ANALYSIS OF AN NPN COMMON-EMITTER AMPLIFIER

Transcription

1 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

2 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: The amplification factor (β) The base-to-emitter resistance (r π ) 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 (R B ) (if necessary) in order to adjust the operating point (Q) to be approximately one-half the value of V CC o To apply a 1 khz, 20mV (p-p) sine wave to the amplifier s input and to measure both the input voltage (V in ) and the output voltage (V O ) 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

3 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

4 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

5 Figure 4: A simplified cross-section of an NPN transistor 1 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 (I C ) dependency on the base current (I B ) and the collector-to-emitter voltage drop (V CE ). 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, < 5

6 I B can be controlled by adjusting the current source (labeled I1 in Figure 5), V CE can be changed by changing V CC since they are equal, and the value of I C is measured. To graph the changes in I C as a result of the values of I B and V CE, I1 is set at 0A, and I C is measured as the value of V CE is changed from 0 to 15V. The resulting graph is in Figure 6, and shows a very small rise in the value of I B as V CE increases. Note the values of I B never climb above 0.25nA, which is so small as to be effectively 0A. I C vs. V CE 2.50E E E-10 I C (A) 1.00E-10 IB = 0A 5.00E E E-11 V CE (V) Figure 6: I C vs. V CE for the values of I B equal to 0A Next, the value of I B is changed to be 10μA. Data is again taken and plotted on the same graph as the previous data. This process is repeated increasing I B in steps of 10μA up to 60μA. The resulting graph is below. 6

7 I C vs. V CE 1.40E E-02 I C (A) 1.00E E E E E-03 IB = 0A IB = 10 ua IB = 20uA IB = 30uA IB = 40uA IB = 50uA IB = 60uA 0.00E E-03 V CE (V) Figure 7: I C vs. V CE in 10μA steps of I B Figure 7 shows that the value of I C depends mostly on the value of I B and less on the value of V CE. This means that in order to accurately predict the value of I C in a circuit both the values of I C and V CE must be known. Also, note that the relationship between V CE and I C 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 I C is mostly dependent on I B, the next logical step is to find what that relationship is. Note from Figure 7 that I B is given in terms of μa (10-6 A), and I C is given it terms of ma (10-3 A), and that the value of I C increases a predictable amount for every 10μA increase in I B. 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 Solving for β results in: I = β I (1) C B 7

8 I C β = (2) I B Using Figure 5 to find the value of β for the transistor, I C can be measured and recorded while V CC is held constant and I B 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. I B (μa) β Table 1: Chart used to display β for different values of I B Input Characteristics of a Transistor: When analyzing a circuit which includes a transistor it is also useful to know how the value of I B changes according to the value of the base-to-emitter voltage drop (V BE ). Knowing that the turn-on voltage for a diode is ~0.6V, it makes sense that the value of I B would be very close to zero until V BE reaches ~0.6V and then it would increase at the same rate that I C is increased. The graph would look similar to Figure 8. I B vs. V BE 7.00E E E E-05 IB (A) 3.00E E E E E E E E E E E E-01 V BE (V) Figure 8: Graph of I B vs. V BE for a NPN transistor 8

9 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 I C to the maximum value of V CE. 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 I B in a circuit (see Equation 3). In this example, it appears that I B 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 I B on V CE discussed earlier, the green line is not horizontal but has a slightly positive slope. I V V CC BE B = (3) RB 9

10 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 I C and V CE. The Q point is defined as a point on the DC load line that indicates the values of I C and V CE for the amplifier at rest. Ideally, the value of V CE at this point should be as close as possible to one-half of V CC. In the example above, the V CE value for point Q appears to be at 7V, which is within 10% of the half of the V CC value of 15V. Projecting the point Q to the y-axis gives the value of I C at approximately 2.1mA, which is about ½ of the value marked at I Cmax. When the values for V CE and I C 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 V CE and I C 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 operation 2 2 From: Course Notes: Unit 8. Electronic Fundamentals I. November 25, < 10

11 If the Q point is not optimally placed it can be changed by adjusting the value of R B. That will change the value of I B which, as discussed above, is the value which primarily affects I C, which in turn affects the value of V CE. To move the Q point towards cutoff, R B should be raised, and to move Q closer to saturation, R B 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

12 supply and flows to the top of the circuit where it splits and travels in two directions. The first branch (I B ) flows through the base resistor (R B ) and into the base of the transistor. The second branch (I C ) flows through the collector resistor (R C ) 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, I C will not pass through the base-to-collector junction, instead it will continue straight through the transistor to the emitter. I B, however, will flow from the base to the emitter. This means that the emitter current (I E ) will be equal to the sum of I B and I C, which yields a useful relationship for this circuit: I = I + I (4) E B C 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

13 Starting with the input loop equation: V + V + V = 0 V CC CC = V RB RB + V BE BE (5) Using Ohm's Law the term V RB can be replaced: V = I R (6) V B = I R (7) R B B Substituting Equation 7 into Equation 5 results in the final input loop equation: V = I R + V (8) CE B B BE A similar process is used to find the output loop equation, which ultimately yields: V = I R + V (9) CC C C CE The following values in the input and output loop equations are known or can be solved for: o V CC, R C, and R B which are set by the user o V BE, which is known to be the turn on voltage of a diode, equal to ~0.6V o I Bmax can be solved for using Ohm's Law o With the value for I Bmax and V CC, a DC Load Line can be drawn and used to find the values of I C and V CE 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

14 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 (i phc ) 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 R B and the other into the base of the transistor. However, the resistance in the transistor is so small in comparison to R B that the vast majority of the current will travel into the base and the current into R B can be considered zero. 14

15 Remember that the AC current is superimposed on the DC current. So for the positive half cycle the base current is: i = i + I (10) B phc B 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 V CC (i C ). That output current will be equal to the input current times the amplification factor: i C = 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 i (12) B nhc 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 = i + i (13) E B C 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

16 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, i B 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 i B travels through the transistor, which has a base-to-emitter resistance called r π. The collector current is dependent on the value of current across r π. The value of r π can be found by using this equation: r ΔV = BE π (14) ΔI BQ 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 ΔV BE will be equal to the peak-to-peak voltage of the AC signal and ΔI BQ equals the corresponding sweep in the base current. Once these values are known the AC voltage amplification factor (A v ) can be found. By definition, it will be equal to the output voltage (v out ) divided by the input voltage (v in ). v out A V = (15) vin 16

17 By Ohm's law Equation 15 can be written as: A V ic RC = (16) i B r π Substituting Equation 11 yields: A V β ib RC = (17) i r B π And finally results in the final equation for the estimation of the voltage amplification factor: A V β RC = (18) r π Which makes it possible to predict the value of v out using Equation 15 to an accuracy of about 20%. Note that, in Figure 14, i c is flowing in the opposite direction as i b, so v out will be 180 out of phase with v in. 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 (r π ), 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 (I B ), from 0 to 60µA in steps of 10µA, while simultaneously stepping up V1 from 0 to 15V in 0.01V increments. 17

18 Figure 15: Test circuit used to obtain output characteristics of a 2N3904 NPN transistor Figure 16: Output characteristics I C vs. V CE in steps of I B 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 (I C ) first. Therefore, we set V1 to 10V and I1 to 10µA and ran the simulation (see Figure 17). 18

19 Figure 17: Obtaining I C for I B =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 I B equal to 10µA below. β 1.556mA = 10μA β = We repeated the steps described above for I B equal to 20, 30, 40, 50 and 60µA, while keeping V1 constant at 10V. The resulting values for β were recorded in Table 2. I B (μa) β Table 2: Transistor's amplification factors 19

20 Looking at the data it became apparent that there is a relationship of some sort between the value of I B 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 I B and β is logarithmically increasing. β According to I B β I B (μa) Figure 18: Chart displaying the relationship between β and I B 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 (I B ) and the base-to-emitter voltage (V BE ). 20

21 Figure 19: Input characteristics of an NPN transistor We calculated the base-to-emitter resistance (r π ) for I B equal to 20, 30, 40, 50 and 60µA using Equation 14. A sample of this calculation for I B = 20µA can be viewed below. The remaining resistances are available in Table 3. r π r π 0.026V = 20μA = 1300Ω I B (μa) r π (Ω) 10 omitted

22 I B (μa) r π (Ω) 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. r π According to I B r π IB (μa) Figure 20: Graphical representation of the relationship between I B and r π PSpice Analysis of a Fixed-Biased Common Emitter Transistor Amplifier with R B =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 (I BQ ), the collector current (I CQ ), the emitter current (I EQ ), the amplification factor ), the collector-toemitter voltage (V CEQ ) and the voltage drop across the R L resistor (V RLQ ). 22

23 Figure 21: Fixed-biased common emitter amplifier with R B =1MΩ We knew that Q point is the point where the load line intersects the I BQ 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, I Cmax and V CEmax, respectively (see below). 15V 0.6V I C max = 3. 3kΩ I C max = 4. 36mA V CEmax = V CC = 15V (V CE will be a maximum when both R B and R C equal to zero.) Using Equation 3, we calculated I BQ below. 15V 0.6V I BQ = 1MΩ I BQ = 14.4μA 23

24 Knowing I Cmax, V CEmax and I BQ, 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 Q- point (see Figure 22). 4.36mA 2.4mA Q 14.4uA LOAD LINE 6.9V 15V Figure 22: Output characteristics of a CE amplifier when I BQ =14.4µA Projecting the Q-point to the vertical and the horizontal axis, we found the collector-toemitter voltage at Q (V CEQ ) to be equal to 6.9V. The collector current at Q (I CQ ) equaled to 2.4mA. 6.9V was quite close to the expected value for V CEQ, which should generally be approximately half of the input voltage (V CC ). In this case, that would ideally be 7.5V. Referring to the calculations below, we found the emitter current at Q (I EQ ), the transistor s amplification factor at Q ) and the voltage across R C at Q (V RCQ ) by using Equations 4, 2, and 6. I EQ = 14.4μ A mA 24

25 I EQ = mA Q = 2.4mA 14.4μA Q = V RCQ = 2.4mA 3. 3kΩ V RCQ = 7. 92V PSpice Analysis of a Fixed-Biased Common Emitter Transistor Amplifier with R B =680kΩ: In this part of the laboratory exercise, we repeated the steps described above with the only difference being the magnitude of R B now reduced to 680kΩ. This modified circuit is shown in Figure 23. Figure 23: Fixed-biased common emitter amplifier with R B =680kΩ 25

26 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 V CEQ that is now equal to 2.2V. The magnitude of I CQ, on the other hand, increased (to 3.8mA). As these numbers demonstrate, decreasing R B 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 mA 3.8mA Q 21.2uA LOAD LINE 2.2V 15V Figure 24: Output characteristics of a CE amplifier when I BQ =21.2µA I BQ =21.2μA = I CQ =3.8mA I EQ =3.8212mA V CEQ =2.2V V RLQ =12.54V Table 4: Characteristics of a CE amplifier when I BQ =21.2µA 26

27 PSpice Analysis of a Fixed-Biased Common Emitter Transistor Amplifier with R B =1.5MΩ: In the preceding section, we reduced the size of R B 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 R B =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, V CEQ increased to 10.3V, which is approximately 37% above what it should be and I CQ dropped down to 1.5mA. As expected, increasing R B 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

28 4.36mA 1.5mA Q 9.6uA LOAD LINE 10.3V 15V Figure 26: Output characteristics of an NPN transistor with IBQ =9.6µA I BQ =9.6μA = I CQ =1.5mA I EQ =1.5096mA V CEQ =10.3V V RLQ =4.95V Table 5: Characteristics of a CE amplifier when I BQ =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 V in, 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 V CEQ equals to 7.537V, which is approximately half of V CC, as we expected. Additionally, we obtained a printout of the AC-input waveform that is displayed in 28

29 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

30 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 V BEQ. 30

31 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 V in superimposed to the collector-to-emitter voltage V CEQ (in other words, is an amplified version of the voltage seen in Figure

32 Figure 31: Simulated AC signal at the collector of the transistor Figure 27 As the next step, we calculated the voltage amplification factor (A V ) 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 = Then, using Equation 14, we obtained the base-to-emitter resistance (see the calculation below). r π 0.026V = 14.32μA r π = 1.815kΩ Further, we substituted into Equation 16 and calculated the gain as shown below. (157.96)(3.3kΩ) A V kΩ A V

33 Finally, we re-calculated A V using Equation 15. As can be seen below, the difference between the two values was approximately 11%. 9.75V 4.67V A V = V A V = % difference = 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 R B and R C. Resistor Expected value Actual value R C 3.3kΩ 3.25kΩ R B 1.0MΩ 1.0MΩ Table 6: Expected and actual resistances Then, using a 2N3904 small-signal transistor, the R C and R B 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 (V CEQ ). It equaled to 8.74V, which was within the acceptable ±20% proximity of the 7.5V ideal value. Assuming that V BE equaled 0.6V, we calculated V CB, I BQ, V RB and V RC using the Kirchhoff s voltage law using the measured values for R C, R B and V CE. 33

34 V CB = V CE V BE V CB = 8.74V 0.6V V CB = 8.14V V CC = I B R B + V BE 15V = I BQ (1.0MΩ) + 0.6V I BQ =14.4µA V RB = V CC - V BE V RB = 15V 0.6V V RB = 14.4V V RC = V CC V CE V RC = 15V 8.74V V RC = 6.26V Then, using the Ohm s law, we found the collector current (I CQ ): V I CQ = R I 6.26V = CQ kω I CQ = 1.926mA Finally, as shown in the calculations below, we found the emitter voltage (V EQ ) by applying Kirchhoff s current law and we found the transistor s transfer function by substituting for I CQ and I BQ into Equation 2. RC C 34

35 I EQ = I BQ + I CQ I EQ = 14.4µA mA I EQ = mA 1.926mA β = 14.4μA β = All the above calculated values are summarized in Table 7. I BQ =14.4μA = I CQ =1.926mA I EQ =1.9404mA V CEQ =8.74V V RCQ =6.26V V RB = 14.4V V CB = 8.14V Table 7: Calculated characteristics of a the CE amplifier in Figure 21 when I BQ =14.4µA Next, we measured the V CEQ, V BEQ, V CBQ, V RB, and V RC. 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 I BQ and I CQ yielded the same results as the calculation above. The list of the voltage measurements can be viewed in Table 8. V CEQ =8.74V V BEQ =0.6V V RB =14.302V V RC =6.2V V CBQ =8.09V Table 8: Measured characteristics of the CE amplifier in Figure 21 when I BQ =14.4µA 35

36 Calculated Values Measured Values % Difference I BQ =14.4μA I BQ =14.4μA 0.00 I CQ =1.926mA I CQ =1.926mA 0.00 I EQ =1.9404mA I EQ =1.9404mA 0.00 V CEQ =8.74V V CEQ =8.74V 0.00 V BEQ =0.6V V BEQ =0.6V 0.00 V CBQ = 8.14V V CBQ =8.09V 0.61 V RB = 14.4V V RB =14.302V 0.68 V RC =6.26V V RC =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 (R 1 and R 2 ) across the function generator in Figure 32. The purpose of R 1 and R 2 in the circuit is to allow for the adjustment of the input voltage (V in ) if needed. We supplied power to the circuit and measured the collector-to-emitter voltage (V CEQ ) to be 8.89V, which was an acceptable value as we expected it to be 7.5 ±2V. If V CEQ 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 R B until it was approximately 7.5V. 36

37 Figure 32: Common emitter amplifier from Figure 27 with the addition of R 1 and R 2 for adjusting V in Next, we set the wave generator (V gen ) to a frequency of 1000Hz and an amplitude of 20mV (p-p) and applied it to the circuit. Using as oscilloscope, the input voltage (V in ) was measured across the R 2 resistor to be 19.89mV (p-p), and the output voltage (V out ) 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 (A V ), as shown below. A V 19.89V ( p p) = 4.156mV ( p p) = Vin and Vout According to Time Vin and Vout (Volts) Time in Seconds Vin Vout Figure 33: Graph showing V in and V out of Figure 32 37

38 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 = 100(%) = 17.7% 254 Figure 34: Oscilloscope screen shot showing V out and V in according to time. Note that the scale for V in (Ch 1) is 5 mv/square, and the scale for V out (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 V in (Ch 1) is 5 mv/square, and the scale for V out (Ch 2) is 1V/square. What stands out the most in this graph is the 180 phase shift between V out and V in. 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 (r π ) 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

39 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 (see Table 2). As the base current was increased to 60μA, β rose logarithmically to (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 (r π ) also varied according to the value of I B. When I B was equal to 20μA, r π was equal to 1300Ω (see Table 3), and it decreased logarithmically to equal 433Ω when I B 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

40 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 (A V ) 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 V CC. 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

41 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 V out and V in 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

42 References: Course Notes: Unit 7. Electronic Fundamentals I. November 25, < Course Notes: Unit 8. Electronic Fundamentals I. November 25, < Npn BJT cross section.png. Answers.com. November 25, < Alexander, Charles K. and Matthew N. O. Sadiku. Fundamentals of Electronic Circuits. Boston: McGraw-Hill Higher Education, Brain, Marshall. "How Semiconductors Work". How Stuff Works. April 25, November 23, < Driscoll, Frederick F. and Robert F. Coughlin. Solid State Devices and Applications. Englewood Cliffs: Prentice-Hall, Inc., Nave, R. Silicon Crystal Structure. Hyperphysics. November 25, < Nave, R. NPN Common Emitter Amplifier. Hyperphysics. November 25, < Sedra, Adel S. and Kenneth C. Smith. Microelectronic Circuits. New York: Oxford University Press, Wang, Ruye. Semiconductor Devices. E84 Lecture Notes. April 25, November 25, < 42