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| university:courses:engineering_discovery:lab_10 [18 Oct 2016 20:51] – [Procedure] Jonathan Pearson | university:courses:engineering_discovery:lab_10 [03 Jan 2018 19:37] (current) – [Observations and Conclusions] Doug Mercer |
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| ===== Class A NPN Common-Emitter Amplifier ===== | ===== Class A NPN Common-Emitter Amplifier ===== |
| {{ analogTV>VideoNumberHere}} | {{ analogTV>5179788276001}} |
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| ==== Introduction ==== | ==== Introduction ==== |
| The inputs and outputs of Class A amplifiers can be either DC- or AC-coupled. When DC-coupled, the DC voltage of the source that is feeding the amplifier input must match the DC bias voltage required by the amplifier input, and similarly, the DC voltage of the output must match the DC voltage requirement of the input of whatever the amplifier is feeding. Often several Class A amplifier stages are DC-coupled together in a cascade, and the inter-stage DC levels are designed to be compatible with one another. AC-coupling is simpler in that each stage can have its own bias configuration, independent of the other stages, but AC-coupled systems do not pass DC, which is often a requirement. In this lab we build a simple AC-coupled CE amplifier stage and use it to amplify a voltage and drive an AC-coupled 1 KΩ load. | The inputs and outputs of Class A amplifiers can be either DC- or AC-coupled. When DC-coupled, the DC voltage of the source that is feeding the amplifier input must match the DC bias voltage required by the amplifier input, and similarly, the DC voltage of the output must match the DC voltage requirement of the input of whatever the amplifier is feeding. Often several Class A amplifier stages are DC-coupled together in a cascade, and the inter-stage DC levels are designed to be compatible with one another. AC-coupling is simpler in that each stage can have its own bias configuration, independent of the other stages, but AC-coupled systems do not pass DC, which is often a requirement. In this lab we build a simple AC-coupled CE amplifier stage and use it to amplify a voltage and drive an AC-coupled 1 KΩ load. |
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| The gain of a CE amplifier can be precisely calculated using circuit analysis, but it can also be estimated by simply inspecting the circuit. In the simplest CE amplifiers, a single emitter resistor R<sub>E</sub> is connected between the emitter and a voltage lower that that connected to R<sub>C</sub>, often the system ground. This resistor forms part of the transistor's bias network and also plays a part in determining the gain of the amplifier. As we learned in the "Introduction to Transistors" lab, we can estimate the emitter and collector currents to be approximately equal to each other. If R<sub>E</sub> is connected between the emitter and system ground, the signal voltage across R<sub>E</sub> will be a close replica of the voltage applied to the base, only shifted down by approximately 0.7 VDC -- the standard base-emitter drop we learned about in the "Introduction to Transistors" lab. This signal voltage produces a current in R<sub>E</sub> that is approximately equal to v<sub>i</sub>/R<sub>E</sub>, where v<sub>i</sub> is the input voltage to the amplifier. This is the emitter signal current i<sub>e</sub>, which is approximately equal to the collector signal current i<sub>c</sub>, which flows through R<sub>C</sub>. From this we can see that the collector signal voltage v<sub>c</sub>, which is also the output voltage v<sub>o</sub> is equal to -i<sub>c</sub>R<sub>C</sub>. Combining these results we get -v<sub>o</sub> = v<sub>i</sub>[R<sub>C</sub>/R<sub>E</sub>], and the gain v<sub>o</sub>/v<sub>i</sub> = -R<sub>C</sub>/R<sub>E</sub>. The minus sign indicates the 180 degree phase inversion that exists between the input and output signal voltages. In the lab we will see how to set the bias point as required by the load, and how loading affects the overall amplifier gain. We will see that the large output resistance of the CE amplifier causes an undesirable voltage drop when the amplifier is driving a load, and why another low-output resistance stage is often added to CE amplifiers that drive moderate to heavy loads. | The gain of a CE amplifier can be precisely calculated using circuit analysis, but it can also be estimated by simply inspecting the circuit. In the simplest CE amplifiers, a single emitter resistor R<sub>E</sub> is connected between the emitter and a voltage lower than that connected to R<sub>C</sub>, often the system ground. This resistor forms part of the transistor's bias network and also plays a part in determining the gain of the amplifier. As we learned in the "Introduction to Transistors" lab, we can estimate the emitter and collector currents to be approximately equal to each other. If R<sub>E</sub> is connected between the emitter and system ground, the signal voltage across R<sub>E</sub> will be a close replica of the voltage applied to the base, only shifted down by approximately 0.7 VDC -- the standard base-emitter drop we learned about in the "Introduction to Transistors" lab. This signal voltage produces a current in R<sub>E</sub> that is approximately equal to v<sub>i</sub>/R<sub>E</sub>, where v<sub>i</sub> is the input voltage to the amplifier. This is the emitter signal current i<sub>e</sub>, which is approximately equal to the collector signal current i<sub>c</sub>, which flows through R<sub>C</sub>. From this we can see that the collector signal voltage v<sub>c</sub>, which is also the output voltage v<sub>o</sub> is equal to -i<sub>c</sub>R<sub>C</sub>. Combining these results we get -v<sub>o</sub> = v<sub>i</sub>[R<sub>C</sub>/R<sub>E</sub>], and the gain v<sub>o</sub>/v<sub>i</sub> = -R<sub>C</sub>/R<sub>E</sub>. The minus sign indicates the 180 degree phase inversion that exists between the input and output signal voltages. In the lab we will see how to set the bias point as required by the load, and how loading affects the overall amplifier gain. We will see that the large output resistance of the CE amplifier causes an undesirable voltage drop when the amplifier is driving a load, and why another low-output resistance stage is often added to CE amplifiers that drive moderate to heavy loads. |
| ==== Objective ==== | ==== Objective ==== |
| To design, build, and test a CE amplifier, using a 2N3904 NPN transistor, with a loaded voltage gain of -5 and input resistance of at least 1 KΩ that is capable of driving an AC-coupled 1 KΩ load with a 2 V<sub>P-P</sub> sine wave. To verify that the amplifier Q-point and gain are close to their designed values. To observe the effects of loading on a CE amplifier output. To understand and be able to calculate CE amplifier voltage gain, power gain, and efficiency. Following completion of this lab you should be able to give the definition of a Class A amplifier and explain the basic operation of a CE amplifier, explain what a Q-point is, explain how output loading affects the overall voltage gain of a CE amplifier, and calculate the amplifier voltage gain, power gain, and efficiency of a CE amplifier. | To design, build, and test a CE amplifier, using a 2N3904 NPN transistor, with a loaded voltage gain of -5 and input resistance of at least 1 KΩ that is capable of driving an AC-coupled 1 KΩ load with a 2 V<sub>P-P</sub> sine wave. To verify that the amplifier Q-point and gain are close to their designed values. To observe the effects of loading on a CE amplifier output. To understand and be able to calculate CE amplifier voltage gain, power gain, and efficiency. Following completion of this lab you should be able to give the definition of a Class A amplifier and explain the basic operation of a CE amplifier, explain what a Q-point is, explain how output loading affects the overall voltage gain of a CE amplifier, and calculate the amplifier voltage gain, power gain, and efficiency of a CE amplifier. |
| <m>A_V ≈ -{470 Ω}/{57 Ω + 5.7 Ω} ≈ -7.5</m> | <m>A_V ≈ -{470 Ω}/{57 Ω + 5.7 Ω} ≈ -7.5</m> |
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| Accounting for the voltage divider loss incurred by the 470 Ω amplifier output resistance and the 1 KΩ loar resistance, we can calculate the loaded gain as | Accounting for the voltage divider loss incurred by the 470 Ω amplifier output resistance and the 1 KΩ load resistance, we can calculate the loaded gain as |
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| <m>{A_V}(loaded) ≈ -7.5({1 kΩ}/{1470 Ω}) ≈ -5.1</m> | <m>{A_V}(loaded) ≈ -7.5({1 kΩ}/{1470 Ω}) ≈ -5.1</m> |
| * Loaded CE amplifier gain depends on its output resistance and load resistance | * Loaded CE amplifier gain depends on its output resistance and load resistance |
| * Power gain of a CE amplifier can be calculated knowing its voltage gain, input resistance, and load resistance | * Power gain of a CE amplifier can be calculated knowing its voltage gain, input resistance, and load resistance |
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