ECE 3110 - Full Lab Report 2
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BJT Common-Emitter Circuit Voltage Gain
October 31, 2023
ECE 3110 – 003
Electrical Engineering Laboratory III
ABSTRACT:
This experiment is intended to show how altering the bias of the circuit and the resistor
values inside it can alter the voltage gain of a
C
E
bipolar transistor circuit. The function of the
emitter bypass capacitor
C
E
will also be demonstrated. Furthermore, it will be illustrated how
the Q-point is determined by the base resistors and how the Q-point impacts the relationship
between the input and output wave shapes.
INTRODUCTION:
One common electronic circuit architecture for amplifying tiny signals is the common-
emitter (
C
E
) transistor amplifier. It is commonly applied for small-signal AC and DC
amplification, and it makes use of a bipolar junction transistor (BJT). When there is no input
signal for the transistor, the common-emitter circuit’s DC function includes setting a stable
operating point or Q-point. The DC values of collector current (
I
C
) and collector-emitter
voltage (
V
CE
) are determined by this Q-point. The transistor is kept in the active zone and
prepared to amplify signals thanks to DC biasing. The
C
E
amplifier also amplifies weak AC
impulses and sends them to its base terminal. The transistor operates differently when the input
signal is placed on the DC biasing point (Q-point). The ratio of the AC output voltage to the AC
input voltage, or
V
O
V
I
, determines the voltage gain of the
C
E
amplifier. By modifying the
DC biasing resistors, the Q-point can be changed, which in turn affects the transistor’s operating
voltage and current (
I
C
and
V
CE
). On the transistor’s characteristic curves, if the Q-point
is moved excessively to the left, the transistor may enter cutoff or saturation states, which would
restrict linearity and signal amplification. To guarantee that the transistor runs in its active region
and enables efficient signal amplification with ideal gain and linearity while preventing
saturation or cutoff, the Q-point must be adjusted correctly.
Changing the values of the emitter and collector resistors has an impact on the circuit’s
voltage gain. A BJT’s collector current (
I
C
) is determined by the transistor’s
, temperature,
and circuit components. However, selecting biasing resistors so that the quiescent (DC) collector
current stays consistent, independent of transistor temperature, is necessary for successful
amplifier design. The collector current and voltage gain are kept reasonably constant by choosing
the appropriate biasing network and resistors. The emitter resistor, RE, and the values of the bias
resistors, R1 and R2, are the essential components that stabilize the circuit against variations in
the transistor.
The stability of the circuit in response to variations in the transistor’s
is influenced by
essential components, namely the emitter resistor (
R
E
) and the bias resistors (
R
1
and
R
2
). To preserve the high gain of the common emitter (
C
E
) circuit, a common practice is
to bypass the emitter resistor with a sizable capacitor. The introduction of an emitter resistor
diminishes the AC voltage gain at the AC component (
i
C
) is obliged to pass through RE to
reach ground. Consequently, RE hinders
i
C
, and the negative feedback voltage across RE
diminishes the AC gain. To counteract this undesirable AC response while retaining effective
control over
I
C
(collector current) and the stabilizing impact of RE, a capacitor is connected
across RE to short-circuit the AC current (
i
C
) around RE to ground. This capacitor, usually
denoted as
C
E
, is typically of substantial magnitude, such as 10
F or greater, facilitating the
majority of the AC current to flow through RE to ground.
I
C
still traverses through
R
E
as
the capacitor functions as an open circuit for DC, thereby maintaining the stabilizing effect of
R
E
. The investigation in this laboratory will explore the impact of the absence of
C
E
,
revealing that the voltage gain is determined by the ratio of
R
L
(total AC load resistance) to
R
E
.
OVERVIEW / BACKGROUND:
Before conducting the lab, review the dc analysis methods for determining a common-
emitter circuit’s Q-point and how changing the ratio between two base resistors (
R
1
and
R
2
) affects the load line’s motion and base voltage. As prelab, LTSpice simulations were
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performed of the circuit in Figure 1, using the resistor values listed in Table 1, both with and
without
C
E
. Assume sinusoidal input VS = 0.02Vp-p at 1 kHz, the transistor is a 2N3904, and
all capacitors are 22
F or greater. The simulated results were recorded in Table 1.
Important formulas:
A
V
=
V
O
−
pp
V
S
−
pp
, which was used to determine the voltage gain.
=
I
C
I
B
V
CE
=
V
CO
−
V
EO
I
C
≈ I
E
=
V
EO
R
E
V
CE
(max) =
V
CC
I
C
(
max
)=
V
CC
R
E
+
R
C
Table 1: List of Resistors for Prelab and Simulated Data
With
C
E
Without
C
E
R
C
(
)
R
E
(
)
V
OMin
(mV)
V
OMax
(mV)
V
OMin
(mV)
V
OMax
(mV)
3.9 k
1 k
-915.3412
756.27914
-19.886941
20.218734
3.9 k
200
-265.35281
218.92653
-86.002651
89.18741
3.9 k
10 k
-32.985416
25.446516
-1.894901
2.183624
1 k
1 k
-422.890132
345.56552
-37.23240
39.28542
20 k
1 k
-15.448301
446.70943
-7.287621
33.058711
Figure 1: Common-Emitter circuit with emitter bypass capacitor
EXPERIMENTAL PROCEDURE and RESULTS:
Equipment used:
-
1 NPN Transistor, 2N3904
-
Curve Tracer function on Waveforms
-
1 Decade Resistance Box
-
NI-ELVIS workstation
-
Resistors (2.2k, 220, 16k)
-
Capacitors (22
F)
-
Lab Kits
-
AD2
Part 1 Procedures: During this part, the proper Q-point was established. Using the curve
tracer, a Q-point closest to
V
CE
= 2.5 and
I
C
= 1 mA was determined which was
V
CE
=
2.5096 and
I
C
= 1.4256 mA. Additionally,
I
b
= 0.01064 mA and
V
bE
= 0.69988 and
=
1.4256
0.01064
= 133.984. Figure 2 shows the screenshot of the determined Q-point on the
transistor characteristics graph.
Figure 2: Screenshot of Q-point found on Waveforms
Using Figure 3 as reference, the circuit was constructed with
V
CC
=
5
V
. The DMM was used
to measure
V
CO
and
V
EO
and got a measurement of
V
CO
= 3.115V and
V
EO
= 0.184
V.
V
CE
was determined as
V
CO
−
V
EO
= 3.115 – 0.184 = 2.931 V and
I
C
≈ I
E
=
V
EO
R
E
=
0.184
220
=
0.83636
mA. Figure 4 portrays the transistor characteristic with the determined Q-
point using the DMM and drawn load line.
.
Figure 3: Common-Emitter Circuit for Q-point measurement
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Figure 4: Output Characteristic plot (hand drawn) with Q-point.
Part 2 Procedures: This part of the lab was performed to determine the AC measurements
and the voltage gain versus the collector resistor (
R
C
) with and without the emitter bypass
capacitor (
C
E
)). First, the circuit was constructed using Figure 5 as reference.
Section 1 & 2 of Part 2 Procedures:
V
O
and
V
S
were measured with the
oscilloscope.
R
C
was replaced with the resistance decade box and set to 3.9 k
. Again, the
oscilloscope was used to measure
V
O
, which was the same as the previous step.
V
O
was
continuously measured while
R
C
was varied from 1 k
to 20 k
and
V
S
was determined
to be 20.292 mV. The
V
O
measurements, output wave shape, and voltage gain was recorded
into Table 2. Voltage gain was determined by
A
V
=
255.041
20.292
= 12.5685 and was repeated for
each measurement.
Once the data was collected, the emitter bypass capacitor,
C
E
, was
removed and the same procedures were repeated. The data was recorded in Table 2.
Figure 5: The Common Emitter Circuit with added capacitors and input signal VS
provided by the function generator and the voltage divider, RS1 and RS2.
Table 2:
V
O
vs
R
C
R
C
()
With
C
E
Without
C
E
V
O
−
pp
Output
shape
A
V
V
O
−
pp
Output
shape
A
V
1k
255.041
sine
12.569
88.426
sine
4.358
2.2k
546.671
sine
26.940
190.282
sine
9.476
6.8k
-
clipped
-
-
clipped
-
10k
-
clipped
-
-
clipped
-
15k
-
clipped
-
-
clipped
-
20k
-
clipped
-
-
clipped
-
Figure 6:
R
C
=
¿
1k
,
R
E
= 220, with
C
E
Figure 7:
R
C
=
¿
6.8 k
,
R
E
= 220, with
C
E
Figure 8:
R
C
=
¿
20 k
,
R
E
= 220, with
C
E
Figure 9:
R
C
=
¿
1k
,
R
E
= 220, without
C
E
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Figure 10:
R
C
=
¿
6.8k
, R
E
= 220, without
C
E
Figure 11:
R
C
=
¿
20k
, R
E
= 220
, without
C
E
1000
2200
6800
10000
15000
20000
0
5
10
15
20
25
30
Plot 1: RC vs Av with CE
RC ()
Av (voltage Gain) (V)
Figure 12: Plot of
R
C
vs
A
V
with
C
E
1000
2200
6800
10000
15000
20000
0
1
2
3
4
5
6
7
8
9
10
Plot 2: RC vs Av without CE
RC (
Av (voltage Gain) (V)
Figure 13: Plot of
R
C
vs
A
V
without
C
E
Table 2’s findings demonstrate that, as predicted, the signal at the collector terminal in the
table and graphs above with
C
E
present is stronger than it is in the absence of
C
E
.
This is
due to the fact that when
C
E
is present in the circuit, the resistor
R
E
is bypassed. For the
values of
R
C
shown in the table, distortion displayed as clipping occurs very early on because
the Q-point value is no longer centered along the DC load line.
Section 3 & 4 of Part 2 Procedures:
R
C
was set to 3.9k
using a fixed resistor
instead of a resistance decade box and the emitter bypass capacitor,
C
E
,
was connected using
Figure 5 as reference.
R
E
was replaced with the resistance decade box and set to 1 k
.
V
O
was measured using the oscilloscope with ranging
R
E
values from 100 to 10k.
V
S
was measured to be 20.292 mV. The determined data was recorded in Table 3. After the data was
collected, the emitter bypass capacitor,
C
E
,
was removed and repeated the steps with data
recorded in the table.
Table 3:
V
O
vs.
R
C
R
E
()
With
C
E
Without
C
E
V
O
−
pp
Output
shape
A
V
V
O
−
pp
Output
shape
A
V
100
711.976
sine
35.087
390.311
sine
19.923
220
557.031
sine
27.451
190.349
sine
9.380
470
457.914
sine
22.566
91.743
sine
4.521
1k
365.122
sine
17.993
44.508
sine
2.193
4.7k
134.561
sine
6.631
12.224
sine
0.602
6.8k
99.901
sine
4.923
9.991
sine
0.492
10k
73.073
sine
3.601
6.553
sine
0.323
Figure 14:
R
C
=
¿
2.2k
, R
E
= 220
, with
C
E
Figure 15:
R
C
=
¿
2.2k
, R
E
= 1k
, with
C
E
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Figure 16:
R
C
=
¿
2.2k
, R
E
= 6.8k
, with
C
E
Figure 17:
R
C
=
¿
2.2k
, R
E
= 220
, without
C
E
Figure 18:
R
C
=
¿
2.2k
, R
E
= 1k
, without
C
E
Figure 19:
R
C
=
¿
2.2k
, R
E
= 6.8k
, without
C
E
100
220
470
1000
4700
6800
10000
0
5
10
15
20
25
30
35
40
Plot 3: RE vs Av with CE
RE ()
Av (Voltage Gain) (V)
Figure 20: Plot of
R
E
vs
A
V
with
C
E
100
220
470
1000
4700
6800
10000
0
5
10
15
20
25
Plot 4: RE vs Av without CE
RE ()
Av (Voltage Gain) (V)
Figure 21: Plot of
R
E
vs
A
V
without
C
E
Higher voltage gains are associated with lower values for
R
E
,
as Table 3 above
illustrates because
C
E
bypasses the
R
E
resistor, which increases the voltage gain. Lower
voltage gain results from increasing the resistance of
R
E
, both with and without
C
E
.
Final section of Part 2 Procedures:
R
E
was set to 1k
and
C
E
was connected.
The input signal frequency was varied from 10 Hz to 50 kHz and the oscilloscope was used to
see how the magnitude of the output signal varied. The results were recorded into Table 4.
Table 4: Frequency Response of
V
O
−
pp
and Voltage Gain,
A
V
Frequency (Hz)
10
100
1k
10k
25k
50k
V
O
−
pp
59.215
108.321
360.965
538.212
538.435
540.794
A
V
2.918
5.338
17.789
26.523
26.534
26.651
Changes to the frequency appear to make large changes to the voltage gain up until 10
kHz. From that point forward the voltage gain remains steady. This indicates that this range falls
at the midrange frequency level, where the voltage gain remains at a steady level. The coupling
capacitor,
C
2
, has high reactance at lower frequencies which means only a small part of the
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signal will be output and it can’t be effectively bypass
R
E
,
which means increased resistance
and a lower voltage gain as it does when
C
E
is not present in the circuit.
Post Lab Questions:
1.
In certain instances, the findings of the comparison between the simulations of the
circuits and the experimental data were extremely similar, but not always. This could be
the result of variances in different physical components or human mistake in the
experiment or simulation.
2.
When
C
E
is present in the circuit, the resistor
R
E
is bypassed, resulting in a greater
voltage gain. Without the bypass capacitor,
C
E
,
there is more resistance on the output
voltage and therefore resulting in a lower voltage gain.
CONCLUSION: As anticipated, the introduction of the
C
E
capacitor resulted in an amplified
output voltage. This amplification occurred as a consequence of bypassing the
R
E
resistor.
Elevations in the
R
C
resistor’s resistance contributed to higher voltage gains, reaching a point
of distortion manifested as clipping at 6.8 k
. Conversely, increases in the
R
E
resistor led to
reduced voltage gains for the same reason observed without the bypass capacitor (
C
E
). This
reduction is attributed to the heightened resistance affecting the output voltage. Additionally, the
frequency of the input signal significantly influences the voltage gain, as it impacts the reactance
of the coupling capacitor
C
2
.
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- with the aid of a diagram describe a simple common emitter circuit with npn bipolar transitor and discus the relationship between base emitter voltage and base currentarrow_forwardA 4.7 zener diode is connected to the base resistor. The left side of the resistor sits at about 12V. The right side of the biasing resistor is connected to the Zener and npn transistor. The Zener bias is 5mA. The base current needed is 80uA. What is the appropriate resistor for this circuit? (In kohms) Iarrow_forwardQuestion 2 (a) To achieve correct operation when using a Bipolar Transistor how should the two p-n junctions be biased?arrow_forward
- 05: Choose the correct answer: 1. The current in FET flows between terminals. (Drain and gate Drain and source Gate and soarce None of these) 2. The threshold voltage is in enhancement type metal oxide semiconductor field effect transistor. (Positive Negative Zero Positive) 3. The drift current density effected by (An electric filed • Concentration gradient in holes • Concentration gradient in free electrons All of these) 4. The band gap between conduction and valance band in an insulator is (Low Very low High Moderate) 5. Operating point of a transistor must be independent of (Biasing Amplification Variations in parameters Filtration) 6. When a transistor operating at the central of the load line is declining, the current gain will change the Q-point (down up nowhere of the load line) 7. A bipolar junction transistor when used as a switch, operates in (cut-off and active region active and saturation region cut-off and saturation region all of these) 8. Inversion layer in an MOS…arrow_forwardIn a properly Biased NPN transistor most of the electrons from the emitter A. recombine with holes in the base, B. recombine in the emitter its self, C. pass through the base to the collector, D. are stopped by the junction barrier. In transistor, the relation among emitter current, base current and Collector Current is: A. Ic=IB+IE, B. Ig=IB+Ic, C. IE=Ic-IB, D. I=Ig-Ic. • Transistor is known as bipolar transistor because it has: A. Electrons, В. Holes, C. Electrons and holes, D. None of the above. In NPN or PNP transistor: A. Ic> IE, B. IE >Ic, C. IE = Ic, D. None of the above. The base of an NPN transistor is: A. Heavily doped, B. Lightly doped, C. Medium Doped, D. Doped by pentavalent material. In a transistor the value of a (B) is 100, the value of (a) is: A. 0.01, В. О.1, C. 0.99, D.1.arrow_forwardQuestion: 8. With the emitter bias shown, a reasonable assumption for troubleshooting work is that the VcC a. base voltage =+1 V +15 V b. emitter voltage = +5 V RC 3.9 k2 c. emitter voltage =-1 V d. collector voltage = Vcc RB 68 kQ %3D RE 7.5 kQ VEE -15 Varrow_forward
- What do each of the three terms in the MOSFET threshold voltage equation represent? ☐ built-in voltage ☐ voltage needed to achieve flat-band voltage due to band bending in semiconductor ☐ voltage across oxide ☐ breakdown voltagearrow_forwardtwo applications for a Bipolar Junction Transistor and talk about one in detailarrow_forwardYou have been given a Silicon-based BJT transistor with the leads labelled as shown in the figure: Plastic Body TO-92 1. Emitter 2. Base 3. Collector 123 Given that this transistor is either a PNP or NPN variant, create a step-by-step procedure for using a DMM with "diode check" feature to identify whether (a) it's an NPN or PNP (b) it's a PNP or NPN that has failed in some way (e.g. failed as open, or as a short)arrow_forward
- This multiple choice questions from electronics lab. Pic one is questions && pic two laws. Ic:current of collecter transistor. Ib:current base Ie:of emitter for transistor. I HOPE to help me,thx advanced.???arrow_forwardA transistor current is given below. The Zener diode breakdown voltage is 5.3V as shown. Take base to emitter voltage drop to be 0.6V. The value of the current gain ß is ..... 4.7kQ2 0.5mA ↓ 5.3V www 41 + 0.6- www 22092 47092arrow_forward
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