached to this question can be found in the attached file below.

So from the given information ;we are to determine the  point that  has the largest Q value at section a–a

In order to do that; we will work hand in hand with the image attached below.

From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.

We also have block partitioned into different point segments . i,e A,B,C, D

For point A ;

Let Q be the moment of the Area A;

SO ; Q_A = Area \times y_1

where ;

y_1 = (6 - \dfrac{1.5}{2})

y_1 = (6- 0.75)

y_1 = 5.25 \  in

Q_A =(L \times B)  \times y_1

Q_A =(6 \times 1.5)  \times 5.25

Q_A =47.25 \ in^3

For point B ;

Let Q be the moment of the Area B;

SO ; Q_B = Area \times y_2

where ;

y_2 = (6 - \dfrac{1.5 \times 3}{2})

y_2= (6 - \dfrac{4.5}{2}})

y_2 = (6 -2.25})

y_2 = 3.75 \ in

Q_B =(L \times B)  \times y_1

Q_B=(6 \times 4.5)  \times 3.75

Q_B = 101.25 \ in^3

For point C ;

Let Q be the moment of the Area C;

SO ; Q_C = Area \times y_3

where ;

y_3 = (6 - \dfrac{1.5 \times 2}{2})

y_3 = (6 - 1.5})

y_3= 4.5 \  in

Q_C =(L \times B)  \times y_1

Q_C =(6 \times 3)  \times 4.5

Q_C=81 \ in^3

For point D ;

Let Q be the moment of the Area D;

SO ; Q_D = Area \times y_4

since there is no area about point D

Area = 0

Q_D =0 \times y_4

Q_D = 0

Thus; from the foregoing ; point B where Q_B = 101.25 \ in^3  has the largest Q value at section a–a

3 0
4 years ago
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