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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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What historical event allowed both aerospace fields to make enormous strides<br> forward? *
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4 years ago
1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0
3 years ago
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