Lab6_ Common-base Amplifier

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Toronto Metropolitan University *

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Apr 3, 2024

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Course Title: Electronic Circuits I Course Number: ELE 404 Semester/Year W2024 Instructor: Prof. Fei Yuan Assignment/ Lab Number: Lab 6 Assignment/Lab Title: Common-base Amplifier Submission Date: Mar. 23, 2024 Due Date: Mar. 24, 2024 Student LAST Name Student FIRST Name Student Number Section Signature* El-Hage Ali 501167729 12 Pope Justin 501153591 12

Lab Report (Common-base Amplifier) By: Justin Pope & Ali El-Hage Date of Preparation: March 13, 2024 Table of Contents 1. Introduction…………..…………..………… .... …………..…………..……R2 2. Objectives..…………..…………..…………..…………..…………..………R2 3. Circuit Under Test..…………..…………..…………..…………..…………R2 4. Experimental Results..…………..………..…………..…………..……...…R4 5. Conclusions and Remarks..………….………..…………..…………..……R5 6. Appendix: Pre-Lab and TA Copy of Results…………………………...…R7 Page R1

1. Introduction The Common-base Amplifier Lab 6 report is presented below. The experiments were conducted on Wednesday, March 13, 2024. The pre-lab report and the TA copy of the experimental results are placed at the end of the report as an Appendix. 2. Objectives The objective of this lab is to learn how to bias a Bipolar-Junction Transistor (BJT) into active mode and test a Common-Base (CB) amplifier. We will also learn a technique for experimental evaluations for both input and output resistances on the amplifier. 3. Circuit under Test Figure 1 shows the schematic of the transistor circuit constructed; The circuit consists of a 2N3904 BJT. We will use the same quiescent voltage and current values as those we found for the CE amplifier in Table P1(a) . Figure 1. Common-Base (CB) amplifier. . Page R2

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Figure 2 shows the schematic of the two-port representation of an amplifier. Figure 2. Two-port representation of an amplifier. Figure 3 uses the amplifier represented by the two-port box of Figure 2 to drive a test load, R t, out , and is fed by a signal source whose Thevenin voltage and resistance are v S and R S , respectively. The signal source however, is not directly connected to the input port of the amplifier, but through another test resistor, R t, in . Figure 3. An amplifier energizing a test load, R t, out , through a signal source in series with a test resistance, R t, in . Page R3

4. Experimental Results In this lab, we built the circuit just like Figure 1 kind of like what lab 6 was about but this time, the signal generator is now connected to the emitter of the BJT transistor. We then measured the input voltage and output voltage (both in RMS) and also calculated the decibels for both input and output nodes. Once the values were found, we calculated the voltage gain using both the voltage and decibel values. We then removed the load resistance causing an open circuit which then led to an infinite load resistance. We then followed the same procedure as before. Now using the technique to measure the input and output resistances, we placed a test resistor from the source to the base of the BJT.. We then measured the voltage from the beginning of the resistor (V t ) and measured the voltage at the end of the test resistor(V i ). With this information, we can then calculate the input resistance. To calculate the output resistance, we placed the test resistor where the load voltage is located. Measured the output voltage with the test resistor and without the load resistor. Taking both output voltages in mind to then calculate the output resistance. Graph E2. Input (yellow) and output (green) voltage waveforms of the CB amplifier of Figure 1, with R S = 50Ω and R L = 10kΩ. Table P1(a) . Quiescent voltages and currents of the CB amplifier. V B [V] V C [V] V E [V] I B [mA] I C [mA] I E [mA] 2.56 8.11 1.86 0.0082 1.23 1.24 Page R4

Table E2(a) . Input and output AC voltages and gain of the CB amplifier, with R L = 10 kΩ V i [Vrms] V o [Vrms] A v [V/V] V I [db] V o [db] A v [db] 0.037 0.413 11.162 -20 -5.66 14.3 Table E2(b) . Input and output AC voltages and gain of the CB amplifier, with R L = ∞. V i [Vrms] V o [Vrms] A vo [V/V] V I [db] V o [db] A v [db] 0.078 0.603 7.73 -19.93 -1.72 17.58 Table E3 . Parameters of the CB amplifier for determining its input resistance. R t,in [kΩ] V t [Vrms] V i [Vrms] R i [kΩ] 0.1 0.127 0.1036 0.442 Table E4 . Parameters of the CE amplifier for determining its output resistance. R t,out [kΩ] (i.e., the load) V o [Vrms] without load (i.e., A vo V i ) V o [Vrms] with load R o [kΩ] 10 1.933 1.64 1.79 5. Conclusions and Remarks This section answers the questions raised in the lab manual. The questions have been repeated in italic bold font, for the ease of reference. C1. Compare the calculated and measured AC parameters of the CB amplifier, and calculate the percent errors from: ?% = ?𝑎????𝑎??? ?𝑎???−??𝑎????? ?𝑎??? ?𝑎????𝑎??? ?𝑎??? × 100 Complete Table C1. Comment on the magnitudes of errors and provide reasons for discrepancies. Page R5

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Table C1. Calculated and measured AC parameters for the CB amplifier of Figure 1. A v [V/V] A vo [V/V] R i [kΩ] R o [kΩ] Calculated Values (from Table P1(b)) 264.9 169.8 0.0207 5.6 Measured Values (from Tables E2, E3, and E4) 11.162 7.73 0.442 1.79 Percent Error, e% 95.8 95.3 2035 68 Our calculated values are significantly different from the measured values given a very large percentage error. After creating the Common-Base amplifier in Figure 1 , we were able to successfully achieve the respective graphs for both V i and V o . However, the lab required us to set the signal generator to produce a 1 kHz asymmetrical sinusoidal signal with the smallest magnitude. Then gradually increase the magnitude of V s until either V o starts to get distorted or clippled, or until V i is about to exceed 10 mV peak-to-peak. Due to the deficiency of the oscilloscope, the maximum peak-to-peak voltage that we were able to achieve was 1.29V. Our theoretical values were under the assumption that our experimental values would have a peak-to-peak value between 5mV - 10mV, hence, being the reasoning behind our large percentage errors. If we were able to achieve a peak-to-peak value closer to that of what was calculated in our theoretical values, then the percentage error would be smaller. Page R6

C2. Based on the measured results, calculate the current gain A i and power gain A p of the CB amplifier. The current gain is defined as the ratio of the output current i o to the input current i i (see Figure 3 to identify those currents). Also, the power gain is defined as the ratio of the power that the amplifier delivers to the load to the power that the amplifier draws from the signal source, i.e., through its input port. Therefore, A = A v A i . Find Current Gain: 𝐴 𝑖 = 𝑖 ? 𝑖 𝑖 = 𝑉 ? 𝑅 𝑖 𝑅 ? 𝑉 𝑖 = (1.16)(442) (1790)(0.07) = 4. 09 Find Power Gain: 𝐴 ? = 𝐴 ? 𝐴 𝑖 = (11. 162)(4. 09) = 45. 67 Therefore, the current gain is 4.09, and the power gain is 45.67. 6. Appendix – Pre-Lab and TA Copy of Results Page R7

Lab 6: Common-Base (CB) Amplifier By: Justin Pope Date of Preparation: March 8, 2024 P1: Simulation of the Circuit of Figure 1 using Multisim Figure 1. Common-Base (CB) amplifier. Table P1(a). Quiescent voltages and currents of the CB amplifier. V B [V] V C [V] V E [V] I B [mA] I C [mA] I E [mA] 2.56 8.11 1.86 0.0082 1.23 1.24 Common-Base (CB) Amplifier Pre-Lab 1

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Lab 6: Common-Base (CB) Amplifier By: Justin Pope Date of Preparation: March 8, 2024 P2: Find A vo (no load) 𝐴 𝑣𝑜 = 𝑉 𝑜 𝑉 𝑖 = −β β+1 ( ) 𝑅 𝐶 ? 𝑒 = −150 151 ( ) 5600 21 ( ) =− 264. 9 Find A v (with load) 𝐴 𝑣 = 𝑉 𝑜 𝑉 𝑖 = −β β+1 ( ) 𝑅 𝐶 || 𝑅 𝐿 ? 𝑒 ( ) = −β β+1 ( ) 1 ? 𝑒 ( ) 𝑅 𝐶 𝑅 𝐿 𝑅 𝐶 +𝑅 𝐿 ( ) = −150 150 ( ) 1 21 ( ) (5600)(10000) 5600+10000 ( ) =− 169. 8 Find R i 𝑅 𝑖 = 𝑅 𝐸 || ? 𝑒 = (1500)(21) 1500+21 = 20. 7Ω Find R o 𝑉 ? = 0, 𝐼 𝐸 = 0 𝑅 𝑜 = 𝑅 𝐶 = 5. 6𝑘Ω Table P1(b). Parameters of the CB amplifier of Figure 1. A vo [V/V] A v [V/V] R i [kΩ] R o [kΩ] -264.9 -169.8 0.0207 5.6 Common-Base (CB) Amplifier Pre-Lab 2

Lab 6: Common-Base (CB) Amplifier By: Justin Pope Date of Preparation: March 8, 2024 Graph P2. Source, input, and output voltage waveforms of the CB amplifier of Figure 1, with R S = 50Ω and R L = 10kΩ. Common-Base (CB) Amplifier Pre-Lab 3

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