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Common-Emitter (CE) Amplifier Introduction In this lab you will bias a Bipolar-Junction Transistor (BJT) in the active mode, and also tests a Common-Emitter (CE) amplifier. Moreover, you will learn a technique for experimental evaluation of the input and output resistances of an amplifier. For this lab, you will use the 2N3904 NPN BJT. Pre-lab Assignment P1. For the transistor circuit of Figure 1, calculate the node voltages 𝑉𝑉𝐡𝐡, 𝑉𝑉𝐢𝐢, and 𝑉𝑉𝐸𝐸, as well as the branch currents 𝐼𝐼𝐡𝐡, 𝐼𝐼𝐢𝐢, and 𝐼𝐼𝐸𝐸. Note that the resistances are different for different lab sections and, therefore, you must consult the document posted on D2L to know the resistance values specific to your designated lab section. Otherwise, assume that 𝑉𝑉𝐢𝐢𝐢𝐢 = 15 𝑉𝑉, 𝛽𝛽 = 150, 𝑉𝑉𝐡𝐡𝐡𝐡,π‘œπ‘œπ‘œπ‘œ = 0.7 𝑉𝑉, and 𝑉𝑉𝐢𝐢𝐢𝐢,𝑠𝑠𝑠𝑠𝑠𝑠 = 0.3 𝑉𝑉, and ignore the Early effect. Based on the calculated node voltages and branch currents, first establish the fact that the transistor is in the active mode and then determine the AC parameters π‘”π‘”π‘šπ‘š, π‘Ÿπ‘Ÿπ‘’π‘’, and π‘Ÿπ‘Ÿπœ‹πœ‹. Complete Table P1. Show all the work. Figure 1. Transistor circuit. Table P1. Quiescent voltages and currents and AC parameters of the transistor circuit of Figure 1. 𝑽𝑽𝑩𝑩[𝑽𝑽] 𝑽𝑽π‘ͺπ‘ͺ[𝑽𝑽] 𝑽𝑽𝑬𝑬[𝑽𝑽] 𝑰𝑰𝑩𝑩[π’Žπ’Žπ’Žπ’Ž] 𝑰𝑰π‘ͺπ‘ͺ[π’Žπ’Žπ’Žπ’Ž] 𝑰𝑰𝑬𝑬[π’Žπ’Žπ’Žπ’Ž] π’ˆπ’ˆπ’Žπ’Ž[π’Žπ’Žπ’Žπ’Ž] 𝒓𝒓𝒆𝒆[π’Œπ’Œπ›€π›€] 𝒓𝒓𝝅𝝅 [π’Œπ’Œπ›€π›€] R1 RC RE VCC R2 VE VB VC IC IB IB P2. Using the results of Step P1, and assuming that 𝑉𝑉𝐢𝐢𝐢𝐢 = 15 𝑉𝑉, 𝑅𝑅𝑆𝑆 = 50 Ξ©, and 𝑅𝑅𝐿𝐿 = 10 π‘˜π‘˜Ξ©, manually calculate the no-load voltage gain 𝐴𝐴𝑣𝑣𝑣𝑣, voltage gain 𝐴𝐴𝑣𝑣 = π‘£π‘£π‘œπ‘œ/𝑣𝑣𝑖𝑖, input resistance 𝑅𝑅𝑖𝑖, and output resistance π‘…π‘…π‘œπ‘œ of the Common-Emitter (CE) amplifier of Figure 2. Complete Table P2. Show all the work. Table P2. Parameters of the CE amplifier of Figure 2. 𝑨𝑨𝒗𝒗𝒗𝒗[𝑽𝑽/𝑽𝑽] 𝑨𝑨𝒗𝒗[𝑽𝑽/𝑽𝑽] for 𝑹𝑹𝑳𝑳 = 𝟏𝟏𝟏𝟏 π’Œπ’Œπ›€π›€ π‘Ήπ‘Ήπ’Šπ’Š[π’Œπ’Œπ›€π›€] 𝑹𝑹𝒐𝒐[π’Œπ’Œπ›€π›€] Figure 2. Common-Emitter (CE) amplifier based on the circuit of Figure 1. R1 RC RE VCC R2 vE vB vO Ro vC RS vI vS + - Signal Source RL Ri C1 C2 C3 RE2 VCC Common-Emitter Amplifier 3 P3. Simulate the CE amplifier of Figure 2, assuming that the transistor is the 2N3904, 𝑉𝑉𝐢𝐢𝐢𝐢 = 15 𝑉𝑉, 𝑅𝑅𝑆𝑆 = 50 Ξ©, 𝑅𝑅𝐿𝐿 = 10 π‘˜π‘˜Ξ©, 𝐢𝐢1 = 𝐢𝐢2 = 10 πœ‡πœ‡πœ‡πœ‡, and 𝐢𝐢3 = 100 πœ‡πœ‡πœ‡πœ‡. Also, assume 𝑣𝑣𝑆𝑆 to be a 1- kHz symmetrical sinusoidal voltage. Choose the magnitude of 𝑣𝑣𝑆𝑆 in such a way that 𝑣𝑣𝑂𝑂 features the maximum swing without noticeable distortion. Although doable by trial and error, this can be done systematically by first calculating the maximum permissible swing of 𝑣𝑣𝑂𝑂, and then dividing the mentioned swing by 𝐴𝐴𝑣𝑣, to find the maximum permissible swing of 𝑣𝑣𝐼𝐼. The swing of 𝑣𝑣𝑆𝑆 is then close to that of 𝑣𝑣𝐼𝐼, since 𝑅𝑅𝑆𝑆 is typically much smaller than 𝑅𝑅𝑖𝑖. Present the waveforms of 𝑣𝑣𝑆𝑆, 𝑣𝑣𝐼𝐼, and 𝑣𝑣𝑂𝑂, for three cycles (periods) as Graph P3. Make sure that 𝑣𝑣𝐼𝐼 and 𝑣𝑣𝑂𝑂 are in-phase (in terms of their zero crossings). Otherwise, check your simulation model and parameters. Also, verify that your manual calculation of 𝐴𝐴𝑣𝑣 (reported in Table P2) agrees well with the gain indicated by the waveforms of Graph P3. Otherwise, check your calculations and/or your simulation model. Graph P3. Source, input, and output voltage waveforms of the CE amplifier of Figure 2, with 𝑹𝑹𝑺𝑺 = πŸ“πŸ“πŸ“πŸ“ 𝛀𝛀 and 𝑹𝑹𝑳𝑳 = 𝟏𝟏𝟏𝟏 π’Œπ’Œπ›€π›€. t t vI vO 0 0 t vS 0 Common-Emitter Amplifier 4 P4. It is useful and a common practice to represent an amplifier (irrespective of its internal circuitry, number of transistors, etc.) with a two-port equivalent circuit such as the one shown in Figure 3 below. Thus, the input port of the amplifier is assumed to effectively appear to the outside world as a resistance, 𝑅𝑅𝑖𝑖, the input resistance of the amplifier. Also, the output port of the amplifier is viewed from the outside as a Thevenin circuit whose Thevenin voltage 𝐴𝐴𝑣𝑣𝑣𝑣𝑣𝑣𝑖𝑖 is a magnified copy of the input voltage, where the magnification factor 𝐴𝐴𝑣𝑣𝑣𝑣 is the open-circuit (or no-load) voltage gain of the amplifier, and whose Thevenin (or series) resistance, π‘…π‘…π‘œπ‘œ, is the output resistance of the amplifier. For further information see Chapter 1, Section 1.5, Sedra-Smith textbook, 6th or 7th Editions. Figure 3. Two-port representation of an amplifier. Now consider the circuit of Figure 4 in which an amplifier represented by the two-port box of Figure 3 drives a test load, 𝑅𝑅𝑑𝑑,π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ, and is fed by a signal source whose Thevenin voltage and resistance are 𝑣𝑣𝑠𝑠 and 𝑅𝑅𝑠𝑠, respectively. However, the signal source is not directly connected to the input port of the amplifier, but through another test resistor, 𝑅𝑅𝑑𝑑,𝑖𝑖𝑖𝑖, as Figure 4 illustrates. Prove that the input resistance, 𝑅𝑅𝑖𝑖, of the amplifier can be found from 𝑅𝑅𝑖𝑖 = 𝑅𝑅𝑑𝑑,𝑖𝑖𝑖𝑖 οΏ½ 𝑣𝑣𝑖𝑖 𝑣𝑣𝑑𝑑 βˆ’ 𝑣𝑣𝑖𝑖 οΏ½ Also, prove that the output resistance, 𝑅𝑅𝑂𝑂, of the amplifier can be found from π‘…π‘…π‘œπ‘œ = 𝑅𝑅𝑑𝑑,π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ οΏ½ 𝐴𝐴𝑣𝑣𝑣𝑣𝑣𝑣𝑖𝑖 π‘£π‘£π‘œπ‘œ βˆ’ 1οΏ½ Figure 4. An amplifier energizing a test load, 𝑹𝑹𝒕𝒕,𝒐𝒐𝒐𝒐𝒐𝒐, through a signal source in series with a test resistance, 𝑹𝑹𝒕𝒕,π’Šπ’Šπ’Šπ’Š. Ro ii Signal Source Rs Rt,in Amplifier vo + - Avovi Ro vi Ri RL ii io + - + - Input Signal Amplifier vL + - vO + vo + - + - Avovi Ro vi Ri + - ii io Signal Source vs Rs Rt,in + - vt Rt,out Amplifier Signal vL - Common-Emitter Amplifier 5 Experiments and Results E1. Construct the transistor circuit of Figure 1, using the BJT 2N3904 as the transistor. Figure 5 helps you identify the pins of the transistor. Set your bench-top power supply to ensure that the supply voltage is 𝑉𝑉𝐢𝐢𝐢𝐢 = 15 𝑉𝑉. Then, using your multimeter in the DC voltage measurement mode, measure the node voltages 𝑉𝑉𝐡𝐡, 𝑉𝑉𝐢𝐢, and 𝑉𝑉𝐸𝐸. Then, using the resistances and the measured voltages, calculate the device currents 𝐼𝐼𝐡𝐡, 𝐼𝐼𝐢𝐢, and 𝐼𝐼𝐸𝐸. Complete Table E1. Figure 5. Terminals of the NPN BJT 2N3904. Table E1. Measured terminal voltages and currents of the BJT in the circuit of Figure 1. 𝑽𝑽𝑩𝑩 [𝑽𝑽] 𝑽𝑽π‘ͺπ‘ͺ[𝑽𝑽] 𝑽𝑽𝑬𝑬[𝑽𝑽] 𝑰𝑰𝑩𝑩[π’Žπ’Žπ’Žπ’Ž] 𝑰𝑰π‘ͺπ‘ͺ[π’Žπ’Žπ’Žπ’Ž] 𝑰𝑰𝑬𝑬 [π’Žπ’Žπ’Žπ’Ž] Common-Emitter Amplifier 6 E2. Use the circuit you built in Step E1 as the kernel and evolve it to the CE amplifier of Figure 2, with 𝑅𝑅𝐿𝐿 = 10 π‘˜π‘˜Ξ©. Use 10-Β΅F electrolytic capacitors as the coupling capacitors 𝐢𝐢1 and 𝐢𝐢2, and a 100-Β΅F electrolytic capacitor as the bypass capacitors 𝐢𝐢3. Be careful with the polarity of the capacitors. Commonly, it is the negative terminal of the capacitor that is identified by a marking on the case of the capacitor. Schematically, the negative terminal corresponds to the curved plate of the capacitor’s symbol (see Figure 2). Set Channel 1 and Channel 2 of your oscilloscope to the DC-coupled mode and use them to monitor voltages 𝑣𝑣𝑖𝑖 and π‘£π‘£π‘œπ‘œ, respectively. Then, set the signal generator to produce a 1-kHz symmetrical sinusoidal signal with the magnitude you found in Step P3. The idea is that you should go for the maximum input signal magnitude that results in an undistorted (or sinusoid-looking) output voltage. As long as the amplifier works fairly linearly, that is, as long as its output voltage waveform looks very much like a sinusoid, your subsequent measurements will be valid. If the output signal appears to be distorted or, even worse, clipped, reduce the signal amplitude until this is no longer the case. Capture the waveforms of 𝑣𝑣𝐼𝐼 and 𝑣𝑣𝑂𝑂, for about three cycles, and save the captured waveforms as Graph E2. If the waveforms are noisy by any chance (for example, due to their small magnitudes), use the oscilloscope’s β€œAcquire” button to capture the β€œAverage” of the waveforms. Next, set the multimeter to the AC voltage measurement mode and measure the rms values of 𝑣𝑣𝐼𝐼 and 𝑣𝑣𝑂𝑂. Then press the β€œdB” button of the multimeter, to measure 𝑣𝑣𝐼𝐼 and 𝑣𝑣𝑂𝑂 in dB. From the measurements, determine the voltage gain 𝐴𝐴𝑣𝑣 = π‘£π‘£π‘œπ‘œ/𝑣𝑣𝑖𝑖, in both V/V and dB. Remember that the value of a voltage 𝑉𝑉 in dB is defined as 𝑉𝑉 [𝑑𝑑𝑑𝑑] = 20log (𝑉𝑉/𝑉𝑉𝑅𝑅), where 𝑉𝑉𝑅𝑅 is a reference voltage (for example, 0.7746 V rms) identified by the datasheet of the meter. The voltage gain in dB is therefore simply the output voltage in dB minus the input voltage in dB, i.e., 𝐴𝐴𝑣𝑣 [𝑑𝑑𝑑𝑑] = 20log (π‘‰π‘‰π‘œπ‘œ/𝑉𝑉𝑖𝑖) = 20log (π‘‰π‘‰π‘œπ‘œ/𝑉𝑉𝑅𝑅) βˆ’ 20log (𝑉𝑉𝑖𝑖/𝑉𝑉𝑅𝑅). Record the measured values of 𝑣𝑣𝐼𝐼 and 𝑣𝑣𝑂𝑂, as well the corresponding voltage gains (in V/V and dB) in Table E2(a). Table E2(a). Input and output AC voltages and gain of the CE amplifier, with 𝑅𝑅𝐿𝐿 = 10 π‘˜π‘˜Ξ©. π‘½π‘½π’Šπ’Š [𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] 𝑽𝑽𝑢𝑢[𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] 𝑨𝑨𝒗𝒗[𝑽𝑽/𝑽𝑽] 𝑽𝑽𝑰𝑰[𝒅𝒅𝒅𝒅] 𝑽𝑽𝑢𝑢[𝒅𝒅𝒅𝒅] 𝑨𝑨𝒗𝒗 [𝒅𝒅𝒅𝒅] Remove the 10-π‘˜π‘˜Ξ© load and repeat the AC voltage measurements used for Table E2(a). These measurements will give us the no-load voltage gain, 𝐴𝐴𝑣𝑣𝑣𝑣. Complete Table E2(b). Table E2(b). Input and output AC voltages and gain of the CE amplifier, with 𝑅𝑅𝐿𝐿 = ∞. π‘½π‘½π’Šπ’Š [𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] 𝑽𝑽𝑢𝑢[𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] 𝑨𝑨𝒗𝒗𝒗𝒗[𝑽𝑽/𝑽𝑽] 𝑽𝑽𝑰𝑰[𝒅𝒅𝒅𝒅] 𝑽𝑽𝑢𝑢[𝒅𝒅𝒅𝒅] 𝑨𝑨𝒗𝒗𝒗𝒗 [𝒅𝒅𝒅𝒅] Common-Emitter Amplifier 7 E3. Let us now measure the input resistance of the CE amplifier built in Step E2, using the method discussed in Step P4. To that end, set the multimeter in the AC voltage measurement mode to measure voltages in terms of their rms values. Then, bring the 10-π‘˜π‘˜Ξ© load back into the circuit. Next insert a test resistance, which you called 𝑅𝑅𝑑𝑑,𝑖𝑖𝑖𝑖 in Step P4, between the output terminal of the signal generator and the input terminal of the amplifier (refer to Figure 4). Choose a value for 𝑅𝑅𝑑𝑑,𝑖𝑖𝑖𝑖 that is close to the value of 𝑅𝑅𝑖𝑖 you expect the amplifier to have (based on your analysis in Step P2). For instance, if 𝑅𝑅𝑖𝑖 to be measured is expected to be about 7 kΩ, then 𝑅𝑅𝑑𝑑,𝑖𝑖𝑖𝑖 should also be of the same order of magnitude (e.g., about a few kΩ). Measure voltages 𝑣𝑣𝑑𝑑 and 𝑣𝑣𝑖𝑖, and based on the measured values, calculate 𝑅𝑅𝑖𝑖. Complete Table E3. Table E3. Parameters of the CE amplifier for determining its input resistance. 𝑹𝑹𝒕𝒕,π’Šπ’Šπ’Šπ’Š [π’Œπ’Œπ›€π›€] 𝑽𝑽𝒕𝒕[𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] π‘½π‘½π’Šπ’Š[𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] π‘Ήπ‘Ήπ’Šπ’Š[π’Œπ’Œπ›€π›€] E4.Next, replace the input test resistance by a short link, to bring the amplifier back to the state it had in Step E2. Then, replace the 10-π‘˜π‘˜Ξ© load resistance with a test resistance, which you called 𝑅𝑅𝑑𝑑,π‘œπ‘œπ‘œπ‘œπ‘œπ‘œ in Step P4, whose value is of the same order of magnitude as the value you expect for 𝑅𝑅𝑂𝑂 (based on your manual calculations in Step P2). Then, record the no-load and loaded rms output voltages of the amplifier. Based on the measured values, calculate 𝑅𝑅𝑂𝑂 and complete Table E4. Table E4. Parameters of the CE amplifier for determining its output resistance. 𝑹𝑹𝒕𝒕,𝒐𝒐𝒐𝒐𝒐𝒐 [π’Œπ’Œπ›€π›€] (i.e., the load) 𝑽𝑽𝑢𝑢[𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] without load (i.e., π‘¨π‘¨π’—π’—π’—π’—π’—π’—π’Šπ’Š) 𝑽𝑽𝒐𝒐[𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽] with load 𝑹𝑹𝒐𝒐[π’Œπ’Œπ›€π›€] Common-Emitter Amplifier 8 Conclusions and Remarks C1. Compare the calculated and measured quiescent (DC) voltages of the CE amplifier, and calculate the percent error from the following expression: 𝑒𝑒% = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 βˆ’ π‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘š 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 π‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘š 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 Γ— 100 Complete Table C1 and comment on the magnitude of, and reasons for, the errors. Table C1. Calculated and measured (DC) voltages in the transistor circuit of Figure 1. 𝑽𝑽𝑩𝑩[𝑽𝑽] 𝑽𝑽π‘ͺπ‘ͺ[𝑽𝑽] 𝑽𝑽𝑬𝑬[𝑽𝑽] Calculated values (from Table P1) Measured values (from Table E1) Percent error, 𝑒𝑒% C2. Compare the calculated and measured AC parameters of the CE amplifier, and calculate the percent errors. Complete Table C2. Comment on the magnitudes of errors and provide reasons for discrepancies. Table C2. Calculated and measured ac parameters for the CE amplifier of Figure 2. 𝑨𝑨𝒗𝒗[𝑽𝑽/𝑽𝑽] 𝑨𝑨𝒗𝒗𝒗𝒗[𝑽𝑽/𝑽𝑽] π‘Ήπ‘Ήπ’Šπ’Š[π’Œπ’Œπ›€π›€] 𝑹𝑹𝒐𝒐[π’Œπ’Œπ›€π›€] Calculated Values (from Table P2) Measured Values (from Tables E2, E3, and E4) Percent Error, 𝑒𝑒% C3. Based on the measured results, calculate the current gain 𝐴𝐴𝑖𝑖 and power gain 𝐴𝐴𝑝𝑝 of the CE amplifier. The current gain is defined as the ratio of the output current π‘–π‘–π‘œπ‘œ to the input current 𝑖𝑖𝑖𝑖 (see Figure 3 and Figure 4 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, 𝐴𝐴𝑝𝑝 = 𝐴𝐴𝑣𝑣𝐴𝐴𝑖𝑖. C4. Explain the effect of resistance 𝑅𝑅𝐸𝐸2 of the CE amplifier on β€’ The voltage gain β€’ The input resistance β€’ The output resistance β€’ The maximum magnitude of 𝑣𝑣𝑖𝑖 before the output voltage exhibits distortions.

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A transistor is an active three-terminal semiconductor device invented in 1925. The transistor replaced the use of vacuum tubes as electronically controlled switches. Thus, transistors pave the way to miniaturized electronic devices that use integrated circuits. There are two types of transistors: Bipolar Junction Transistors (BJT) and Field Effect Transistors (FET). In this Lab, the BJT transistor is studied. The typical pin out of a BJT transistor is shown in the Figure below. The three terminals are called emitter(E), base(B), and collector(C). In a BJT transistor, the base current can be used to control the collector current, which can be used to build amplifiers and switches...

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