Frequency Response of CE Amplifier in Mechatronics Engineering, Study Guides, Projects, Research of Electronics

Information on the frequency response of a Common Emitter Amplifier (CEA) in the context of an Electronics 2 lab experiment at Sana'a University. the concept of frequency response, its importance, and how capacitors affect it. It also presents methods to establish the low and high frequency responses and plots the global frequency response. the objectives, circuit components, and an overview of the topic.

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2021/2022

Uploaded on 10/15/2022

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Experiment NO.1 FREQUENCY RESPONSE OF CE AMPLIFIER
Program: Mechatronics Engineering
Subject: Electronics 2 Date:
Subervisor: Eng.Mohammed AL-Yemeni
Experiment 1: FREQUENCY RESPONSE OF CEAMPLIFIER
Name: ID:
Sana'a University
Faculty of Engineering
Mechatronics Engineering
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Program: Mechatronics Engineering

Subject: Electronics 2 Date:

Subervisor: Eng.Mohammed AL-Yemeni

Experiment 1: FREQUENCY RESPONSE OF CEAMPLIFIER

Name: ID: Sana'a University Faculty of Engineering Mechatronics Engineering

OBJECTIVES

 In this LAB, we will focus on this important feature of amplifiers. First of all, the notion of frequency response is detailed along with some basic related concepts and we will present how to quantify it.  We will understand which component affects the frequency response and how.  In the rest of LAB a method to establish the low and high frequency responses is presented.  These results will finally be synthesized in the conclusion to plot the global frequency response of a Common Emitter Amplifier.

Circuit components/Equipment:

 BJT NPN Transistor.  Regulated power Supply (0-30V).  Function Generator.  Resistors Capacitors.  Bread Board  Connecting Wires  DMM

Overview:

As such for any electronic circuit, the behavior of amplifiers is affected by the frequency of the signal on their input terminal. This characteristic is known as the frequency response. Frequency response is one of the most important property of amplifiers. In the frequency range that amplifiers have been designed for, they must deliver a constant and acceptable level of gain. The frequency response depends directly on the components and the architecture chosen for the design of the

The light blue curve is called the asymptotic representation while the dark blue curve is the real frequency response of the circuit. In Figure 1 , two different cutoff frequencies can be distinguished : flc for “low cutoff” and fhc for “high cutoff”. The quantity fhc-flc is called the bandwidth and represents the frequency range where the gain is above the - 3 dB plateau. EFFECT OF THE CAPACITORS Let’s consider a Common Emitter Amplifier (CEA) which configuration is shown in Figure 2 :

The structure around the BJT transistor consists of a voltage divider network (R1 and R2), a load (RL), coupling capacitors (C1 and C3) and a bypass capacitor C2. What is important to keep in mind is that capacitors have a property called reactance that is an equivalent of the resistance. The reactance (XC) of capacitors depends on the frequency and the value of the capacitor, it satisfies the following formula : Independently of the value of the capacitor, when the frequency is low, XC tends to be high. Near DC signals, capacitors behave therefore as open circuits.

as an open circuit and the transistor is not affected at all. However, if the frequency increases, more signal passes through them instead of going in the base branch of the transistor, therefore decreasing the voltage gain. A very important formula is given in Equation 3 and links the cutoff frequency of a RC filter : LOW FREQUENCY RESPONSE

Input Coupling Capacitor The CE amplifier has three capacitors, to understand how each capacitor affects the frequency response, we will investigate each one individually. The critical frequency can be calculated using the following equation: Thevenin’s Theorem can be used to derive the resistance as seen by the capacitor Cin. Rin is a resistance from R1, R2, and the transistor AC resistance in parallel with each other. You can calculate Rin with the following formula: The total resistance as seen from the source voltage is:  Output Coupling Capacitor The coupling capacitor Cout is affected by the load resistor Rload and Rc, therefore:  Emitter Bypass Capacitor The equivalent resistance seen by the emitter capacitor is given by: The resistor Re is not included in the equation because it is very large when compared to rth in the above equation, recall when two resistors are in parallel, the total resistance is closer to the smaller value.

We can note that the coupling capacitors are not represented since they behave as short circuits at high frequencies. Moreover, the emitter branch is shorten to the ground for the same reason applying to the bypass capacitor. The internal capacitor CBC is converted via Miller’s theorem into the equivalent Cin and Cout capacitors. Moreover, this theorem states that

  • The total input capacitance of this circuit is
  • the total input resistance is
  • The numerical application to our lab gives
  • The high cutoff frequency of the input is therefore From the output point of view, the high cutoff frequency is simply given by the filter (RC//RL)Cout with

Procedure

  1. Connect the circuit as shown in Figure above.
  2. Apply the bias voltage Vcc and check the dc bias voltages at test points.
  3. Apply an input sine wave signal of 10mV, 1 0 kHz from the function generator.
  4. Observe the output in DSO. Calculate the corresponding gain and compare with the designed values.
  5. Vary the frequency of the input signal and tabulate the output signal gain for different frequencies.
  6. Plot the Frequency vs Gain (dB) using semilog sheet and calculate the bandwidth of the given amplifier from the plot.