EE-221L Basic Electronics Lab Lab-07

Lab 07
BJT as Switch and Emitter-Follower Amplifier
Name: laiba munib , shaheer bin tariq, Student ID: lm09327, st09213
S.M.Asad Ali

7.1 Objective
To investigate the operation of BJT common-collector (emitter-follower) amplifier and design a circuit
for LED control using BJT as a switch.

Note: Pre-lab (printed or hand-written) should be done on a separate paper (use A4 sheet) and will
be submitted at the start of Lab within first 10 minutes.

7.2 Pre-Lab Task

Task 1: To derive expressions for designing BJT switch controlling LED.

i. For the circuit given above derive the expressions for the base resistor 𝑅𝐵 and the current
limiting resistor 𝑅 using KCL.

Rb =( Vcc – Vbe)/Ib

R = (Vcc – Vled – Vce(sat))/Ic

ii. Given that the forward voltage 𝑉𝐿𝐸𝐷 and current through LED is 2V and 20mA respectively,
what will be the value of collector current 𝐼𝐶 .
Since current is 20mA flowing into the collector, Ic = 20mA.

7.3 Introduction to Emitter-Follower Common Collector Amplifier

When the output of BJT amplifier is taken from the emitter terminal of the transistor as shown in Figure
1, the network is referred to as an emitter-follower. This configuration is actually a common-collector
configuration because the collector is directly connected to supply and this will be grounded for AC
analysis. The emitter-follower configuration is frequently used for impedance-matching purposes.

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Figure 1: Emitter-follower amplifier circuit

For the circuit shown in Figure 1, if BJT is replaced by its re model, the AC equivalent of the network
can be represented by Figure 2. Here, R’ is the equivalent impedance R1 || R2.

Figure 2: AC equivalent of the emitter-follower network

7.4 Characteristic Parameters of Emitter-Follower CC Amplifier

Using the network of Figure 2, the characteristic parameters of emitter-follower circuit are given as:
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 CC amplifier (under no-load) can be calculated using:
RE
Av =
RE + re
Where, transistor AC dynamic resistance re shown in model can be calculated as
VT 26mV
re = =
I E I E (mA)

• AC Input Impedance, Zi, is that of the amplifier as seen by the input signal. For CC amplifier, it
can be given as
Z i = R1 || R2 ||  ( RE + re )

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• AC Output Impedance, Zo, for CE amplifier can be approximated to

Z o = re

Task 2: To determine DC operating point of transistor for emitter-follower

amplifier
1. Implement the circuit shown in Figure 3 using 2N3904 BJT model and supply of 10V.
2. Note down the measured values of resistances used in your implemented circuit.
3. Measure the DC operating values of the circuit as given in Table 1.
Table 1: DC operating point of BJT in CC amplifier circuit
Measured Measured Measured
Parameters Parameter Parameter
Value Value Value

VCC 10V Base Voltage: VB 2.27V Base Current: IB 7.64 uA

Emitter Voltage:
R1 33k ohm 1.7v Collector Current: IC 1.03 mA
VE
Collector Voltage:
R2 10 k ohm 10V Emitter Current: IE 1.47 mA
VC
Base-Emitter
RE 1k ohm 0.64V DC Current Gain: β 134.8
Voltage: VBE

Figure 3: Emitter-Follower CC amplifier circuit for implementation

Task 3: To calculate the characteristic parameters of emitter-follower

amplifier
1. Now, using the measured values of Table 1, calculate the characteristic parameters of amplifier
listed in Table 2.
Table 2: Characteristic parameters of CC amplifier circuit
Parameters Calculated Values
Transistor AC dynamic resistance re 17.6 ohms
AC Voltage Gain Av 0.94
Input Impedance Zi 7.26 k ohm
Output Impedance Zo 17.6

Task 4: To experimentally determine characteristic parameters of emitter-

follower CC amplifier
1. Complete the amplifier implementation, and apply AC input signal Vsig of 1Vrms (1.414V peak) at
frequency of 1k Hz at no load condition through coupling capacitor C1. Measure the output voltage

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through C2 at no-load. Make sure there is no distortion or clipping of waveform at the output
terminal emitter.

• Amplifier Voltage Gain: Av

1. Obtain the waveforms of input and output voltage. Note down their measured values in Table 3.
Determine the gain of amplifier.
2. Observe the phase difference between the two and assign corresponding sign (+/-) to the gain.
3. What are your comments about the gain of amplifier? Keeping the formula of gain in terms of
circuit parameters, do you think it is easily adjustable? Justify the name of this CC amplifier as
emitter-follower.

1. Yellow- Input
Purple - output

2. The yellow waveform (input) and purple waveform (output) seem to be in phase.
Since the phase difference is approx 0°(given that Av ~ 1), the amplifier does not invert the
signal.
This means the gain has a positive sign (+).

3. Since the voltage gain is slightly less than 1, this amplifier primarily functions as a buffer rather
than providing amplification. This is a key characteristic of a common-collector (CC)
amplifier, also known as an emitter-follower.
The gain depends on circuit parameters such as the transistor’s properties and resistor values.
While it is not easily adjustable like in a voltage amplifier, it can be modified by changing the
emitter resistance.
The output voltage follows the input voltage with nearly the same amplitude (slightly less due
to the voltage drop across the transistor). The Emitter-Follower (Common Collector) amplifier
is called so because the output voltage at the emitter "follows" the input voltage at the base

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with nearly the same amplitude but slightly reduced due to the base-emitter voltage drop
(VBE≈0.6V−0.7V). Its Av can be changed with the changing of emitter resistor.

• Amplifier Input Impedance: Zi

1. To determine the input impedance of this amplifier, connect an input measurement resistor,
Rx=10kΩ, as shown in Figure 4. Apply the AC input signal of 2Vrms at frequency of 1 kHz but
this time through this known resistance. Measure Vi which is the signal after this resistor Rx input
to amplifier. Note down in Table 3.

Figure 4: Circuit for determining input impedance

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

Vsig
Vi = Zi
Z i + Rx
Solve the given equation for Zi.
Z i = ViRx / (Vsig - Vi)
2. Determine the value of input impedance. Note down in Table 3.

Table 3: Experimentally determined values of CC amplifier characteristic parameters

Experimentally Required Measured
Characteristic Parameter
Determined Values Measurements Values
Input Voltage: Vi 2.92 V
CC Amplifier Voltage Gain:
0.95 Output Voltage
Av 2.76 V
(Unloaded): Vo
Measurement Resistor:
10 k ohm
Rx
CC Amplifier Input
7520 Source Voltage: V Sig 5.67V pk-pk
Impedance: Zi
Input Voltage: Vi 2.34 V pk-pk
Output Voltage 300mV(peak to
Unloaded: Vo peak)
CC Amplifier Output
22.3 Load Resistor: RL 98ohms
Impedance: Zo
Output Voltage Load:
240mV
VL

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• Amplifier Output Impedance: Zo

1. Remove the input measurement resistor Rx.
2. Apply AC input signal (100 mV rms at 1 kHz) directly. Measure the unloaded output voltage Vo.
3. Now, connect load resistor of 100 Ω and measure the corresponding loaded output voltage VL. Note
it down in Table 3.
4. When loaded, the output voltage can be expressed as
RL
VL = Vo
Z o + RL

Solve the above for output impedance Zo.

Zo = (VoRL - VLRL) / VL

5. Calculate the output impedance to complete Table 3.

7.5 Transistor as a Switch

Transistor can be used as a switch to control the ‘on’ and ‘off’ states of a load. The behavior of switch
is represented in Figure 5. When a switch is off, it can be replaced by open circuit and when it is fully
on, it can be replaced by short-circuit.

Figure 5: Characteristics of ideal switch

For implementation of switch by transistor, the transistor operates in saturation region and cut-off
region. In cut-off region, both the junctions remain reversed-biased so it acts as “fully-OFF” and in
saturation region, both the junction remain forward-biased so it acts as “fully-ON”. In common-emitter
configuration, the voltage across collector-emitter terminal is equal to the supply voltage Vcc in cut-off
region and in saturation region, collector-emitter voltage remains approximately equal to zero (0.3 V).
Therefore, the base terminal can be used to control the operation of transistor switch that can control
the state of load connected at collector terminal. The collector and emitter form the switch terminals
and the base is the switch handle. In other words, the small base current can be made to control a much
larger current between the collector and emitter. For example, the circuit of Figure 6 can be modified
to control an electric motor. The BJT switch only needs to handle the small base current due to the
current amplification provided by the transistor. If high current motors are directly switched on and off,
mechanical switch contacts will eventually wear out from switching the high current, causing the switch
to fail. However, the BJT can operate nearly indefinitely as the switch as it has no mechanism that
causes it wear out. The loads that require a few milliamps can be directly controlled by connecting them
at collector terminal. For devices that require more power like large motors, the transistor switch can
be used to control relay (as relay driver) that controls motor.

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Figure 6: Transistor as switch

When EBJ is forward biased, let VBE be 0.7 V. We know that when collector voltage is lower than the
base voltage by 0.4V, the transistor enters saturation region. For that edge of saturation (EOS) point,
we can take VC equal to 0.3V. For a transistor that operates deeper into saturation region, we take V CE_sat
equal to 0.2V.
VCC − VCE _ Sat
Ic Sat =
RC
Using minimum value of current gain βmin, corresponding edge-of-saturation base current IB (EOS)
required to saturate the transistor is given by
I C _ Sat
I B _ Sat =
 min
In saturation region, the transistor can be forced to operate at any desired β. The ratio of collector (in
saturation) to base current is therefore, called βforced. Ratio of IB to IB_Sat is called overdrive factor. This
can be used to calculate required base current and resistance RB.
VBB − VBE = I B RB

Task 5: To design circuit for implementing BJT switch to control an LED

To design a circuit that utilizes transistor as a switch for controlling on and off status of Red LED,
consider the circuit shown in Figure 7. Here, in series with LED, there is a current limiting resistor 𝑅.
The LED forward voltage is 2V and has forward current rating of 20mA. The LED with current limiting
resistor will act as load across which 5V should appear when the switch is on and 0V when switch is
off.

Figure 7: Implementation of transistor switch to control LED

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Find the value of current limiting resistor 𝑅 and base resistor 𝑅𝐵 with βmin of 100 and an overdrive
factor of at least 10. Show your calculations. Hint: Use equations derived in Task 1.

Working:
R = (5-2-0.2)V/(2mA)
R = 140ohm

Ib = 20mA/100 =0.2mA

Ibsat = Ib * 10 = 2mA
Rb = (5-0.7)/2mA = 2.15k ohm

Task 6: To implement and investigate operation of BJT switch for LED

control
1. Construct the circuit designed in Task 5 using 2N3904 BJT.
2. Identify the modes of operation of transistor when voltage is applied at base terminal through
𝑹𝑩 and when the applied voltage at this terminal is 0V. Don’t change the voltage 𝑉𝐶𝐶 applied at
load. Observe the status of LED Measure the voltage across the load (LED + Resistor) and transistor
switch in each case along with base current and collector current.
Mode of Operation ( )@ Vcc=5V ( )@ Vcc=0V
𝑽𝑳𝑶𝑨𝑫 = 𝑽(𝑳𝑬𝑫+𝑹) 5.0V 0V
𝑰𝑪 15.86 mA 0A
𝑰𝑩 1.82 mA 0A

3. Now, instead of controlling transistor switch through constant supply at base terminal, control it
through a pulse signal with 50 % duty cycle having amplitude of 5V. Use the function generator to
generate 1Hz square wave of 0-5V (5V peak to peak and offset of 2.5V). Why is offset added?
As the signal is pure AC, it would also give negative values for 5V pk-pk, which could turn off
the transistor completely. On adding 2.5V offset, it is ensured that the values oscillates between
0 and 5V, ensuring the proper working of the LED i.e. turning ON at 5V pk-pk and OFF at 0V
cutoff. 50% duty cycles makes the LED turn on for half of the cycle.

4. Initially set frequency of 1Hz and then increase it. Does the transistor switch keep changing its
status at higher frequencies? You can verify this by observing glow of LED and obtaining voltage
waveform across transistor switch / load.

At low frequency, LED works as expected, as the frequency keeps on increasing, we notice that

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LED keeps on blinking faster and faster, to the point (at around 900 Hz), it keeps on blinking so
fast as if it is turned on in a continuous state, though it keeps getting dimmer as high frequency is
approach (above 1k Hz). It is initially because increasing the frequency reduced the time period
between the pk-pk values, making the blinking faster, but at certain high frequency, it will
eventually appear as a continuous dim light due to persistence of vision. At high frquency , it gets
so dim, as if it is turned off, due to switching limitations.

7.6 Post Lab Task

Task 7: To analyze the performance of BJT as amplifier and switch
1. Discuss the application of emitter-follower circuit as voltage buffer.

The emitter-follower circuit, which is a common-collector (CC) configuration, is widely used as a

voltage buffer because It has a voltage gain close to 1, isolates the input voltage from the output
voltage, providing a high input impedance and low output impedance.

Applications:

• Impedance Matching: matches impedence with b/w two circuits ensuring that the power
transfer is maximum.
• Signal Isolation: Prevents one circuit from interfering with another.
• Signal Conditioning: Ensures voltage transfer without attenuation or distortion.

2. Compare characteristic parameters of CE amplifier and CC amplifier.

common-emitter (CE) amplifier, provides high voltage gain ( 50 to 500) with moderate input
impedance and high output impedance, the common-collector (CC) amplifier offers almost no voltage
gain, but provides excellent impedance matching with high input impedance and very low output
impedance. While CE amplifiers are best for applications requiring significant voltage amplification,
CC amplifiers are best as buffers, ensuring that signals are transferred between stages with minimal
loss and distortion. the CE amplifier introduces a 180-degree phase shift between input and output,
while the CC amplifier maintains 0-degree phase shift, meaning the output follows the input directly.

Assessment Rubric
Lab 07
BJT as Switch and Emitter-Follower Amplifier

Name: Student ID:

Points Distribution

LR1 LR 4 LR 6 LR 10 LR 11
Task No. Circuit Layout Data Collection Calculation Analysis Design

Task 1 - - - - /8
Task 2 /4 /10 - - -

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Task 3 - - /8 - -
Task 4 /4 /12 /4 /8 -
Task 5 - - - - /8
Task 6 /4 /8 - /8 -
Task 7 - - - /10 -
/12 /30 /12 /26 /16
CLO
Mapped
CLO 1 CLO 1 CLO 1 CLO 1 CLO 2

Affective Domain Rubric Points CLO

Mapped
AR 4 Report Submission /4 CLO 4

CLO Total Points Points Obtained

1 80
2 16
4 04
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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