Lab 3: Amplifier Circuits
►3.1◄
LAB – 3
AMPLIFIER CIRCUITS
Experiment 3.1:
Transistor amplifier
(Small-signal Low-frequency voltage amplifier)
Experiment 3.2:
op-amp feedback amplifier
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.2◄
EXPERIMENT – 3.1
COMMON – EMITTER AMPLIFIER
3.1.1 OBJECTIVE
a. To sketch a common – emitter amplifier circuit and explain the operation of the circuit.
b. To analyze the amplifier circuit to determine input resistance output resistance, voltage gain,
current gain and power gain.
c. To design a CE amplifier Circuit for the given specifications.
d. To observe with an oscilloscope, the transient signal voltages of the input and output of the
amplifier.
e. To measure the voltage gain of the amplifier over and range of frequencies and plot the
frequency response curve.
f. To determine the values of lower and upper 3-dB frequencies and 3-dB bandwidth.
g. To trouble shoot a non-operational CE amplifier.
(i) To make a dynamic test which will determine whether the ac amplifier is operating
properly?
(ii) To consider dc voltage and resistance norms at test points in the amplifier which is
operating properly, and to draw inferences as to the nature of the trouble from the
voltage and resistance measurements in a defective amplifier.
3.1.2 HARDWARE REQUIRED
a. Power supply
:
Variable regulated low voltage dc source
b. Equipments
:
AFO, CRO, DMM
c. Resistors
:
d. Capacitors
:
e. Semiconductors
:
BC 107 (or equivalent)
f. Miscellaneous
:
Breadboard and wires.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.3◄
3.1.3 PRE LAB QUESTIONS
1. Calculate the base bias voltage for the circuit shown below when no signal source is present,
and when signal source is directly connected.
Vcc=12V
R1
33k
Rc
2.2k
C2
C1
Q1
rs
600
Vs
RL
33k
R2
15k
RE
2.7k
CE
Fig. (a)
2. Calculate the transistor collector voltage for the circuit of problem – 1 with C2 present, and with
RL directly connected.
3. The CE circuit shown in problem – 1 has the following transistor parameters: hie=1KΩ, hfe=85,
hoe=2µS. Calculate Zi, ZO & AV.
4. For the circuit in problem–1, recalculate Zi &AV when the bypass capacitor is removed from RE.
5. For a CE circuit with voltage–divider bias, a bypass emitter resistor, a capacitor–coupled signal
source, and a capacitor-coupled load,
(i)
Sketch the dc equivalent circuit and write expressions for the dc voltages and currents.
(ii)
Sketch the ac equivalent circuit and approximate h-parameter equivalent circuit and
write expressions for Zi, Zo, Av, Ai and AP.
(iii)
Draw ac and dc load lines on the output characteristics of CE configuration and explain
their significances.
6. The CE circuit has the following component values and parameter values: hie = 1.4kΩ, hfe = 55,
RL = 47K, VCC = 12V, rs= 600Ω and ZO = 3.9KΩ. Sketch the circuit and determine suitable
resistor and capacitor values. Also, calculate Zi, Zo, AV, Ai and AP for the circuit designed.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.4◄
3.1.4 THEORY
Amplifier is an electronic circuit that is used to raise the strength of a weak signal.
The process of raising the strength of a weak signal is known as amplification. One importance
requirement during amplification is that only the magnitude of the signal should increase and there
should be no change in signal shape. The transistor is used for amplification. When a transistor is
used as an amplifier, the first step is to choose a proper configuration in which device is to be used.
Then the transistor is biased to get the desired Q-point. The signal is applied to the amplifier input
and gain is achieved.
3.1.4.1 CE amplifier operation
Consider a CE amplifier circuit as shown in fig. 3-1-1
Vcc
R1
Rc
C2
C1
Q1
RL
Vs
R2
RE
CE
Fig. 3-1-1(a) CE amplifier circuit
(b) Waveforms for CE amplifier
When the capacitors are regarded as ac short circuits, it is seen that the circuit input terminals are
the transistor base and emitter, and the output terminals are the collector and the emitter. So, the
emitter terminal is common to both input and output, and the circuit configuration is termed
Common –Emitter (CE).
From the voltage waveforms for the CE circuit shown in Fig. 3-1-1 (a) it is seen that
there is a 180o phase shift between the input and output waveforms. This can be understood by
considering the effect of a positive going input signal. When VS increases in a positive direction, it
increases the transistor VBE. The increase in VBE raises the level of IC, thereby increasing the drop
across Rc, and thus reducing the level of the VC. The changing level of VC is capacitor-coupled to
the circuit output to produce the ac output voltage, VO. As VS increases in a positive direction, VO
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.5◄
goes in a negative direction. Similarly, When VS changes in a negative direction, the resultant
decrease in VBE reduces the IC level, thereby reducing VRC, and producing a positive going output.
The circuit in Fig. 3-1-1(a) has input impedance (Zi) and output impedance (ZO).
These can cause voltage division of the circuit input and output voltages. So, for most transistor
circuits Zi and ZO are important parameters. The circuit voltage amplification (AV), or voltage gain,
depends on the transistor parameters and on resistor RC and RL.
3.1.4.2 CE amplifier circuit elements and their functions
(i)
Biasing circuit: The resistances R1, R2 and RE form the biasing and stabilization circuit. The
biasing circuit must establish a proper operating point, otherwise a part of the negative halfcycle of the signal may be cut-off in the output.
(ii)
Input capacitor, C1: An electrolyte capacitor C1 is used to couple the signal to the base of the
transistor. If it is not used, the signal source resistance, rs will come across R2 and thus change
the bias. C1 allows only ac signal to flow but isolates the signal source from R2
(iii) Emitter bypass capacitor, CE: An Emitter bypass capacitor, CE is used parallel with RE to
provide low reactance path to the amplified ac signal. If it is not used, then ac amplified ac
signal following through RE will cause a voltage drop across it, thereby reducing the output
voltage.
(iv) Coupling capacitor, C2: The coupling capacitor, C2 couples one stage of amplification to the
next stage. If it is not used, the bias conditions of the next stage will be drastically changed
due to the shunting effect of RC. This is because RC will come in parallel with the upper
resistance
R1
of the biasing network of the next stage, thereby altering the biasing conditions
of the latter. In short, the coupling capacitor C2 isolates the dc of one stage from the next
stage, but allows the passage of ac signal.
3.1.4.3 CE amplifier circuit currents
(i)
Base current
iB = IB +ib
where IB = dc base current when no signal is applied, ib = ac base when ac signal is applied
and iB = total base current
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
(ii)
►3.6◄
Collector current
iC = IC+ic
where IC = zero signal collector current, ic = ac collector current when ac signal is applied
and iC = total collector current
(iii) Emitter Current
iE = IE + ie
where IE = Zero signal emitter current, Ie = ac emitter current when ac signal is applied
and iE = total emitter current
It is useful to keep in mind that IE = IB + IC and ie = ib +ic
Also, IE ≈ IC and ie ≈ ic
3.1.4.4 CE amplifier frequency response
The voltage gain of an amplifier varies with signal frequency. It is because
reactances of the capacitors in the circuit changes with signal frequency and hence affects the output
voltage. The curve between voltage gain and signal frequency of an amplifier is known a frequency
response. Figure 3-1-2 shows the frequency response of a typical CE amplifier.
Fig. 3-1-2 Frequency response of CE amplifier
It is clear that the voltage gain drops off at low (< fL) and high (> fH) frequencies whereas it is
uniform over mid-frequency range (fL to fH).
(i)
At low frequencies (< fL), the reactance of coupling capacitor is quite high and hence very
small part of signal will pass from amplifier stage to the load. Moreover, CE cannot shunt the
RE effectively because of its large reactance at low frequencies. These two factors cause a
falling of voltage gain at low frequencies.
(ii)
At high frequencies (> fH), the reactance of C2 is very small and it behaves as a short circuit.
This increases the loading effect of amplifier stage and serves to reduce the voltage gain.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.7◄
Moreover, at high frequency, capacitive reactance of base-emitters junction is low which
increases the base current. These reduce the current amplification factorβ. Due to these two
reasons, the voltage gain drops off at high frequency.
(iii) At mid frequencies (fL to fH), the voltage gain of the amplifier is constant. The effect of
coupling capacitor C2 in this frequency range is such as to maintain a uniform voltage gain.
Thus, as the frequency increases in this range, reactance of CC decreases which tend to
increase the gain. However, at the same time, lower reactance means higher almost cancel
each other, resulting in a uniform fain at mid-frequency.
3.1.4.5 CE amplifier analysis
The first step in ac analysis of CE amplifier circuit is to draw ac equivalent circuit by
reducing all dc sources to zero and shorting all the capacitors. Fig. 3-1-3 shows the ac equivalent
circuit.
rs
Q1
Vs
R1
Rc
R2
RL
Fig. 3-1-3 AC equivalent circuit for CE amplifier
The next step in the ac analysis is to draw h-parameter circuit by replacing the transistor in the ac
equivalent circuit with its h-parameter model. Fig. 3-1-4 shows the h-parameter equivalent circuit
for CE circuit.
rs
Vs
B
+
Vi
R1
Zi .
R2
C
R3
hfeib
1/hoe
Zb .
-
-
.
Zc
E
E
Vo
Rc
RL
.
Zo
+
Fig. 3-1-4 h-parameter equivalent circuit for CE amplifier
The typical CE circuit performance is summarized below:
Device input impedance, Zb = hie
(3-1-1)
Circuit input impedance, Z i = R1 R2 Z b
(3-1-2)
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
Device output impedance, Z C =
1
hoe
(3-1-3)
Circuit output impedance, Z O = RC Z C ≈ RC
h fe
Circuit voltage gain,
AV = −
Circuit current gain,
Ai =
Circuit power gain,
AP = AV x Ai
hie
►3.8◄
(3-1-4)
( RC RL )
(3-1-5)
h fe RC RB
(3-1-6)
( RC + RL )( RC + hie )
(3-1-7)
3.1.4.6 CE amplifier circuit design
Design of CE circuit normally commences with a specification of supply voltage,
minimum voltage gain, frequency response, source impedance, load impedance, stability factor and
Vcc
Q-point.
Rc
R1
C2
C1
Q1
RL
Vs
R2
RE
CE
Fig. 3-1-5 CE amplifier circuit (redrawn from fig. 3-1-1(a))
Selection of IC, RC and RE
h fe
From eq. 3-1-5, AV = −
( RC RL )
hie
For satisfactory transistor operation, Ic should not be less than 500µA. A good minimum Ic
to aim for is 1mA.
The VCE should typically be around 3V to ensure that the transistor operates linearly and to
allow a collector voltage swing of ±1V which is usually adequate for small-signal amplifier
Note: RC should normally be very much larger than RL, so that RL has little effect on
voltage gain.
Select VE = 5V for good bias stability in most circumstances.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.9◄
Note: When VE>>VBE, VE will be only slightly affected by any variation in VBE (due to
temperature change or other effects)
Once VE, VCE and Ic are selected, VRC is determined as VRC = VCC – VCE – VE
Then, RC and RE are calculated as RC =
VRC
V
and RE = E
IC
IC
Selection of bias resistors
As discussed in lab-1, experiment-1.1, section-1.1, selection of voltage divider
current (I2) as IC/10 gives good bias stability and reasonably high input resistance. The bias resistors
are calculated as
R2 =
V − VB
VB
and R1 = CC
I2
I2
Selecting R2 = 10RE gives I2 = IC/10 the precise level of I2 can be calculated as I2 = VB/R2 and this
can be used in the equation for R1.
Selection of bypass capacitor, CE
Basically the capacitor values are calculated at the lowest signal frequency that the
circuit is required to amplify. This frequency is the lower cut-off frequency, fL.
Choose X CE =
hie
at fL for CE calculation to give the smallest value for the bypass capacitor.
1 + h fe
Selection of coupling capacitors, C1 and C2
The coupling capacitors C1 and C2 should have a negligible effect on the frequency
response of the circuit. To minimize the effects of C1 and C2, the reactance of each coupling
capacitor is selected to be approximately equal to one-tenth of the impedance in series with it at the
lowest operating frequency of the circuit (fL).
X C1 =
Z i + rs
10
X C3 =
Z O + RL
10
Usually, RL >> ZO and often Zi >> rS, so that ZO and rS can be omitted in the above equations.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.10◄
Design problem
(i)
Design a single stage CE transistor amplifier using BC107 transistor with Vcc = 15V,
VCEQ = 5V, VE = 3V, RL = 47KΩ and fL = 100Hz.
(ii)
Determine Zi, ZO, AV, Ai and AP for the CE circuit designed in problem (i).
Procedure
Given VCC = 15V, VCE = 5V, VE = 3V, RL = 47kΩ and fL = 100Hz.
The data sheet of BC107 transistor shows: hie = 3kΩ and hFE=190
Selection of RC
RC << RL so that RL will have little effect on the circuit voltage gain.
Select RC =
R L 47 K
=
= 4.7 KΩ (Standard value)
10
10
Selection of RE
RE =
VE VE
≈
IE
IC
Where I C =
∴ RE =
V RC VCC − VCE − V E (15 − 5 − 3)V
=
=
= 1.4mA
4 .7 K
RC
RC
3V
= 2.14 KΩ (use a standard 2.2 kΩ)
1.4mA
Selection of R1 and R2
Selection of voltage divider current I2 as IC/10 gives good bias stability and reasonably high
input resistance
Selecting R2 = 10 RE gives I2 = IC/10
i.e., R2 = 10 × 2 KΩ = 22 KΩ (standard value)
and I 2 =
∴ R1 =
I C 1.4mA
=
= 140 µA
10
10
VCC − VB 15 − (VBE + VE ) 15 − (0.7 + 5)
=
=
= 66.43KΩ (use standard 68kΩ)
I2
140 µA
140 µA
Selection of C1 and C2
The coupling capacitors C1 and C2 should have negligible effect on the frequency response of
the circuit. So, the reactance of each coupling capacitor is selected to be approximately equal to
1/10th of the impedance in series with it at the lowest operating frequency for the circuit.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
X C1 ≈
Z i R1 R2 hie 68K 22 K 3K
=
=
= 254
10
10
10
∴ C1 =
1
1
=
= 6 µF
2πf L X C1 2 × π × 100 × 254
X C2 ≈
RL 47 K
=
= 4 .7 K Ω
10
10
∴ C2 =
►3.11◄
1
1
=
= 0.34 µF (use a standard 0.33µF)
2πf L X C 2 2 × π × 100 × 4.7 K
Selection of CE
hie
3 KΩ
X CE =
=
= 15.71
1 + h fe 1 + 190
∴ CE =
1
1
=
= 101.36µF (use a standard 100µF)
2πf L X CE 2 × π × 100 × 15.71
Calculation of Zi, ZO, AV, Ai and AP
Input impedance, Z i = R1 R2 hie = 68 K 22 K 3K = 2.54 KΩ
Output impedance, ZO = RC = 4.7kΩ
h fe
AV = −
Current gain,
Ai =
Power gain,
AP = AV x Ai = 270.61 X 37.23 = 10K
hie
( RC RL ) = −
190
(4.7 K 47 K ) = 270.61
3K
Voltage gain,
h fe RC RB
( RC + RL )( RC + hie )
=
190 × 4.7 K × (68 K 22 K )
(4.7 K + 47 K )(4.7 K + 3K )
= 37.23
Vcc=15V
R1
68K
Rc
4.7K
C2
0.33uF
C1
Q1
6uF
Vs
R2
22K
RL
47K
RE
2.2K
CE
4.7K
Fig. 3-1-6 CE amplifier circuit with design values of components
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.12◄
3.1.5 TROUBLE SHOOTING A CE AMPLIFIER
When you are faced with having to trouble shoot a circuit, the first thing you need is
a schematic with the proper dc and signal voltages labeled. You must know what the correct
voltages in the circuit should be before you can identify an incorrect voltage.
Vcc=15V
R1
68K
0Vdc
C1
6uF
Vs
Rc
4.7K
C 5Vdc
5.7Vdc
B
C2
0Vdc
0.33uF
Q1
E 3Vdc
R2
22K
RE
2.2K
RL
47K
CE
4.7K
Fig. 3-1-7 CE amplifier with correct voltages indicated
Instability
After the circuit has been constructed, the power supply should be set to the
appropriate voltage and then connected and switched on. An oscilloscope should be connected
to monitor the output of the amplifier to check that the circuit is not oscillating. If the circuit is
oscilloscope the oscillations must be stopped before proceeding further.
Amplifier instability can be the result of incorrect design or poor circuit layout. It can
also be caused by feedback along the conductors from the power supply to the circuit. To
stabilize an unstable amplifier, commence by connecting a 0.01µf decoupling capacitor from the
positive supply line to ground. Where a plus–minus supply is used, connect capacitor from each
supply line to ground.
If the circuit is still unstable, small shunt capacitors should be connected from the
transistor collect terminal to ground, or between collector and base.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.13◄
Vcc
decoupling
capacitor
C1
Rc
R1
shunt
capacitor
C2
Q1
Fig. 3-1-8 CE amplifier circuit with decoupling capacitors and small shunt capacitors to eliminate
oscillations in an unstable
DC Voltage measurements
Once it is established that the circuit is stable the next step is to measure the dc
voltage levels at all transistor terminals. A digital multimeter (DMM) should be used IF the dc
vulgates are not satisfactory; they must be corrected before proceeding further.
Vcc=15V
R1
68K
Rc
4.7K
C 5Vdc
5.7Vdc
B
Q1
E 3Vdc
R2
22K
RE
2.2K
Fig. 3-1-9 Equivalent dc voltages for the CE
amplifier designed
The measured dc voltage levels should at least show that
(i)
The EB junction of the transistor should be forward biased. The VBE may vary from 0.65
to 0.75V (for silicon transistor).
(ii)
The VCB may range from approximately half the supply voltage to almost the full battery
voltage.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.14◄
Inferences from DC voltage measurements
DC voltage readings are used to draw inferences of proper or improper junctioning in
transistor circuits. To shown this, we will assume certain abnormal voltages in the circuit of Figure
3-1-9 and analyze the possible causes of these voltages.
1. VC = VCC = +15V. Possible troubles could be (a) an open in the emitter circuit, (b) an open
in the base circuit, (c) base-emitter short circuit, (d) base short-circuited to ground.
2. VC=0V. Possible troubles include (a) open collector circuit, and (b) collector short-circuited
to ground.
3. VE = VC collector to emitter short-circuited
4. VE=0V. Possible troubles include (a) there is no current following in the emitter, or (b) the
emitter is short-circuited to ground.
Dynamic test
When satisfactory dc levels are established throughout the circuit, dynamic test may
proceed. Suppose that there is no signal at the output terminals of the amplifier for a specified signal
input. The dynamic signal-tracing method can be used to determine what is wrong with the circuit.
The procedure is as follows. A sine-wave signal no larger than the amplifier can handle is injected
into the input terminals of the amplifier and observed at these terminals with an oscilloscope. If the
observed signal is normal, the oscilloscope probe is moved to point B (base) of the amplifier. (Refer
Fig.3-1-7) The sine-wave signal at this point should be approximately the same as at the input
terminals if the amplifier input is normal. If there is no signal at the base, two possible reasons exist.
The first is that the capacitor C1 is open. The second is that the base terminal is short circuited to
ground. An open capacitor may readily be found by connecting a 0-1µf capacitor across C1 and
observing with oscilloscope the output signal. If an output signal appears, this indicates that
capacitor is open. If no signal appears at output terminals, the oscilloscope probe is connected to the
base. No signal indicates a short circuit in the base circuit.
We may also determine if capacitor C2 is open by signal tracing. Assume that the
input circuit, including C1, is found to be operating correctly, but that there is no signal at the output
terminals of the amplifier. The oscilloscope probe is then connected directly to the collector of
transistor. If normal signal appears at the collector, but non exists at the output terminals, we know
that C2 is open.
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.15◄
Resistance measurements
Resistance measurements in transistor circuits, always made with power turned off,
are helpful in determining defective components. The resistance measured at the base B to G should
be 22kΩ. The resistance measured at the emitter E to G should be 2.2kΩ. The resistance from
collector C to G would be the sum of RC, R1 and R2; in this case 94.7kΩ. These value, then, are the
standard or norm for the circuit of Fig. 3-1-10.
Vcc=0V
Rc
4.7K
R1
68K
C
B
Q1
E
DMM
R2
22K
RE
2.2K
Fig. 3-1-10 Resistance measurements in transistor circuit
The most obvious resistor defects are opens, which can be spotted very easily. For example, if the
resistance measured from E to G is infinite (∞), then either RE is open, or the connective wiring is
open. So, the resistance checks and continuity checks will reveal where the defect is.
3.1.6 EXPERIMENT
1. DC voltage measurements
1.1 Assemble the dc equivalent of the CE amplifier circuit you have designed, as shown in fig. 3-19. Use the BC 107 transistor, or equivalent.
1.2 Measure Q point and other transistor terminal voltages and currents as per the procedure given
in experiment 1.1. Tabulate the readings in table 3-1-1. Compare the measured values with the
calculated values.
Voltage and Current quantities
Measured values
Calculated values
Table 3-1-1 DC Voltage measurements
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.16◄
2. Transient voltage measurements
2.1 Feed 100mV (peak-to-peak) sinusoidal signal at 1KHz frequency as the input signal Vs to the
CE circuit shown in Fig. 3-1-6.
2.2 Observe the input and output voltages simultaneously on a CRO. Note down the amplitude,
frequency and phase difference between the two voltages in the table 3-1-2. Compute the gain
of the amplifier circuit and compare it with the calculated value.
2.3 Plot on a linear graph the transient voltage of the input & output of the amplifier circuit
Particulars
Amplitude
Time period
Frequency
(volts)
(msec)
(Hz)
Input Voltage
Output Voltage
Table 3-1-2 Transient Voltage measurements
3. Frequency response curve measurements
3.1 In the above assembled circuit, keep the magnitude of the source same, ie.,100mV and decrease
the frequency from 1KHz and measure voltage gain of the amplifier at each frequency. Now
increase the frequency from 1 KHz to 1MHz and measure the voltage gain of the amplifier at
each frequency. Take at least 5 readings on either side of the 1 KHz frequency. Tabulate the
reading in table 3-1-3.
3.2 Plot on a semi log graph sheet the frequency response (voltage gain vs frequency) curve using
the above measurements.
3.3 From the plot, determine the values of (a) Mid band voltage gain, AV(mid), (b) Lower cut-off
frequency, (c) Upper cut-off frequency and (d) Bandwidth.
Input Voltage, VS = _______ mV
Voltage Gain, VO
Signal Frequency
Output Voltage
(Hz)
(volts)
(
VO
)
VS
Voltage Gain, dB
20 log10 (
VO
)
VS
Table 3-1-3 Frequency response curve measurements
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.17◄
4. Resistance measurements
4.1 In the above assembled circuit, keep the magnitude and frequency of the source same, ie.,
100mVpp at 1KHz frequency.
4.2 Connect a potentiometer Rin (variable resistance) in series with the circuit input terminal and the
signal source, as shown in fig. 3-1-11
4.3 Connect a two-channel CRO to simultaneously monitor the input and output signal voltage
Vcc=15V
waveforms.
R1
68K
C1
6uF
Rc
4.7K
C2
0.33uF
R4
Q1
10k
R5
Vs
R2
22K
RE
2.2K
CE
4.7K
50k
Fig. 3-1-11 Experimental CE circuit to measure input and output impedances
4.4 Adjust the POT until a new output signal VO, equal to one-half the original measured value of
VO is obtained. Now, remove Rin from the circuit and measure its resistance using DMM. The
measured value in ohms equals the input impedance, Zi.
4.5 To measure the output impedance ZO of the amplifier, connect a potentiometer Rout to the
output circuit.
4.6 Adjust the POT until a new output signal VO, equal to one-half the original measured value of
VO is obtained. Now, remove Rout from the circuit and measure its resistance using DMM. The
measured value in ohms equals the output impedance, Zo. Tabulate the readings in table 3-1-4.
Particulars
Measured Value
Calculated Value
Input Impedance, Zi
Output Impdance, ZO
Table 3-1-4 Resistance Measurements
EC0222 Electronic Circuits Lab Manual
Experiment 3.1: Common Emitter Amplifier
►3.18◄
3-1-7 Post–Lab questions
1. Check your understanding by answering these questions.
(a) The voltage gain of the amplifier in Fig. 3-1-11 is 50. When CE is opened the gain of the
amplifier should ___________ (increase, decrease, remain the same)
(b) The circuit of fig. 3-1-11 must amplify sine-wave signals in the frequency range 20 to 20,000
Hz. The design value of RE is 2200Ω. The highest value XC of CE which will act a good bypass
for this amplifier is _____ Ω at _______ Hz. The value of CE is _______ µF.
(c) In an audio amplifier the collector-to-base must be _______ (forward, reverse) biased.
(d) The ac signal voltage measured at the base of a CE amplifier is 50mV. The output signal voltage
measured at the collector is 2.5V. The voltage gain of the amplifier is ________
(e) The rms voltage measured at the collector of the amplifier in Fig. 3-1-11 is 4.6V without load.
When a 250Ω load is connected across the output, the rms voltage measured at the collector
load is 2.3V. The output impedance of the circuit is _________ Ω.
(f) A CE amplifier has a gain of 50, an input impedance of 1000Ω, and an output impedance of
200Ω. The power gain of this amplifier is _________.
(g) The decibel power gain of the amplifier in question 1(f) is __________ dB.
(h) A sine wave injected into the base of the transistor Fig. 3-1-6 results in a normal output. When
the generator leads are moved to the input, no signal appears in the output. The most probable
cause of trouble is a ____________.
(i) In the circuit shown in Fig. 3-1-9, a technician measures 10V at collector. VCC = 12V. IC is ___.
(j) If RC in Fig. 3-1-9 is open, a resistance check from C to G will measure _________Ω.
(k) If the emitter is short circuited to base in Fig. 3-1-9, a resistance check from the base to ground
will show approximately _________Ω.
(l) For the conditions in question (k), the voltage at the collector will measure __________V
(m) If R1 in Fig. 3-1-9 is open, the voltage measured from E to G will be _________V
2. How do coupling capacitors C1 and C2 affect the frequency response? Why?
3. What is the effect on the amplifier performance of omitting RE?
4. What is the effect on input impedance of removing bypass capacitor CE?
5. (a) What is the phase relationship between the input and output signals of a CE amplifier?
(b) Was this relationship confirmed by the results of your experiments? Explain how.
6. Is the output impedance of a CE amplifier a fixed quantity? Confirm your answer by referring
specifically to any substantiating data in this experiment.
EC0222 Electronic Circuits Lab Manual