EE/CE-211L Basic Electronics Lab Lab-06

Lab 06
Common Emitter Transistor Amplifier
Name: Rafay Ahmed, Syed Saad, Shaheer Bin Student ID: rk08919, si09370, st09213
Tariq

6.1 Objective
To investigate the performance of single-stage common emitter BJT amplifier

6.2 Introduction to BJT Amplifiers

One of the most useful applications of a transistor is amplification. With an amplifier, the output
sinusoidal signal can be achieved greater than the applied input sinusoidal signal. When BJT is operated
in active mode, changes in base-emitter voltage gives rise to change in collector current. Voltage
amplification is obtained utilizing this effect.

Figure 1: Two-port network representation of amplifier showing its parameters

Biasing (application of DC voltage) is required to make BJT operate in active mode at desired operating
point before AC signals are applied for amplification. The superposition theorem is applicable for the
analysis, permitting the separation of the analysis of the DC and AC responses of the system. In other
words, one can make a complete DC analysis of a system before considering the AC response. Once
the DC analysis is complete, the AC response can be determined using a complete AC analysis.

6.3. Pre-Lab Task

Task 1:
1. What is the significance of small signal analysis in transistor amplifiers?
2. Why are Coupling and Bypass Capacitors used? Discuss their behavior during AC and DC input
signals.
3. List down the advantages and applications of CE Amplifiers.

ANSWERS FOR PRELAB:

1. Significance of Small Signal Analysis in Transistor Amplifiers

Small signal analysis is crucial in transistor amplifiers because it allows us to study how the amplifier
responds to weak input signals. The key points are:

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 Linear Operation: Transistors are nonlinear devices, but for small variations around a bias
point (Q-point), they can be approximated as linear.
 Gain Calculation: Helps determine voltage gain, current gain, and power gain.
 Impedance Estimation: Helps calculate input and output impedances, which are critical for
impedance matching in circuits.
 Frequency Response Analysis: It allows us to understand the behavior of the amplifier
across different frequencies.
 Noise and Stability Considerations: Small signal analysis helps in designing stable
amplifiers with minimal distortion.

2. Coupling and Bypass Capacitors – Role and Behavior

Coupling Capacitors:

 Purpose: Block DC and allow AC signals to pass.

 Behavior in AC Signals: Acts as a high-pass filter, passing AC signals to the next stage.
 Behavior in DC Signals: Blocks DC to prevent shifting of bias conditions in the next stage.
 Application: Used between amplifier stages to prevent DC level shifts.

Bypass Capacitors:

 Purpose: Provide a low-impedance path for AC signals, improving gain.

 Behavior in AC Signals: Shunts AC signals around emitter resistance, increasing gain by
preventing negative feedback.
 Behavior in DC Signals: Acts as an open circuit; has no effect on DC biasing.
 Application: Used in Common Emitter amplifiers to enhance gain.

3. Advantages and Applications of Common Emitter (CE) Amplifiers

Advantages:

1. High Voltage Gain: Provides significant amplification of the input signal.

2. Moderate Input Impedance & High Output Impedance: Suitable for signal processing
applications.
3. Good Frequency Response: Can operate over a wide range of frequencies.
4. Phase Inversion: Output signal is 180° out of phase with the input.
5. Low Cost & Simplicity: Simple circuit design with minimal components.

Applications:

1. Audio Amplifiers: Used in microphones and speaker systems.

2. RF Amplifiers: Used in radio and TV transmitters/receivers.
3. Sensor Signal Conditioning: Used in preamplifier circuits for sensors.
4. Oscillators & Signal Processing: Forms the basis of many signal generation circuits.
5. Digital Logic Circuits: Used in switching circuits for logical operations.

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6.4. Common Emitter Transistor Amplifier

Common emitter amplifier configuration is widely used. It provides large voltage gain (typically ten to
hundreds) and moderate input and output impedance. CE amplifier circuit is shown in Figure 2.

Figure 2: Common-emitter BJT amplifier circuit

Using a DC source and biasing configuration, the BJT is first biased to establish operating point in
active region. Then, AC signal Vs is applied at input port through a capacitor. The source resistance is
Rs here which will be neglected in our analysis for now. So, applied input to amplifier Vi is source
voltage. The output voltage Vo is measured from collector terminal. The capacitors C1 and C2 are called
coupling capacitors. The emitter resistor is bypassed through C3 bypass capacitor for AC signal.

For a particular DC operating point, when AC input signal vi is superimposed on DC voltage VBE, the
corresponding instantaneous curve of base-emitter voltage is shown with instantaneous base current.
The corresponding instantaneous collector-emitter voltage is represented in ic-vce curve which shows
large variation in collector voltage by small changes in applied input AC voltage. The amplification is
obtained at output node i.e. collector. From this, you can see the importance of locating Q point of BJT.
If VCE is very small, the BJT will enter saturation region as output voltage swings across it and if closer
to Vcc, it will enter cut-off region. The location of bias point should allow swing in both directions
equally, otherwise, output waveform peaks (positive or negative) will be clipped.

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Figure 3: Graphical representation of AC signal components superimposed on DC

DC analysis has been covered in previous labs. For AC analysis, equivalent models are used. Here, for
CE configuration re model will be used as shown in Figure 4.

Figure 4: BJT AC equivalent model for CE configuration

Moreover, the AC equivalent of complete transistor network is obtained by:

 Setting all dc sources to zero and replacing them by a short-circuit equivalent
 Replacing all capacitors by a short-circuit equivalent
 Removing all elements bypassed by the short-circuit equivalents introduced by steps 1 and 2

Using the equivalent model of BJT, the AC equivalent of the network shown in Figure 2 is drawn in
Figure 5.

Figure 5: AC equivalent of CE amplifier network shown in Figure 2

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6.5. Characteristic Parameters of Common Emitter Amplifier

The characteristic parameters of an amplifier include its voltage gain Av, input impedance Zi, output
impedance Zo and current gain Ai.

VT 26mV
Transistor AC dynamic resistance re shown in model can be calculated as re  
I E I E (mA)

Vo
 AC Voltage Gain Av is defined as Av 
Vi
Here, Vo and Vi are output and input voltage which can be measured as rms, peak or peak-to-peak.
From model, the AC voltage gain of a CE amplifier (under no-load) can be calculated using:
 RC
Av 
RE  re
With bypass capacitor, use RE =0 in the above equation which results in
 RC
Av 
re
 AC Input Impedance, Zi, is that of the amplifier as seen by the input signal. For CE amplifier, it can
be given as
Z i  R1 || R2 ||  ( RE  re )  R1 || R2 ||  (re )
 AC Output Impedance, Zo, for CE amplifier can be approximated to
Z o  RC

Task 2: To perform DC analysis for CE BJT Amplifier

1. For the common-emitter amplifier circuit shown in Figure using BJT model 2N3904, perform DC
analysis to complete Table 1. The supply voltage VCC is 20 V. Before, performing DC analysis
identify the BJT bias configuration and verify the conditions that should be satisfied for of the
chosen analysis. Use DC Gain β equal to 300 in your calculations if needed.
The circuit component values are given here:
RC =3kΩ, RE=1kΩ, R1=33kΩ, R2=10kΩ, C1= C2=15μF and C3=100μF.

Table 1: DC analysis of Common-Emitter BJT amplifier

Resistances Measured* DC Parameters Calculated Measured

VB 4.6v 4.53v
RC 2.93k ohm
VE 3.9v 3.81v
IE 3.9mA 3.5mA
RE 991 ohm
VC 8.3V 8.76v
3.8mA
IC 3.9mA
R1 32.7k ohm
IB .015mA 16.4 uA
R2 9.8 k ohm DC Gain β 300 (Given) 232

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2. Now, implement the circuit and verify the DC analysis by measuring the values for all the DC
parameters listed in Table 1.

Task 3: To perform AC analysis for Common Emitter BJT Amplifier

For the same circuit, now perform AC analysis and calculate the parameters listed in Table 2. Use
measured values in your calculations for AC analysis and fill in second column.

Table 2: AC analysis of Common-Emitter BJT amplifier

Characteristic Parameters Calculated Values Measured Results
Transistor AC Dynamic
6.66ohms 7.43
Resistance: re
Input Impedance: Zi 1.76k 1403 ohms
Output Impedance: ZO 3000 2930

AC Voltage Gain: AV -466.5 -394.35

Task 4: To implement Common Emitter Amplifier Circuit

1. In the implemented circuit, now apply AC signal of magnitude 50mV pk sine wave at 1k Hz through
coupling capacitor of 15μF.
2. Use bypass capacitor of 100μF at emitter and measure the output voltage through another coupling
capacitor of 15μF as shown in Figure 2. This is the unloaded output of amplifier as there isn’t any
load resistor connected at the output terminal of amplifier. Write down your observations in Table
3.
3. Observe the output voltage waveform. Analyze the magnitude and phase of this amplifier output
voltage with respect to the applied AC input signal.

Vs Across C1

Vo / C2

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Magnitude Analysis

 The peak-to-peak voltage ) is 16.2V).

 The RMS voltage is 6.80V, which aligns with a sinusoidal signal behavior.
 The frequency of the signal is 1.00001 kHz, indicating that the amplifier is processing an AC
signal with a frequency of approximately 1 kHz.
 The waveform appears distorted, suggesting possible non-linearity or rectification effects.

Phase Analysis:

 A typical common-emitter amplifier introduces a 180° phase shift between input and
output.
 However, the observed waveform does not resemble a standard sinusoidal output. Instead, it
appears to be rectified or clipped, meaning that parts of the waveform may be missing or
altered.

Task 5: To experimentally determine the characteristic parameters of CE

amplifier
The CE amplifier characteristic parameters discussed in Section 6.5 will now be experimentally
determined and compared with the ones calculated in Task 3.

 Amplifier Voltage Gain: Av

1. Using the measured values of input and output voltages of the amplifier, determine the magnitude
of amplifier voltage gain and note it down in Table 2.
2. Keeping the phase difference in mind, assign its sign (+/-).

 Amplifier Input Impedance: Zi

1. To determine the input impedance of this amplifier, connect an input measurement resistor,
Rx=1kΩ, as shown in Figure 6. Apply the same input signal (50mVpk at 1 kHz) but this time
through this known resistance. Measure Vi i.e., the signal after this resistor Rx. Note down in Table
3.

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Figure 6: Circuit for determining input impedance

2. Mathematically, for this circuit this voltage Vi can now be expressed as

Vsig
Vi  Zi
Z i  Rx
Solve the given equation for Zi.
Zi 
3. Determine the value of input impedance. Note down in Table 2.

Table 3: Data table for experimental determination of CE amplifier characteristic parameters

Characteristic Parameter Required Measurements Measured Values

For CE Amplifier Voltage Input Voltage: Vi 56mV

Gain Output Voltage (Unloaded): Vo 7.2V

Measurement Resistor: Rx 0.99k

For CE Amplifier Input
Source Voltage: VSig
Impedance 64 mV
Input Voltage: Vi 35.7mV

Output Voltage Unloaded: Vo 7.2V

For CE Amplifier Output
Load Resistor: RL 2.934k
Impedance
Output Voltage Load: VL 3.16V

 Amplifier Output Impedance: Zo

1. Remove the input measurement resistor Rx.
2. Apply AC input signal (50mV pk at 1 kHz) directly. Measure the unloaded output voltage Vo.
3. Now, connect load resistor of 3kΩ (use measured resistance) and measure the corresponding loaded
output voltage. Note it down in Table 3.
4. When loaded, the output voltage can be expressed as
RL
VL  Vo
Z o  RL

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Solve the above for output impedance Zo.

Zo 

5. Calculate the output impedance to complete Table 2.

Task 6: To investigate the performance of CE amplifier

1. Using the data collection of Table 2, compare the estimated and measured characteristic parameters.
What are your comments about the approximations considered in the analysis?
There are small differences in the estimated and measured values of re, Zi, and Av. And the
values of Zo are very close together.

The small differences arise from factors such as transistor variations and the non-ideal behavior
of capacitor

2. Now, in the actual CE amplifier circuit where there is no output load or input measurement
resistance connected in the circuit having the applied input of 50mVpk sinewave at 1k Hz, obtain
the output voltage waveform and observe this output as you gradually increase the input signal
voltage from 50mV to 200mV. Analyze the changes you observe in amplifier output voltage
waveform.
As the input voltage rises from 50mV to 200mV, the output waveform also increases in
amplitude. Initially, with a small input signal, the output remains largely undistorted, maintaining
the expected amplified sinewave shape. However, as the input voltage continues to rise, the
waveform peaks begin to flatten, indicating saturation and clipping. This occurs because the
transistor in the amplifier reaches its operational limits, preventing linear amplification. At higher
input levels, the output voltage swing becomes restricted by the supply voltage and biasing
conditions, resulting in distortion

3. Try measuring the output before the coupling capacitor connection at output node. Justify the need
of coupling capacitor?
In a CE amplifier, the coupling capacitor serves to block DC components while allowing the
AC signal to pass through to the next stage or measuring device, such as an oscilloscope. If the
output is measured before this capacitor, the signal includes its DC bias. Because oscilloscopes
usually display signals relative to ground, the presence of the collector’s DC bias causes the
entire waveform to appear shifted upward.

4. If you increase the resistor at collector terminal to 5kΩ, what changes do you observe in the output
voltage waveform? Justify your analysis. Hint: Observe the DC operating point Vc too to relate.

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After increasing the collector resistor to 5kΩ, the output signal exhibits greater distortion. Instead
of maintaining a smooth sine wave, the waveform shows signs of clipping and is more square

6.6. Post Lab Task

Task 7: To calculate current of the CE amplifier
1. Find the current gain of the amplifier implemented in lab when it is loaded.
2. From voltage gain and current gain, one can say that the output AC power is greater than the input
AC power. How can you justify this when conservation of energy dictates that over time the total
power output, Po, of a system cannot be greater than its power input, Pi, and that the efficiency
defined by efficiency cannot be greater than 1?

Answer

1)

2)

Although the amplifier increases the output power, energy conservation still holds
since the extra power comes from the DC supply (VCC). The transistor controls the

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power transfer rather than creating energy, ensuring that efficiency never exceeds
100%.

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Assessment Rubric
Lab 06
Common Emitter Transistor Amplifier

Name: Student ID:

Points Distribution

LR1 / LR2 LR 4 LR 6 LR 10
Task No. Simulation / Circuit Data Collection Calculation Analysis
36 16 24 20
Task 1 - - - /4
Task 2 /8 /4 /8 -
Task 3 - - /8 -
Task 4 /8 /4 - -
Task 5 /12 /4 /4 -
Task 6 /8 /4 - /8
Task 7 - - /4 /8
CLO Mapped CLO 1 CLO 1 CLO 1 CLO 1
Total

Affective Domain Rubric Points CLO

Mapped
AR 4 Report Submission /4 CLO 4

CLO - LDL Total Points Points Obtained

CLO 1 – (Psy-3) 96
CLO 4 – (Aff-3) 4
Total 100

For description of different levels of the mapped rubrics, please refer the provided Lab Evaluation
Assessment Rubrics and Affective Domain Assessment Rubrics.

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