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Course: Electronic Circuit Devices Lab No: 05 Title: Single Stage Common Emitter Amplifier Design CID: - Date

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Department of Electronics & Telecommunication Engineering

Course: Electronic Circuit Devices

Lab No: 05

Title: Single Stage Common Emitter Amplifier Design

CID: ________________

Date: _______________

S.
Name SID Initials Marks
No.
1

Faculty/Lab Engineer

Comments
LABORATORY EXERCISE - 4

SINGLE STAGE COMMON EMITTER AMPLIFIER DESIGN

Introduction
1. This laboratory execise is meant to enble the students to make a preliminary design of voltage divider
bias BJT transistor configuration and to observe the ac operation of common-emitter amplifier.

Objective
2. This laboratory will enable the students to achieve the following:
2.1. To design a Voltage-Divider bias.
2.2. To demonstrate the ac operation of the common-emitter (CE) amplifier.
2.3. To demonstrate the effects that a bypass capacitor has on amplifier voltage gain (Av).

Conduct of Lab
3. The students are required to work in groups of three to four; each student must attempt to understand the
the use of a simple set up for measuring the parameters of the given transistor.
4. In case some aspect of the lab experiment is not understood the students are advised to seek help from
the
teacher, the lab technicians or the assigned Teaching Assistant (TA).

Equipment required
5. The following equipment would be used in this experiment:

5.1. Instruments
5.1.1. DMM
5.2. Components
• Resistors
o As required
• Transistor
o 2N3904
• Supplies
o DC power supply

SMALL SIGNAL Common Emitter (CE) AMPLIFIER

After you have completed part-1; do not remove your circuit instead reconnect the circuit with coupling and
bypass capacitors as shown in fig 5.2. Now you have connected a classical CE amplifier.

The input signal to a CE amplifier is applied across the base-emitter junction of the device. The Output from
this circuit is taken from the collector terminal of the transistor. The CE amplifier is one of the most commonly
used BJT amplifier configurations. The amplifier has a relatively high voltage gain, a relatively high current
gain, and a voltage phase shift of 1800 between its input (base) and output (collector) terminals. In this part, you
will observe the ac operation of the CE amplifier. You will also observe the effects of the emitter bypass
capacitor, a component used to increase the voltage gain of the CE amplifier.
Figure 4.1

PROCEDURE

A. Construct the circuit shown in fig. 4.1.

B. Apply power to the circuit and adjust R1a to zero.

C First do the DC analysis of the circuit with R1a set to zero and note down calculated and measured

values (you will find measure values by operation point analysis).

IC = calculated_______________Measured__________,

Ib = calculated_______________Measured___________

VCEQ = calculated_______________Measured___________

D Draw the equivalent small signal circuit using the π Model. Derive the relationship for the voltage gain;
assume that ro is infinite and would not affect the circuit. Also write the expression for the voltage gain in
the space provided below:
AV = _____________________________________________
Small Signal Equivalent Circuit and derivation of AV (Use Space below)
E. Apply a 1-kHz, 10-m V pp ac signal to the input of the amplifier.

Note: You should set the signal generator so that you have 10 m VPP at the base terminal of the
transistor (measured with respect to ground). Do transient analysis with end time upto 10ms step size of
0.001ms and start time of 0ms.

F. Measure and record the peak-to-peak output voltage at the collector terminal of the transistor.

VOUT = _________________________V pp.

G Using the values measured in steps d and e above; calculate the voltage gain of the amplifier.

Av = _________________________

I It was stated in the introduction, that the CE amplifier produces an 1800 Voltage phase Shift from
input (base) to output (collector). So give proof of that concept and draw input and output waveforms as
displayed infornt of you with exact values of amplitudes and frequency.
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k. Remove the emitter bypass capacitor and for the same Vin that is 10mVPP note down the output voltage

Vout = ______________ Vpp

Av = _______________

l. What happened to Av when the bypass capacitor was removed? Why? Derive the expression for A V
with the emitter bypass removed and check your practical answer with the calculation that you get for the
gain with emitter bypass capacitor removed.

[AV] with RE = ________________________________

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