Solving Common Base Amplifier Problems: Homework Statement & Solutions

In summary, the conversation discusses the problem statement and solution for a circuit, specifically looking at the calculation of R_out, R_in, and A_v. The speaker also brings up confusion over the use of approximate and exact equations, the omission of certain circuit elements, and multiple equations for each characteristic. They express frustration with finding errors and the time it takes to complete the homework.
  • #1
hisotaso
27
0

Homework Statement


The problem statement, given solutions, and my small signal drawing are attached


Homework Equations


Rout, Rin, Av


The Attempt at a Solution


To begin with, our text seems to alternate between using approximate equations and exact derived equations without stating why at any given point. Also, circuit elements are frequently ignored as you will see below.

Looking at my small signal drawing, I don't understand why r is not being considered in any of the calculations. By my calculations, r and R4 are the same size: r = β0/gm = 100 kΩ and R4 is given as 100 kΩ. I see that RI is only 50 Ω. Should r not be considered in the parallel combination in the equation for Rout?

Also, Rc is not considered at all. I know it is not given, does that mean I am supposed to assume there isn't a bias resistor on the collector? Or would I normally need to try and solve for Rc somehow?

The problem is, I have found several errors in the text and in the given solutions so far, and the text gives up to three different equations for each characteristic at times (the derivation and up to two different approximations). This is making me doubt everything I do, and making my homework take forever.

Regards

 

Attachments

  • 14.37 Problem statement.jpg
    14.37 Problem statement.jpg
    30.1 KB · Views: 368
  • 14.37 Small Signal.jpg
    14.37 Small Signal.jpg
    31.9 KB · Views: 373
  • 14.37 Solution.png
    14.37 Solution.png
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  • #2
In 'real life' you would say:

A_v = R_L/(R_l + 25mV/25uA) = 200e3/1050 = 190, R_in = 50 + 25mV/25uA = 1050 ohms, R_out = infinity (this is not including R_L in the computation), A_i = +1.

But that obviously would not satisfy the problem issuer!

So comfort yourself in the thought that on a job you would probably never use all the variables & constants given in your answer sheet ...
 
  • #3
Deleted
 

What is a Common Base Amplifier?

A Common Base Amplifier is a type of transistor amplifier circuit where the input signal is applied to the base terminal of the transistor, and the output is taken from the emitter terminal. It is commonly used for high-frequency applications due to its high input and output impedances.

What are the advantages of a Common Base Amplifier?

Some advantages of a Common Base Amplifier include high voltage gain, low input and output impedances, and a wide frequency response. It also has good stability and low distortion, making it useful for amplifying weak signals.

What are the disadvantages of a Common Base Amplifier?

One disadvantage of a Common Base Amplifier is that it has a low input impedance, which can cause loading effects on the input signal. It also has a low power gain, making it less suitable for applications that require high power amplification.

How does a Common Base Amplifier work?

A Common Base Amplifier works by using a transistor to amplify the input signal. The input voltage causes a small current to flow through the base-emitter junction, which then controls the larger current flowing through the collector-emitter junction. The amplified output signal is taken from the emitter terminal.

What are some common applications of a Common Base Amplifier?

Common Base Amplifiers are commonly used in radio frequency (RF) circuits, high-frequency amplifiers, and impedance matching circuits. They are also used in some types of audio amplifiers and in certain types of sensors and detectors.

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