Download Analysis on an NPN Common-Emitter Amplifier

ANALYSIS OF AN NPN COMMON-EMITTER AMPLIFIER
Experiment Performed by:
Michael Gonzalez
Filip Rege
Alexis Rodriguez-Carlson
Report Written by:
Filip Rege
Alexis Rodriguez-Carlson
November 28, 2007
Objectives:
In completing these experiments, our main objective was to familiarize ourselves with
NPN transistors in general and their use in a common emitter amplifier in specific. To that end
we had the following goals:
o To use PSpice to find the following input (base) and output (collector)
characteristics of a 2N3904 NPN small-signal bipolar transistor:
ƒ
The amplification factor (β)
ƒ
The base-to-emitter resistance (rπ)
o Use PSpice to simulate the DC and AC performance of a midpoint biased
common-emitter (CE) amplifier
o To wire a basic CE amplifier circuit and to adjust the value of the base resistor
(RB) (if necessary) in order to adjust the operating point (Q) to be approximately
one-half the value of VCC
o To apply a 1 kHz, 20mV(p-p) sine wave to the amplifier’s input and to measure
both the input voltage (Vin) and the output voltage (VO) to determine the circuit’s
AC voltage gain.
Discussion of Theory:
The Diode and the Semiconductor:
Basically, an NPN transistor can be thought of as two diodes with a shared anode region.
A diode is a component which allows electrical current to flow through it in only one direction.
The symbol for a diode (see Figure 1) reflects this directionality. Current can flow in the
2
direction of the arrowhead, but not against it. The input side is referred to as the anode, and the
output side is called the cathode.
Figure 1: Circuit symbol for a diode
When a battery or other voltage source is connected to the diode there are two possible
outcomes. If the voltage is more positive on the cathode side then no current is conducted. This
diode is “reverse biased”. If, however, the voltage is more positive on the anode side (making
the diode forward biased) current will flow across the diode. When an ideal diode is reverse
biased its model is an open circuit which passes no current. When an ideal diode is forward
biased in a circuit it can be imagined as a short circuit which passes all current.
Figure 2: Ideal models of forward and reversed biased diodes.
3
In actuality, a reverse biased diode will conduct a small current of about 10μA, and a
forward biased diode will not start conducting current until the potential across it reaches about
0.6V. This is referred to as the turn-on voltage.
When applying an AC voltage to a diode, the diode will be forward biased for half of the
AC cycle and reverse biased for the other half. When analyzing a circuit with an AC voltage as
the input, it is useful to do two separate analyses, one for the positive half-cycle (when the
voltage is greater than 0V) and one for the negative half-cycle (when the voltage is smaller than
0V). Then, the models shown in Figure 2 will apply as expected.
As stated before, a NPN transistor can be thought of as two diodes connected at the
anode. This description would indicate that no current could flow since at any given time one of
the diodes would be reverse biased. However, a transistor has 3 terminals, a collector (the
cathode of one of the diodes), an emitter (the cathode of the other), and a base, which is an input
where the two diodes meet at their anodes. The symbol for an NPN transistor is below in Figure
3. The arrowhead points towards the emitter.
Figure 3: Circuit symbol for a transistor
A more accurate description of the NPN transistor’s physical shape is to envision a
sandwich made of two slices of N-type semiconductor with a single slice of P-type
semiconductor in the middle, as in Figure 4 below. Note that the three different regions of the
transistor are different sizes.
4
Figure 4: A simplified cross-section of an NPN transistor1
Output Characteristics of a Transistor:
In order to determine the output characteristics of an NPN transistor, a simple circuit as
seen in Figure 5 can be used. This circuit is used only to find the characteristics of an NPN
transistor in order to use that information in another circuit. The first output characteristic is the
collector current’s (IC) dependency on the base current (IB) and the collector-to-emitter voltage
drop (VCE). Another characteristic is the transistor’s amplification factor (β), which is the
relationship between the collector current and the base current.
Figure 5: Circuit used to find the input characteristics of a transistor
1
From “Npn BJT cross section.PNG”. Answers.com. November 25, 2007.
<http://www.answers.com/topic/npn-bjt-cross-section-png>.
5
IB can be controlled by adjusting the current source (labeled I1 in Figure 5), VCE can be
changed by changing VCC since they are equal, and the value of IC is measured. To graph the
changes in IC as a result of the values of IB and VCE, I1 is set at 0A, and IC is measured as the
value of VCE is changed from 0 to 15V. The resulting graph is in Figure 6, and shows a very
small rise in the value of IB as VCE increases. Note the values of IB never climb above 0.25nA,
which is so small as to be effectively 0A.
IC vs. VCE
2.50E-10
2.00E-10
1.50E-10
I C (A) 1.00E-10
IB = 0A
5.00E-11
0.00E+00
0
2
4
6
8
10
12
14
16
-5.00E-11
VCE (V)
Figure 6: IC vs. VCE for the values of IB equal to 0A
Next, the value of IB is changed to be 10μA. Data is again taken and plotted on the same
graph as the previous data. This process is repeated increasing IB in steps of 10μA up to 60μA.
The resulting graph is below.
6
IC vs. VCE
1.40E-02
1.20E-02
1.00E-02
IB = 0A
IB = 10 uA
8.00E-03
IB = 20uA
I C (A) 6.00E-03
IB = 30uA
IB = 40uA
4.00E-03
IB = 50uA
IB = 60uA
2.00E-03
0.00E+00
0
2
4
6
8
10
12
14
16
-2.00E-03
VCE (V)
Figure 7: IC vs. VCE in 10μA steps of IB
Figure 7 shows that the value of IC depends mostly on the value of IB and less on the
value of VCE. This means that in order to accurately predict the value of IC in a circuit both the
values of IC and VCE must be known. Also, note that the relationship between VCE and IC is only
linear in the middle of the graph. In the far left of the graph, which is called the saturation
region, and on the far right of the graph, which is called the cutoff region, the relationship is not
clear and the transistor’s performance is not predictable.
Having found that IC is mostly dependent on IB, the next logical step is to find what that
relationship is. Note from Figure 7 that IB is given in terms of μA (10-6 A), and IC is given it
terms of mA (10-3 A), and that the value of IC increases a predictable amount for every 10μA
increase in IB. This indicates a proportional relationship between the two. The proportionality
constant is called transistor’s amplification factor (β) (in specification sheets, the amplification
factor is often referred to as “hfe”). The relationship can be described mathematically as
I C = β ×I B
Solving for β results in:
7
(1)
β=
IC
IB
(2)
Using Figure 5 to find the value of β for the transistor, IC can be measured and recorded while
VCC is held constant and IB is changed in increments of 10μA. Then, a chart such as the one
below can be filled in. The value of β in all of these cases should be within approximately 10%
of each other.
IB (μA)
10
20
30
40
50
60
β
Table 1: Chart used to display β for different values of IB
Input Characteristics of a Transistor:
When analyzing a circuit which includes a transistor it is also useful to know how the
value of IB changes according to the value of the base-to-emitter voltage drop (VBE). Knowing
that the turn-on voltage for a diode is ~0.6V, it makes sense that the value of IB would be very
close to zero until VBE reaches ~0.6V and then it would increase at the same rate that IC is
increased. The graph would look similar to Figure 8.
IB vs. VBE
7.00E-05
6.00E-05
5.00E-05
4.00E-05
IB (A)
3.00E-05
2.00E-05
1.00E-05
0.00E+00
0.00E+00
1.00E-01
2.00E-01
3.00E-01
4.00E-01
5.00E-01
6.00E-01
7.00E-01
V BE (V)
Figure 8: Graph of IB vs. VBE for a NPN transistor
8
The DC Load Line:
Figure 9 is a copy of the graph from Figure 7 with some additional information on it.
The line in red is referred to as the DC Load Line. This line is created by drawing a line from
the maximum value of IC to the maximum value of VCE. These values may be found using
traditional circuit analysis techniques for any circuit which utilizes a NPN transistor as a
common emitter amplifier (CE Amplifiers will be discussed in a later section).
Figure 9: The graph from Figure 7 showing the DC load line and the operating point, Q
The line in green represents the measured or calculated value for IB in a circuit (see
Equation 3). In this example, it appears that IB is equal to approximately 14μA, so the green line
has been drawn slightly below the halfway mark between 10μA and 20μA. Note that, because of
the slight dependence of IB on VCE discussed earlier, the green line is not horizontal but has a
slightly positive slope.
IB =
9
VCC − VBE
RB
(3)
The main purpose of a CE amplifier is to amplify an AC voltage. An amplifier without
an AC input is known as a quiescent amplifier, or an amplifier which is resting. If the amplifier
is resting, it has constant values of IC and VCE. The Q point is defined as a point on the DC load
line that indicates the values of IC and VCE for the amplifier at rest. Ideally, the value of VCE at
this point should be as close as possible to one-half of VCC. In the example above, the VCE value
for point Q appears to be at 7V, which is within 10% of the half of the VCC value of 15V.
Projecting the point Q to the y-axis gives the value of IC at approximately 2.1mA, which is about
½ of the value marked at ICmax.
When the values for VCE and IC at the Q point are half of their maximum values the circuit is said
to be midpoint biased. When an AC voltage is applied both VCE and IC will oscillate around their
Q point values, so when the Q point is centered they can both have the largest possible
amplitudes above and below their DC values. The relationship between the Q point and the
amplifier operation can be seen below in Figure 10.
Figure 10: Optimum amplifier operation2
2
From: “Course Notes: Unit 8”. Electronic Fundamentals I. November 25, 2007.
<http://technology.niagarac.on.ca/courses/elnc1126/Notes/Unit8.pdf>.
10
If the Q point is not optimally placed it can be changed by adjusting the value of RB.
That will change the value of IB which, as discussed above, is the value which primarily affects
IC, which in turn affects the value of VCE. To move the Q point towards cutoff, RB should be
raised, and to move Q closer to saturation, RB should be lowered.
A Transistor in a CE Amplifier:
Depending on how a transistor is placed into a circuit it can have different characteristics
and functions. For the purposes of this experiment, the relevant configuration is one known as a
common emitter (CE) amplifier. As the name implies, the emitter is the terminal which is
common to both the base and the collector. The transistor’s purpose in the circuit is to amplify
an AC voltage. In the CE configuration, the base is the input of the transistor and the collector is
the output of the transistor. Figure 11 shows the circuit schematic of a very simple CE
amplifying circuit.
Figure 11: A simple CE Amplifier circuit. The path of the DC current is in blue.
The capacitor in Figure 11 keeps the DC current from entering the AC voltage source.
When following the DC current path through the circuit, the current originates at the DC voltage
11
supply and flows to the top of the circuit where it splits and travels in two directions. The first
branch (IB) flows through the base resistor (RB) and into the base of the transistor. The second
branch (IC) flows through the collector resistor (RC) and then into the collector of the transistor.
The base-to-collector junction is reverse biased with respect to the DC voltage supply,
while the base-to-emitter junction is forward biased. Under these conditions, IC will not pass
through the base-to-collector junction, instead it will continue straight through the transistor to
the emitter. IB, however, will flow from the base to the emitter. This means that the emitter
current (IE) will be equal to the sum of IB and IC, which yields a useful relationship for this
circuit:
I E = I B+ I C
(4)
Without any additional information, more properties of the circuit in Figure 11 can be
found. By drawing the input and output loops (see Figure 12) and using Kirkoff’s Voltage Law
(which states that the sum of the voltage drops around a loop will equal zero) the input and
output loop equations can be found.
Figure 12: The circuit from Figure 11 with the input and output loops.
12
Starting with the input loop equation:
− VCC + VRB + VBE = 0
VCC = VRB + VBE
(5)
Using Ohm's Law the term VRB can be replaced:
V = I×R
(6)
V RB = I B × R B
(7)
Substituting Equation 7 into Equation 5 results in the final input loop equation:
VCE = I B RB + VBE
(8)
A similar process is used to find the output loop equation, which ultimately yields:
VCC = I C RC + VCE
(9)
The following values in the input and output loop equations are known or can be solved
for:
o VCC, RC, and RB which are set by the user
o VBE, which is known to be the turn on voltage of a diode, equal to ~0.6V
o IBmax can be solved for using Ohm's Law
o With the value for IBmax and VCC, a DC Load Line can be drawn and used to find
the values of IC and VCE
o With the above information, all of the voltage, current, and resistance values for
the quiescent amplifier in Figure 11 can be found.
This is sufficient information to predict the DC performance of a CE Amplifier. The next two
sections will describe how to predict the circuit’s AC performance.
13
AC Current and the CE Amplifier:
Since the main purpose of a CE amplifier is to amplify an AC voltage, it is important to
know how a transistor behaves in the presence of an AC voltage. For example, in Figure 13, the
AC voltage travels through the circuit differently depending on whether the voltage source is in
its positive or negative half cycle.
Figure 13: AC current (red and green) superimposed on the DC current (blue) through a CE
Amplifier.
During the positive half cycle, which is in red above, the AC current (iphc) flows through
the capacitor (recall that a capacitor is seen as a short circuit by AC current and as an open
circuit by DC current; also, a DC voltage source is seen as a short circuit to an AC current.) and
splits into two branches. One branch goes through RB and the other into the base of the
transistor. However, the resistance in the transistor is so small in comparison to RB that the vast
majority of the current will travel into the base and the current into RB can be considered zero.
14
Remember that the AC current is superimposed on the DC current. So for the positive half cycle
the base current is:
i B = i phc + I B
(10)
Just as with DC current, the AC current traveling into the base of the transistor will
control the AC current flowing from the power supply VCC (iC). That output current will be
equal to the input current times the amplification factor:
iC = β × i B
(11)
Looking to the negative half cycle, in green in Figure 13, note that the base current will
equal the DC base current minus AC current, so:
i B = I B − inhc
(12)
As with DC current, the value of the emitter current at any given time will be equal to the
collector current added to the base current, so:
i E = i B + iC
(13)
The Hybrid Pi Model:
In order to determine the magnitude and shape of the AC voltage output from the CE
amplifier, the hybrid pi model can be used. The hybrid pi model takes advantage of the fact that
to an AC voltage the transistor in a CE Amplifier will act as a current controlled current source.
The circuit in Figure 14 is the AC equivalent circuit. This circuit is never to be built, it is simply
a model used to determine the output voltage.
15
Figure 14: The hybrid pi model AC circuit equivalent.
Figure 14 shows how the AC current travels through the circuit. As discussed in the
previous section, iB splits between the transistor and the base resistor, but since the base resistor’s
resistance is so large in comparison to the transistor, it can be assumed that all of iB travels
through the transistor, which has a base-to-emitter resistance called rπ. The collector current is
dependent on the value of current across rπ.
The value of rπ can be found by using this equation:
rπ =
ΔVBE
ΔI BQ
(14)
The changes in the base-to-emitter voltage and the base current are due to the sinusoidal
shape of the AC current as discussed above. Therefore, the value of ΔVBE will be equal to the
peak-to-peak voltage of the AC signal and ΔIBQ equals the corresponding sweep in the base
current.
Once these values are known the AC voltage amplification factor (Av) can be found. By
definition, it will be equal to the output voltage (vout) divided by the input voltage (vin).
AV =
16
vout
vin
(15)
By Ohm's law Equation 15 can be written as:
AV =
iC × RC
i B × rπ
(16)
Substituting Equation 11 yields:
AV =
β × i B × RC
i B × rπ
(17)
And finally results in the final equation for the estimation of the voltage amplification factor:
AV =
β × RC
rπ
(18)
Which makes it possible to predict the value of vout using Equation 15 to an accuracy of about
20%. Note that, in Figure 14, ic is flowing in the opposite direction as ib, so vout will be 180° out
of phase with vin.
This completes the collection of tools necessary to analyze this type of circuit.
Procedure and Results:
Modeling the Output Characteristics of an NPN Transistor in PSpice:
In order to find the output characteristics of an NPN transistor such as the amplification
factor (β) and the base-to-emitter resistance (rπ), we started by creating the circuit shown in
Figure 15 in PSpice. This simple circuit was assembled using four components: DC voltage
supply (V1), DC current supply (I1), NPN transistor (Q2N3904), and analog ground (0). The
plot of the output characteristics of this transistor (in Figure 16) was obtained by increasing the
value of the current supply, representing the base current (IB), from 0 to 60µA in steps of 10µA,
while simultaneously stepping up V1 from 0 to 15V in 0.01V increments.
17
Figure 15: Test circuit used to obtain output characteristics of a 2N3904 NPN transistor
Figure 16: Output characteristics IC vs. VCE in steps of IB of an NPN transistor
In order to calculate the transistor’s amplification factor β for every one of the six base
currents (I1), we needed to find the corresponding values of the collector current (IC) first.
Therefore, we set V1 to 10V and I1 to 10µA and ran the simulation (see Figure 17).
18
Figure 17: Obtaining IC for IB=10µA by setting V1=10V and I1=10µA
Using 1.556mA for the collector current, we used Equation 2 to calculate the amplification factor
for IB equal to 10µA below.
β =
1.556mA
10 μA
β = 155.6
We repeated the steps described above for IB equal to 20, 30, 40, 50 and 60µA, while keeping V1
constant at 10V. The resulting values for β were recorded in Table 2.
IB (μA)
β
10
155.60
20
168.85
30
174.83
40
177.80
50
179.20
60
179.67
Table 2: Transistor's amplification factors
19
Looking at the data it became apparent that there is a relationship of some sort between
the value of IB and the value of β. In order to examine this relationship we used Microsoft Excel
to graph the points in Table 2. The graph showed that the relationship between IB and β is
logarithmically increasing.
β According to IB
185.00
180.00
175.00
170.00
β
165.00
160.00
155.00
150.00
0
10
20
30
40
50
60
70
IB (μA)
Figure 18: Chart displaying the relationship between β and IB
Modeling the Input Characteristics of an NPN Transistor in PSpice:
To obtain the input characteristics of an NPN transistor, we used the same circuit that we
used for modeling the output characteristics of an NPN transistor (Figure 15). But this time, we
set V1 to a constant 10V and the DC current sweep range from 0 to 60µA in 0.1µA increments.
The resulting plot (in Figure 19) displays the relationship between the base current (IB) and the
base-to-emitter voltage (VBE).
20
Figure 19: Input characteristics of an NPN transistor
We calculated the base-to-emitter resistance (rπ) for IB equal to 20, 30, 40, 50 and 60µA
using Equation 14. A sample of this calculation for IB = 20µA can be viewed below. The
remaining resistances are available in Table 3.
rπ =
0.026V
20μA
rπ = 1300Ω
IB (μA)
rπ (Ω)
10
20
omitted
1300
30
867
40
650
50
520
21
IB (μA)
rπ (Ω)
60
433
Table 3: Base-to-emitter resistances
As with β, we used Excel to create a graph to give a visual representation of this
relationship. The relationship appeared to be a logarithmically decreasing.
rπ According to IB
1400
1200
1000
800
rπ
600
400
200
0
0
10
20
30
40
50
60
70
IB (μA)
Figure 20: Graphical representation of the relationship between IB and rπ
PSpice Analysis of a Fixed-Biased Common Emitter Transistor Amplifier with RB=1MΩ:
At this time, we were interested in finding the characteristics of the operation point (Q) of
a fixed-biased common-emitter (CE) circuit (in Figure 21), such as the base current (IBQ), the
collector current (ICQ), the emitter current (IEQ), the amplification factor (β@Q), the collector-toemitter voltage (VCEQ) and the voltage drop across the RL resistor (VRLQ).
22
Figure 21: Fixed-biased common emitter amplifier with RB =1MΩ
We knew that Q point is the point where the load line intersects the IBQ line (see Figure 9).
Therefore, we used Ohm's law and first calculated the points of intersection of the load line with
the vertical and the horizontal axes, ICmax and VCEmax, respectively (see below).
I C max =
15V − 0.6V
3.3kΩ
I C max = 4.36mA
VCEmax = VCC = 15V
(VCE will be a maximum when both RB and RC equal to zero.)
Using Equation 3, we calculated IBQ below.
I BQ =
15V − 0.6V
1MΩ
I BQ = 14.4μA
23
Knowing ICmax, VCEmax and IBQ, we plotted the load line on the printout of the output
characteristics of an NPN transistor obtained from PSpice earlier (Figure 16) and located the Qpoint (see Figure 22).
4.36mA
Q
2.4mA
14.4uA
LOAD
LINE
15V
6.9V
Figure 22: Output characteristics of a CE amplifier when IBQ =14.4µA
Projecting the Q-point to the vertical and the horizontal axis, we found the collector-toemitter voltage at Q (VCEQ) to be equal to 6.9V. The collector current at Q (ICQ) equaled to
2.4mA. 6.9V was quite close to the expected value for VCEQ, which should generally be
approximately half of the input voltage (VCC). In this case, that would ideally be 7.5V.
Referring to the calculations below, we found the emitter current at Q (IEQ), the transistor’s
amplification factor at Q (β@Q) and the voltage across RC at Q (VRCQ) by using Equations 4, 2,
and 6.
I EQ = 14.4μA + 2.4mA
24
I EQ = 2.414mA
β@Q =
2.4mA
14.4μA
β @ Q = 166.67
V RCQ = 2.4mA × 3.3kΩ
VRCQ = 7.92V
PSpice Analysis of a Fixed-Biased Common Emitter Transistor Amplifier with RB=680kΩ:
In this part of the laboratory exercise, we repeated the steps described above with the
only difference being the magnitude of RB now reduced to 680kΩ. This modified circuit is
shown in Figure 23.
Figure 23: Fixed-biased common emitter amplifier with RB =680kΩ
25
Figure 24 displays the location of the operating point for this circuit. It has shifted to the
left (towards saturation) resulting in a notable decrease in VCEQ that is now equal to 2.2V. The
magnitude of ICQ, on the other hand, increased (to 3.8mA). As these numbers demonstrate,
decreasing RB leads to increase in the base current as well as the collector current and decrease in
collector-to-emitter voltage. The characteristics for this particular circuit were found using the
same methods as for the last circuit and are summarized in Table 4.
4.36mA
21.2uA
Q
3.8mA
LOAD
LINE
15V
2.2V
Figure 24: Output characteristics of a CE amplifier when IBQ =21.2µA
IBQ =21.2μA
β@Q =179.25
ICQ =3.8mA
VCEQ =2.2V
IEQ =3.8212mA
VRLQ =12.54V
Table 4: Characteristics of a CE amplifier when IBQ =21.2µA
26
PSpice Analysis of a Fixed-Biased Common Emitter Transistor Amplifier with RB=1.5MΩ:
In the preceding section, we reduced the size of RB to 680kΩ. In this section we
increased it to 1.5MΩ. The modified circuit is shown in Figure 25.
Figure 25: Fixed-biased common emitter amplifier with RB =1.5MΩ
As can be seen in Figure 26, this time the Q-point moved to the right towards the cutoff
region. In other words, VCEQ increased to 10.3V, which is approximately 37% above what it
should be and ICQ dropped down to 1.5mA. As expected, increasing RB results in reducing the
base current as well as the collector current and this leads to increase in collector-to-emitter
voltage. The characteristics for this CE amplifier are summarized in Table 5.
27
4.36mA
Q
1.5mA
9.6uA
LOAD
LINE
10.3V
15V
Figure 26: Output characteristics of an NPN transistor with IBQ =9.6µA
IBQ =9.6μA
β@Q =156.25
ICQ =1.5mA
VCEQ =10.3V
IEQ =1.5096mA
VRLQ =4.95V
Table 5: Characteristics of a CE amplifier when IBQ =9.6µA
AC Analysis of the Fixed-Biased Common Emitter Amplifier in PSpice:
We modified the circuit in Figure 21 by adding the coupling capacitor C1 and the sine
wave generator Vin, as shown in Figure 27. The wave generator was set to frequency of 1000Hz
and amplitude of 20mV(p-p). The simulation yielded the outputs shown in Figure 28. It can be
seen in this figure that VCEQ equals to 7.537V, which is approximately half of VCC, as we
expected. Additionally, we obtained a printout of the AC-input waveform that is displayed in
28
Figure 29. Reading the peak values of the sine wave in this figure confirms that the amplitude is
20mV(p-p), as expected.
Figure 27: A common emitter amplifier
Figure 28: Simulation of the common emitter amplifier from Figure 27
29
Figure 29: Simulated AC-input signal of the CE amplifier in Figure 27
In order to see what the input voltage was at the base of the transistor, we moved the
voltage marker from point (A) to point (B) (see Figure 27) and ran the simulation again. The
resulting waveform (in Figure 30) shows that the amplitude of the input voltage at this point
remains 20mV(p-p) but it is superimposed to the DC base-to-emitter voltage VBEQ.
30
Figure 30: Simulated AC signal at the base of the transistor in Figure 27
To find the input voltage at the collector of the transistor, we moved the voltage marker
from point (B) to point (C) (see Figure 27) and ran the simulation one more time. The waveform
of the input voltage at this point is shown in Figure 31. Upon a closer look, the graph reveals
that the input voltage at the collector is an amplified version of Vin superimposed to the
collector-to-emitter voltage VCEQ (in other words, is an amplified version of the voltage seen in
Figure 30.
31
Figure 31: Simulated AC signal at the collector of the transistor Figure 27
As the next step, we calculated the voltage amplification factor (AV) below using the
values obtained from Figure 28. First, we calculated the transistors’ amplification factor (β) by
substituting into Equation 2.
β=
2.262mA
14.32μA
β = 157.96
Then, using Equation 14, we obtained the base-to-emitter resistance (see the calculation below).
rπ =
0.026V
14.32 μA
rπ = 1.815kΩ
Further, we substituted into Equation 16 and calculated the gain as shown below.
AV ≅
(157.96)(3.3kΩ)
1.815kΩ
AV ≅ 287
32
Finally, we re-calculated AV using Equation 15. As can be seen below, the difference between
the two values was approximately 11%.
AV =
9.75V − 4.67V
0.020V
AV = 254
%difference =
287 − 254
× 100(%) = 11.5%
287
Laboratory Test of a Fixed-Biased Common Emitter Amplifier:
We proceeded to build the circuit in Figure 21 in the laboratory so we could verify the
simulated results to the actual values. We first measured and recorded in Table 6 the actual
resistances of the RB and RC.
Resistor
Expected value
Actual value
RC
3.3kΩ
3.25kΩ
RB
1.0MΩ
1.0MΩ
Table 6: Expected and actual resistances
Then, using a 2N3904 small-signal transistor, the RC and RB resistors, and a bench power supply,
we assembled the circuit in Figure 21. We used a bench digital multimeter (DMM) to measure
the collector-to-emitter voltage (VCEQ). It equaled to 8.74V, which was within the acceptable
±20% proximity of the 7.5V ideal value. Assuming that VBE equaled 0.6V, we calculated VCB,
IBQ, VRB and VRC using the Kirchhoff’s voltage law using the measured values for RC, RB and
VCE.
33
VCB = VCE – VBE
VCB = 8.74V – 0.6V
VCB = 8.14V
VCC = IB RB + VBE
15V = IBQ (1.0MΩ) + 0.6V
IBQ =14.4µA
VRB = VCC - VBE
VRB = 15V – 0.6V
VRB = 14.4V
VRC = VCC – VCE
VRC = 15V – 8.74V
VRC = 6.26V
Then, using the Ohm’s law, we found the collector current (ICQ):
I CQ =
I CQ =
V RC
RC
6.26V
3.25kΩ
ICQ = 1.926mA
Finally, as shown in the calculations below, we found the emitter voltage (VEQ) by
applying Kirchhoff’s current law and we found the transistor’s transfer function by substituting
for ICQ and IBQ into Equation 2.
34
IEQ = IBQ + ICQ
IEQ = 14.4µA + 1.926mA
IEQ = 1.9404mA
β=
1.926mA
14.4μA
β = 133.75
All the above calculated values are summarized in Table 7.
IBQ =14.4μA
β@Q =133.75
ICQ =1.926mA
VCEQ =8.74V
IEQ =1.9404mA
VRCQ =6.26V
VRB = 14.4V
VCB = 8.14V
Table 7: Calculated characteristics of a the CE amplifier in Figure 21 when IBQ =14.4µA
Next, we measured the VCEQ, VBEQ, VCBQ, VRB, and VRC. The voltage measured across
the base-to-emitter junction of the transistor was found to be equal to 0.6V, which was what we
assumed for the calculations above. Therefore, re-calculating IBQ and ICQ yielded the same
results as the calculation above. The list of the voltage measurements can be viewed in Table 8.
VCEQ =8.74V
VRB =14.302V
VBEQ =0.6V
VRC =6.2V
VCBQ =8.09V
---------------------
Table 8: Measured characteristics of the CE amplifier in Figure 21 when IBQ =14.4µA
35
Calculated Values
Measured Values
% Difference
IBQ =14.4μA
IBQ =14.4μA
0.00
ICQ =1.926mA
ICQ =1.926mA
0.00
IEQ =1.9404mA
IEQ =1.9404mA
0.00
VCEQ =8.74V
VCEQ =8.74V
0.00
VBEQ =0.6V
VBEQ =0.6V
0.00
VCBQ = 8.14V
VCBQ =8.09V
0.61
VRB = 14.4V
VRB =14.302V
0.68
VRC =6.26V
VRC =6.2V
0.01
Table 9: Comparison of the measured and calculated values
Verifying AC Characteristics of a Fixed-Biased Common Emitter Amplifier in a Laboratory:
We built the circuit shown in Figure 32 using a 2N3904 small-signal NPN transistor, one
1kΩ and one 100Ω resistor, and a bench power supply. When comparing Figure 32 to Figure 27,
the only difference is the addition of two resistors (R1 and R2) across the function generator in
Figure 32. The purpose of R1 and R2 in the circuit is to allow for the adjustment of the input
voltage (Vin) if needed. We supplied power to the circuit and measured the collector-to-emitter
voltage (VCEQ) to be 8.89V, which was an acceptable value as we expected it to be 7.5 ±2V. If
VCEQ had been more than 9.5V or less then 5.5V we would have adjusted its value by slightly
decreasing or slightly increasing the resistance of RB until it was approximately 7.5V.
36
Figure 32: Common emitter amplifier from Figure 27 with the addition of R1 and R2 for
adjusting Vin
Next, we set the wave generator (Vgen) to a frequency of 1000Hz and an amplitude of 20mV(p-p)
and applied it to the circuit. Using as oscilloscope, the input voltage (Vin) was measured across
the R2 resistor to be 19.89mV(p-p), and the output voltage (Vout) was measured across the
collector-to-emitter terminal of the transistor and found to equal to 4.156V(p-p). Using these
numbers and Equation 15, we calculated the voltage gain (AV), as shown below.
AV =
19.89V( p − p )
4.156mV( p − p )
= 208.95
Vin and Vout According to Time
2.5
2
Vin and Vout (Volts)
1.5
1
0.5
Vin
0
Vout
-0.5
-1
-1.5
-2
-2.5
-0.0006
-0.0004
-0.0002
0
0.0002
0.0004
0.0006
Time in Seconds
Figure 33: Graph showing Vin and Vout of Figure 32
37
Comparing the voltage gain calculated from the measured values with the one calculated
from the PSpice values (as seen below), we concluded that the two numbers were within 17.7%
of each other.
%difference =
254 − 208.95
× 100(%) = 17.7%
254
Figure 34: Oscilloscope screen shot showing Vout and Vin according to time. Note that the scale
for Vin (Ch 1) is 5 mV/square, and the scale for Vout (Ch 2) is 1V/square
To get a clearer idea of the shape of the input voltage, this screen shot of the oscilloscope
was taken. Note that the scale for Vin (Ch 1) is 5 mV/square, and the scale for Vout (Ch 2) is
1V/square. What stands out the most in this graph is the 180° phase shift between Vout and Vin.
Conclusion:
In performing this experiment our primary goal was to become more familiar with NPN
transistors and their use in common emitter amplifiers. In order to do that, we used PSpice to
find the amplification factor (β) and the base-to-emitter resistance (rπ) of a 2N3904 NPN smallsignal bipolar transistor as well as to examine the effect that the value of the base resistor had on
the operating point. We also used PSpice to simulate the AC performance of a midpoint biased
38
common emitter amplifier. We then confirmed our simulated findings by building a midpoint
biased common emitter amplifier and found its amplification factor and base-to-emitter
resistance to compare to the simulated values. Our last step was to build the circuit in Figure 32
and to compare it’s AC and DC performance to that of the PSpice simulation of the similar
circuit in Figure 27.
When performing the PSpice simulations, we found that the amplification factor varied
depending on the value of the base current. When the base current was equal to 10μA, β was
equal to 155.6 (see Table 2). As the base current was increased to 60μA, β rose logarithmically
to 179.67 (see Figure 18). This is in contrast to the relationship between the base current and the
base-to-emitter voltage which showed an exponential increase in base current as the base-toemitter voltage increased (see Figure 19). The value of the base-to-emitter resistance (rπ) also
varied according to the value of IB. When IB was equal to 20μA, rπ was equal to 1300Ω (see
Table 3), and it decreased logarithmically to equal 433Ω when IB was equal to 60μA (see Figure
20).
The effect that the base current had on both the base-to-emitter resistance and β confirms
that it is important to know the characteristics of the operating point in order to analyze a circuit
containing a transistor. Our next exercise displayed the effect that the value of the base resistor
has on the location of the operating point. When the base resistor’s value was equal to 1MΩ the
operating point was centered between the cutoff and the saturation region (see Figure 22).
However, when we lowered the value of the base resistor to 680kΩ the operating point shifted
toward the saturation region (see Figure 24), and when we increased the base resistor to 1.5MΩ
the operating point shifted closer to the cutoff region. This confirmed the theoretical relationship
between the base resistor and the operating point, which states that in order to move the
39
operating point towards cutoff, the base resistor should be made larger, and to move the
operating point closer to saturation, the base resistor’s value should be decreased.
With this information we were able move on to an AC analysis of a common emitter
amplifier by simulating the circuit in Figure 25 in PSpice. By graphing the voltage at different
points in the circuit we found that the AC signal into the base was superimposed on the DC
signal (see Figure 30) and that it was that combined signal which was amplified by the transistor
and output at the collector (see Figure 31). The exact voltage amplification factor (AV) was
calculated using Equation 15 and then compared to the estimation found using Equation 16.
These two values were found to be within 12% of each other which confirmed that Equation 16
is an acceptable method of determining the approximate voltage gain for a CE amplifier.
Having completed the simulations we moved on to using practical circuits in the
laboratory. By building the circuit in Figure 25 and measuring the common-to-emitter voltage
we were able to calculate values of every voltage and current in the circuit. To see how good our
analysis was, we then measured each of the voltage values and used those measured values to
calculate the exact currents. The measured and calculated values were very close to one another
(see Table 9), with the largest difference being in the value of the voltage across the base
resistor, where the measured value was .68% different than the calculated value. This confirmed
that this method of analysis provides extremely accurate results for a common emitter amplifier.
To complete this experiment we built the circuit in Figure 32 and checked to see that the
collector-to-emitter voltage was within 2V of ½ VCC. Since the collector-to-emitter voltage
value of 8.89V is within 2V of 7.5V we were able to move on to the AC analysis of the circuit.
By using the oscilloscope to monitor both in input voltage signal and the output voltage signal
we were able to obtain the graph seen in Figure 33. We used the peak-to-peak values for the
40
input and output voltages in that graph to calculate the voltage amplification factor using
Equation 15 and found that it was within 18% of the simulated value from the PSpice simulation
of Figure 27. This difference is more than likely due to the fact that in PSpice all of the
components are assumed to be ideal while in reality our components are not ideal. Had we
wanted to estimate the value of the voltage gain using Equation 16 we would have had to
measure the collector and base current as well at the base-to-emitter voltage.
In Figure 34 we noted that the phase shift between Vout and Vin is 180°. This confirmed
the prediction of the voltage from the hybrid pi model discussed in the theory section of this
report. This means that a common emitter amplifier is very similar to an inverting amplifier built
with an operational amplifier, with one of the main differences being that an operational
amplifier has almost infinite impedance while the transistor has a very small impedance.
41
References:
“Course Notes: Unit 7”. Electronic Fundamentals I. November 25, 2007.
<http://technology.niagarac.on.ca/courses/elnc1126/Notes/Unit7.pdf>.
“Course Notes: Unit 8”. Electronic Fundamentals I. November 25, 2007.
<http://technology.niagarac.on.ca/courses/elnc1126/Notes/Unit8.pdf>.
“Npn BJT cross section.PNG”. Answers.com. November 25, 2007.
<http://www.answers.com/topic/npn-bjt-cross-section-png>.
Alexander, Charles K. and Matthew N. O. Sadiku. Fundamentals of Electronic Circuits. Boston:
McGraw-Hill Higher Education, 2007.
Brain, Marshall. "How Semiconductors Work". How Stuff Works. April 25, 2001. November
23, 2007. <http://computer.howstuffworks.com/diode.htm>.
Driscoll, Frederick F. and Robert F. Coughlin. Solid State Devices and Applications.
Englewood Cliffs: Prentice-Hall, Inc., 1975.
Nave, R. “Silicon Crystal Structure”. Hyperphysics. November 25, 2007.
<http://hyperphysics.phy-astr.gsu.edu/hbase/solids/sili2.html#c1>.
Nave, R. “NPN Common Emitter Amplifier”. Hyperphysics. November 25, 2007.
<http://230nsc1.phy-astr.gsu.edu/hbase/electronic/npnce.html>.
Sedra, Adel S. and Kenneth C. Smith. Microelectronic Circuits. New York: Oxford University
Press, 2004.
Wang, Ruye. “Semiconductor Devices”. E84 Lecture Notes. April 25, 2006. November 25,
2007. <http://fourier.eng.hmc.edu/e84/lectures/ch4/ch4.html>.
42